source_id
int64 1
4.64M
| question
stringlengths 0
28.4k
| response
stringlengths 0
28.8k
| metadata
dict |
|---|---|---|---|
71,184 | Over on Photography , a question was asked as to why (camera) lenses are always cylindrical . Paraphrasing slightly, one of the answers and follow-up comments asserted that quantum effects are significant and that you need to understand QED in order to understand lens design. My intuition tells me that this probably isn't the case, and that modern lens design can be handled via classical optical theory. Is my intuition correct, or are quantum effects really significant in modern lenses? | Stan Rogers' answer on photography.SE seems to be claiming that QED is not just sufficient but also necessary to explain the the effect of the lens's shape. This is wrong. Ray optics suffices at ordinary magnifications, and even at high magnifications, classical wave optics suffices. Let's say you use a rectangular lens rather than a cylindrical one. First off, the shape of the lens won't matter at all unless you have the aperture all the way open; on any slower setting, the approximately circular shape of the diaphram will be the determining factor. Assuming that you do have the aperture all the way open, the main effect, which is purely a geometrical optics (ray optics) effect will be as follows. You will have a certain depth of field. If object point A is at the correct distance to produce a pointlike image, then this point is still a point regardless of the rectangular shape of the lens. However, if object point B is at some other distance, we get a blur as the image of that point. The blur occurs because there is a bundle of light rays, and the bundle has some finite size where it intersects the film or chip. Since the lens is rectangular, this bundle is pyramidal, and the blur will be a rectangular blur rather than the usual circular one. For example, say you're photographing someone's face with a starry sky in the background. You focus on the face. The stars will appear as little fuzzy rectangles. At very high magnifications (maybe with a very long lens that's effectively a small telescope), it's possible that you would also see diffraction patterns. In the example of the face with the starry background, suppose that we change the focus to infinity, putting the face out of focus. Wave optics would now predict that (in the absence of aberrations), the diffraction pattern for a star would be a central (order 0) fringe surrounded by a ring (first-order fringe) if you used a circular aperture, but a rectangular aperture would give a different pattern (more like a rectangular grid of fringes). In practice, I don't think a camera would ever be diffraction-limited with the aperture all the way open. Diffraction decreases as the aperture gets wider, while ray-optical aberrations increase, so aberration would dominate diffraction under these conditions. Quantum effects are totally irrelevant here. Stan Rogers says: It's difficult to explain without launching into a complete explanation of quantum electrodynamics, but all of the light that reaches the sensor "goes through" all of the lens, at least in a sense, even if we're just talking about a single photon. A photon doesn't take just one path (unless you make the mistake of trying to figure out which path it took), it takes all possible paths. Weird, but true. This is an OK description of why QED suffices for a description of the phenomenon, but it's completely misleading in its implication that QED is necessary. The word "photon" can be replaced with the word "ray" wherever it occurs in this quote, and the description remains valid. | {
"source": [
"https://physics.stackexchange.com/questions/71184",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27112/"
]
} |
71,263 | There is a popular belief that wet skin burns or tans faster. However, I've never heard a believable explanation of why this happens. The best explanation I've heard is that the water droplets on the skin act as a lens, focusing the sunlight onto your skin. I don't see how this would affect an overall burn, because the amount of sunlight reaching the skin is the same (ignoring reflection). Is this 'fact' true, and if so, what causes it? | I don't know of any research to find out if skin sunburns faster when wet, though someone did a comparable experiment to find out if plants can be burnt by sunlight focussed through drops of water after the plants have been watered. You need to be clear what is being measured here. The total amount of sunlight hitting you, and a plant, is unaffected by whether you're wet or not. The question is whether water droplets can focus the sunlight onto intense patches causing small local burns. The answer is that under most circumstances water droplets do not cause burning because unless the contact angle is very high they do not focus the sunlight onto the skin. Burning (of the plants) could happen if the droplets were held above the leaf surface by hairs, or when the water droplets were replaced by glass spheres (with an effective contact angle of 180º). My observation of water droplets on my own skin is that the contact angles are less than 90º, so from the plant experiments these droplets would not cause local burning. The answer to your question is (probably) that wet skin does not burn faster. I would agree with Will that the cooling effect of water on the skin may make you unaware that you're being burnt, and this may lead to the common belief that wet skin accelerates burning. | {
"source": [
"https://physics.stackexchange.com/questions/71263",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/17427/"
]
} |
71,292 | Why does a cork ball float to the side of a glass as illustrated in the following GIF? What is the physical phenomenon behind this observation and why does it happen? | What seems to be happening is that capillary effects in the presence of gravity create a situation in which the cork being maximally decentralized in the glass corresponds to a minimum energy configuration. My guess is that the cork is non-wetting , and therefore surrounded by a water surface that bends down in the proximity of the cork, thereby creating a minute overall increase of water level in the glass. With the cork resting against the edge of the glass, this water level increase gets minimized. If all of this is right, the effect should disappear if the cork is replaced by a floating material that is water wet. | {
"source": [
"https://physics.stackexchange.com/questions/71292",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/26264/"
]
} |
71,503 | The ideal black-body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. Many objects approximate ideal blackbodies and have radiation curves very similar in shape and continuity to that of an ideal black-body (often minus some emission and absorption lines from the atoms in an object, such as radiation curves seen from stars). I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (or perhaps the degree of quantization is so small the radiation curves look continuous). | perhaps the degree of quantization is so small the radiation curves look continuous Yes, this is the reason. The correspondence principle says that quantum mechanics has to become classical in the appropriate limit. One way to obtain an appropriate limit is with large numbers of particles. As you increase the number of particles in a material many-body system, you get more and more ways of putting together combinations of states for your material object. The density of states of the object grows very quickly (roughly exponentially) with the number of particles. Therefore the number of possible transitions between states also grows very rapidly. The number of particles in a tungsten lightbulb filament is something like Avogadro's number. The exponential of Avogadro's number is really, really big. | {
"source": [
"https://physics.stackexchange.com/questions/71503",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/25680/"
]
} |
71,582 | My understanding of orbital mechanics is very limited, but as I understand geostationary satellite, they stay in place by having an orbital speed corresponding to the spot they're orbiting over. So my basic intuition tells me that it's not possible to have a geostationary satellite over the poles, since they're effectively standing still, and to my knowledge, a satellite must keep a certain velocity to keep from getting pulled down by gravity. Am I missing something? | Your understanding is correct. There cannot be a geostationary satellite at the poles, basically because it would have to be at rest, which cannot happen as it would get pulled by the earth's gravity and eventually crash to the surface. In fact, there cannot be a geostationary satellite anywhere else, except above the equator(in an equatorial orbit). This is fairly easy to prove. Imagine that you wanted a satellite directly above the place where you are right now, lets say 500 km away. Now we know that the earth is rotating, so the place 500 km directly above you will also move in a circle.This circle has its center somewhere on the rotational axis of the earth(Not necessarily coinciding with the center of the earth). If you want your satellite to move on that path, It will require a centripetal force continuously acting towards that center. Now if a satellite is purely under the influence of earth's gravity, there is a force acting on it directed towards the earth's center. Now if this force were to act as the centripetal force for the motion we want for our satellite, the orbit's center would have to coincide with the earth's center, leading us to the fact that a geostationary orbit has to necessarily be an equatorial orbit! P.S. I'm sorry I couldn't provide you with a 3D diagram, but Wikipedia has a great diagram for a geostationary orbit: | {
"source": [
"https://physics.stackexchange.com/questions/71582",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27265/"
]
} |
71,611 | In Heaviside-Lorentz units the Maxwell's equations are: $$\nabla \cdot \vec{E} = \rho $$
$$ \nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$
$$ \nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$
$$ \nabla \cdot \vec{B} = 0$$ From EM Lagrangian density: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} - J^\mu A_\mu$$ I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it? | Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations
$$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$ $$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$ are already identically satisfied. To prove them, just use the definition of the electric field $$\vec{E}~:=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t},$$ and the magnetic field $$\vec{B}~:=~\vec{\nabla}\times\vec{A}$$ in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$. The above is more naturally discussed in a manifestly Lorentz-covariant notation .
OP might also find this Phys.SE post interesting. Thus, to repeat, even before starting varying the Maxwell action $S[A]$, the fact that the action $S[A]$ is formulated in terms the gauge $4$-potential $A^{\mu}$ means that the source-free Maxwell equations are identically satisfied. Phrased differently, since the source-free Maxwell equations are manifestly implemented from the very beginning in this approach, varying the Maxwell action $S[A]$ will not affect the status of the source-free Maxwell equations whatsoever. | {
"source": [
"https://physics.stackexchange.com/questions/71611",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21270/"
]
} |
71,620 | If it's really hot inside, but cooler outside; what is the best way to place a single fan to try and cool a room down? I always assumed it would be better pointing inwards (and this thread suggests the same). However; today I had a thought - if the room had a bad smell in it, we would probably expect a fan blowing air out would remove the smell faster than a fan blowing inwards. Is a smell really any different to heat? Would it be more efficient to remove heat by blowing the fan a different way to a smell? | From a purely temperature point of view, not human perceived level of hotness, it is better to point the fan outward. This is because the fan motor will dissipate some heat, and when the air is blown outwards, this heat goes outside. This is all assuming the room has enough ventilation cracks and the like that the pressure inside is still effectively the same as the pressure outside regardless of what the fan is doing. Human-perceived hotness is quite different because humans are a heat source themselves and have a built-in evaporative cooling system. Air flow will help with the cooling process and remove heat from the area around the body. A human sitting in a chair in the room with the fan blowing in will feel cooler than with the fan blowing out due to the higher motion of the air in the room. If the point is to make you in the room feel cooler, blow the air in. The extra power from the fan motor is a minuscule effect in the overall scheme of a normal room in a house and the kind of airflow such a fan would create. Worrying about the fan motor power is really nitpicking, but can be significant for things like cooling chassis of electronics. Another issue is where the air comes from that enters the room if the fan blows outward. If it is coming from other parts of the same house that are also hot, then that may technically be the most efficient for bringing down the temperature in the whole house, but less useful for just the room in question. Added The original question was about blowing air "in" or "out" with a fan. That implied the fan was in a window or such so that inside air would be on one side and outside air on the other. The more the fan is inside the room, the less effective it will be. Just moving air around inside the room does nothing to cool it. In fact, the extra power from the fan actually heats the room, although very slightly. This can still be useful if the point is to make a human feel cooler. However, to actually cool the room, the hot air in the room must be swapped for the cooler air outside. With a single fan you only get to force this in one direction, and the other happens through open windows, doorways, etc. In that sense, the direction of the fan is irrelevant (ignoring the tiny extra power of the fan itself). Cool air will come in, and warm air will go out. It is best to place the fan in a window or the like where there is a direct connection between the inside and outside air. For best effect, this portal should be sealed around the outside of the fan so that air can't just loop around the fan and not contribute to the overall movement. If the fan can't be placed right at the inside/outside interface, then it will rapidly become less useful as it is moved into the room. 20 cm (8 inches) inside from a window is enough to make a difference. In that case, blowing the air out is better. That is because the exhaust air of the fan has is in a tighter stream and therefore faster and stays together for a short distance. It if exists the room within this short distance, then a good fraction of the air moved by the fan is still moved outside the room. Again, this effect diminishes rapidly with distance. 20 cm might still be somewhat effective if the fan has a considerably larger diameter than that. If you can make a duct so that all the air moved by the fan is forced to go outside, then the efficiency increase greatly. However, the longer the duct, the more resistance to air movement it creates, and the less overall air the fan moves. Usefulness goes down due to the fan moving less air, even though all the moved air goes outside. | {
"source": [
"https://physics.stackexchange.com/questions/71620",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27278/"
]
} |
71,661 | You can try it with your own uncooked spaghetti if you want; it almost always breaks into three when you snap it. I am asking for a good physical theory on why this is along with evidence to back it up. Or, a reference to a good study previously done on this would also be satisfactory. The math behind everything would be a great bonus if it exists. My hypothesis is that a strand of dry spaghetti has two main pressure areas being put on it due to how brittle and how high the length to thickness ratio is, and that they snap at almost the same time due to the manufacturing process of the strand causing it to have a very consistent breaking point. Also that vibrations from a first break could influence the causing of another. I'm not sure how this could be tested, though. | The breaking of dry spaghetti was discussed in a 2005 Phys. Rev. Lett. by French physicists Audoly and Neukirch. Bottom line is that elastic (flexural) waves propagating along the spaghetti cause local increases in curvature leading to multiple breaking points: abstract to article . In essence, your assumption "that vibrations from a first break could influence the causing of another" is correct. This work earned both authors the 2006 Ig Nobel prize for physics . | {
"source": [
"https://physics.stackexchange.com/questions/71661",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/26597/"
]
} |
71,823 | It seems to me that there is no such thing as time. There is only movement in the universe and we compare our own movement to a different object to have a sense of time. It can be a clock or a atomic vibration. Does this view of time work within the current framework of phsics? Do physicists have an explanation/proof about time's existence? | It's easy to get mixed up between time and the flow of time , and I think you've done this in your question. Take time first: since 1905 we describe any event as a spacetime point and label it with four co-ordinates ($t$, $x$, $y$, $z$). Saying that time doesn't exist means we can ignore the time co-ordinate and label everything by just it's spatial co-ordinates ($x$, $y$, $z$), which is contradiction with observations. The time co-ordinate obviously exists and be used to distinguish events that happen at the same place but at different times. Now consider the flow of time : actually this is a tough concept, and relativity makes it tougher. We all think we know what we mean by the flow of time because we experience time passing. To take your example of movement, we describe this as the change of position with time, $d\vec{r}/dt$, where we regard time as somehow fundamental. I'm guessing that this is what you're questioning i.e. whether the flow of time is somehow fundamental. I don't think there is a good answer to this. To talk about the flow of time you'd have to ask what it was flowing relative to. In relativity we can define the flow of co-ordinate time relative to proper time, $dt/d\tau$, and indeed you find that this is variable depending on the observer and in some circumstances (e.g. at black hole event horizons) co-ordinate time can stop altogether. But then you'd have to ask whether proper time was flowing. You could argue that proper time is just a way of parameterising a trajectory and doesn't flow in the way we think time flows. At this point I'm kind of stuck for anything further to say. If I interpret your question correctly then you do have a point that just because we observe change of position with time (i.e. movement ) this doesn't necessarily mean time is flowing in the way we think. However I'm not sure this conclusion is terribly useful, and possibly it's just semantics. | {
"source": [
"https://physics.stackexchange.com/questions/71823",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27310/"
]
} |
72,368 | Why do most metals (iron, tin, aluminum, lead, zinc, tungsten, nickel, etc.) appear silver or gray? What makes copper and gold have different colors? What atomic characteristics determine the color? | Why do most metals appear silver in color, with gold being an exception? It is hardly surprising that the answer to this question relies heavily on quantum theory, but most people will be surprised to hear that the full answer brings relativistic considerations into the picture. So we are talking quantum relativistic effects. The quantum bit of the story tells us that the colour of metals such as silver and gold is a direct consequence of the absorption of photons by d electrons. This photon absorption results in d electrons jumping to s orbitals. Typically, and certainly for silver, the 4d→5s transition has a large energy separation requiring ultraviolet photons to enable the transition. Therefore, photons with frequencies in the visible band have insufficient energy to be absorbed. With all visible frequencies reflected, silver has no colour of its own: it's reflective, an appearance we refer to as 'silvery'. Now the relativistic bit. It is important to realize that electrons in the s orbitals have a much higher likelihood of being in the neighborhood of the nucleus. Classically speaking, being close to the nucleus means higher velocities (cf speed of inner planets in solar system with that of the outer planets). For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity). This shifts the light absorption (for gold primarily due to the 5d→6s transition) from the ultraviolet down to the lower frequency blue range. So gold tends to absorb blue light while it reflects the rest of the visible spectrum. This causes the yellowish hue we call 'golden'. Reflectivity as function of wavelength. Purple/blue light corresponds to 400 - 500 nm, the red end of the visible spectrum to about 700 nm. See: the color of gold , relativistic quantum chemistry . | {
"source": [
"https://physics.stackexchange.com/questions/72368",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/23795/"
]
} |
72,641 | Why is there a size limitation on human/animal growth? Assuming the technology exists for man to grow to 200 feet high, it's pretty much a given that the stress on the skeletal structure and joints wouldn't be possible to support the mass or move...but WHY is this? if our current skeletal structures and joints can support our weight as is, wouldn't a much larger skeletal structure do the same assuming it's growing in proportion with the rest of the body? And why wouldn't a giant person be able to move like normal sized humans do? (I'm honestly thinking Ant Man, or even the non-biological sense of mechs/gundams/jaegers)...I'm just having a hard time grasping why if it were possible to grow to gigantic sizes or create giant robots, why it then wouldn't be possible for them to move. | The following fact lies at the heart of this and many similar issues with sizes of things: Not all physical quantities scale with the same power of linear size. Some quantities, like mass, go as the cube of your scaling - double every dimension of an animal, and it will weigh eight times as much. Other quantities only go as the square of the scaling. Examples of this latter category include Muscle strength (a longer muscle can exert no more force than a shorter one of equal cross sectional area), Heart pumping ability (the heart is not solid but rather hollow, so the amount of muscle powering it goes as the surface area), The compression/tension that can be safely transmitted by a bone (material strength is intrinsic and independent of size, so the pressure that can be supported is constant, so the force - cross sectional area times pressure - that can be supported goes as the square of size), and The ability to exchange material and heat the the environment (single cells for example have a hard time growing large because their metabolism goes as the cube of the size, but their ability to transport nutrients across their outer membranes only scales as the area of those membranes), at least to a first approximation. You could also come up with other quantities that scale differently with size. As a result, simply scaling up an organism will undo the balance that has been achieved for that particular size. Its muscles will likely be too weak, its bones will likely break, and it will generate so much internal heat (if it is warm blooded) that the only equilibrium achievable given its comparatively small surface area would be at a high enough temperature to denature many proteins. For a completely non-biological example, consider the fact that airplanes cannot be made arbitrarily large, and in fact different sizes of planes have very different shapes and engineering requirements. The surface area of the wings does not scale the same way as the total mass, and the stresses and pressures the material needs to withstand will not stay constant as you enlarge the plane. | {
"source": [
"https://physics.stackexchange.com/questions/72641",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27641/"
]
} |
73,403 | Final edit : I think I pretty much understand now (touch wood)! But there's one thing I don't get. What's the physical reason for expecting the correlation functions to be independent of the cutoff? I.e. why couldn't we just plump for one "master Lagrangian" at the Planck scale and only do our integration up to that point? Perhaps it has something to do with low energy experiments not being influenced by Planck scale physics. Maybe it's because there isn't any fundamental scale, i.e. that $\Lambda$ must be arbitrary in a QFT approximation, for some reason. I'll award the bounty to anyone who can explain this final conundrum! Cheers! $$***$$ Apologies if this question is too philosophical and vague! I've been thinking about QFTs and continuum mechanics, and reading about their interpretation as effective theories. In these theories we have natural cutoffs at high momentum (small scales). We make the assumption ($\star$) that the large scale physics is decoupled from the small-scale. Therefore we hope that our predictions are independent of the cutoff (after some renormalization if necessary). Why is the assumption ($\star$) so reasonable? I guess it seems observationally correct, which is powerful empirical evidence. But could it not be the case that the small scale physics had ramifications for larger scale observations? In other words, would it be reasonable to expect that the predictions of a TOE might depend on some (Planck scale) cutoff? This question may be completely trivial, or simply ridiculous. Sorry if so! I'm just trying to get a real feel for the landscape. Edit : I'd like to understand this physically from the purely QFT perspective, without resorting to analogy with statistical physics. It might help if I rephrase my question as follows. In the Wilsonian treatment of renormalization we get a flow of Lagrangians as the energy scale $\Lambda$ changes. For a renormalizable theory we assume that there's a bare Lagrangian independent of $\Lambda$ in the limit $\Lambda \to \infty$. We calculate with this quantity, by splitting it into physical terms and counterterms. I think these counterterms come from moving down the group flow, but I'm not quite sure... But why do we care about (and calculate with) the bare Lagrangian , rather than one at some prescribed (high) energy scale (say the Planck scale)? I don't really understand the point of there existing a $\Lambda\to \infty$ limit. | This is a very interesting question which is usually overlooked. First of all, saying that "large scale physics is decoupled from the small-scale" is somewhat misleading, as indeed the renormalization group (RG) [in the Wilsonian sense, the only one I will use] tells us how to relate the small scale to the large scale !
But usually what people mean by that is that if there exists a fixed-point in the RG flow, then some infrared (IR) [large scale] physics is independent of the details at small scale [ultraviolet (UV)], that is it is universal. For instance, the behavior of the correlation functions at long distance is independent of the bare parameters (to fix the setting, say a scalar field with bare parameters $r_\Lambda, g_\Lambda$ for the quadratic and quartic interaction and $\Lambda$ is the (for now) finite UV cut-off). But one should not forget that a lot of physical quantities are non-universal. For example, the critical value of $r_\Lambda$ (at fixed $g_\Lambda$ and $\Lambda$) to be at the critical point is not universal. And this is a physical quantity in condensed-matter/stat-phys, the same way that $\Lambda$ also has a physical meaning. The point of view of the old-school RG (with conterterms and all that) is useful for practical calculations (beyond one-loop), but make everything much less clear.
In the spirit of high-energy physics with a QFT of everything (i.e. not an effective theory), one does not want a cut-off, because it has no meaning, the theory is supposed to work at arbitrary high-energy. This mean that we should send $\Lambda$ to infinity. And here comes another non-trivial question : what do we mean by $\Lambda\to\infty$ ? The perturbative answer to that is : being able to send $\Lambda\to\infty$ order by order in perturbation in $g$. But is it the whole answer to the question ? Not really. When we say that we want $\Lambda\to\infty$, it means that we want to define a QFT, at a non-perturbative level, which is valid at all distance, and we want this QFT to be well-defined, that is defined by a finite number of parameters (say two or three). And in fact, this non-perturbative infinite cut-off limit (that I will call the continuum limit) is much more difficult to take. Indeed, having a theory described in the limit $\Lambda\to\infty$ by a finite number of parameter means that the RG flows in the UV to a fixed point. In the same way, the RG has to flow in the IR to another fixed point in order to be well controlled. This implies that very few QFTs in fact exist in the continuum limit, and that some QFTs which are perturbatively renormalizable ($\Lambda\to\infty$ order by order in perturbation in $g$) are not necessarily well defined in the continuum limit ! For instance, some well known QFTs in dimension four (such as scalar theories or QED) do not exist in the continuum limit ! The reason is that even if these theories are controlled by a fixed point in the IR (at "criticality", which for QED means at least electrons with zero masses), it is not the case in the UV, as the interaction grows with the cut-off. Therefore one has to specify the value an infinite number of coupling constants (even "non-renormalizable") to precisely select one RG trajectory. One of the QFTs which exists in the continuum limit is the scalar theory in dimension less that four (say three). In that case, at criticality, there exists one trajectory which is controlled by a fixed point in the UV (the gaussian fixed point) and in the IR (the Wilson-Fisher fixed point). All (!) the other trajectories are either not well defined in the UV (critical theories but with otherwise arbitrary coupling constants) or in the IR (not a critical theory). One then sees why this $\Lambda\to\infty$ limit is less and less seen as important in the modern approach to (effective) QFTs. Unless one wants to describe the physics at all scale by a QFT, without using a fancy up-to-now-unknown theory at energies above $\Lambda$. Nevertheless, this idea of controlling a QFT both in the IR and the UV is important if you want to prove that General relativity is (non-perturbatively) renormalizable (i.e. can be described at all scales by few parameters) in the asymptotic safety scenario : if there is a non trivial UV fixed point, then there exists a trajectory from this fixed point to the gaussian fixed point (which is, I think, Einstein gravity), and you can take the continuum limit, even though the perturbative $\Lambda\to\infty$ does not exists. Reference : Most of this is inspired by my reading of the very nice introduction to the non-perturbative RG given in arXiv 0702.365 , and especially by the section 2.6 "Perturbative renormalizability, RG flows, continuum limit, asymptotic freedom and all that". | {
"source": [
"https://physics.stackexchange.com/questions/73403",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/10833/"
]
} |
73,564 | My five-year old daughter was asking about astronauts the other day and why they float in space. After me showing her a few bits on the kids section on the NASA web site I started explaining about the planets and how the sun was a ball of gas and you couldn't stand on it. I then had to explain to her what a gas was, using ice, water and steam as examples for the three main states. A few hours later she asked me if the bubbles from the bubble-bath in the bath water were a liquid and if they were why didn't the gravity make them run off her hands like water. That had me stumped. I'm not a scientist but I've been doing my best to try and figure this out and explain it to her but I'm not fully sure, either. Firstly, are bubbles from bubble-bath a liquid or have I made a bad assumption? I thought it might be down to friction but I assume bubble-bath is a detergent and shouldn't encounter much friction. I then thought it might be because the bubbles had a large surface area relative to their weight and would suffer wind resistance as a result. My best guess now is that it's because the bubbles were more viscous than the water. Are any of these right, if not can someone explain it to me in simple terms so I can tell her, please? UPDATE Wow, thanks for all the responses! I'm usually on Stackoverflow and I don't think that in two and half years on there I've asked anything that's generated this level of response.
Thank you to everyone who took the time to answer or comment - I've read all that's been written here. I explained this to my daughter last night saying that the bubble bath liquid was sticky and as a result would stick to things and slow down gravity's pull. I drew a bubble and explained about the liquid stopping the air in the middle getting out and by standing opposite and holding hands both I explained how we were like the liquid of the bubble squeezing the air in the middle which she understood. I expect that if she's asking questions like this now I'll be on here again soon asking something else. I've certainly learnt a lot! Thanks again. | The reason fluids flow off your hand while solids don't, is that fluids can change shape and solids can't. The molecules in a fluid want to stay together, but they don't care about the shape they're in, so gravity will cause them to spread out over your hand and flow off the sides. Solids can't change shape so they just stay on top off your hand, held in place by friction. A bubble is a thin sphere of a water/soap mixture filled with air. The water/soap mixture has surface tension. This means that the molecules are pulling on each other to try and reduce the size of the bubble. But the air inside the bubble has air pressure. If the bubble gets smaller, the air pressure increases, pushing back on this thin layer of water and soap. This will result in a stable situation: the surface tension is pulling inwards, and the air pressure is pushing outwards, resulting in a specific size and shape. If the bubble somehow got smaller the air pressure would restore its size, and if it got bigger the surface tension would. If the bubble is deformed to something other then a sphere, the surface tension and air pressure are no longer regular and equal, and they will keep pulling and pushing until they are again, which, again, makes the bubble a sphere. So in a sense, a bubble is behaving as if it was a solid , because it has a rigid shape and size . The bubble can't spread out over your hand and flow off the sides, because it wants to maintain its shape and size. And the bubble as a whole doesn't move as easily because of adhesion to your hand (the fluid-counterpart of friction). If you blow against the bubble or tilt your hand, the airflow or the gravity will overpower the adhesion, and the bubble as a whole will slide of your hand. It will never spread out and flow off unless you pop it, at which point there is no bubble to speak of any more, but just the water/soap mixture, which is a fluid. In summary, a bubble has a somewhat rigid shape because of the combination of surface tension and air pressure. This means it can't flow, but only move as a whole. Adhesion between the bubble and your hand prevents the bubble from simply sliding off your hand. I'm not great at this, but here's my attempt to phrase it as to be understandable for a child: If something flows, it has to change shape. Fluids flow because they don't care about what shape they are. Solids, like a die, don't flow because they do want to be in a specific shape. A die is always a cube. Because of this, the die can only move as a whole. The die doesn't fall off your hand because there is friction between the die and your hand. Just like a piece of rubber, or a strip of anti-slip, on a table. A bubble is a ball with air inside and a thin layer of water on the outside. Everything is made up of tiny things called 'molecules' (let's not get ahead of ourselves here) . The molecules in a solid hold each other very tight, that's why solid things can't change shape. The molecules in a liquid pull on each other, but they don't hold each other. Because the molecules are pulling on each other, the water in the bubble wants to get smaller. But, the air inside the bubble also has molecules. Air is a gas. The molecules in a gas don't hold each other at all, they just wan't to get as far away from each other as possible. So the molecules in the air inside the bubble want the bubble to get bigger. If the molecules in the air are pushing harder than the molecules in the water are pulling, the bubble gets bigger. If the molecules in the water are pulling harder, then the bubble gets smaller. After a while, the bubble will become exactly so big that the molecules in the air are pushing just as hard as the molecules in the water are pulling. Now if the bubble becomes smaller, the air molecules will push it out again. If the bubble becomes bigger, the water molecules will push it in again. So the bubble can't change shape. You can see this in a balloon (thanks to Bobson) . Take an empty balloon. It is very small because the rubber is pulling the balloon together, and there is no air in the balloon to push it out. Now if you inflate the balloon, more and more air will get inside. So the air will push out harder and harder, making the balloon bigger. If you poke the balloon, you can feel the air pushing against your finger. And if you take your finger away again, the air pushes the balloon back into shape. This is exactly the same as in a bubble. Except the water will 'break' much easier then the rubber in the balloon. So you can't really poke it. So just like the die, the bubble and the balloon want to be in a specific shape. This means the bubble can only move as a whole. The die couldn't slide off your hand because of friction. With the bubble something similar is happening: Hold your hands in a cup and throw some water in. Now open your hands. The water flowed off your hands, but some of the water is still sticking to your hand. This is because the molecules in the water and the molecules in your hand are pulling on each other too. It's called adhesion. Because of this adhesion between the water at the bottom of the bubble and your hand, the bubble can't slide off your hand, just like the die. | {
"source": [
"https://physics.stackexchange.com/questions/73564",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27882/"
]
} |
73,593 | Only because Rep is unitary, so saves positive-definite norm (for possibility density), Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and It is the representation of the global group of relativistic symmetry, yes? | First, note that in physics, we consider unitary representations $U$ of the Poincare group acting on the Hilbert space $\mathcal H$ of the theory because we are interested in a precise formulation of the concept of Poincare transformations acting on the quantum mechanical states of the theory as symmetries (since the laws of physics should be inertial frame-invariant); and by Wigner's theorem, we choose these symmetries to be realized by unitary operators. These observations are related to your #1 and #3 and I think they should be kept conceptually distinct from the notion of a state that represents a single particle state. Second, since such quantum field theories are supposed to allow for the emergence of states of particles, and in particular should account for states in which there is a single elementary particle, we expect that there is some subset $\mathcal H_1$ of the Hilbert space of the theory corresponding to states "containing" a single elementary particle. Given these observations, let's rephrase your question as follows: What properties do we expect that the action of the representation $U$ will have when its domain is restricted to the subspace $\mathcal H_1$? In particular, we would like to justify the following statement The restriction of the unitary representation $U$ acting on $\mathcal H$ to the single-particle subspace $\mathcal H_1$ is an irreducible representation of the Poincare group acting on $\mathcal H_1$. This requires justifying two things: The restriction maps $\mathcal H_1$ into itself. The restriction is irreducible. I think that the justification of the first property is pretty intuitive. If all we are doing is applying a Poincare transformation to the state of the system, namely we are just changing frames, then the number of particles in the state should not change. It would be pretty strange if you were to, for example, boost or rotate from one inertial frame into another and find that there are suddenly more particles in our system. The irredicibility requirement means that the only invariant subspace of the single particle subspace $\mathcal H_1$ is itself and $\{0\}$. The physical intuition here is that since we are considering a subspace of the Hilbert space in which there is a single elementary particle, expect that there is no non-trivial subspace of $\mathcal H_1$ in which vectors of this subspace are simply "rotated" into one another. If there were, then the particle would not be "elementary" in the sense that the non-trivial invariant subspace would represent the states of some "more elementary" particle. When it really comes down to it, however, I'm not sure if there is some more fundamental justification for why the restriction of $U$ to $\mathcal H_1$ is irreducible aside from the decades of experience we've now had with particle physics and quantum field theory. | {
"source": [
"https://physics.stackexchange.com/questions/73593",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
73,654 | Cosmic rays can have energies going into the $10^{20}$ eV domain. Asteroids and meteoroids originating in the solar system are probably limited in their speed because they all started out from the same lump of matter having more or less the same speed, but what about rocks from other galaxies ? Could they reach relativistic speeds in relation to the solar system? Are there astronomical events that could be interpreted as a collision between such an asteroid and a planet or star? Just a mole of hydrogen atoms with that kind of energy would have three times the energy of the Chicxulub impact event, so I hope there's a reason for them not existing! | First, the speed of other galaxies isn't too helpful. For example, the radial velocity of the Andromeda galaxy relatively to us is 300 km/s, i.e. 0.1% of the speed of light only. Moreover, internally, everything in that galaxy moves by pretty much the same speed and is confined to the vicinity of that galaxy which makes us pretty sure that no piece will reach us before Andromeda galaxy will. More importantly, macroscopic systems in outer space aren't moving with the same huge speeds near the speed of light as the cosmic rays essentially because of the second law of thermodynamics. When cosmic rays are accelerated to high speeds, we may treat them statistically and the high energy of these elementary particles may be assigned a high temperature. But the physical systems with many degrees of freedom prefer to evolve to the most likely, high-entropy configurations. That's why the excess energy (e.g. in a supernova) tends to divide between the elementary particles chaotically. In particular, the individual particles' kinetic energy results from velocities that have a random direction. At these huge temperatures (kinetic energies per particle), the atoms are unbound (well above the ionization energy) and macroscopic matter doesn't exist. So the likelihood that a large object will move towards the Earth at a near-luminal velocity is as unlikely as the possibility that the numerous atoms in the large objects are assigned velocities with the exactly equal direction although the directions are being chosen from an isotropic, random distribution. After some time for "thermalization" (interactions between atoms are allowed to change the system with the conservation laws' being the only absolutely constraints), the greater object you consider, the less likely it is that all the atoms in that object will be doing the same thing. This is a form of the second law of thermodynamics. The previous paragraph prevents the creation of "coherent macroscopic cosmic rays", macroscopic objects that would move in the same direction, from the thermal chaos of hot environments such as the supernova. But even if some astrophysical process managed to eject a chunk of matter at these high speeds, the previous paragraph will still guarantee that we won't receive it on Earth. Instead, the individual atom of that speedy object would still have some residual mutual velocities so the object would split into individual atoms and we would observe cosmic rays only once again. I could summarize the situation in this way: to shoot a large object by a huge speed from a very distant celestial body to the Earth requires one to have the same and huge radial velocities of all atoms but almost vanishing relative transverse velocities. The likelihood of that goes to zero exponentially in any thermal environment. If one assumes that the source of the initial speed is not thermal, then one must accept that the projectiles will derive their speed from their broader environments – speeds of astrophysical bodies that already exist – and those are simply of order 0.1 percent of the speed of light as in the Andromeda example or lower. These modest speeds boil down to the inhomogeneities during the structure formation and, ultimately, to inflation, or to the planetary speeds derivable for a star. Whenever you want to locally violate the modesty of the speeds, e.g. by a gravitational collapse, you don't "shoot" any new particles before the collision and when the collision happens, the excess energy of the collision also inevitably creates a high temperature and we're back to the previous paragraph. So near-luminal macroscopic bodies aren't observed. I would even go further and despite my being a believer that life in the Universe is extremely rare, I would say that an object moving by a near-luminal speed would prove the existence of an advanced civilization. I should be a bit careful: the gravitational slingshot is a process that allows the speed to be enhanced even naturally. But even if the source of the speed were a gravitational slingshot and the speed would be really high, like 99.9999% of the speed of light, chances would be high that some intelligence was behind the optimization of the gravitational slingshot because it's extremely unlikely for such an outcome to occur naturally. | {
"source": [
"https://physics.stackexchange.com/questions/73654",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27232/"
]
} |
73,705 | I'm making several assumptions, not sure if any are correct: there is a black hole at the center of a galaxy the black hole is eating the galaxy Eventually the galaxy will be gone, right? Has this been observed? Do we know what happens afterwards? Posting here since astronomy got merged into physics | Assumption #1 is quite correct - there is a very large ("supermassive") black hole in the center of our galaxy. Assumption #2, however, is false . Black holes are no better at drawing in distant objects than any other thing in space with the same mass would be. If you collapsed the Sun into a black hole right now, the Earth's orbit would not change. That is because gravity is gravity - it doesn't matter that we are being pulled by a star or by a black hole; all that matters is the mass of the object doing the pulling. So yes, anything that falls into the black hole in the center of the galaxy is lost. But there isn't terribly much falling in. True, the large number of stars at the center tend to be moving in all sorts of random directions, and some inadvertently get sent on trajectories into the black hole, but most objects in the galaxy are unaffected. In fact, our Solar System is held in its orbit far more by the mass of all the stars near the central black hole than by the black hole itself, as these collectively outweigh the black hole by quite a lot. Part of this misconception I think is due to artists' renditions of accretion disks where all the material inescapably spirals into the black hole in the center. However, the reason this material spirals inward is that gas can transfer angular momentum via collisions. Stuff moving in a nearly circular orbit has too much tangential velocity to just "fall" in. What happens is that material further away in the disk is moving slower, and frictional drag slows down the stuff toward the inner part of the disk, allowing its orbit to decay. Basically, angular momentum (essentially the tendency to have a tangential rather than radial velocity) is transferred outward as matter is dragged inward. With stars, this process basically cannot happen, simply because they are not colliding with one another. Our Sun will continue for many billions of years in a stable orbit around the center of the Milky Way, black hole or no. Stars do scatter gravitationally (a process called dynamic relaxation) so there is an analog to the viscosity of an accretion disc, but it's much smaller and slower. | {
"source": [
"https://physics.stackexchange.com/questions/73705",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28084/"
]
} |
73,867 | I've been studying quantum mechanics and quantum field theory for a few years now and one question continues to bother me. The Schrödinger picture allows for an evolving state, which evolves through a unitary, reversible evolution (Schrödinger’s equation, represented by the Hamiltonian operator) and an irreversible evolution (wave function collapse, represented by a projection operator). The Heisenberg picture holds the states constant and evolves the operators instead. It provides an equivalent representation of the unitary evolution on operators, but I haven't yet seen an equivalent Heisenberg representation of wave function collapse. Is there any accepted explanation for how to represent state collapse in the Heisenberg picture? Thanks! | Well, I think you said the answer yourself when you used the words "projection operator." In the Heisenberg picture the operators get projected down to a subspace at the time of the collapse. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state. Forget about pictures for a second, the physical thing is the full matrix element \begin{equation}
\langle \psi,t_1 | U(t_1,t_2) \mathcal{O}(t_2) U(t_2,t_1) | \psi,t_1 \rangle
\end{equation} Knowledge of the hamiltonian is buried inside of the time evolution operator $U$. The Schrodinger picture amounts to grouping the $U$ with the state so that $|\psi(t)\rangle =U(t,t_*)|\psi(t_*)\rangle $, the Heisenberg picture amounts to grouping the $U$ with the operator so that $\mathcal{O}(t)=U(t,t_*)\mathcal{O}(t_*)U(t_*,t)$. This is clearly an artificial split and nothing can ever depend on your choice of picture: if you express things in terms of the full matrix element the difference between the pictures always amounts to a different way of grouping terms. How do we describe collapse? There is some special time $t_c$, the collapse time, at which something non-unitary happens. We cannot use $U$ to evolve past $t_c$. Or in other words, the relationship
\begin{equation}
U(t_2,t_1)=U(t_2,t_c)U(t_c,t_1)
\end{equation}
is no longer true for $t_2>t_c>t_1$. We need to include a projection operator, as you said in your question:
\begin{equation}
U(t_2,t_1)=U(t_2,t_c)N_c P_c U(t_c,t_1)
\end{equation}
where $P_c$ is the operator that projects us down onto the collapsed subspace, and where $N_c$ is a normalization factor so that the state is correctly normalized after collapse. The projection operator will by hermitian and satisfies $P_c^2=P_c$, although the full operator that is being applied at $t_c$, namely the combination $N_c P_c$, is not a projection operator. So let's say we want to evaluate the physical matrix element, we have to include this projection operator \begin{equation}
\langle \psi,t_1 | U(t_1,t_c) N_c^* P_c U(t_c,t_2) \mathcal{O}(t_2) U(t_2,t_c) N_c P_c U(t_c,t_1) | \psi,t_1 \rangle
\end{equation} So again we have a choice of how we group things. We could group things in a Schrodinger way so that \begin{array}
\ |\psi(t_c+\epsilon)\rangle &=& N_c U(t_c+\epsilon,t_c)P_c U(t_c,t_1)|\psi,t_1\rangle \\
& =& N_c P_c |\psi,t_c\rangle + O(\epsilon)
\end{array} This is the 'state collapse.' At $t_c$ the state changes so that it is projected down onto a subspace. Or, we could group things in a Heisenberg way, so that \begin{array}
\ \mathcal{O}(t_c+\epsilon) &=& U(t_c+\epsilon,t_c) N_c^* P_c U(t_c,t_1) \mathcal{O(t_1)} U(t_1,t_c)N_c P_c U(t_c,t_c+\epsilon)\\
&=& |N_c|^2 P_c \mathcal{O}(t_c) P_c + O(\epsilon)
\end{array} This is "the operator being projected onto a subspace." The state is the same, but the operator now includes a projection piece that cancels out the part of the state that is no longer physical. EDIT # 1: I previously said $U(t_2,t_1)=U(t_2,t_c)P_c U(t_c,t_1)$, which is incorrect. The basic point still stands but the math was technically wrong. Whoops! I was right the first time. Thanks to Bruce Connor for making me rethink through this point. I was confused because I thought wanted the transformation rule $P_c U P_c$, which is how you would project the time evolution operator to the collapsed subspace. But that is not what we want here: the time evolution operator is special. The point is that you project down to the subspace (say a position eigenstate) at $t_c$, then you evolve normally from there. In particular you are allowed to evolve out of the subspace. For example, after we observe a particle at position $x$ the particle is allowed to evolve a probability to be at $x'$. You don't want to force the evolution to stay in the subspace, that's what the second $P_c$ would have done. EDIT # 2: Sorry for all the edits, this is a little more subtle to get exactly right than I originally thought. You aren't just projecting the state down to a subspace, you are projecting the state and then rescaling it so that it has the correct normalization. | {
"source": [
"https://physics.stackexchange.com/questions/73867",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28137/"
]
} |
73,959 | I'm just starting to learn physics and I have a question (that is probably stupid.) I learned that energy levels that the bound electron can have are discrete. I also learned that when an electron transitions from one level to another, a photon with a specific wavelength (energy) is created and released into the wild. My question is this: are there holes in the electromagnetic spectrum, values of frequency that the created photon can't possibly have? | No, there aren't any holes like that in the EM spectrum. There are other ways of creating photons than by having electrons bound in atoms transition from one level to another. (For example, you can create pretty much any frequency of photon you want by accelerating a free electron.) | {
"source": [
"https://physics.stackexchange.com/questions/73959",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28160/"
]
} |
74,569 | As a planet moves through the solar system, a bow shock is formed as the solar wind is decelerated by the magnetic field of the planet. Presumably the creation of this shock wave would cause drag on the planet, certainly in the direction of orbit but possibly rotation as well. Is there an estimate for the amount of drag on the Earth as it orbits the Sun? Based on the drag, how long would it take before the orbital velocity slows to the point that we spiral slowly into the Sun? Would any planets fall into the Sun prior to the Sun expanding into a Red Giant, gobbling them up? | This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$ . You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$ . Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$ . Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$ . The time it would take for this to make a significant dint the the Earth's orbital velocity ( $30\ \mathrm{km/s}$ ) is of the order of $10^{15}\ \mathrm{yr}$ . I think we're safe. For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $ , which for Mercury is about 0.1. So the end result is not much different for Mercury. This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in: Note that the number of digits displayed in the final column is ludicrous. :) | {
"source": [
"https://physics.stackexchange.com/questions/74569",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/6634/"
]
} |
74,570 | Trivial thought ... Materials may be broadly superconductive, diamagnetic, paramagnetic, ferromagnetic. An object is magnetized by repetitive motion of a magnetic field across it's surface Say a field of strength 1T were to be moved across a steel cylinder. Could the field created in the cylinder be greater than 1T? What if the same field were to be applied to a more strongly ferromagnetic material? | This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$ . You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$ . Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$ . Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$ . The time it would take for this to make a significant dint the the Earth's orbital velocity ( $30\ \mathrm{km/s}$ ) is of the order of $10^{15}\ \mathrm{yr}$ . I think we're safe. For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $ , which for Mercury is about 0.1. So the end result is not much different for Mercury. This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in: Note that the number of digits displayed in the final column is ludicrous. :) | {
"source": [
"https://physics.stackexchange.com/questions/74570",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/5265/"
]
} |
74,578 | Why does a large drive sprocket require more power to accelerate at the same rate as a smaller drive sprocket if the overall gear ratio is the same (with all other components such as the chain the same between systems)? As an example, we could compare 11/33 to 13/39 (same 1/3 ratio). Why does the 13 tooth sprocket need more power to accelerate at the same rate as the 11 tooth setup? Can we agree that the top sprocket will need more torque to accelerate an atv at the same rate as the same atv with the bottom sprocket? And assuming the overall gear ratio is the same. The moment of inertia differences of the sprockets are insignificant. Assume the moments of inertia are equal. | This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$ . You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$ . Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$ . Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$ . The time it would take for this to make a significant dint the the Earth's orbital velocity ( $30\ \mathrm{km/s}$ ) is of the order of $10^{15}\ \mathrm{yr}$ . I think we're safe. For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $ , which for Mercury is about 0.1. So the end result is not much different for Mercury. This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in: Note that the number of digits displayed in the final column is ludicrous. :) | {
"source": [
"https://physics.stackexchange.com/questions/74578",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/26382/"
]
} |
74,593 | I'm trying to get a sense of how much energy a $27$ horsepower engine outputs. $$27\ \mathrm{hp} = 20133\ \mathrm{W}$$ Potential energy can be calculated as $$E = mgh$$ where $g \approx9.8\ \mathrm{m\ s^{-2}}$ on earth. $$\frac{20133\ \mathrm{W}}{9.8\ \mathrm{m\ s}^{-2}} \approx 3000\ \mathrm{kg\ s^{-1}}$$ So for example if you use a $27\ \mathrm{hp}$ engine to pump hydraulic fluid. Would it output as much fluid as a $3000\ \mathrm{kg}$ stone block pushing down a piston at $1\ \mathrm{m\ s^{-1}}$ ? | This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$ . You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$ . Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$ . Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$ . The time it would take for this to make a significant dint the the Earth's orbital velocity ( $30\ \mathrm{km/s}$ ) is of the order of $10^{15}\ \mathrm{yr}$ . I think we're safe. For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $ , which for Mercury is about 0.1. So the end result is not much different for Mercury. This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in: Note that the number of digits displayed in the final column is ludicrous. :) | {
"source": [
"https://physics.stackexchange.com/questions/74593",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/7953/"
]
} |
74,625 | This popular question about "whether an AC circuit with one end grounded to Earth and the other end grounded to Mars would work (ignoring resistance/inductance of the wire) " was recently asked on the Electronics SE. (Picture edited from the one in the above link) Though I respect the AC/DC experts there, I think (with the exception of the top answer) they are all wrong. My issue is that they all assume that AC requires a complete circuit in order to function. However, my understanding is that a complete circuit is necessary for DC, but not AC. My intuitive understanding is that AC is similar to two gas-filled rooms with a pump between them - the pump couldn't indefinitely pump gas from one room to another without a complete circuit (DC) , but it could pump the gas back and forth indefinitely (AC) . In the latter case, not having a complete circuit just offers more resistance to the pump (with smaller rooms causing a larger resistance) . Is my understanding correct - can AC circuits really function without a complete loop? More importantly, what are the equations that govern this ? If larger isolated conductors really offer less AC-resistance than smaller AC conductors, how is this resistance computed/quantified? Would its "cause" be considered inductance, or something else? | You're actually hitting on a very famous concept here that revolutionized physics!! Your understanding is almost wholly correct and your analogy is a good one - excellent reasoning - the only thing missing is radiation from the system. This latter lack is mostly irrelevant for the level of question you have been thinking about: but I'll address that below. So mostly you have answered your own question. I'll break this answer into three parts by: Giving an overview of the fundamental physics, which is all to do with generalized current flows that comprise a conduction current component (what most of the answers you have seen on Electronics SE have centred on) as well as the displacement current . This latter component hasn't been discussed yet, and it is all important; Presenting a "canonical" system to "play with" and a discussion of the solution of its general equations, which, in the general case must be solved numerically; Discussing the general solution in approximate cases: these will allow you to see how and to what degree a "circuit" description is applicable. Here I can answer your questions about what AC resistance (more correctly, impedance) means in this case. Theoretical Overview Your thinking can be made more precise by saying that: AC circuits must always be closed, either by conductive current paths OR by displacement current paths . Conductive current paths are represented by the flow lines of the conduction term $\mathbf{J}$ (electric current density vector) and displacement current paths are the flow lines of the displacement current term $\partial_t \mathbf{D}$ (here $\mathbf{D}$ is the electric displacement vector) in Ampère's law : $$\nabla\wedge\mathbf{H} = \mathbf{J} + \partial_t \mathbf{D}$$ and the concept of "closed" is defined by the fulfilling of all fields by the electric charge continuity equation : $$0 = \nabla\cdot\mathbf{J} + \partial_t \rho$$ where $\rho$ is the electrical charge density: the Ampère and Gauss electric laws imply the continuity equation (take the divergence of each side of Ampère's law and then apply Gauss's law $\nabla.\mathbf{D} = \rho$). Look carefully at the derivation of the continuity equation because this is precisely the mathematical encoding of your idea "similar to two gas-filled rooms with a pump between them". The above just says the net flux $(\nabla \cdot \mathrm{J}) \, \delta x, \delta y\, \delta z \times \delta t$ out of the small volume $\, \delta x, \delta y\, \delta z$ in time $\delta t$ is the amount $\delta\, \rho \, \delta x, \delta y\, \delta z$ that the charge in that little volume falls by in time $\delta t$. " What goes in either must come out or stays in: nothing goes missing " - it's that simple. This is not just like your idea: if you think about it carefully, it IS your idea, so you should have confidence in the powerful and simple principles you are reasoning by. It works for charges, masses in fluids and all kinds of continuum mechanics problems. So you're dead right in your question: there is no need for a closed conduction path, but instead we have the more general concept of the total current - conduction and displacement - contributing to the continuity equation rather than only the conduction current. Where the conduction current leaves off at the plates of a capacitor, the displacement current "takes over" to make sure that the continuity equation stays fulfilled. This generalized fulfilling of the continuity law through the postulating of the displacement current term was James Clerk Maxwell's stupendous achievement. To see what he did will help strengthen your understanding. I have drawn the "planet capacitor" system (not to scale!) above and some rough field lines of both $\mathbf{J}$ and the displacement current $\partial_t \mathbf{D}$. Exactly as you are thinking with your pump analogy, the field lines of $\mathbf{J}$ end at the planet surfaces and charge alternately gathers there or is drained from there repeatedly with each AC cycle. The problem that Maxwell dealt with was that Ampère's law $\nabla \wedge\mathbf{H} = \mathbf{J}$ (this is the before-Maxwell form) had up until then been inconsistent. We can see this three ways: If we take the divergence of both sides of the before-Maxwell Ampère law we get $\nabla\cdot\mathbf{J} = 0$ (the divergence of a curl for a twice-differentiable vector field is nought) and this gainsays what happens in the figure above: a divergenceless vector field must have closed flow lines, whereas here they end at the planet's surfaces. Suppose we think of surface $\Sigma_1$ bounded by its boundary path $\Gamma_1 = \partial \Sigma_1$ (if you've not seen this notation before, $\partial$ here stands for "boundary of"). There most certainly is a nonzero flux of $\mathbf{J}$ through $\Sigma_1$; suppose we work it out as $\oint_{\Gamma_1} \mathbf{H} \cdot \mathrm{d}\mathbf{r}$; this is equal to the flux of $\mathbf{J}$ through surface $\Sigma_1$ by Stoke's theorem . (If you've not already done so, you should learn the equivalence of the differential and surface integral forms of the Ampère law, afforded by Stokes's theorem. Now, we continuously deform surface $\Sigma_1$ into $\Sigma_1^\prime$ and we must get the same answer if our definition of the flux of $\mathbf{J}$ through a loop is going to be meaningful - if it depends on the surface bounded by the loop, we haven't got a unique definition. But of course the flux through $\Sigma_1^\prime$ is nought , so we have an inconsistency in the before-Maxwell Ampère law. This inconsistency is actually the same one as in point 1, aside from being stated in an integral rather than differential form; Experimentally if we consider the loop $\Gamma_2 = \partial \Sigma_2$ bounding the surface $\Sigma_2$, there is most certainly found to be a nonzero magnetic field circulating around the loop, even though the flux of $\mathbf{J}$ through $\Sigma_2$ is clearly nought. To clear all these inconsistencies up , there clearly needs to be another term added to the right hand side of the before-Maxwell Ampère law whose divergence is equal to $\partial_t \rho$: then the divergence of the new law will give the continuity equation. Now, by Gauss's law for electricity $\nabla \cdot \mathbf{D} = \rho$ we see that $\partial_t \mathbf{D}$ most certainly is a field whose divergence is equal to $\partial_t \rho$ and so it will fix all these problems up (aside from possibly point 3 above: we have to do the experiment first). So that's just what Maxwell did: he defined the "displacement current" $\partial_t \mathbf{D}$ and added it to the right hand side of Ampère's law. Moreover, the now displacement-current-kitted Maxwell equations foretold electromagnetic waves, and the rest is history! Now, it's important to understand that $\partial_t \mathbf{D}$ is only ONE possible term that will do the job, because the divergence operator is many to one. Indeed we can add any term of the form $\nabla \wedge \tilde{\mathbf{H}}$ for any twice differentiable vector field to $\partial_t \mathbf{D}$ to get a field $\partial_t \mathbf{D} + \nabla \wedge \tilde{\mathbf{H}}$ that will do the job of clearing up inconsistencies 1 and 2 above just as well as $\partial_t \mathbf{D}$ does: again, recall that the divergence wipes out the curl. So the displacement current was a hunch on Maxwell's part: he postulated that there was no further mysterious $\tilde{\mathbf{H}}$ field. The ultimate test is the experimental one: so, with reference to inconsistency 3, if we work out the flux of $\partial_t\mathbf{D}$ through $\Sigma_2$ it is indeed found experimentally to be equal to $\oint_{\Gamma_2} \mathbf{H} \cdot \mathrm{d} \mathbf{r}$, not $\oint_{\Gamma_2} (\mathbf{H} - \tilde{\mathbf{H}}) \cdot \mathrm{d} \mathbf{r}$ for some other field $\tilde{\mathbf{H}}$. Wherever there is an AC electric field in space around an AC circuit, there are source charges repeatedly bunching up and gathering and then draining somewhere in the circuit and the displacement current measures this "compressible" part of the conduction current flow that leads to these oscillating excess charge densities. When people speak of "parasitic" or "stray" capacitance degrading a circuit's performance, they're really speaking about unforeseen or unavoidable displacement current paths ad indeed they are often using fancy words to hide the following real meaning: "there's an oscillating electric field outside this circuit, it's complicated and we can't model all of the circuit's in all of its EM field complexity so we don't really know how to hinder the displacement current that's degrading the performance". Before I go on: your comment that LC circuits are closed owing to a capacitor's closing the circuit is a little bit inconsistent with your description of the Earth and Mars plugged into the wall. They are really very alike: you can "continuously deform" one into the other: just bring the two balls at the end of your wire together and shrink the loop down to less than wavelength lengths. The "closed" AC circuit is just Earth and Mars flattened out into disks and brought near to one another so that the electric field between them is essentially "electrostatic" - i.e. the magnetic field induced by the displacement current becomes negligible. Yes the capacitor is a circuit "closing" in the sense that it conducts a displacement current $\partial_t \mathbf{D}$ in the space between its plates but it is not a conductive "closing". The Canonical System and its General Mathematical Description I suggest that canonical system you need is the loaded electric dipole antenna : For simplicity you have a system that is symmetric about the $x-z$ plane, so here you have two "small" (i.e. with dimensions much smaller than a wavelength at frequency in question) conducting balls as shown in my drawing at $\pm\mathbf{r}_0 = (0, \pm y_0, 0)$ linked to a "small" (in the same sense) AC source somewhere on the $x$-axis through conductors that bear a current density $\mathbf{J}(x, y, z)$ to "fill" and "drain" the balls of charge, as in your own conception of the system. Let $\Gamma$ stand for the thread like volume of the conductor linking the source and top ball in the picture, $\Gamma^\prime$ for its mirror image in the $x-z$ plane. Let us represent all assumedly sinusoidally varying with time quantities by phasors (i.e. the positive frequency part of each quantity) so that the charge stored in the top ball as a function of time (i.e. as it is repeatedly filled and drained of charge) is $+Q\,e^{-i\,\omega\,t}$ and that stored in the lower ball $-Q\,e^{-i\,\omega\,t}$. Now we can write down the full electrodynamic solution of Maxwell's equations for this system; that for the electric potential (in Lorenz gauge ) is the retarded wave arising from the free charge in the system, i.e. the first component to the electrical potential arises from the charge stored in the balls: $$\phi_B(\mathbf{r}) = \frac{Q}{4\,\pi\,\epsilon_0} \left(\frac{e^{i\,k\,|\mathbf{r} - \mathbf{r}_0|}}{|\mathbf{r} - \mathbf{r}_0|}-\frac{e^{i\,k\,|\mathbf{r} + \mathbf{r}_0|}}{|\mathbf{r} + \mathbf{r}_0|}\right)\tag{1}$$ and those for the magnetic vector potential: $$A_x(\mathbf{r}) = \frac{\mu_0}{4\,\pi} \int_\Gamma J_x(\tilde{\mathbf{r}}) \left(\frac{e^{i\,k\,|\mathbf{r} - \tilde{\mathbf{r}}|}}{|\mathbf{r} - \tilde{\mathbf{r}}|}-\frac{e^{i\,k\,|\mathbf{r} + \tilde{\mathbf{r}}|}}{|\mathbf{r} + \tilde{\mathbf{r}}|}\right) \,\mathrm{d}^3 \tilde{r}\tag{2}$$ $$A_y(\mathbf{r}) = \frac{\mu_0}{4\,\pi} \int_\Gamma J_y(\tilde{\mathbf{r}}) \left(\frac{e^{i\,k\,|\mathbf{r} - \tilde{\mathbf{r}}|}}{|\mathbf{r} - \tilde{\mathbf{r}}|}+\frac{e^{i\,k\,|\mathbf{r} + \tilde{\mathbf{r}}|}}{|\mathbf{r} + \tilde{\mathbf{r}}|}\right) \,\mathrm{d}^3 \tilde{r}\tag{3}$$ $$A_z = 0\tag{4}$$ where, of course, $\omega$ is the angular frequency in question and $k=\frac{2\pi}{\lambda} = \frac{c}{\omega}$ is the wavenumber for the wavelength $\lambda$ at that frequency. Note: Here we have cancelled the $e^{-i\,\omega\,t}$ multiplying phase function from all quantites: we simply put it back as a multiplier for all quantities at the end of the calculation and take the real part to find the actual, real-valued field variation at ny point with position vector $\mathbf{r}$; In the phasor notation, time derivatives get replaced by multiplications by $-i\omega$; We only have to integrate over the volume $\Gamma$ to get the vector magnetic potential, the equations above automatically take care of the contributions from the corresponding, mirror image points on $\Gamma^\prime$; Once we are done accounting for all charge and currents, we infer the electric and magnetic fields from the above equations (and that for $\phi_C$ defined further dow) as: $$\mathbf{E} = - \nabla (\phi_B + \phi_C) + i\omega\mathbf{A}\tag{5}$$
$$\mathbf{B} = \nabla \wedge \mathbf{A}\tag{6}$$ We must also write down conditions describing the electrical connexion between the conductor and the balls: $$\int_S \mathbf{J} \cdot \hat{\mathbf{n}}\,\mathrm{d}S = -i\,\omega\,Q(t)\tag{7}$$
$$\int_{S^\prime} \mathbf{J} \cdot \hat{\mathbf{n}}\,\mathrm{d}S = +i\,\omega\,Q(t)\tag{8}$$ where the surface integrals are done over the end faces $S$ and $S^\prime$ of the conductors where they feed into the balls. Now I'd like to say that's the whole story, but here is where it gets rather complicated. We don't actually know the current distribution $\mathbf{J}$. We can infer it analytically in simple cases as I do below, but it in general there is a complicated feedback loop from the electromagnetic field calculated from $(5)$ and $(6)$ back onto the magnetic potential equations $(2)$ and $(3)$. The electric field in the conductors must be consistent with the condition: $$\mathbf{J} = \sigma \mathbf{E}\tag{9}$$ where $\sigma$ is the conductor's conductivity. Moreover, in general there are propagation delays, so the current no longer flows like an incompressible flow of fluid in a pipe; unmatched charge actually gathers at different points along the conductor in accordance with the continuity equation: its as though there were little charge storing balls along the conductor's whole length: $$i\,\omega\,\rho(x, y, 0) = \nabla\cdot \mathbf{J}\tag{10}$$ and this excess charge arising from the "bunched up" currents adds yet another component to the electric potential!: $$\phi_C(\mathbf{r}) = \frac{-i}{4\,\pi\,\omega\,\epsilon_0} \int_\Gamma\nabla\cdot \mathbf{J}(\tilde{\mathbf{r}})\left(\frac{e^{i\,k\,|\mathbf{r} - \tilde{\mathbf{r}}|}}{|\mathbf{r} - \tilde{\mathbf{r}}|}-\frac{e^{i\,k\,|\mathbf{r} + \tilde{\mathbf{r}}|}}{|\mathbf{r} + \tilde{\mathbf{r}}|}\right)\mathrm{d}^3 \tilde{r}\tag{11}$$ So you're stuck with a complicated problem that must in general be solved numerically. The following procedure works in principle: We specify the problem in terms of the sinusoidally varying charge on the balls $Q e^{-i\,\omega\,t}$; We assume a constant current density through the wire, inferring the right value from equations (7)and (8); Calculate the electric and magnetic fields from this assumed current using equations (1) through (6); Calculate the corrected to the current density in the wire through equation (9) Calcuate the corrected charge density on the wire through equation (10); Replace the constant current density in step 2 with the corrected current density and charge density just calculated then repeat steps 2 through 5. We simply iterate around the loop comprising steps 2 through 6 until the current densities are consistent with Maxwell’s equations. This procedure actually works numerically. So you could in principle study the whole “continuous deformations” of the system all the way from one where we have a small, much smaller than a wavelength conducting ring with the two balls near together, corresponding to the electrostatic circuit with an electrostatic capacitor (comprising the two balls) to the full Mars and Earth plugged in system! Approximations to the General Description and a "Circuit" Description of the System For intuition, let’s look at a few limiting cases. The short electric dipole (Hertzian dipole). The two balls form a very "unclosed" (conductively) circuit and they are on the ends of short conductors (much much shorter than $\lambda$). This problem has an exact solution: the only important term in the general analysis above is the electric potential $\phi_B$ arising from the charges on the balls as the AC source repeatedly fills them up with charge and drains them. The conductors are so short that the integrals in equations (2), (3) and (4) is negligible and the magnetic potential is negligible everywhere. The electric field is simply: $$\mathbf{E}(\mathbf{r}) = - \nabla \phi_B(\mathbf{r})\tag{11}$$ where $\phi_B(\mathbf{r})$ is given by equation (1). Now $-i\,\omega\,\epsilon_0\,\mathbf{E}(\mathbf{r})$ is the displacement current that "completes the circuit" through freespace. You will also find that this expression reduces to the electrostatic analysis of a capacitor comprising two balls near to one another for low frequencies, and, because the whole thing is very small compared to the wavelength, the radiated power is very small. But it is not nought, and if you work out the potential difference between the two balls, you will find that it is not exactly 90 degrees out of phase with the current flowing from the source, so there is a capacitor plus a small resistance - the radiation resistance. Equation (11) is worked out on the Wikipedia page for the dipole antenna , where there the concept of the charge as a function of time is replaced by current in the conductors, so that we replace $I_0$ in the Wiki page by $-i\,\omega\,Q$ (the time derivative of the charge on the balls). The second approximate model is the same as the above, but we take account of the magnetic vector potential from the current flowing in the wires. This analysis is similar - it holds for a short conductor that is a bit longer than the first case, but still very short compared to a wavelength. This model is qualitatively very like the first: almost wholly describing an electrostatic capacitor with a tiny radiation resistance term and a small amount of radiated power. Now let's look at the system's "circuit description". Let's first suppose that our frequency is very low - say well less than $10^{-3}\mathrm{Hz}$ - and the wavelength is therefore much more than the distance from Earth to Mars. Then the first approximation given by equation (11) is applicable here. The system is a giant electrostatic circuit and Earth and Mars form the "plates" of a capacitor. Suppose for simplicity we replace them by conducting balls. With Mars and Earth well separated, the potential difference between them is approximately: $$\Delta V = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{r_e} + \frac{1}{r_m}\right)\tag{12}$$ where $r_e$ and $r_m$ are Earth's radius and Mars's radius, respectively. So the capacitance of the two planet system is: $$C_{e-m} = 4\pi\epsilon_0\frac{r_e\,r_m}{r_e + r_m}\tag{13}$$ which I reckon to be about a quarter of one millifarad ($r_e = 6371\mathrm{km}$, $r_m = 3390\mathrm{km}$). So at $10^{-4}\mathrm{Hz}$ and a one volt peak to peak source, your current is going to be about 0.15 microamps peak to peak. This is the electrostatic current. There is another component of the current which I discuss below. Notice how the capacitance, at least in the large separation limit, does NOT depend on the separation between the planets. The separation does, however, limit the upper frequency that such a simple model will be applicable to. At these frequencies, of course, the next level of sophistication is the use of equation (11) to calculate the radiation resistance. Using the expression for the radiation resistance on the Wiki page : $$R_{rad} = \frac{2\pi}{3}\sqrt{\frac{\mu_0}{\epsilon_0}} \left(\frac{\Delta L}{\lambda}\right)^2\tag{14}$$ where $\Delta L$ is the Earth-Mars distance (say $2.25\times10^8\mathrm{km}$) we get about $4\Omega$ radiation resistance ($\lambda = 3\times10^9\mathrm{km}$ at $10^{-4}\mathrm{Hz}$). So, for a one volt peak to peak source, we're going to be radiating about 63mW. So, from a circuit standpoint, at $10^{-4}\mathrm{Hz}$, the system looks roughly like that below: You will need very thick conductors to ensure that the resistance of the conduction path is small comparer to the radiation resistance given that the conductors have to reach all the way from Earth to Mars. At higher frequencies, where the Earth-Mars distance is many wavelengths, we in general have very complicated antenna behavior and it is hard to think in circuit terms: the full analysis of described above applies. However, there is one configuration shown below which can be analysed roughly in circuit terms, and this is shown below: The source is place a long way from both planets and the conductors form a parallel pair such that the distance between the wires increases slowly with axial distance, so that, locally, the distance between them can be treated as constant. It is easiest conceptually if the conductors are thin walled pipes. The wires now behave as TEM (transverse electromagnetic) waveguides (they can also support higher order modes, but TEM modes make the system most like a distributed circuit and will be present alone in the steady state, harmonic case. TEM (as well as higher order modes) propagate when the system's cross section is strictly translationally invariant (i.e. does not vary along the axial direction $z$) - hence this is only an approximation in this case. The electromagnetic fields on the wires for TEM modes have the same form as the electrostatic / magnetostatic fields calculated for two dimensional problems, but multiplied by functions of $z$ and time that give the total field wave behavior. To understand what this means, electro- / magentostatic fields are gradients of scalar potentials i.e. $\mathbf{E} = \nabla_\perp \psi_E = \partial_x \psi_E \hat{\mathbf{x}} + \partial_y \psi_E \hat{\mathbf{y}} =\mathbf{E}_\perp$, $\mathbf{H} = \nabla_\perp \psi_H= \partial_x \psi_H \hat{\mathbf{x}} + \partial_y \psi_H \hat{\mathbf{y}} =\mathbf{H}_\perp$: to see that TEM modes exist, we substitute fields of the form $\mathbf{E}_\perp E_z(z, t)$ and $\mathbf{H}_\perp H_z(z, t)$ into Faraday's and Ampère's laws, thus proving that they do fulfill Maxwell's equations as long as: $$\hat{\mathbf{z}}\wedge \mathbf{E}_\perp = \mathbf{H}_\perp\tag{15}$$
$$\partial_z E_z = -\mu_0 \partial_t H_z\tag{16}$$
$$\partial_z H_z = -\epsilon_0 E_z\tag{17}$$ so that both $E_z$ and $H_z$ fulfill the wave equation $\partial_t^2 E_z = c^2 \partial_z^2 E_z$ (where $c$ is the freespace lightspeed and $c^2 \epsilon_0 \mu_0 = 1$), so that their solutions are dispersionless waves of the general form: $$E_z(z, t) = f_+(z - c\,t) + f_-(z + c\,t)\tag{18}$$
$$H_z(z, t) = \sqrt{\frac{\epsilon_0}{\mu_0}}\left(f_+(z - c\,t) - f_-(z + c\,t)\right)\tag{19}$$ Here $f_+$ and $f_-$ are arbitrary twice differentiable functions. Equation (15) means $\partial_x \psi_E = \partial_y \psi_H$ and $\partial_y \psi_E = -\partial_x \psi_H$, i.e. the Cauchy Riemann relationships and so there is a really neat and compact way of doing TEM waveguide analysis wherein we can define a complex potential $\Psi(\zeta) = \Psi(x + i\,y) = \psi_E(x, y) + i\,\psi_H(x, y)$ which is a holomorphic function of the complex variable $\zeta = x + i\,y$ and the electric and magnetic vector fields can be interpreted as the complex numbers: $$\mathbf{E}_\perp = \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^*\tag{20}$$
$$\mathbf{H}_\perp = -i \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^*\tag{21}$$ so that the whole field variation as a function of $\zeta$ (encoding the transverse co-ordinate as a complex number) $z$ (axial position) and $t$ (time) is: $$\mathbf{E}(\zeta,\,z,\,t)= \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^* \left(f_+(z - c\,t) + f_-(z + c\,t)\right)\tag{22}$$
$$\mathbf{H}(\zeta,\,z,\,t) = -i \sqrt{\frac{\epsilon_0}{\mu_0}} \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^*\left(f_+(z - c\,t) - f_-(z + c\,t)\right)\tag{23}$$ In our parallel hollow conductor example, the complex potential can be shown to be: $$\Psi(\zeta) = \frac{i\,q}{2\,\pi\,\epsilon_0} \log\left(\frac{\zeta + a}{\zeta - a}\right)\tag{24}$$ where $q$ is the peak charge per unit length on either wire at the cross section in question (one bears charge $+q$ per unit length, the other $-q$) and where the branch points of the potential at $\zeta = \pm a$ are sited such that: $$a = \frac{1}{2}\sqrt{w^2 - d_c^2}\tag{25}$$ where $w$ is the distance between the centres of the conductors and $d_c$ is the diameter of the conductors and the conductor surfaces themselves are the contours $\mathrm{Re}(\Psi) = \pm V$ where the electric potentials at the conductor surfaces in volts are: $$V = \frac{q}{\pi\,\epsilon_0} \operatorname{arcosh}\left(\frac{w}{d_c}\right)\tag{26}$$ Thus the capacitance and inductance per unit length (found by calculating the flux of $\mathbf{D}$ and $\mathbf{B}$ through the vertical and horizontal axes respectively in the picture below) of this system are: $$C(z) = \frac{\pi\,\epsilon_0}{\operatorname{arcosh}\left(\frac{w(z)}{d_c}\right)}\tag{27}$$
$$L(z) = \frac{\mu_0}{\pi}\operatorname{arcosh}\left(\frac{w(z)}{d_c}\right)\tag{28}$$ In the notation of (18) and (19), the charge per unit length $q$ and the current $I$ implied by the continuity equation is: $$q(z, t) = q_+(z - c\,t) + q_-(z + c\,t)\tag{29}$$
$$I(z, t) = c\,q_+(z - c\,t) - c\,q_-(z + c\,t)\tag{30}$$ and some details of the field are shown below: Further details of the theory of TEM transmission lines can be found on the Wikipedia page for transmission lines . Given the TEM transmission lines, the whole system, from a circuit standpoint, can be approximated like the drawing below: so that the system is "discretized" so that each each length $\Delta z \ll \lambda$ of is represented by the lumped inductance $L(z) \Delta z$ (see (27)), the capacitance $C(z) \Delta z$ (see (28)), any ohmic resistance of the conductor $R \Delta Z$, where $R$ is the resistance per unit length. The discretized $LCR$ circuits are concatenated in the ladder circuit above. The end of the ladder is loaded with the Earth-Mars capacitance of (13) as well as the frequency dependent radiation resistance. However, in general, (14) will not work as it is only valid when the Earth-Mars distance is much smaller than the wavelength. In general, one must use (11) with (1) and then calculate the total radiated power from the two planets to get the general radiation resistance value. I noted above that we assume the transmission lines spread slowly from the source. Other approximations in this model are: The Earth / Mars radiated field will couple back into the transmission lines, leading to waves on the lines not modeled by the ladder circuit. Again, the general numerical procedure would be needed; In this case, the ladder circuit and TEM waves only model the steady state AC behavior. Transients at switch on mean that higher order modes other than the TEM modes will propagate along the transmission lines. Higher order modes are especially important in this system for understanding transients: a TEM mode is one where the disturbance over the whole cross section of a transmission line is in-phase, and nonzero time is definitely needed to establish this highly nonlocal condition over the huge distances within this system. Lastly, to plot the field lines, I have used the Mathematica code to plot the electric and magnetic field lines: The equipotential lines are circles of the form: $$\left(x - a\,\coth\rho\right)^2 + y^2 = \left(\frac{a}{\sinh\rho}\right)^2$$ where: $$\rho = \cosh\left(\frac{2\,\pi\,\epsilon_0\,V}{q}\right)$$ and $V$ is the potential of the line in question. The electric field lines are orthogonal circles of the form: $$\left(x - a\,\cot\theta\right)^2 + y^2 = \left(\frac{a}{\sin\theta}\right)^2$$ where $\theta$ is an angle, the so called "stream function". | {
"source": [
"https://physics.stackexchange.com/questions/74625",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/175/"
]
} |
74,682 | First, I shall say that I am familiar with the intuitive idea that a spinor is like a vector (or tensor) that only transforms "up to a sign" when acted on by the rotation group. I have even rotated a plate on my palm to explain this to my fiancee! I have also looked at spinors as mathematical objects, such as the 2d subspace of complex 3-space such that $X·X = 0$, and feel that I understand this well also. I am confused by spinors in physics. Are they still vectors that are isotropic (inner product with themselves is 0)? In what vector space? Normally states are vectors in infinite dimensional spaces! Every attempt to find literature which pins down specifically what a spinor is (in physical terms) seems to assume that one is already well acquainted with the idea. Take, for example, the Dirac equation. I can see that the solutions are four-component wavefunctions, which then splits into two parts. Is this a spinor? Why? What vector space do these solutions live in?
I believe I've heard that the answer has something to do with representation theory, perhaps of the Poincare group? I am also familiar with the basics there, so don't hesitate to explain in terms of representations. | At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes. Ingredients: The essential ingredients to this explanation are: A physical "system" which evolves in and whose "events" happen in some space $\mathcal{U}$ (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is always a linear (wontedly it's Minkowsky spacetime) space wherein stuff happens: let's call $\mathcal{U}$ the "scene" of where stuff we want to talk about happens; A connected Lie group $\mathfrak{G}$ which represents the co-ordinate transformations the a system can undergo: in physics these are all linear transformations $\mathcal{U}\to\mathcal{U}$ of the scene $\mathcal{U}$ . Wontedly in physics, $\mathfrak{G} = SO^+(1,3)$ (the identity connected component of the Lorentz group comprising all rotations and boosts of "physical space", sometimes called the "proper, orthochronous Lorentz group" (proper = unimodular determinant =1, i.e. "does not invert space" and orthochronous = does not switch the time direction) or the Poincaré group; A cover of $\mathfrak{G}$ ; this is almost always (I've never seen it not so) the universal cover $\tilde{\mathfrak{G}}$ of $\mathfrak{G}$ as explained in my article "Lie Group Homotopy and Global Topology" on my website here ; A vector space $\mathcal{V}$ which can be, for example, a quantum state space, possibly infinite dimensional and its group $GL(\mathcal{V})$ of linear endomorphisms, i.e. bijetive, linear maps $\phi:\mathcal{V}\to\mathcal{V}$ . Informally, $GL(\mathcal{V})$ is the group of invertible matrices acting on $\mathcal{V}$ . Most importantly: this space is different from the physical "scene" $\mathcal{U}$ . The scene is $\mathcal{U}$ spacetime all around us, the state space $\mathcal{V}$ is a Hilbert space of quantum states. And, actually, although we talk about a "linear" state space $\mathcal{V}$ , we're a bit sloppy: sure, all quantum states are linear superpositions of the basis for $\mathcal{V}$ but they are always of unit magnitude: the probabilities of a measurement's "collapsing" the state into one of the basis vectors must all sum up to one - "we have to end up in some state". So, if we're being precise, we take heed that we are actually talking about the unit sphere within $\mathcal{V}$ as the state of quantum states. This state space is very different in character from spacetime, where there is no obligation for 4-positions of events to be unit magnitude; Representations $\rho : \mathfrak{G}\to GL(\mathcal{V})$ , $\tilde{\rho}:\tilde{\mathfrak{G}}\to GL(\mathcal{V})$ of both $\mathfrak{G}$ and its cover $\tilde{\mathfrak{G}}$ , respectively. Recall that a representation of a Lie group $\mathfrak{G}$ is a homomorphism from from $\mathfrak{G}$ to $GL(\mathcal{V})$ , i.e. a transformation that "respects the group product" so that, given $\gamma,\,\zeta\in\mathfrak{G}$ , we have $\rho(\gamma\,\zeta)=\rho(\gamma)\,\rho(\zeta)$ . And, as discussed above, the linear transformations of the form $\rho(\gamma),\,\tilde{\rho}(\tilde{\gamma}) \in GL(\mathcal{V})$ for $\gamma \in\mathfrak{G}$ and $\tilde{\gamma} \in\tilde{\mathfrak{G}}$ must be unitary so that the transformed quantum states stay normalised . So we can see that $GL(\mathcal{V})$ is very different from $\mathfrak{G}$ or $\tilde{\mathfrak{G}}$ : Lorentz boosts most assuredly are not unitary! We say that $\mathfrak{G}$ or $\tilde{\mathfrak{G}}$ "act on the state space $\mathcal{V}$ through the respresentations $\rho$ , $\tilde{\rho}$ ". I'm using here more of the mathematician's description of a representation, because here (i'm not always so fussed) I believe it is clearer than the physicists because we need to take heed that there are two different classes of representations in our discussion: those whereby the group of co-ordinate transformations $\mathcal{G}$ act on the state space $\mathcal{V}$ and those whereby its cover $\tilde{\mathcal{G}}$ acts on $\mathcal{V}$ . Baking Instructions: Wigner's Theorem and Why Covers Are Interesting Why are we interested in covers at all? After all, elements of $\tilde{G}$ are not the "physical" co-ordinate transformation. This is where we meet our baking oven for our ingedients: Wigner's theorem . Clearly, when our scene $\mathcal{U}$ undergoes a co-ordinate transformation, then the transformations wrought on the quantum state has to preserve inner products in the quantum state space so that the state stays properly normalised. From this assumption alone, i.e. one does NOT have to assume linearity , Wigner proved that the when the scene $\mathcal{U}$ undergoes a "symmetry" (a Lorentz transformation), the state space must undergo a "projective homomorphism" $\sigma$ , i.e. if $\gamma,\,\zeta$ are two Lorentz transformations, then the state space transformation corresponding to their product is: $$\sigma(\gamma\,\zeta) = \pm \sigma(\gamma)\,\sigma(\zeta)\tag{1}$$ The fact that we don't get exactly a homomorphism is why we are interested in covers: the image of the representation $\tilde{\rho}(\tilde{\mathfrak{G}})$ (recall that this is a group of unitary operators in $GL(\mathcal{V})$ acting on the state space) of the cover $\tilde{\mathfrak{G}}$ contains both the transformations that fulfill the genuine $+$ -sign homomorphism in (1) (which are simply unitary operators in the image $\rho(\mathfrak{G})$ of the co-ordinate transformation group $\mathfrak{G}$ ) AND those that flip the sign. So if we allow representations of the cover, we get every possible unitary transformation (even without an assumption of linearity - this automatically follows) that can be wrought on the state space $\mathcal{V}$ when the scene $\mathcal{U}$ is transformed. Here's the punchline. Quantum states that transform by the transformations belonging to the image $\rho(\mathfrak{G})$ of $\mathfrak{G}$ under the genuine homomorphism $\rho$ are called vectors . Quantum states that transform by the transformations belonging to the image $\tilde{\rho}(\tilde{\mathfrak{G}})$ of the cover $\tilde{\mathfrak{G}}$ under the "projective homomorphism" $\tilde{\rho}$ are called spinors . The above can be intuitively thought of as follows: in quantum mechanics, a global phase $e^{i\phi}$ multiplying a system's quantum state does not affect measurements we make on the system. So quantum systems "don't care whether a homomorphism is genuine or projective". The universal (only) cover of the Lorentz group $SO(1,3)$ is the group $SL(2,\,\mathbb{C})$ . So "spinors" transform by a representation of $SL(2,\,\mathbb{C})$ . Vectors transform by a representation of $SO(1,3)$ . The word "spinor" can be pretty vague in my experience: it can refer to the transformation in $SL(2,\,\mathbb{C})$ rather than the quantum state that is transformed by it, and people often speak of the unit quaternions as "spinors": Roger Penrose's "Road To Reality" Chapter 11 simply defines a spinor as something that takes a negative sign when rotated through $2\,\pi$ and comes back to its beginning point after a rotation through $4\,\pi$ . This is actually a pretty good definition, for that is exactly how elements of a representation of $SL(2,\,\mathbb{C})$ act on the state space $\mathcal{V}$ , and is the essential difference between how elements of a representation of $SO(1,3)$ act on state spaces. Forget about "quantities with direction" as a vector's definition: in physics the word "vector" always talks about how something transforms when our scene $\mathcal{U}$ undergoes a symmetry. Remember this is pretty near to the word's literal meaning vehor (transliterated as vector) literally means "I am borne" or "I am carried" in Latin, so it's all about how the "vector" is borne, either by a transformation in physics or as a pathogen in biology (the word's original English meaning). | {
"source": [
"https://physics.stackexchange.com/questions/74682",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28464/"
]
} |
75,222 | I would like to know where gravity gets its energy to attract physical bodies? I know that the law of conservation states that total energy of an isolated system cannot change. So gravity has to be getting its energy from somewhere, or else things like hydropower plants wouldn't be able to turn the power of the falling water into a spinning rotor. Just to be clear, Lets create an example: Lets say we have two objects with equal mass close to each other. So gravity does its job and it pulls each other closer, this gets turned into kinetic energy. This is where I'm lost. According to the law of conservation energy can't be created or destroyed and the kinetic energy comes from the gravitational pull so where does the gravitational pull gets its energy. If that energy isn't being recycled from some where else then that means you have just created energy, therefore breaking the law of conservation. | According to the conservation of energy, we cannot create or destroy energy, we can only transform it from one form to the other. So this justifies that gravity doesn't have an infinite source of energy which never runs out! So it must be getting this energy from somewhere else, right? Let's take the example of a ball dropped from some height. Gravity of the earth pulls it downward, doing work on the ball and giving it kinetic energy. The question you ask is where did it get this energy from? Go back a step and think about how this ball ended up at such a height? You lifted it up with your arms, and put it on that height. Your arms did work against gravity , spent some energy to put that ball on that height. Where did that spent energy go? This was given to gravity! When you do work against gravity, you store energy in the gravitational field as gravitational potential energy , which then gravity uses to do work on that object. In case of hydro power-plants, the sun is giving energy to the water at sea level, to evaporate and rise(in effect doing work against gravity ), which ultimately ends up in dams at a higher height, and then falls converting that initial solar energy to electricity! | {
"source": [
"https://physics.stackexchange.com/questions/75222",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28768/"
]
} |
75,229 | Here's an interesting "proof" that there is no such thing as magnetism. I know the answer but I love this so much I had to ask it here. It's a great way to confuse people! As we all know, $$\nabla \cdot\vec{B} =0$$
Using the divergence theorem, we find $$ \iint_S \vec{B} \cdot \hat{n} \, dS = \iiint_V \nabla \cdot \vec{B} \, dV = 0$$
Since $\vec{B}$ has zero divergence, there exist a vector function $\vec{A}$ such that $$\vec{B} = \nabla \times \vec{A}$$
Combining the last two equations, we get $$\iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$
Applying Stokes' theorem, we find $$\oint_C \vec{A} \cdot \hat{t} \, ds = \iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$
Therefore, $\vec{A}$ is path independent and we can write $\vec{A} = \nabla \psi$ for some scalar function $\psi$.
Since the curl of the gradient of a function is zero, we arrive at:
$$\vec{B} = \nabla \times \nabla \psi = 0,$$
which means that all magnetic fields are zero, but that can't be! Can you see where we went wrong? | Note that $\partial V=S$, so that $$\tag{1} C~=~\partial S~=~\partial^2V~=~\emptyset$$ is the empty set. (Topologically, the boundary of a boundary is empty, or equivalently, the boundary operator $\partial^2=0$ squares to zero.) On the other hand, the circulation $$\tag{2} \Gamma~=~\oint_{C=\emptyset}\vec{A}\cdot d\vec{r}~=~0$$ along the empty curve $C=\emptyset$ vanishes identically for any vector field $\vec{A}$. In particular, one can not conclude from (2) that the magnetic potential $\vec{A}$ should be a gradient field. | {
"source": [
"https://physics.stackexchange.com/questions/75229",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
75,363 | Wave equations take the form: $$\frac{ \partial^2 f} {\partial t^2} = c^2 \nabla ^2f$$ But the Schroedinger equation takes the form: $$i \hbar \frac{ \partial f} {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$ The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation? And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions? | Actually, a wave equation is any equation that admits wave-like solutions, which take the form $f(\vec{x} \pm \vec{v}t)$. The equation $\frac{\partial^2 f}{\partial t^2} = c^2\nabla^2 f$, despite being called " the wave equation," is not the only equation that does this. If you plug the wave solution into the Schroedinger equation for constant potential, using $\xi = x - vt$ $$\begin{align}
-i\hbar\frac{\partial}{\partial t}f(\xi) &= \biggl(-\frac{\hbar^2}{2m}\nabla^2 + U\biggr) f(\xi) \\
i\hbar vf'(\xi) &= -\frac{\hbar^2}{2m}f''(\xi) + Uf(\xi) \\
\end{align}$$ This clearly depends only on $\xi$, not $x$ or $t$ individually, which shows that you can find wave-like solutions. They wind up looking like $e^{ik\xi}$. | {
"source": [
"https://physics.stackexchange.com/questions/75363",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28823/"
]
} |
75,471 | As a child, I imagined this device, which may seem to rotate indefinitely. I have two questions. Is this perpetual motion machine already known? If it is, could you please give some references? What is the exact mechanism that makes it stop? By this, I mean an explanation, not simply "because it would break law of energy conservation". Of course energy conservation is true and this machine cannot work indefinitely. But for any known (presumed) perpetual mobile, there was an explanation, usually based on showing that the force generating the motion is balanced by another force . For example, Stevin obtained the laws of the inclined plane from the perpetual motion device I would like to make some observations. I don't try to convince anyone that it will move forever because I don't believe in breaking the energy conservation . It is true that great physicists like Bohr, Kramers and Slater admitted the possibility , and nowadays some who think that there may be energy exchanges between parallel worlds in MWI believe, but I don't. But I don't consider an enough explanation simply to refer to the energy conservation. I am interested in an explanation showing exactly how the magnetic forces making it rotate, are balanced. If the forces are balanced, only then friction will make it slow down and stop. I don't think that we can explain only by referring to friction, which in principle can be made as small as needed. There has to be a balance of forces. Why spending time trying to understand or explain something that admittedly can't work? Well, even though perpetual motion machines cannot actually work, I think they may be interesting as puzzles. | You do not need to invoke friction. The magnetic forces are in equilibrium by themselves so if you place the magnets in that configuration, they will not spontaneously begin to move. The reason is that there is a corresponding force on the magnets when they are vertical that matches the ones you've already drawn. Let me make a simple model. First of all, start by upping the game and including two big magnets, which can only make it better: If red is a north pole, then each rotating magnet, when horizontal, has a north pole repelling its backside north pole and a north pole attracting its front south pole. Focusing on the big magnets for the moment, the fact that their north poles face each other suggests that we can trade them for a pair of anti-Helmholtz coils. This means the important character of their field is its quadrupolar nature, and we can approximate the magnetic field as
$$\mathbf B=\frac{B_0}a\begin{pmatrix}x\\y\\-2z\end{pmatrix},$$
where the $z$ axis goes from one big magnet to the other, $a$ is some characteristic length, and $B_0$ is some characteristic field strength. Now, for the little magnets, I think it is uncontroversial to model them as point dipoles. If $\theta$ is the angle the wheel spoke makes with the $x$ axis (with the wheel in the $x,z$ plane) then each magnet is a dipole with moment
$$\mathbf m=m
\begin{pmatrix}-\sin(\theta)\\0\\\cos(\theta)\end{pmatrix}
\text{ at }
\mathbf r=R
\begin{pmatrix}\cos(\theta)\\0\\\sin(\theta)\end{pmatrix}.$$
With this, the potential energy of each spoke magnet is
$$U=-\mathbf m\cdot\mathbf B=3\frac R a mB_0\sin(\theta)\cos(\theta)=\frac32\frac Ra mB_0\sin(2\theta).$$ To see how this behaves, here is a colour plot of the energy, with negative energy in red and positive energy in blue. You can see there is a gradient pointing up on the right and down on the left. However, these are matched by clockwise gradients when the spokes are vertical. A single magnet will settle on the lower left or the upper right ; a pair of magnets will settle on that diagonal. For a symmetrical wheel with three or more magnets, the total potential energy is flat at zero,
$$U=\sum_{k=1}^n\frac32\frac Ra mB_0\sin\left(2\left(\theta_0+\frac{2\pi}{n}k\right)\right)
=\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta}\sum_{k=1}^n(e^{2\pi i/n})^k\right]
=\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta+2\pi i/n}\frac{1-(e^{2\pi i/n})^n}{1-e^{2\pi i/n}}\right]
=0
$$
and there is no resultant magnetic force. | {
"source": [
"https://physics.stackexchange.com/questions/75471",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/730/"
]
} |
75,619 | A popular assumption about black holes is that their gravity grows beyond any limit so it beats all repulsive forces and the matter collapses into a singularity . Is there any evidence for this assumption? Why can't some black holes be just bigger neutron stars with bigger gravity with no substantial difference except for preventing light to escape? And if neutrons collapse, can they transform into some denser matter (like quark-gluon plasma) with strong interaction powerful enough to stop the gravity? In this video stars are approaching supermassive black hole in the center of our galaxy in a fraction of parsec. The tidal force should tear them apart, but it doesn't. Can there be some kind of repulsive force creating limits for attractive forces? | A popular assumption about black holes is that their gravity grows beyond any limit so it beats all repulsive forces and the matter collapses into a singularity. [...] Is there any evidence for this assumption? It's not an assumption, it's a calculation plus a theorem, the Penrose singularity theorem. The calculation is the Tolman-Oppenheimer-Volkoff limit on the mass of a neutron star, which is about 1.5 to 3 solar masses. There is quite a big range of uncertainty because of uncertainties about the nuclear physics involved under these extreme conditions, but it's not really in doubt that there is such a limit and that it's in this neighborhood. It's conceivable that there are stable objects that are more compact than a neutron star but are not black holes. There are various speculative ideas -- black stars,
gravastars, quark stars, boson stars, Q-balls, and electroweak stars. However, all of these forms of matter would also have some limiting mass before they would collapse as well, and observational evidence is that stars with masses of about 3-20 solar masses really do collapse to the point where they can't be any stable form of matter. The Penrose singularity theorem says that once an object collapses past a certain point, a singularity has to form. Technically, it says that if you have something called a trapped lightlike surface, there has to be a singularity somewhere in the spacetime. This theorem is important because mass limits like the Tolman-Oppenheimer-Volkoff limit assume static equilibrium. In a dynamical system like a globular cluster, the generic situation in Newtonian gravity is that things don't collapse in the center. They tend to swing past, the same way a comet swings past the sun, and in fact there is an angular momentum barrier that makes collapse to a point impossible. The Penrose singularity theorem tells us that general relativity behaves qualitatively differently from Newtonian gravity for strong gravitational fields, and collapse to a singularity is in some sense a generic outcome. The singularity theorem also tells us that we can't just keep on discovering more and more dense forms of stable matter; beyond a certain density, a trapped lightlike surface forms, and then it's guaranteed to form a singularity. Why can't some black holes be just bigger neutron stars with bigger gravity with no substantial difference except for preventing light to escape? This question amounts to asking why we can't have a black-hole event horizon without a singularity. This is ruled out by the black hole no-hair theorems, assuming that the resulting system settles down at some point (technically the assumption is that the spacetime is stationary). Basically, the no-hair theorems say that if an object has a certain type of event horizon, and if it's settled down, it has to be a black hole, and can differ from other black holes in only three ways: its mass, angular momentum, and electric charge. These well-classified types all have singularities. Of course these theorems are proved within general relativity. In a theory of quantum gravity, probably something else happens when the collapse reaches the Planck scale. Observationally, we see objects such as Sagittarius A* that don't emit their own light, have big masses, and are far too compact to be any stable form of matter with that mass. This strongly supports the validity of the above calculations and theorems. Even stronger support will come if we can directly image Sagittarius A* with enough magnification to resolve its event horizon. This may happen within 10 years or so. | {
"source": [
"https://physics.stackexchange.com/questions/75619",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28954/"
]
} |
75,945 | Alternatively, why does the force created by blowing out air feel so much stronger than the force created by sucking in air? Ok, so forget the human factor involved in blowing out candles . Consider a vacuum cleaner with a suction end and a blower end. Anyone who has tried it out notices the blower end creates a much stronger force than the suction end, despite the discharge being (more or less) equal at both ends. Why does this happen? | In a blower, the air is directed along the axis of the blower as it exits, creating a high-pressure narrow cone. Exit pressure can also be multiple times of atmospheric pressure. At a sucker entry, the low-pressure zone is fed by a much wider angle of atmospheric air at atmospheric pressure. Additionally, the underpressure can at most be 1x atmospheric pressure. therefore the inflow has an upper limit to its velocity. | {
"source": [
"https://physics.stackexchange.com/questions/75945",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/11484/"
]
} |
76,650 | My (just completed) PhD involved a considerable amount of research involved with the detection of solar UV radiation. This generated quite a bit of interest, especially when I was conducting my experiments outside. A friend's 6 year old was most fascinated, but could not grasp the concept of UV radiation, primarily as she could not 'see' it. She understood the importance of protecting oneself from too much UV (this is a big thing in Australian schools). On reflection, it occurred to me that a lot of people don't truly grasp what UV radiation actually is, despite knowing of the risks involved with overexposure. Conversely, I know and understand the technical side of it, but struggle to put it in simpler terms. What is a simple and meaningful way of explaining UV radiation? | First, try and see if you can get the 6 year old to think about "what if there are colors we can't see"? Explain to her that the color we see is the color of "light". Now, show her a remote control, and press some button. There's an IR bulb up front, ask her if it flashes when you press the button (it shouldn't). Now, use a phone camera to look at the IR bulb, most phone cameras will show white light when the button is pressed. Explain to her that the light coming from the remote is "invisible", in the sense that it's of a color we can't see. However, the camera can see it because the camera sees slightly more "colors" than we can, and when it tries to display it it shows it as white. Explain to her that this is "infrared" light, a light that is "more red than red itself". Whenever someone turns on the TV, a light signal is sent to the TV. (You may want to explain that this light has some "bending" capabilities, but that's not entirely necessary). This ought to get her past the mental block when it comes to "light that isn't light". Mentioning that some animals see more/less colors than we do helps. Now, talk about the spectrum: Explain that the light that we can see is a very small portion of the kinds of light that actually exist. The spectrum is what she sees when she looks at a rainbow, but it really doesn't "stop" at red or purple; she just can't see it. If you wish, you can then talk about radio waves, and how they are light that can easily "bend" (i.e. diffract). Talk about X-rays, which is light that can pass through skin but not bones. This can actually lead to an interesting side track where you explain how an X-ray is nothing but a photograph with a different kind of light. Once you reach here, it's easy to explain UV. Mention that while the sun emits a lot of visible light, it's not limited to the visible spectrum and emits a significant amount (much less, but not negligible) of UV and IR as well. You can actually extend this to sound as well, talk about how there are sounds we can't hear. For that matter, sounds just outside your hearing range will be clearly audible to most six year olds. If you can generate increasing frequencies from your computer (It's actually possible for our vocal cords to work in the inaudible ranges, but it takes some practice to get that to work so it's just easier to use a computer), you can both show here that different people/ages have different frequency ranges 1 , and that there are sounds that even she can't hear. (to do the latter you may want to set up a microphone and have it show the amplitude on the screen or something). Similarly, you can go to lower frequencies (and show the transition from invisibly fast vibrations but audible sounds to visible vibrations and inaudible sounds in a string instrument or possibly a rubber band). It's a good opportunity to explain how a dog whistle works, too. The concept of there being light that we can't see and sound that we can't hear is a really amazing one when one hears of it first. I certainly was intrigued by it when I learned about this as a child. 1. This may not be so easy and may not be desirable, see Cleonis' comment below | {
"source": [
"https://physics.stackexchange.com/questions/76650",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
76,721 | I'd like some clarification regarding the roles of active and passive diffeomorphism invariance in GR between these possibly conflicting sources. 1) Wald writes, after explaining that passive diffeomorphisms are equivalent to changes in coordinates, "This 'passive' point of view on diffeomorphisms is, philosophically, drastically different from the above 'active' viewpoint, but, in practice, these viewpoints are really equivalent since the components of the tensor $\phi*T$ at $\phi(p)$ in the coordinate system $\{y^\mu\}$ in the active viewpoint are precisely the components of $T$ at $p$ in the coordinate system $\{x'^\mu\}$ in the passive viewpoint." (Wald, General Relativity, Appendix C.1) 2) Gaul and Rovelli write, "General relativity is distinguished from other dynamical field theories by its invariance under active diffeomorphisms. Any theory can be made invariant under passive diffeomorphisms. Passive diffeomorphism invariance is a property of the formulation of a dynamical theory, while active diffeomorphism invariance is a property of the dynamical theory itself ." (Gaul and Rovelli, http://arxiv.org/abs/gr-qc/9910079 , Section 4.1) It seems that Wald is saying that there is no mathematical difference between the two, and that both imply the same physical consequences. It seems that Gaul and Rovelli, however, are saying that only active diffeomorphism invariance has physical meaning that may influence the dynamics of the theory. Can anyone explain? | I think the best approach is to try to understand a concrete example: Let's look at a piece of the Euclidean plane coordinatized by $x^a=(x, y); a=1,2$ in a nice rectangular grid with Euclidean metric. Now suppose we define a transformation $$X(x,y)=x(1+\alpha y^2) $$ $$Y(x,y)=y(1+\alpha x^2) $$ $\alpha$ is just a constant, which we will take as 5/512 - for the sake of being able to draw diagrams. A point P with coordinates $(x,y)=(8,8)$ is mapped to a point P' with coordinates $(X,Y)=(13,13)$. Passive View Here we don't think of P and P' as different points, but rather as the same point and $(13,13)$ are just the coordinates of P in the new coordinate system $X^a$. In the picture, the blue lines are the coordinate lines $x^a=$ const and the red lines are the coordinate lines $X^a=$ const. Metric components on our manifold $g_{ab}(x)$ get mapped to new values $$h_{ab}(X)={\frac{\partial x^c}{\partial X^a}}{\frac{\partial x^d}{\partial X^b}} g_{cd}(x) \ \ \ (1) $$ This represents the same geometric object since $$ h_{ab}(X)dX^a\otimes dX^b = g_{ab}(x)dx^a\otimes dx^b$$ Active View One description of the active view that is sometimes used is that points are "moved around" (in some sense perhaps it's better to think just of an association between points, "moving" implies "with respect to some background"). So in our example, we'd think of the point P as having been "stretched out" to the new location P'. (These locations are with respect to the old $x$ coordinate system). The old (blue) $x=$ constant coordinate lines get dragged along too, into the red lines shown in the diagram. So the point P retains its old coordinate values $(8,8)$ in its new location, i.e $(X,Y)=(8,8)$. The metric is also dragged along (see for example Lusanna ) according to: $$h_{ab}(X)|_{P'} \ dX^a \otimes dX^b = g_{ab}(x)|_{P}\ dx^a \otimes dx^b \ \ \ (2)$$ So the old Euclidean metric $dx^2+dy^2$ becomes $dX^2+dY^2$, i.e. still Euclidean in the new $(X,Y)$ chart - nothing has changed. So, for example, the angle between the red vectors $\frac{\partial}{\partial X}$, $\frac{\partial}{\partial Y}$ is still 90 degrees, as it was for the blue vectors $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ ! My guess is that this is what Wald means by the physical equivalence - in this example a Euclidean metric remains Euclidean. Now, if we look at the red vectors from the point of view of the blue frame, they sure don't look orthogonal*, so from the blue point of view, it can only be a new metric in which the red vectors are orthogonal. So active diffeomorphisms can be interpreted as generating new metrics. Now suppose we have a spacetime - a manifold with metric for which the Einstein tensor $G_{\mu\nu}$ vanishes. Applying an active diffeomorphism, we can generate the drag-along of the Einstein tensor by a rule analogous to (2). As we have discussed, if we compare the dragged along metric with the old one in the same coordinates, we see we have a spacetime with a new metric. Moreover, the new spacetime must also have vanishing Einstein tensor - by the analog of (2), the fact that it vanished in the old system means it vanishes in the new system and hence our newly created Einstein tensor vanishes too (if a tensor vanishes in one set of coordinates it vanishes in all). In this respect, the invariance of Einstein's equations under active diffeomorphisms is special. If we take, for example, the wave equation in curved spacetime $$(g^{\mu\nu}{\nabla}_{\mu}{\nabla}_{\nu}+\xi R)\phi(x) = 0 $$ then active diffeomorphisms don't naturally take solutions to solutions - they change the metric, and the metric in this equation is part of the background, and fixed. By contrast, in Einstein's equations, the metric is what you're solving for so active diffeomorphism invariance is built in. *Just compute the vectors $\frac{\partial}{\partial X}, \frac{\partial}{\partial Y}$ in terms of $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and test their orthogonality using the original Euclidean metric. | {
"source": [
"https://physics.stackexchange.com/questions/76721",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
76,748 | So the air pressure on earth remains relatively constant, right? Is there air gained or lost through transitions of any sort? e.g. plumes going out into space, earth gaining new air particles with gravity from air in the solar system. I was curious to know whether in general the particles that brushed up against the dinosaurs, and against Homo erectus 's face, are the same as the wind today, ebbing and flowing as pressure variation directs it. Is this true? :) I suppose air goes into trees, and into mammals, and recycles around. It seems like the recyclability rate must be very very high. Is this logic correct? It's all very fascinating! | The atmosphere of the Earth is mainly composed of nitrogen (N 2 , 78%) and oxygen (O 2 , 21%) molecules, which together make up about 99% of its total volume. The remaining 1% contains all sorts of other stuff like argon, water and carbon dioxide, but let's ignore those for now. As you probably know, the oxygen we breathe is produced by plants from water and carbon dioxide as a byproduct of photosynthesis . Conversely, animals (including humans) use the oxygen to burn organic compounds (like sugars, fats and proteins) back into water and carbon dioxide, obtaining energy in the process. So do many bacteria and fungi, too, and some oxygen also gets burned in abiotic processes like wildfires and the oxidization of minerals. The result is that oxygen cycles pretty rapidly in and out of the atmosphere. According to the Wikipedia article I just linked to, the average time an oxygen molecule spends in the atmosphere before being burned or otherwise removed from the air is around 4,500 years.
The most recent known Homo erectus fossil dates from about 143,000 years ago, so the probability that a particular oxygen molecule hitting your face today has been around since that time is roughly $\exp(- 143000 / 4500) = \exp(-31.78) \approx 1.58 \times 10^{-14}$, i.e. basically zero. Of course, the oxygen atoms used for respiration don't disappear anywhere: they just become part of the water and carbon dioxide molecules. Those that end up in carbon dioxide usually get photosynthesized back into free oxygen pretty soon, unless they happen to get trapped in a carbonate sediment or something like that. The oxygen atoms that end up in water, on the other hand, may spend quite a long time in the oceans before being recycled back into the air; if I'm not mistaken, the total amount of oxygen in the hydrosphere is about 1000 times the amount in the atmosphere, so the mean cycle time should also be about 1000 times longer. Still, eventually, even the oxygen in the oceans gets cycled back into the atmosphere. Thus, while the oxygen molecules you breathe might not have been around for more than a few thousand years, the atoms they consist of have been around since long before the dinosaurs. How about nitrogen, then? Perhaps a bit surprisingly, given how inert nitrogen generally is, it's also actively cycled by the biosphere . Unfortunately, the actual rate at which this cycling occurs seems to be still poorly understood, which makes estimating the mean cycle times difficult. If I'm reading these tables correctly, the annual (natural) nitrogen flux in and out of the atmosphere is estimated to be somewhere between 40 and 400 teragrams per year, while the total atmospheric nitrogen content is about 4 zettagrams . This would put the mean lifetime of a nitrogen molecule in the atmosphere somewhere between 10 million and 100 million years, well above the time since Homo erectus first appeared (about 1.8 million years ago). Thus, it seems that most of the air molecules around you have probably been around since the days of Homo erectus , and some of them might even have been present during the age of the dinosaurs , which ended about 66 million years ago. | {
"source": [
"https://physics.stackexchange.com/questions/76748",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/29228/"
]
} |
76,842 | Ok, so entropy increases... This is supposed to be an absolute statement about entropy. But then someone imagines a box with a 10 particle gas, and finds that every now and then all particles are in the left. Conclusion, the 2nd law holds only in a statistical sense. But then Szilard comes up with a thought experiment with only one particle and a piston that can compress to the left or right. The apparent loss of entropy when finding the particle in the left half is compensated by the very bit of information that indicates where the particle is. So perhaps the 2nd Law actually holds in an absolute sense, except that... Is there consensus on the absolute vs statistical nature of the 2nd Law, or is it subject to interpretation?
Can the issue be settled within a classical setting, or does one have to go quantum? Addendum: (as per Ben Crowell's request) Here is the paper Szilard, L., 1929, “On the Decrease of Entropy in a Thermodynamic System by the Intervention of Intelligent Beings” , Zeitschrift fur Physik 53: 840–856. English translation in The Collected Works of Leo Szilard: Scientific Papers, B.T. Feld and G. Weiss Szilard (eds.), Cambridge, Massachusetts: MIT Press, 1972, pp. 103–129. | Giving a full answer to this one takes quite a bit of information, so I'll first give a few references and then summarise how they all fit in. References Relevant Physics SE Questions Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution? Theoretical proof forbidding Loschmidt reversal? Perpetual motion machine of the second kind possible in nano technology? Review Papers There are two review papers describing the concepts I am about to talk about: Sevick, E. M.; Prabhakar, R.; Williams, Stephen R.; Bernhardt, Debra Joy, "Fluctuation Theorems", Annual Rev. of Phys. Chem., 59, pp. 603-633 (this one is paywalled). E. T. Jaynes, "Gibbs vs Boltzmann Entropies", Am. J. Phys. 33, number 5, pp 391-398, 1965 as well as many other of his works in this field Charles Bennett, "The Thermodynamics of Computing: A Review", Int. J. Theoretical Physics, 21 , 12, 1982 And a remarkable experiment that actually BUILDS AND TESTS the Maxwell Daemon. Shoichi Toyabe; Takahiro Sagawa; Masahito Ueda; Eiro Muneyuki; Masaki Sano (2010-09-29). "Information heat engine: converting information to energy by feedback control". Nature Physics 6 (12): 988–992. arXiv:1009.5287. Bibcode:2011NatPh...6..988T. doi:10.1038/nphys1821 . "We demonstrated that free energy is obtained by a feedback control using the information about the system; information is converted to free energy, as the first realization of Szilard-type Maxwell’s demon." Now Your Question Now to your question. You are quite right in your conclusion about the second law's statistical nature: ... But then someone imagines a box with a 10 particle gas, and finds that every now and then all particles are in the left. Conclusion, the 2nd law holds only in a statistical sense ... and indeed various fluctuation theorems (see the "Fluctuation Theorem" Wikipedia page as well as the "Fluctuation Theorems" review paper I cited above) quantify the probability of observing deviations of a given "severity" from the second law. For the reason you clearly understand, the smaller the system, the less meaningful it becomes to describe it in terms of "macroscopic" properties such as temperature, pressure and so forth (indeed these quantities can be construed to be a parameter of a statistical population , which have less and less relevance for smaller and smaller sample sizes from that population). So I think the most meaningful version of the second law to address for this question is Carnot's classic macroscopic statement that it is "impossible to build a perpetual motion machine of the second kind". A particular property of such a perpetual motion machine is its periodicity in its interactions with its surroundings: it undergoes a periodic cycle and when it comes back to its beginning point, both it and the surrounding world are in the same state. So the impossibility of the second kind perpetual motion machine talks about "not winning in the long term": you might make small conversions of the heat in a uniform thermodynamic temperature system into useful work in the short term by dint of fluctuations, but in the long term you cannot. Ultimately this is an experimental fact and is thought to be owing to the boundary conditions of the universe. The Szilard Engine and Maxwell Daemons: Information is Physical Let's look at the Szilard engine and Maxwell Daemon first: the latter was conceived by Maxwell to illustrate that the second law was "just statistical" and it does seem to thwart the second law, as does the Szilard engine. Indeed they do win in the short term, but in the long term they do not. The full resolution to the problem is discussed in detail in Bennett's paper that I cited above, and the reason they do not is Landauer's Principle : the idea that the merging of two computational paths or the erasing of one bit of information always costs useful work, an amount given by $k_B\,T\,\log 2$, where $k_B$ is Boltzmann's constant and $T$ the temperature of the system doing the computation. Bennett invented perfectly reversible mechanical gates ("billiard ball computers") whose state can be polled without the expenditure of energy and then used such mechanical gates to thought-experimentally study the Szilard Engine and to show that Landauer's Limit arises not from the cost of finding out a system's state (as Szilard had originally assumed) but from the need to continually "forget" former states of the engine. Probing this idea more carefully, as also done in Bennett's paper: One can indeed build the Maxwell Daemon with simple finite state machines in the laboratory, as described in the Nature paper I cited. As the Daemon converts heat to work, it must record a sequence of bits describing which side of the Daemon's door (or engine's piston, for an equivalent discussion of the Szilard engine) molecules were on. For a finite memory machine, one needs eventually to erase the memory so that the machine can keep working. However, "information" ultimately is not abstract - it needs to be "written in some kind of ink" you might say - and that ink is the states of physical systems. The fundamental laws of physics are reversible, so that one can in principle compute any former state of a system from the full knowledge of any future state - no information gets lost . So, if the finite state machine's memory is erased, the information encoded that memory must show up, recorded somehow, as changes in the states of the physical system making up and surrounding the physical memory. So now those physical states behave just like the computer memory: eventually those physical states can encode no more information, and the increased thermodynamic entropy of that physical system must be thrown out of the system, with the work expenditure required by the Second Law, before the Daemon can keep working. The need for this work is begotten of the need to erase information, and is the ultimate justification for Landauer's principle. The Szilard Engine and Daemon "win" in the short term because they are not truly cyclic: they change the states of memory: the second law prevails when that memory is brought back to its beginning state too. Another Illustration of Non Cyclic Thwarting of the Second Law Another illustration of the importance of true cycles in considering the second law is a "trick" whereby one can extract ALL of the enthalpy of a chemical reaction as useful work IF one has a sequence of cooler and cooler reservoirs that one can use as follows: (1) Lower the reactants down to absolute zero temperature by drawing heat from the reactants into the reservoirs, (2) Let the reaction to go ahead at aboslute zero thus extracting all the reactants' enthalphy as work and then (3) Use the sequence of reservoirs in rising temperature order to bring the reaction products back to the beginning temperature. The point is that some of the enthalpies of formation will now be left in the cold reservoirs and so the system has not been taken through a complete cycle. One can't do this indefinitely: the cool reservoirs will eventually heat up if one does this repeatedly. You might "win" with small amounts of reactants, but you can't do so indefinitely because you are degrading the system: the work needed to restore the cold reservoirs to their beginning state is then the difference between the enthalpy of reaction and the free energy. "Proofs" of the Second Law E. T. Jaynes tried to bring information theory rigorously to thermodynamics and critically examines Boltzmann's concept of entropy. In particular the Boltzmann "stosszahlansatz" (assumption of molecular chaos) can often only be applied once, as later changes to the system leave the states of molecules of a gas correlated, thus begetting the difference between the Gibbs (informational) and Boltzmann ("experimental", i.e. defined only when you have big systems) entropies, with the former unchanged in things like irreversible volume changes, the latter always increasing. So, from an assumption of molecular chaos, one can prove once that the Boltzmann entropy must increase in an irreversible change. But the irreversible change and the correlation between system constituents it begets means that one cannot apply the assumption of molecular chaos again and repeat the proof unless one comes up with an explanation of how the system gets back to a state where the states of all its constituent parts are uncorrelated. See the Jaynes papers in my references: Jaynes does eventually argue that one needs to appeal to experiment to support the large scale second law of thermodynamics. So ultimately it would seem that the statement that the Boltzmann entropy of a system always increases in the long term can only be substantiated experimentally. Why the entropy of a system always increases when physical laws are just as valid with time running backwards is called "Loschmidt's Paradox". There has been a great deal of work to understand this and it's generally agreed that the answer has to do with the "boundary conditions" of the universe - roughly put, the universe was (observed fact) in an exquisitely low entropy state at the big bang, and so the overwhelmingly likeliest history is one where entropy rises with increasing time. But how and why that low entropy state arose is, as I understand it, one of the profound mysteries of modern physics. A good layperson's summary of why we have a second law of thermodynamics, how entropy is to some extent a subjective concept, and the discussion of this profound mystery is to be found in chapter 27 of Roger Penrose's "The Road to Reality". I would highly recommend you look at this reference. | {
"source": [
"https://physics.stackexchange.com/questions/76842",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/29377/"
]
} |
77,196 | We are always told that there are the four fundamental forces or interactions of nature: gravitation, electromagnetism, and the weak and strong forces. We know that gravitation is attractive, that electromagnetism can be attractive or repulsive depending on the electric charge of the interacting particles, and that the strong force is attractive between quarks. But when the weak force is mentioned, the description is always something such as 'responsible for radioactive decay', but there is no mention of whether this force is attractive or repulsive. So my question is: is the weak force/interaction attractive or repulsive? | Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract". In the electroweak case, the charges in question are weak isospin and weak hypercharge. For weak isospin, there are two isospin charges (or flavors), up and down, and their associated anti-charges, anti-up and anti-down. So: up repels up (anti-up repels anti-up) down repels down (anti-down repels anti-down) up attracts down (anti-up attracts anti-down) up attracts anti-up (down attracts anti-down) up repels anti-down (down repels anti-up) For weak hypercharge, there is just one type of charge and its associated anti-charge. So: hypercharge repels hypercharge (anti-hypercharge repels
anti-hypercharge) hypercharge attracts anti-hypercharge Note that electric charge is a certain mixture of weak isospin and weak hypercharge. Since (left-handed) particles carry both weak isospin and weak hypercharge, both must be taken into account to determine which particles attract or repel under the electroweak interaction. | {
"source": [
"https://physics.stackexchange.com/questions/77196",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/17324/"
]
} |
77,197 | Recently, I performed experiments to characterise the ultraviolet-A response in a smartphone camera (with the lens still attached). This question focuses on Figure 2 of my paper "Characterization of a Smartphone Camera's Response to Ultraviolet A Radiation" - please note, I am only posting this link to show the figure. In figure 2, the transmission of UVA through the lens drops significantly at wavelengths of less than about 370nm, not shown in the paper is that several lenses were tested with similar results. However, with the lens in place, UVA radiation from a monochromator at wavelengths as low as 320nm still cause saturation of the silicon-based CMOS image sensor despite the heavy attenuation at wavelengths lower than ~370nm. What is reason for heavily attenuated UV radiation at 320nm causing as strong response as for far less attenuated UV radiation at 380nm? | Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract". In the electroweak case, the charges in question are weak isospin and weak hypercharge. For weak isospin, there are two isospin charges (or flavors), up and down, and their associated anti-charges, anti-up and anti-down. So: up repels up (anti-up repels anti-up) down repels down (anti-down repels anti-down) up attracts down (anti-up attracts anti-down) up attracts anti-up (down attracts anti-down) up repels anti-down (down repels anti-up) For weak hypercharge, there is just one type of charge and its associated anti-charge. So: hypercharge repels hypercharge (anti-hypercharge repels
anti-hypercharge) hypercharge attracts anti-hypercharge Note that electric charge is a certain mixture of weak isospin and weak hypercharge. Since (left-handed) particles carry both weak isospin and weak hypercharge, both must be taken into account to determine which particles attract or repel under the electroweak interaction. | {
"source": [
"https://physics.stackexchange.com/questions/77197",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
77,227 | How do I solve the speed of light in gravitational field? Should I just add gravitational acceleration in speed of light? $$c'=c_0+g(r)t~?$$ | This is a far more complicated question than you (probably) realise, as to answer it requires an understanding of general relativity. In GR the speed of light is locally invariant, that is if you measure the speed of light at your location you'll always get the value $c$. However if you measure the speed of light at some distant location you may find it to be less than $c$. The obvious example of this is a black hole, where the speed of light falls as it approaches the event horizon and indeed slows to zero at the event horizon. The reason we may measure the speed of light at a distant location to be less than $c$ is because, as alexo says in his answer, spacetime is curved by mass/energy. The co-ordinates that you use for measuring spacetime will not match the co-ordinates a distant observer uses, and that's why the two of you measure different values for the speed of light. To calculate the speed of light at some distant point you need to solve Einstein's equations to find out how spacetime curves relative to your co-ordinate system. To show this let's take an example. If you solve Einstein's equations for a spherically symmetric mass you get the Schwarzschild metric : $$ \mbox{d}s^2 = -\left(1-\frac{r_s}{r}\right)c^2~\mbox{d}t^2 + \frac{\mbox{d}r^2}{\left(1-\frac{r_s}{r}\right)} + r^2 (\mbox{d}\theta^2 + \sin^2\theta~ \mbox{d}\phi^2) $$ In this equation $r$ is the distance to the black hole (the radius) and $t$ is time (what you measure on your wristwatch). $\theta$ and $\phi$ are basically longitude and latitude measurements. The quantity $\mbox{d}s$ is called the interval . $r_s$ is the radius of the event horizon. The co-ordinate system strictly speaking is the one used by an observer at infinity, but it's a good approximation as long as you are well outside the event horizon. For light rays $\mbox{d}s$ is always zero, and we can use this to calculate the velocity of the light ray. For simplicity let's take a ray headed directly towards the black hole, so the longitude and latitude are constant i.e. $\mbox{d}\theta$ and $\mbox{d} \phi$ are both zero. This simplifies the above equation to: $$ 0 = -\left(1-\frac{r_s}{r}\right)c^2~\mbox{d}t^2 + \frac{\mbox{d}r^2}{\left(1-\frac{r_s}{r}\right)} $$ The velocity of the light, $v$, is just the rate of change of radius with time, $\mathrm dr/\mathrm d t$, and we get this by a quick rearrangement: $$ \frac{\mathrm{d}r}{\mathrm{d}t} = v = c \left(1-\frac{r_s}{r}\right) $$ The variation of the velocity of light with distance from the black hole looks like: At large distances (large $r$) the velocity tends to 1 (i.e. $c$) but close to the black hole it decreases, and falls to zero at the event horizon. So, to calculate the speed of light in your co-ordinate system solve the Einstein equations to get the metric, set $\mbox{d} s$ to zero and solve the resulting equation - sounds easy, but it rarely is! But, but, but, be absolutely clear what you're calculating. All you're calculating is the speed of light in your co-ordinate system i.e. the result you get applies only to you. Other observers in other places will calculate a different value, and every observer everywhere will find the local speed of light to have the same value of $c$. | {
"source": [
"https://physics.stackexchange.com/questions/77227",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28936/"
]
} |
77,368 | I've seem the notion of bundles, fiber bundles , connections on bundles and so on being used in many different places on Physics. Now, in mathematics a bundle is introduced to generalize the topological product: describe spaces that globally are not products but that locally are. In geometry we use this idea to introduce the notion of vectors into a manifold and so on. Now, what is the connection of this mathematical intuition and the importance that bundles have in Physics? The point is that there are many objects that we naturally see how they fit into Physics: manifolds intuitively can be viewed as abstract spaces where we can put coordinates in a smooth manner and do calculus, so it's very natural that whenever we need coordinates, there'll probably be a manifold involved. Now, with bundles I'm failing to see this intuition. | All of physics has two aspects: a local or even infinitesimal aspect, and a global aspect. Much of the standard lore deals just with the local and infinitesimal aspects -- the perturbative aspects_ and fiber bundles play little role there. But they are the all-important structure that govern the global -- the non-perturbative -- aspect. Bundles are the global structure of physical fields and they are irrelevant only for the crude local and perturbative description of reality. For instance the gauge fields in Yang-Mills theory, hence in EM, in QED and in QCD, hence in the standard model of the known universe, are not really just the local 1-forms $A_\mu^a$ known from so many textbooks, but are globally really connections on principal bundles (or their associated bundles) and this is all-important once one passes to non-perturbative Yang-Mills theory, hence to the full story, instead of its infinitesimal or local approximation. Notably what is called a Yang-Mills instanton in general and the QCD instanton in particular is nothing but the underlying nontrivial class of the principal bundle underlying the Yang-Mills gauge field . Specifically, what physicists call the instanton number for $SU(2)$-gauge theory in 4-dimensions is precisely what mathematically is called the second Chern-class , a " characteristic class " of these gauge bundles_ YM Instanton = class of principal bundle underlying the non-perturbative gauge field To appreciate the utmost relevance of this, observe that the non-perturbative vacuum of the observable world is a "sea of instantons" with about one YM instanton per femto-meter to the 4th. See for instance the first sections of T. Schaefer, E. Shuryak, Instantons in QCD , Rev.Mod.Phys.70:323-426, 1998 ( arXiv:hep-ph/9610451 ) for a review of this fact. So the very substance of the physical world, the very vacuum that we inhabit, is all controled by non-trivial fiber bundles and is inexplicable without these. Similarly fiber bundles control all other topologically non-trivial aspects of physics. For instance most quantum anomalies are the statement that what looks like an action function to feed into the path integral, is globally really the section of a non-trivial bundle -- notably a Pfaffian line bundle resulting from the fermionic path integrals . Moreover all classical anomalies are statements of nontrivializability of certain fiber bundles. Indeed, as the discussion there shows, quantization as such, if done non-perturbatively, is all about lifting differential form data to line bundle data, this is called the prequantum line bundle which exists over any globally quantizable phase space and controls all of its quantum theory. It reflects itself in many central extensions that govern quantum physics, such as the Heisenberg group central extension of the Hamiltonian translation and generally and crucially the quantomorphism group central extension of the Hamiltonian diffeomorphisms of phase space. All these central extensions are non-trivial fiber bundles, and the "quantum" in "quantization" to a large extent a reference to the discrete (quantized) characteristic classes of these bundles. One can indeed understand quantization as such as the lift of infinitesimal classical differential form data to global bundle data. This is described in detail at quantization -- Motivation from classical mechanics and Lie theory . But actually the role of fiber bundles reaches a good bit deeper still. Quantization is just a certain extension step in the general story, but already classical field theory cannot be understood globally without a notion of bundle. Notably the very formalization of what a classical field really is says: a section of a field bundle . The global nature of spinors, hence spin structures and their subtle effect on fermion physics are all enoced by the corresponding spinor bundles . In fact two aspects of bundles in physics come together in the theory of gauge fields and combine to produce higher fiber bundles : namely we saw above that a gauge field is itself already a bundle (with a connection), and hence the bundle of which a gauge field is a section has to be a "second-order bundle". This is called gerbe or 2-bundle : the only way to realize the Yang-Mills field both locally and globally accurately is to consider it as a section of a bundle whose typical fiber is $\mathbf{B}G$, the moduli stack of $G$-principal bundles. For more on this see on the nLab at The traditional idea of field bundles and its problems . All of this becomes even more pronounced as one digs deeper into local quantum field theory, with locality formalized as in the cobordism theorem that classifies local topological field theories. Then already the Lagrangians and local action functionals themselves are higher connections on higher bundles over the higher moduli stack of fields. For instance the fully local formulation of Chern-Simons theory exhibits the Chern-Simons action functional --- with all its global gauge invariance correctly realized -- as a universal Chern-Simons circle 3-bundle . This is such that by transgression to lower codimension it reproduces all the global gauge structure of this field theory, such as in codimension 2 the WZW gerbe (itself a fiber 2-bundle: the background gauge field of the WZW model!), in codimension 1 the prequantum line bundle on the moduli space of connections whose sections in turn yield the Hitchin bundle of conformal blocks on the moduli space of conformal curves. And so on and so forth. In short: all global structure in field theory is controled by fiber bundles, and all the more the more the field theory is quantum and gauge. The only reason why this can be ignored to some extent is because field theory is a compex subject and maybe the majority of discussion about it concerns really only a small little perturbative local aspect of it. But this is not the reality. The QCD vacuum that we inhabit is filled with a sea of non-trivial bundles and the whole quantum structure of the laws of nature are bundle-theoretic at its very heart. See also at geometric quantization . For an expanded version of this text with more pointers see on the nLab at fiber bundles in physics . | {
"source": [
"https://physics.stackexchange.com/questions/77368",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21146/"
]
} |
77,410 | By definition scalar fields are independent of coordinate system, thus I would expect a scalar field $\psi [x]$ would not change under the transformation $x^\mu \to x^\mu + \epsilon^\mu $. Correct? Now when I look in the book "Introduction to QFT" by Peskin and Schroeder they state (in an example) that the scalar field $\psi[x]$ under an infinitesimal coordinate transformation $x^\mu \to x^\mu -a^\mu$ transforms as $\psi[x] \to \psi[x+a] = \psi[x] +a^\mu\partial_\mu\psi[x]$. One possible solution I thought was that a scalar field $f:M\to\mathcal{R}$ from a variety (in this case spacetime) is invariant under coordinate transformations, but that the coordinate representation $\psi[x]$ does change under coordinate transformations (obviously). Is this anywhere near correct? | Here's what's really going on. In classical field theory, a basic set of objects that we often consider are scalar fields $\phi:M\to \mathbb R$ where $M$ is a manifold. Now we can ask ourselves the following question: Is there some natural notion of how a scalar field defined on a given manifold "transforms" under a coordinate transformation? I claim that the answer is yes, and I'll attempt to justify my claim both mathematically, and physically. The bottom line is that we ultimately have to define the way in which fields transform under certain types of transformations, but any old definition will not necessarily be useful in math or physics, so we must make well-motivated definitions and then show that they are useful for modeling physical systems. Mathematical perspective. (manifolds and coordinate charts) Recall that a coordinate system (aka coordinate chart) on an $n$-dimensional manfiold $M$ is a (sufficiently smooth) mapping $\psi:U\to \mathbb R^n$ where $U$ is some open subset of $M$. We can use such a coordinate system to define a coordinate representation $\phi_\psi$ of the scalar field $\phi$ as
\begin{align}
\phi_\psi = \phi\circ\psi^{-1}:V\to\mathbb R
\end{align}
where $V$ is the image of $U$ under $\psi$. Now let two coordinate systems $\psi:U_1\to \mathbb R^n$ and $\psi_2:U_2\to\mathbb R^n$ be given such that $U_1\cap U_2\neq \emptyset$. The coordinate representation of $\phi$ in these two coordinate systems is $\phi_1 = \phi\circ \psi_1^{-1}$ and $\phi_2 = \phi\circ \psi_2^{-1}$. Now consider a point $x\in U_1\cap U_2$, then $x$ is mapped to some point $x_1\in \mathbb R^n$ under $\psi_1$ and to some point $x_2\in \mathbb R^n$ under $\psi_2$. We can therefore write
\begin{align}
\phi(x) &= \phi \circ \psi_1^{-1} \circ \psi_1(x) = \phi_1(x_1) \\
\phi(x) &= \phi \circ \psi_2^{-1} \circ \psi_2(x) = \phi_2(x_2)
\end{align}
so that
\begin{align}
\phi_1(x_1) = \phi_2(x_2)
\end{align}
In other words, the value of the coordinate representation $\phi_1$ evaluated at the coordinate representation $x_1 = \psi_1(x)$ of the point $x$ agrees with the value of the coordinate representation $\phi_2$ evaluated at the coordinate representation $x_2 = \psi_2(x)$ of the same point $x$. This is one way of understanding what it means for a scalar field to be "invariant" under a change of coordinates. If, in particular, the manifold $M$ we are considering is $\mathbb R^{3,1} = (\mathbb R^4, \eta)$, namely four-dimensional Minkowski space, then we could consider the following two coordinate systems:
\begin{align}
\psi_1(x) &= x \\
\psi_2(x) &= \Lambda x+a
\end{align}
where $\Lambda$ is a Lorentz transformation and $a\in \mathbb R^4$, then the coordinate representations $\phi_1$ and $\phi_2$ of $\phi$ are, as noted above, related by
\begin{align}
\phi_1(x) = \phi_2(\Lambda x + a)
\end{align}
If we switch notation a bit and write $\phi_1 = \phi$ and $\phi_2 = \phi'$, then this reads
\begin{align}
\phi'(\Lambda x +a) = \phi(x)
\end{align}
which is the standard expression you'll see in field theory texts. Physical perspective. Here's a lower-dimensional analogy. Imagine a temperature field $T:\mathbb R^2 \to \mathbb R$ on the plane that assigns a real number that we interpret as the temperature at each point on some two-dimensional surface. Suppose that this temperature field is generated by some apparatus under the surface, and suppose that we translate the apparatus by a vector $\vec a$. We could now ask ourselves: What will the temperature field produced by the translated apparatus look like? Well, each point in the temperature distribution will be translated by the amount $\vec a$. So, for example, if the point $\vec x_0$ has temperature $T(\vec x_0) = 113^\circ\,\mathrm K$, then after the apparatus is translated, the point $\vec x_0 + \vec a$ will have the same temperature $113^\circ\,\mathrm K$ as the point $\vec x_0$ before the apparatus was translated. The mathematical way of writing this is that if $T'$ denotes the translated temperature field, then $T'$ is related to $T$ by
\begin{align}
T'(\vec x+\vec a) = T(\vec x)
\end{align}
The a similar argument could be made for a scalar field on Minkowski space, but instead of simply translating some temperature apparatus, we could imagine boosting or translating something producing some Lorentz scalar field, and we would be motivated to define the transformation law of a scalar field under Poincare transformation as
\begin{align}
\phi'(\Lambda x+a) = \phi(x)
\end{align} | {
"source": [
"https://physics.stackexchange.com/questions/77410",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/29669/"
]
} |
77,690 | One of the most useful tools in dimensional analysis is the use of square brackets around some physical quantity $q$ to denote its dimension as
$$[q].$$
However, the precise meaning of this symbol varies from source to source; there are a few possible interpretations and few strict guidelines. What conventions are there, who uses them, and when am I obliged to follow them? | I had an extensive look around, and I turned up four conventions. This included a short poll of google , other questions on this and other sites, and multiple standards documents. (I make no claim of exhaustiveness or infallibility, by the way.) Using $[q]$ to denote commensurability as an equivalence relation. That is, if $q$ and $p$ have the same physical dimension $Q$, one might write $$[q]=[p]=[Q],$$ but no bracketed quantity is ever shown equal to an unbracketed symbol. Thus, if $v$ is a speed one might write $[v]=[L]/[T]$ or $[v]=[L/T]$ or $[v]=[L\,T^{-1}]$ or some equivalent construct. You can see $L$ and $T$ as denoting the dimension or just "some length" and "some time". To see how you would work without evaluating braces, here is a proof that the fine structure constant is dimensionless:
$$
[\alpha]=\left[\frac{e^2/4\pi\epsilon_0}{\hbar c}\right]=\frac{[F\,r^2]}{[E/\omega][r/t]}
=\frac{[F r][\omega t]}{[E]}=\frac{[E]}{[E]}[1]=[1]
,$$
so $\alpha$ and $1$ are commensurable. Some examples are this , this , this , or this . Using $[q]$ to denote the dimensions of a quantity. Thus if the physical quantity $q$ has dimension $Q$, one writes $$[q]=Q.$$ A velocity would then be written as $[v]=L\,T^{-1}$ or its equivalents. This seems to be the leading candidate on Google, closely followed by the convention 1. Some examples are this , this , this , this , and this . This is my personal favourite, as I find that it permits the most flexibility without horribly formalizing the whole business (though I will often skip the actual evaluation of the braces, essentially using convention 1). Using $[q]$ to denote the units of a quantity. Here if $q$ can be written as a multiple of some unit $\text q$, you write $$[q]=\text q.$$ This is contingent on what unit system you choose but different units for the same dimension are of course equivalent. When this approach is used, the notation $\{q\}=q/[q]$ is sometimes used to denote the purely numerical value of the quantity. A speed would be written, for example, as $[v]=\text m\,\text s^{-1}$. This use is endorsed by the NIST Guide to the SI , section 7.1 ,
the IUPAC guide Quantities, units and symbols in physical chemistry , the IUPAP guide Symbols, units, nomenclature and fundamental constants in physics , as well as the ISO standard ISO 80000 -1:2009, section 3.20. (That document is very paywalled, but chapters 0-3 are available for free preview here .) Google results seem relatively scarce, with this and this as examples, although that could simply be poor representation. (There is also this document , which uses the notation $[\text W]=[\text V][\text A]$, but I think this is quite uncommon as well as not very useful.) Using $\operatorname{dim}(q)$ to denote the dimensions of a quantity. This is the notation set as standard by the Bureau International des Poids et Mesures in the SI Brochure (8th edition, chapter 1.3, p. 105). This also sets roman sans-serif as the standard for physical dimensions, so $\mathsf{Q}$ would be the dimension of $q$ and you write $$\operatorname{dim}(q)=\mathsf Q.$$ (To typeset roman sans-serif in TeX or MathJax, use \mathsf ; note that this is distinct from \operatorname , which is used for $\operatorname{dim}$ and would produce $\operatorname{Q}$ through \operatorname{Q} .)
A real-world usage example is thus $\operatorname{dim}(v)=\mathsf L\,\mathsf T^{-1}$ for a velocity. This use is set as standard by ISO 80000-1:2009 , section 3.7, and it is also endorsed by the NIST Guide to the SI, section 7.14 .
(NIST also reproduces the BIPM text in p. 16 of The International System of Units .)
Examples of this online are this , this , this and this ; I note, though, that most examples I found are technical, while pedagogical examples tended to use conventions 1 and 2. (This also feels less common, but that's hard to judge.) I also find it important to add that few academic journals impose standards in this area. As a working physicist in academia, the style guidance of one's chosen journal is often the only style standard one is really obliged to follow. The style manuals of the American Physical Society , the Institute of Physics , Reviews of Modern Physics , Nature Physics and several Elsevier journals have no mention of which convention should be used in their publications. As was made clear in Should we necessarily express the dimensions of a physical quantity within square brackets? , the choice of what the symbol $[q]$ means is entirely a matter of convention. The most important thing is that your usage is consistent . Do not jump conventions within a document. If your work is closely allied to other resources (e.g. textbooks) that use a particular convention, it is best to stick to that, to avoid confusing your students. If you are presenting an exam, use the notations used in your course to avoid confusing your examiner, or - at the very least - define all non-standard notation you use. So, what convention should you use? There is really no requirement to use any one of the above (and you can even make up your own notation, as long as you define it appropriately and don't overdo it ). This is really less of an issue than it looks, as there's actually rather rarely a need to use this notation in print except in pedagogic settings. (That's not to say that professional physicists don't use it in practice: we do use it, often, in everyday life, but it's mostly informal work used on the side to keep calculations straight or as exploratory scaling arguments when starting work on a problem, for example.) If your work is a commercial report, or similar document, and it could potentially have legal repercussions, then you should check whether there is a legal standard you should be using, which will then probably be conventions 3 and 4. Academically, you are typically free to choose the conventions you find most convenient as long as you use them properly and you avoid conflicts with other allied resources. If you are publishing in a journal or as part of a bigger work, you should check if they provide style guidance on this, though as I said journals rarely take stances on this. (You should really be reading the style guidance anyway as part of your submission process, though.) For your informal work, you should use whatever you're most comfortable with! Finally, if you have questions about the typesetting of these notations in LaTeX, you should go to How should I typeset the physical dimensions of quantities? on TeX.SE. | {
"source": [
"https://physics.stackexchange.com/questions/77690",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/8563/"
]
} |
77,730 | The news that physicists have discovered a geometrical object that simplifies a lot our models of quantum physics has recently became viral. For an outsider like me, it is difficult to actually understand the significance of this finding. Is it actually a new model that literaly makes every quantum physics book obsolete, or is it just a tool for a very specific calculation or effect that barely changes anything else in the field? | There was a presentation of the idea at SUSY2013 by Nima Arkani-Hamed which is available on video at http://susy2013.ictp.it/video/05_Friday/2013_08_30_Arkani-Hamed_4-3.html The amplituhedron is a buzzword for a description of a way to solve maximally supersymmetric (i.e. N=4) Yang-Mills theory in 4 dimensions. Ordinary Yang-Mills theory is a generalization of quantum gauge field theories which include electrodynamics and quantum chromodynamics. The supersymmetric extensions have not been found in nature so far. The usual way to calculate scattering amplitudes in quantum field theory is by adding together the effects of many Feynman diagrams, but the number and complexity of diagrams increases rapidly as the number of loops increases and if the coupling is strong the sum converges slowly making it difficult to do accurate calculations. The new solution for Super Yang-Mills uses the observation that the theory has a superconformal invariance in space-time and another dual superconformal invariance in momentum space. This constrains the form that the scattering amplitudes can take since they must be a representation of these symmetries. There are further constraints imposed by requirements of locality and unitarity and all these constraints together are sufficient to construct the scattering amplitudes in the planar limit without doing the sum over Feynman diagrams. The mathematical tools needed are twistors and grassmanians. The answer for each scattering amplitude takes the form of a volume of a high dimensional polytope defined by the positivity of grassmanians, hence the name amplituhedron. The first thing to say about this is that so far it is only applicable to the planar limit of one specific quantum field theory and it is not one encountered in nature. It is therefore very premature to say that this makes conventional quantum field theory obsolete. Some parts of the theory can be generalised to more physical models such as QCD but only for the tree diagrams and the planar limit. There is some hope that the ideas can be broadened beyond the planar limit but that may be a long way off. On its own the theory is very interesting but of limited use. The real excitement is in the idea that it extends in some way to theories which could be physical. Some progress has been made towards applying it in maximal supergravity theories, i.e. N=8 sugra in four dimensions. This is possible because of the observation that this theory is in some sense the square of the N=4 super Yang Mills theory. At one time (about 1980) N=8 SUGRA was considered a candidate theory of everything until it was noticed that its gauge group is too small and it does not have chiral fermions or room for symmetry breaking. Now it is just considered to be another toy model, albeit a very sophisticated one with gravity, gauge fields and matter in 4 dimensions. If it can be solved in terms of something like an amplituhedron it would be an even bigger breakthrough but it would still be unphysical. The bigger hope then is that superstring theory also has enough supersymmetry for a similar idea to work. This would presumably require superstring theory to have the same dual superconformal symmetry as super Yang Mills, or some other even more elaborate infinite dimensional symmetry. Nothing like that is currently known. Part of the story of the amplituhedron is the idea that space, time, locality and unitarity are emergent. This is exciting because people have always speculated that some of these things may be emergent in theories of quantum gravity. In my opinion it is too strong to call this emergence. Emergence of space-time implies that space and time are approximate and there are places such as a black hole singularity where they cease to be a smooth manifold. The amplituhedron does not give you this. I think it is more accurate to say that with the amplituhedron space-time is derived rather than emergent. It is possible that true emergence may be a feature in a wider generalisation of the theory especially if it can be applied to quantum gravity where emergence is expected to be a feature. Having space-time and unitarity as a derived concept may be a step towards emergence but it is not the same thing. For what my opinion is worth I do think that this new way of looking at quantum field theories will turn out to something that generalises to something that is really part of nature. I have advocated the idea that string theory has very large symmetries in the form of necklace algebras so these ideas seem on the right track to me. However I do think that many more advances will be required to work up from super yang mills to sugra and then string theory. They will have to find a way to go beyond the planar limit, generalise to higher dimensions, include gravity and identify the relevant symmetries for string theory. Then there is just the little issue of relating the result to reality. It could be a long road. | {
"source": [
"https://physics.stackexchange.com/questions/77730",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/6913/"
]
} |
77,733 | If we set up an evacuated vessel where we keep one end at a higher temperature than the other, and then introduce a liquid at the warm end, it will evaporate and condense at the cooler end. If we trap the liquid there, all the liquid will move to the cooler end. What limits the speed of this mass transfer? To have example values, let's use water for the liquid, 25 °C for the warm and 4 °C for the cold end. The most simple geometry of the vessel I can think of that disallows the liquid to flow back to the warm end would be a pipe bent into an upturned U shape. The pressure inside the vessel will eventually settle at the vapour pressure of the liquid at the cooler temperature. How would it look during the mass transfer? | There was a presentation of the idea at SUSY2013 by Nima Arkani-Hamed which is available on video at http://susy2013.ictp.it/video/05_Friday/2013_08_30_Arkani-Hamed_4-3.html The amplituhedron is a buzzword for a description of a way to solve maximally supersymmetric (i.e. N=4) Yang-Mills theory in 4 dimensions. Ordinary Yang-Mills theory is a generalization of quantum gauge field theories which include electrodynamics and quantum chromodynamics. The supersymmetric extensions have not been found in nature so far. The usual way to calculate scattering amplitudes in quantum field theory is by adding together the effects of many Feynman diagrams, but the number and complexity of diagrams increases rapidly as the number of loops increases and if the coupling is strong the sum converges slowly making it difficult to do accurate calculations. The new solution for Super Yang-Mills uses the observation that the theory has a superconformal invariance in space-time and another dual superconformal invariance in momentum space. This constrains the form that the scattering amplitudes can take since they must be a representation of these symmetries. There are further constraints imposed by requirements of locality and unitarity and all these constraints together are sufficient to construct the scattering amplitudes in the planar limit without doing the sum over Feynman diagrams. The mathematical tools needed are twistors and grassmanians. The answer for each scattering amplitude takes the form of a volume of a high dimensional polytope defined by the positivity of grassmanians, hence the name amplituhedron. The first thing to say about this is that so far it is only applicable to the planar limit of one specific quantum field theory and it is not one encountered in nature. It is therefore very premature to say that this makes conventional quantum field theory obsolete. Some parts of the theory can be generalised to more physical models such as QCD but only for the tree diagrams and the planar limit. There is some hope that the ideas can be broadened beyond the planar limit but that may be a long way off. On its own the theory is very interesting but of limited use. The real excitement is in the idea that it extends in some way to theories which could be physical. Some progress has been made towards applying it in maximal supergravity theories, i.e. N=8 sugra in four dimensions. This is possible because of the observation that this theory is in some sense the square of the N=4 super Yang Mills theory. At one time (about 1980) N=8 SUGRA was considered a candidate theory of everything until it was noticed that its gauge group is too small and it does not have chiral fermions or room for symmetry breaking. Now it is just considered to be another toy model, albeit a very sophisticated one with gravity, gauge fields and matter in 4 dimensions. If it can be solved in terms of something like an amplituhedron it would be an even bigger breakthrough but it would still be unphysical. The bigger hope then is that superstring theory also has enough supersymmetry for a similar idea to work. This would presumably require superstring theory to have the same dual superconformal symmetry as super Yang Mills, or some other even more elaborate infinite dimensional symmetry. Nothing like that is currently known. Part of the story of the amplituhedron is the idea that space, time, locality and unitarity are emergent. This is exciting because people have always speculated that some of these things may be emergent in theories of quantum gravity. In my opinion it is too strong to call this emergence. Emergence of space-time implies that space and time are approximate and there are places such as a black hole singularity where they cease to be a smooth manifold. The amplituhedron does not give you this. I think it is more accurate to say that with the amplituhedron space-time is derived rather than emergent. It is possible that true emergence may be a feature in a wider generalisation of the theory especially if it can be applied to quantum gravity where emergence is expected to be a feature. Having space-time and unitarity as a derived concept may be a step towards emergence but it is not the same thing. For what my opinion is worth I do think that this new way of looking at quantum field theories will turn out to something that generalises to something that is really part of nature. I have advocated the idea that string theory has very large symmetries in the form of necklace algebras so these ideas seem on the right track to me. However I do think that many more advances will be required to work up from super yang mills to sugra and then string theory. They will have to find a way to go beyond the planar limit, generalise to higher dimensions, include gravity and identify the relevant symmetries for string theory. Then there is just the little issue of relating the result to reality. It could be a long road. | {
"source": [
"https://physics.stackexchange.com/questions/77733",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/16477/"
]
} |
77,894 | When the Hamiltonian is Hermitian(i,e. beyond the effective mass approximation), generally under which conditions the eigenfunctions/wavefunctions are real? What happens in 1D case like the finite quantum well symmetric with respect to the origin might be an example. any general rule? further generalization into 2D case? | All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation,
$$
-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)$$
and therefore for any solution you can construct a second solution by taking the complex conjugate $\psi(\mathbf r)^\ast$. This second solution will either be linearly dependent on $\psi$, in which case $\psi$ differs from a real-valued function by a phase, or linearly independent, in which case you can "rotate" this basis into the two independent real-valued solutions $\operatorname{Re}(\psi)$ and $\operatorname{Im}(\psi)$. For continuum states this also applies, but things are not quite as clear as the boundary conditions are not invariant under conjugation: incoming scattering waves with asymptotic momentum $\mathbf p$, for example, behave asymptotically as $e^{i\mathbf p\cdot \mathbf r/\hbar}$, and this changes into outgoing waves upon conjugation. Thus, while you can still form two real-valued solutions, they will be standing waves and the physics will be quite different. In the second case, when you have a degeneracy, the physical characteristics of the real-valued functions are in general different to those of the complex-valued ones. For example, in molecular physics, $\Pi$ states typically have such a degeneracy: you can choose $\Pi_x$ and $\Pi_y$ states, which are real-valued, have a node on the $x$ and $y$ plane, resp., have a corresponding factor of $x$ and $y$ in the wavefunction, and have zero expected angular momentum component along the $z$ axis, or $\Pi_\pm$ states, which have a complex factor of $x\pm i y$ and no node, and have definite angular momentum of $\pm \hbar$ about the $z$ axis. Thus: you can always choose a real-valued eigenstate, but it may not always be the one you want. | {
"source": [
"https://physics.stackexchange.com/questions/77894",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27216/"
]
} |
78,442 | I am trying to understand intuitively what a phonon is, but for the moment I find it quite difficult (having a limited background in quantum mechanics, an undergraduate course in non-relativistic QM). In fact, I find it hard to formulate good questions, so I hope my questions below make some sense. I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes? Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Are the quantum mechanical frequency and wave vector the same as the frequency and wave vector of the corresponding classical oscillation (vibration in the crystal)? In what sense is it (like) a particle? In that it is always observed or it always interacts at a specific location? | I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes? The reason that phonons are described in terms of normal modes is because the phonon Hamiltonian looks nice in that basis. In other words, the normal mode basis diagonalises the phonon Hamiltonian:
$$H = \sum\limits_{\mathbf{k}} \omega_{\mathbf{k}} a^{\dagger}_{\mathbf{k}} a_{\mathbf{k}},$$ where the bosonic ladder operator $a^{\dagger}_{\mathbf{k}}$ creates a phonon with wavevector $\mathbf{k}$ and oscillation frequency $\omega_{\mathbf{k}}$. These are also the wavevector and oscillation frequency of the corresponding normal mode.$^{\ast}$ A general single-phonon state is a superposition of normal modes, and would be written
$$|\psi_1\rangle = \sum\limits_{\mathbf{k}} f(\mathbf{k}) \,a^{\dagger}_{\mathbf{k}}|0\rangle,$$
where $|0\rangle$ is the vibrational ground state of the lattice. A two-phonon state takes the form
$$|\psi_2\rangle = \sum\limits_{\mathbf{k},\mathbf{k}^{\prime}} f(\mathbf{k},\mathbf{k}^{\prime}) \,a^{\dagger}_{\mathbf{k}}a^{\dagger}_{\mathbf{k}^{\prime}}|0\rangle$$
etc. The functions $f$ can be considered like a "wave-function" in momentum space. However, there is only a limited analogy with the familiar wave functions describing, say, an electron bound to an atomic nucleus. Phonons are not conserved particles, so it is not possible to write down a "single-phonon Hamiltonian" governing the dynamics of $f(\mathbf{k})$. Phonons are collective excitations of a many-body system and must be treated within the quantum many-body formalism, in general. Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Regarding the position-space "wave-function", one can also define the position-space ladder operators (assuming periodic boundary conditions):
$$ a^{\dagger}(\mathbf{x}) = \sum\limits_{\mathbf{k}} e^{-i \mathbf{k}\cdot\mathbf{x}} a_{\mathbf{k}}^{\dagger}.$$
The state $ a^{\dagger}(\mathbf{x})|0\rangle$ describes a single phonon created at the position $\mathbf{x}$. Therefore the position-space "wave-function" of a single-phonon state is given by
$$ \langle 0|a(\mathbf{x}) |\psi_1\rangle = \sum\limits_{\mathbf{k}} e^{i\mathbf{k}\cdot\mathbf{x}} f(\mathbf{k}),$$
which is the Fourier transform of $f(\mathbf{k})$ (up to normalisation factors which I'm ignoring). So we have a nice analogy with the familiar rule for transforming wavefunctions from momentum to position space. The squared modulus of this "wave-function" gives information on the shape in position space of the compression and rarefaction profile throughout the crystal of a longitudinal vibration on average . However, since this is quantum mechanics, the "wave-function" really means the probability amplitude for finding a single phonon at position $\mathbf{x}$. The shape of the wave can only be built up after performing many measurements. For a normal-mode state you will find that the "wave-function" is $\sim e^{i\mathbf{k}\cdot\mathbf{x}}$, which is a plane wave that is indeed delocalised across the entire crystal. However, a more realistic phonon state that might arise, say, if I lightly tap the crystal in a certain position, would be a superposition of more than one frequency. This means that $f(\mathbf{k})$ has a finite width in momentum space, so that the position "wave-function" also has finite width. Of course, as the phonon state evolves over time this wavepacket will spread out as it moves through the crystal. $^{\ast}$In general one would also have to consider polarisation, but let's assume for simplicity that only longitudinal modes are present. * EDIT IN RESPONSE TO COMMENT * Would you say that mathematically there are some analogies between phonons and ordinary particles, but that you don't intuitively think of phonons as particles? Phonons are quasiparticles . They reduce a description in terms of interacting degrees of freedom (lattice ions) to a simpler description in terms of non-interacting collective excitations (phonons). (Of course, when electron-phonon interactions or other non-linearities are taken into account, the phonons cease to be free particles, but the description is still simpler.) Intuitively I think of phonons a lot like photons, which are collective excitations of the electromagnetic field. Phonons are collective excitations of the lattice displacement field. There are two key distinctions between phonons and fundamental particles like electrons. Firstly, phonons are an effective description that only makes sense above a certain length scale, the lattice spacing. If you look so closely that you can resolve the microscopic motion of individual lattice ions, then the description in terms of phonons is meaningless. The other distinction is that phonons are gapless (massless), which means you can create one with an arbitrarily small amount of energy. New electrons can only be created by processes involving energies larger than the electron rest mass. These energies are inaccessible at the low temperatures dealt with by condensed matter physicists. However, such energies are accessible in high energy physics, where one must replace the description of electrons by wave functions to a description in terms of quantum fields. Then electrons are viewed as collective excitations of the Dirac field, which exists at every point in space-time. So in relativistic quantum field theory the distinction between fundamental particles and collective excitations becomes blurred by the formalism. One should bear in mind, though, that an electron is considered a fundamental particle in the Standard Model, while a phonon is really a simplified description of the complicated quantised motion of an enormous number of lattice ions. This is because we know that phonons arise from a more fundamental structure, the crystal lattice, which we can observe directly in X-ray diffraction experiments. On the other hand, no experiment to date has revealed a more fundamental structure from which the electron field emerges. Nevertheless, the tight mathematical correspondence between collective excitations in low-energy condensed matter and fundamental particles at high energy has led some eminent condensed matter physicists (e.g. Laughlin, Wen) to suggest that the fundamental fields of the Standard Model are really effective low-energy (compared to the Planck scale) descriptions of a more fundamental structure of the quantum vacuum. This structure would only become apparent on length scales too small to be resolved with current technology. | {
"source": [
"https://physics.stackexchange.com/questions/78442",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/25794/"
]
} |
78,445 | If you look at the commutation relation of the position and momentum operators (in 1D position space), you get: $$[\hat{x}, \hat{p}_x] = [x,-i \hbar \frac{\partial}{\partial x}] = i \hbar$$ All this says to me is that if you prepare a system in state (A) and measure the position, the system is now in state (B), which is an eigenstate of the position operator. You then measure the momentum of state (B) and now you're in state (C), which is an eigenstate of the momentum operator. Note that these must be consecutive measurements. Alternatively, let's assume you reversed the order of the measurements. Beginning with state (A) again, you measure momentum first and put the system in state (D), an eigenstate of the momentum operator, but not necessarily state (C) [right?]. You then measure position, and put the system in state (E). $$(B) \neq (E)$$
$$(C) \neq (D)$$ That's all HUP says to me. It says nothing about simultaneous measurements of position and momentum — is that even possible? (What operator would that be?) It only says how much the final wavefunctions of two systems that started out the same differ if you perform two measurements in a different order. Where is the uncertainty? You know the position and momentum exactly — right when you measure them. You get some random value weighted by the coefficients of the eigenfunctions in the linear combination that forms $\psi$. So I don't think it's accurate to say you can't "simultaneously know position and momentum to arbitrary precision", because as far as I can tell, you can't even measure the two at the same time. | I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes? The reason that phonons are described in terms of normal modes is because the phonon Hamiltonian looks nice in that basis. In other words, the normal mode basis diagonalises the phonon Hamiltonian:
$$H = \sum\limits_{\mathbf{k}} \omega_{\mathbf{k}} a^{\dagger}_{\mathbf{k}} a_{\mathbf{k}},$$ where the bosonic ladder operator $a^{\dagger}_{\mathbf{k}}$ creates a phonon with wavevector $\mathbf{k}$ and oscillation frequency $\omega_{\mathbf{k}}$. These are also the wavevector and oscillation frequency of the corresponding normal mode.$^{\ast}$ A general single-phonon state is a superposition of normal modes, and would be written
$$|\psi_1\rangle = \sum\limits_{\mathbf{k}} f(\mathbf{k}) \,a^{\dagger}_{\mathbf{k}}|0\rangle,$$
where $|0\rangle$ is the vibrational ground state of the lattice. A two-phonon state takes the form
$$|\psi_2\rangle = \sum\limits_{\mathbf{k},\mathbf{k}^{\prime}} f(\mathbf{k},\mathbf{k}^{\prime}) \,a^{\dagger}_{\mathbf{k}}a^{\dagger}_{\mathbf{k}^{\prime}}|0\rangle$$
etc. The functions $f$ can be considered like a "wave-function" in momentum space. However, there is only a limited analogy with the familiar wave functions describing, say, an electron bound to an atomic nucleus. Phonons are not conserved particles, so it is not possible to write down a "single-phonon Hamiltonian" governing the dynamics of $f(\mathbf{k})$. Phonons are collective excitations of a many-body system and must be treated within the quantum many-body formalism, in general. Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Regarding the position-space "wave-function", one can also define the position-space ladder operators (assuming periodic boundary conditions):
$$ a^{\dagger}(\mathbf{x}) = \sum\limits_{\mathbf{k}} e^{-i \mathbf{k}\cdot\mathbf{x}} a_{\mathbf{k}}^{\dagger}.$$
The state $ a^{\dagger}(\mathbf{x})|0\rangle$ describes a single phonon created at the position $\mathbf{x}$. Therefore the position-space "wave-function" of a single-phonon state is given by
$$ \langle 0|a(\mathbf{x}) |\psi_1\rangle = \sum\limits_{\mathbf{k}} e^{i\mathbf{k}\cdot\mathbf{x}} f(\mathbf{k}),$$
which is the Fourier transform of $f(\mathbf{k})$ (up to normalisation factors which I'm ignoring). So we have a nice analogy with the familiar rule for transforming wavefunctions from momentum to position space. The squared modulus of this "wave-function" gives information on the shape in position space of the compression and rarefaction profile throughout the crystal of a longitudinal vibration on average . However, since this is quantum mechanics, the "wave-function" really means the probability amplitude for finding a single phonon at position $\mathbf{x}$. The shape of the wave can only be built up after performing many measurements. For a normal-mode state you will find that the "wave-function" is $\sim e^{i\mathbf{k}\cdot\mathbf{x}}$, which is a plane wave that is indeed delocalised across the entire crystal. However, a more realistic phonon state that might arise, say, if I lightly tap the crystal in a certain position, would be a superposition of more than one frequency. This means that $f(\mathbf{k})$ has a finite width in momentum space, so that the position "wave-function" also has finite width. Of course, as the phonon state evolves over time this wavepacket will spread out as it moves through the crystal. $^{\ast}$In general one would also have to consider polarisation, but let's assume for simplicity that only longitudinal modes are present. * EDIT IN RESPONSE TO COMMENT * Would you say that mathematically there are some analogies between phonons and ordinary particles, but that you don't intuitively think of phonons as particles? Phonons are quasiparticles . They reduce a description in terms of interacting degrees of freedom (lattice ions) to a simpler description in terms of non-interacting collective excitations (phonons). (Of course, when electron-phonon interactions or other non-linearities are taken into account, the phonons cease to be free particles, but the description is still simpler.) Intuitively I think of phonons a lot like photons, which are collective excitations of the electromagnetic field. Phonons are collective excitations of the lattice displacement field. There are two key distinctions between phonons and fundamental particles like electrons. Firstly, phonons are an effective description that only makes sense above a certain length scale, the lattice spacing. If you look so closely that you can resolve the microscopic motion of individual lattice ions, then the description in terms of phonons is meaningless. The other distinction is that phonons are gapless (massless), which means you can create one with an arbitrarily small amount of energy. New electrons can only be created by processes involving energies larger than the electron rest mass. These energies are inaccessible at the low temperatures dealt with by condensed matter physicists. However, such energies are accessible in high energy physics, where one must replace the description of electrons by wave functions to a description in terms of quantum fields. Then electrons are viewed as collective excitations of the Dirac field, which exists at every point in space-time. So in relativistic quantum field theory the distinction between fundamental particles and collective excitations becomes blurred by the formalism. One should bear in mind, though, that an electron is considered a fundamental particle in the Standard Model, while a phonon is really a simplified description of the complicated quantised motion of an enormous number of lattice ions. This is because we know that phonons arise from a more fundamental structure, the crystal lattice, which we can observe directly in X-ray diffraction experiments. On the other hand, no experiment to date has revealed a more fundamental structure from which the electron field emerges. Nevertheless, the tight mathematical correspondence between collective excitations in low-energy condensed matter and fundamental particles at high energy has led some eminent condensed matter physicists (e.g. Laughlin, Wen) to suggest that the fundamental fields of the Standard Model are really effective low-energy (compared to the Planck scale) descriptions of a more fundamental structure of the quantum vacuum. This structure would only become apparent on length scales too small to be resolved with current technology. | {
"source": [
"https://physics.stackexchange.com/questions/78445",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/960/"
]
} |
78,446 | This is part of a homework question which I've been stuck on for several hours. I've tried googling "lowest point of rope", "lowest point of hanging cable", "lowest point of pulley", and a bunch of other combinations without luck. Cable ABC has a length of 5m. The cable is attached to a wall on the
left at A, and attached to a wall on the right at C, 0.75m above the
vertical position of A (so C is attached at a higher location on the
wall). The distance between the walls is 3.5m. A 100kg sack is hanging
by a pulley on this cable at equilibrium, at B. Find the horizontal
distance x of the pulley from the left wall (neglect the size of the
pulley). Intuitively I think the the pulley would hang at the location where it's closest to the ground (hence "lowest point of loose cable..."). However, I have no idea how to calculate it. We're only allowed scientific calculators (no graphing) and whenever I try to set up an equation it blows up. Once I figure out x it should be relatively easy to calculate the component forces for equilibrium. I tried looking for examples in the textbook and internet for something like this without luck. | ---- 3.5m -----|
---D-------------* <- C
| | /|
| <- 0.75m / |
* <- A / |
|\ | / F
| \|/
| * <- B
E |
_______
| 100kg |
|--| <- x
Length of cable: 5m Here's a text diagram, as best as I could make it Update : Found a hint from the textbook - (3.5 - x)/cos(o) + x/cos(o) = 5 . Not quite sure what to make of it, but it does kinda remind me of an ellipse at a slant... https://math.stackexchange.com/questions/108270/what-is-the-equation-of-an-ellipse-that-is-not-aligned-with-the-axis Update 2 : Upon closer inspection of aufkag's angle-suggestion and the hint from the textbook, I believe he is correct about the angles being equal - the formula calculates the two segments of the rope from the adjacent sides x and 3.5 - x . By the way, how can it be explained or "proved" or what's the law that says the angles between AB and the wall and AC and the wall in a setup like this are equal? Update 3 : (after solved, see comment for aufkag): Added D, E, F. ABD = BAE and CBD = BCF, but can anyone prove or point out the law that says ABD = CBD or BAE = BCF? Anyways, the steps are: o = angle AB and the horizontal or BC and the horizontal
x / cos(o) + (3.5 - x) / cos(o) = 5 (sum of segments of rope is 5)
tension in AB = tension in BC, therefore they share the same "load" of the mass, so we can calculate the tension in just one side
100 * 9.81 / 2 / sin(o) = 687N (approximately - first half of answer)
0.75 + xtan(o) = (3.5 - x)tan(o) (equal lengths for line segment BD)
solve for x to get 1.38m | I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes? The reason that phonons are described in terms of normal modes is because the phonon Hamiltonian looks nice in that basis. In other words, the normal mode basis diagonalises the phonon Hamiltonian:
$$H = \sum\limits_{\mathbf{k}} \omega_{\mathbf{k}} a^{\dagger}_{\mathbf{k}} a_{\mathbf{k}},$$ where the bosonic ladder operator $a^{\dagger}_{\mathbf{k}}$ creates a phonon with wavevector $\mathbf{k}$ and oscillation frequency $\omega_{\mathbf{k}}$. These are also the wavevector and oscillation frequency of the corresponding normal mode.$^{\ast}$ A general single-phonon state is a superposition of normal modes, and would be written
$$|\psi_1\rangle = \sum\limits_{\mathbf{k}} f(\mathbf{k}) \,a^{\dagger}_{\mathbf{k}}|0\rangle,$$
where $|0\rangle$ is the vibrational ground state of the lattice. A two-phonon state takes the form
$$|\psi_2\rangle = \sum\limits_{\mathbf{k},\mathbf{k}^{\prime}} f(\mathbf{k},\mathbf{k}^{\prime}) \,a^{\dagger}_{\mathbf{k}}a^{\dagger}_{\mathbf{k}^{\prime}}|0\rangle$$
etc. The functions $f$ can be considered like a "wave-function" in momentum space. However, there is only a limited analogy with the familiar wave functions describing, say, an electron bound to an atomic nucleus. Phonons are not conserved particles, so it is not possible to write down a "single-phonon Hamiltonian" governing the dynamics of $f(\mathbf{k})$. Phonons are collective excitations of a many-body system and must be treated within the quantum many-body formalism, in general. Could we say that a phonon is a particle whose position wave function extends over the whole crystal? Regarding the position-space "wave-function", one can also define the position-space ladder operators (assuming periodic boundary conditions):
$$ a^{\dagger}(\mathbf{x}) = \sum\limits_{\mathbf{k}} e^{-i \mathbf{k}\cdot\mathbf{x}} a_{\mathbf{k}}^{\dagger}.$$
The state $ a^{\dagger}(\mathbf{x})|0\rangle$ describes a single phonon created at the position $\mathbf{x}$. Therefore the position-space "wave-function" of a single-phonon state is given by
$$ \langle 0|a(\mathbf{x}) |\psi_1\rangle = \sum\limits_{\mathbf{k}} e^{i\mathbf{k}\cdot\mathbf{x}} f(\mathbf{k}),$$
which is the Fourier transform of $f(\mathbf{k})$ (up to normalisation factors which I'm ignoring). So we have a nice analogy with the familiar rule for transforming wavefunctions from momentum to position space. The squared modulus of this "wave-function" gives information on the shape in position space of the compression and rarefaction profile throughout the crystal of a longitudinal vibration on average . However, since this is quantum mechanics, the "wave-function" really means the probability amplitude for finding a single phonon at position $\mathbf{x}$. The shape of the wave can only be built up after performing many measurements. For a normal-mode state you will find that the "wave-function" is $\sim e^{i\mathbf{k}\cdot\mathbf{x}}$, which is a plane wave that is indeed delocalised across the entire crystal. However, a more realistic phonon state that might arise, say, if I lightly tap the crystal in a certain position, would be a superposition of more than one frequency. This means that $f(\mathbf{k})$ has a finite width in momentum space, so that the position "wave-function" also has finite width. Of course, as the phonon state evolves over time this wavepacket will spread out as it moves through the crystal. $^{\ast}$In general one would also have to consider polarisation, but let's assume for simplicity that only longitudinal modes are present. * EDIT IN RESPONSE TO COMMENT * Would you say that mathematically there are some analogies between phonons and ordinary particles, but that you don't intuitively think of phonons as particles? Phonons are quasiparticles . They reduce a description in terms of interacting degrees of freedom (lattice ions) to a simpler description in terms of non-interacting collective excitations (phonons). (Of course, when electron-phonon interactions or other non-linearities are taken into account, the phonons cease to be free particles, but the description is still simpler.) Intuitively I think of phonons a lot like photons, which are collective excitations of the electromagnetic field. Phonons are collective excitations of the lattice displacement field. There are two key distinctions between phonons and fundamental particles like electrons. Firstly, phonons are an effective description that only makes sense above a certain length scale, the lattice spacing. If you look so closely that you can resolve the microscopic motion of individual lattice ions, then the description in terms of phonons is meaningless. The other distinction is that phonons are gapless (massless), which means you can create one with an arbitrarily small amount of energy. New electrons can only be created by processes involving energies larger than the electron rest mass. These energies are inaccessible at the low temperatures dealt with by condensed matter physicists. However, such energies are accessible in high energy physics, where one must replace the description of electrons by wave functions to a description in terms of quantum fields. Then electrons are viewed as collective excitations of the Dirac field, which exists at every point in space-time. So in relativistic quantum field theory the distinction between fundamental particles and collective excitations becomes blurred by the formalism. One should bear in mind, though, that an electron is considered a fundamental particle in the Standard Model, while a phonon is really a simplified description of the complicated quantised motion of an enormous number of lattice ions. This is because we know that phonons arise from a more fundamental structure, the crystal lattice, which we can observe directly in X-ray diffraction experiments. On the other hand, no experiment to date has revealed a more fundamental structure from which the electron field emerges. Nevertheless, the tight mathematical correspondence between collective excitations in low-energy condensed matter and fundamental particles at high energy has led some eminent condensed matter physicists (e.g. Laughlin, Wen) to suggest that the fundamental fields of the Standard Model are really effective low-energy (compared to the Planck scale) descriptions of a more fundamental structure of the quantum vacuum. This structure would only become apparent on length scales too small to be resolved with current technology. | {
"source": [
"https://physics.stackexchange.com/questions/78446",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30087/"
]
} |
78,536 | I was hoping someone could give an overview as to how the Lie groups $SO(3)$ and $SU(2)$ and their representations can be applied to describe particle physics? The application of Lie groups and their representations is an enormous field, with vast implications for physics with respect to such things as unification, but I what specifically made these groups of physical importance and why there study is useful. I have just started studying these two groups in particular, but from a mathematical perspective, I'd very much appreciate understanding some sort of physical motivation. | QuantumMechanic's links turn up a dizzying array of meanings for $SU(2)$ in physics, so your question probably turns out to be too broad for a simple answer. Nonetheless, I do like it, and similar questions that grope for pithy overviews of things, so I'll try to answer it with my non particle physicist's understanding. Probably the "main" meaning of $SU(2)$ you're going to find is as the (or highly nontrivial part of the) gauge group of certain Yang-Mills-kind theories (see Yang-Mills Wiki page ), notably: The $SU(2) \times SU(1)$ gauge group of the electroweak interaction (see Wiki page of this name) . Here the three orthonormal (wrt the Killing form) Lie algebra basis vectors of $SU(2)$ correspond to the three W-bosons. Isospin Symmetry (see Wiki page "Isospin" The group of approximate symmetries leaving invariant the strong interaction Hamiltonian, as formulated by Heisenberg in 1932. The proton and neutron "live in" the fundamental representation of $SU(2)$ (I'm guessing you're a mathematician - so in case you haven't yet picked this up, physicists are wont to mean the vector space that images of group members under the representation act on as the "representation" - it took me a while to grasp this), whereas the three pions live in the adjoint representation of $SU(2)$, i.e. they are transformed by corresponding members of $SO(3)$. The protons and neutrons have spin $1/2$, being spinors, and they can be thought of as basis vectors for a $\mathbb{C}^2$ state space, which is acted on by a group member $\gamma$ simply through $X\in\mathbb{C}^2\mapsto \gamma\,X$. The three neutral pions are basis vectors in an $\mathbb{R}^3$ state space which is acted on by $Y\in\mathbb{R}^3\mapsto\mathrm{Ad}(\gamma)\,Y$. In the case of gauge theories, my understanding of their importance (the ones of the Yang-Mills kind with a finite dimensional structure group) is in this answer . If you're a bit slow on the uptake like me, you may need someone to point out to you that "all we're doing" in constructing a gauge theory is building a fibration on a physical observable theory and the gauge group is nothing more than the fibre bundle's structure group: we put hair on a theory and see what beautiful braids we can make with it. (Yes, I really did need someone to point this out explicitly to me, even though I have a reasonable grasp of fibre bundles!) But why do we do this, i.e. on the surface seem to add complexity, when it would seem the aim of physics to simplify things, not kit them with more hair (complexity)? There are two answers here: There is a known classical gauge theory - Maxwell's electromagnetism with the $U(1)$ symmetry - whose curious gauge symmetry we seek to take to other physics, just as a "suck and see" mathematical physics analogy; There are either (i) experimentally observed continuous symmetries or (ii) conserved quantities in physically observed processes, so we add the fibration as a way to beget these symmetries or conserved quantites in theory. In the case of observed conserved quantities, this procedure works through Noether's theorem , but it is important to understand the implication through Noether's theorem is only one way: a Lagrangian with continuous symmetries implies the same number of conserved quantities, but a conserved quantity doesn't needfully imply a continuous symmetry. Again, it is a suck and see approach - we know one way to force a conserved quantity in a theory - to wit: adding a fibration or gauge symmetry - so we try it and see what happens, and it so happens that experimentally the theories built in this way work rather well (the Standard Model). Other resources that I found helpful - particularly if you haven't already delved into gauge theories - are the following: John C. Baez and John Huerta, "The Algebra of Grand Unified Theories" Gerard 't Hooft's "Lie Groups in Physics" Terrence Tao's blog "What is a gauge?" "Gauge Theory" Wikipedia Page "Introduction to Gauge Theory" Wikipedia Page Summary of Gerard 't Hooft's 1999 Nobel Lecture Relevant chapters in Roger Penrose's "Road To Reality" (I don't have it before me at the moment). I found the first two papers by Baez/Huerta and 't Hooft invaluable here. Like I said, I am not a particle physicist but after reading this I feel I can at least follow many discussions in this field without too much (let's say < 80%) going over my head. Thanks to John Baez and his wonderful literature, I think that withering away in a nursing home is not going to be too bad, if I can still read by then! (this isn't in the offing BTW). I find almost anything written on physics and its relationship with mathematics by Baez, 't Hooft and Penrose well worth reading. There was (likely still is) an excellent introduction to gauge theory on Gerard 't Hooft's webpage but the webpage itself is a bit hard to navigate and I can't find it at the moment - I guess such disorganisation is inevitable for someone as polymathematic as 't Hooft is wanting to share so much varied material. But maybe the deepest, simplest and (for me, the most beautiful) meaning of all for $SO(3)$ and $SU(2)$ is the simple relationship between the two groups, the one being the universal cover of the other (see my answer here) , as was taught to me indirectly by a seven year old boy (the depth of physical meaning, rather than the universal cover property) when I was demonstrating the Dirac belt trick and cup tricks at my daughter's school and he asked the question "can you make a fancier arrangement of ribbons so that you have to spin her [the doll] three times rather than twice to get back to the start?" (I use a doll on a ribbon rather than just a marked card with children because, as social animals, we're hard wired to ken a face, so keeping track of the spins is unmistakable with a doll. Many smallish children of about six years old and over find the belt trick really enthralling, BTW.) I was thoroughly impressed by his question and wished I could answer it better for him. But as far as particles are concerned, the answer is the same: an emphatic no: there are only half integer spins, not spin $1/3$ and so forth, because $SU(2)$ is the universal cover of $SO(3)$. There are only bosons and fermions in the World, and the double cover relationship between $SO(3)$ and $SU(2)$ is why - "a simply connected topological space admits no non trivial coverings" to quote from W. S. Massey, "Algebraic Topology: An Introduction" - so the universal cover is the whole gig! The belt trick works because the evolution of the Serret-Frenet frames along the twisted ribbon encodes a continuous path through $SO(3)$ from the identity to the transformation defined by the doll's orientation in space and so the ribbon precisely encodes the homotopy class of this path . If you can loop it over the doll (deform the path continuously) and undo the twists, the ribbon is still encoding the same homotopy class. The belt trick is a precise physical analogy to the mathematical procedure for building a universal cover. So this humble observation about the relationship between $SO(3)$ and $SU(2)$ explains all the following: There are no other spin $1/3$, or any $1/N$ aside from $1/2$, ribbons realisable in a Dirac belt trick; Spinors and tensors exhaust the list of everything that transforms compatibly with rotations . Actually the idea broadens from the $SO(3)$ with $SO(2)$ relationship to general proper Lorentz transformations: we add boosts to the mix and get the identity connected component of the Lorentz group $O(3,1) \cong PSL(2, \mathbb{C}) \cong \operatorname{Aut}(\mathbb{C})$ (the latter being the group of invertible Möbius transformations) and the double cover of this beast is $SL(2, \mathbb{C})$, so spinors and tensors exhaust the list of everything that transforms compatibly with rotations, boosts and general combinations thereof; and There are only bosons and fermions - i.e. only particles with half integer or whole number spins in the World. Truly I find this simple relationship is a little jewel to behold. There is a footnote in Chapter 17 of the third volume "the Feynman Lectures on Physics" where Feynman says he had been trying to find a simple demonstration that there are only half integer spins and had failed - "We'll have to talk about it with Prof. Wigner, who knows all about such things"!, he ends the footnote. I rather think Feynman, from what I know of his work and sense of humour, would have been delighted to have the explanation suggested to him by a seven year old, were he alive. Lastly, I'd just like to mention how $SU(2)$ and $SO(3)$ show up in my own field of optics and electromagnetism. It's not quite what people wontedly mean by "particle physics" but it is an application in the physics of the photon. The general polarisation state of a one-mode electromagnetic field $\Psi = \left(\begin{array}{c}\psi_+\\\psi_-\end{array}\right)$ can be encoded in two complex amplitudes, one for each circular polarisation's basis state (or, amplitudes of the two Riemann-Silberstein vectors for a given wavevector more as discussed in my answer here ). A lossless polarisation transformer (waveplate, mirror system, and so forth) must impart a general unitary transformation on these two amplitudes, for the sum of their square magnitudes is the wave's power. Often, we're not worried about phase that is common to both polarisation eigenstates, so we can think of the matrix of our polarisation transformer as living in $SU(2)$ rather than $U(2) \cong SU(2) \otimes U(1)$, but the Jones calculus actually handles $U(2)$ as well: $$\Psi \mapsto \Psi^\prime = \mathbf{U}\,\Psi;\;U\in SU(2)$$ In this context, $\mathbf{U}$ is called the transformer's Jones Matrix . We can also represent the polarisation state by the Stokes parameters: $$s_j(\Psi) = \Psi^\dagger \sigma_j \Psi$$
$$\begin{array}{l}
s_0 = \Psi^\dagger\,\Psi = |\Psi|^2\\
s_1 = 2 \operatorname{Re}(\psi_+^*\,\psi_-)\\
s_2 = 2 \operatorname{Im}(\psi_+^*\,\psi_-)\\
s_3 = |\psi_+|^2 - |\psi_-|^2
\end{array}$$ where $\sigma_j$ are the Pauli spin matrices (here $\sigma_j;\,j=1,\,2,\,3$ are the matrices on the Pauli Matrix Wiki page and $\sigma_0$ is the $2\times2$ identity); $i\,\sigma_j;\,j=1,\,2,\,3$ of course span $\mathfrak{su}(2)$ and $i\,\sigma_j;\,j=0,\,1,\,2,\,3$ span $U(2)$. This definition of the Stokes parameters is slightly different to that wontedly given in optics ( e.g. section 1.4 of Born and Wolf, "Principles of Optics" sixth edition ); there is an unimportant sign switch and a renumbering. The Pauli spin matrices $i \sigma_1,\,i \sigma_2,\, i \sigma_3$ are a basis for $\mathfrak{su}(2)$ and $U$ can be written as
$U = \exp(-i \theta \sum \gamma_j \sigma_j/2);\;\theta,\,\gamma_j\in\mathbb{R},\;\sum\gamma_j^2 = 1$. If the system input is $\Psi$, then, after transformation by $U$, its Stokes parameters are transformed by the spinor map: $$s_k = \Psi^\dagger U^\dagger \sigma_k U \Psi =
\Psi^\dagger U^{-1} \sigma_k U \Psi =
- i \Psi^\dagger
\exp\left(i \frac{\theta}{2} \sum_j \gamma_j \sigma_j\right)
i \sigma_k
\exp\left(-i \frac{\theta}{2} \sum_j \gamma_j \sigma_j\right) \Psi
$$ or, alternatively, the unit sphere of Stokes vectors $(s_1,\,s_2,\,s_3)^T$ is transformed by precisely the rotation $\exp(\theta \,\mathbf{H})$ matrix corresponding to $\mathbb{U}$ when the latter is mapped by the standard adjoint representation homomorphism: $$\exp(\theta \,\mathbf{H}) = \exp\left(\theta \left(\begin{array}{ccc}0&\gamma_z&-\gamma_y\\-\gamma_z&0&\gamma_x\\\gamma_y&-\gamma_x&0\end{array}\right)\right)$$ so that we can visualise polarisation state changes as rotations of the unit sphere, as long as we are happy to be blind to the difference between a transformation $\mathbf{U}$ and its negative $-\mathbf{U}$, i.e. we are happy to see only cosets of this homomorphism's kernel. A slight generalisation of this procedure is to use Mueller Calculus(see Wiki page "Mueller calculus" , which is the density matrix notation in disguise and can deal with partially polarised light states, which are classical statistical mixtures of pure quantum states. I describe this aspect of the Mueller Calculus in my answer here . | {
"source": [
"https://physics.stackexchange.com/questions/78536",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/7270/"
]
} |
78,586 | Helium is lighter than air, so it should fly off from Earth.
Is it possible that in the future we will run out helium? | Yes, helium can leave the Earth, and yes, we will run out of helium, but because of different reasons. When you buy a helium balloon and its contents get released, this helium goes into the atmosphere. It isn't gone, and it could in principle be purified out of normal air. However, the total amount of helium in the atmosphere is so small it is technologically not feasible to reclaim it. At some point the technology might be developed, but it is unlikely to be economical. On top of that, helium does also escape from the atmosphere. Since it is so light, it drifts naturally to the upper layers, and there it is easily torn away by the solar wind. However, this process will occur on geological timescales, unless we were to waste so much helium that the total atmospheric content changed appreciably. Keep in mind, though, that even if the helium doesn't leave Earth it is lost to us once it's diluted in the atmosphere. So: yes, we will run out, and yes, it will make everything awful. And yes, you should cringe when you see helium balloons at a childrens' party. | {
"source": [
"https://physics.stackexchange.com/questions/78586",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28172/"
]
} |
78,684 | Every so often, one sees on this site* or in the news † or in journal articles ‡ a statement of the form "we have measured a change in such-and-such fundamental constant" (or, perhaps more commonly, "we have constrained the rate of change of..."). These are often shrouded in controversy, with people often stating quite loudly that statements of the sort are meaningless and even unverifiable, but the end result is often quite confusing. Thus: Is it possible to speak about changes to a physical constant that contains dimensional information? If not, why not? * These are some examples but there's more .$\ \ $† example .$\ \ $‡ example . | The problem is that any statement that a dimensionful physical constant has changed is meaningless if you do not state what it is you are keeping constant . When you do state it, and you work it out in full, it turns out you're measuring the change of a dimensionless physical constant. Suppose, for example, that I claim "the speed of light has increased by 10% since last year". My experimental evidence for this is simple: I have a ruler and a clock, and every day I use them (the exact same ruler and clock) to measure the speed of light. I then calculate the time $T$ - as a quantity, in seconds, given by my clock - taken by light to cross the ruler, and I divide it into the length $L$ of the ruler. Over a year, this quantity $L/T$ increases by 10%. For the metrologically minded, let's take a look at my experimental apparatus: My ruler consists, essentially, of a bunch of atoms in a linear arrangement. For example, I could make my ruler by putting a bunch of hydrogen atoms in a line, separated from each other by one Bohr radius (which can be experimentally determined, at least in principle, as the distance at which the ground state charge density decreases to $e^{-2}$ of the maximum). If my ruler is 1m long, it will have $N=1\:\mathrm m/a_0\approx1.89\times10^{10}$ atoms in it. Moreover, since it's the same ruler, and because I'm lazy, I don't recalibrate it to the SI every day before I measure (which would have catastrophic consequences for my result!). Instead, what I do before I measure is look at all the bonds to ensure they're still one Bohr radius long, and re-count the atoms to ensure the ruler still has the exact same number $N$ of atoms. I should also note that, despite sounding esoteric, this ruler is close to the best possible model for an actual physical ruler made of platinum or whatever, since its length is governed by the same physics (nonrelativistic QM plus electrostatics, a.k.a. chemistry) that governs the size of all everyday objects. It's been distilled into a form that's clearly definable, but the essence of the definition is along the lines of "this ruler here, so long as the ends don't get worn out and it doesn't get bent and there's no monkey business with thermal expansion or whatever". My clock is a caesium clock as per the SI second. That is, it has a bunch of caesium atoms which are placed in a specific state (technically, a coherent superposition of their ground state at and the first hyperfine excited state) which emits microwave radiation. The clock measures this radiation and counts the number of maxima; every 9,192,631,770 cycles it increases the counter by 1 second. With this apparatus, then, I observe the measured speed of light to change. What does this result mean? The easiest interpretation is simple: The speed of light has changed. Light simply travels faster than it did last year. That's pretty mysterious, but then the result is pretty weird too. However, there are other possible interpretations for the result. For example, the size of the hydrogen atoms might have changed. That is, the speed of light is still the same, but for some mysterious reason all my hydrogen atoms are 10% bigger than they were. This is not that crazy at all: the Bohr radius is determined by the Schrödinger equation under the electrostatic force, so their constants $\hbar,m_e$ and $e^2$ determine $a_0$ to be $a_0=\hbar^2/m_ee^2$. If any of those changed - say, all electrons are suddenly 10% lighter, which is as mysterious as light being faster - then I'd see exactly the same result I do observe. It's important to note that this is the reading of my result that would come from a strict application of the current SI system of units, since the SI meter is defined as the distance covered by light in 1s as given by my clock. However, if my ruler has "shrunk" then so have I (as I'm made of atoms) and so has every piece of equipment in my lab and elsewhere on Earth. I could then speak of the "mysterious growth of the SI meter" equally meaningfully. Alternatively, the caesium atoms could be getting progressively more sluggish. They could simply no longer be giving out microwave peaks and troughs as fast as they used to (or as I'd see if I teleported myself to the past), so even though the speed of light is the same, the length of the ruler is the same, and the traversal time is the same, the counter on the clock now reads 10% less seconds than it would have a year ago. These three explanations are all equally mysterious, or equally reasonable, as each other. Moreover, from my experiment I have no way to test which is right, as I have no way of teleporting myself back to the past to compare my (possibly) engorged hydrogens or my (allegedly) sluggish caesiums with their previous versions, in the same way that I cannot set up a race between my (supposedly) faster light and the light of yesteryear. It is clear that something in physics has indeed changed , but you can pan the change into different factors depending on your perspective. As it turns out, of course, the thing in physics which has changed, and which is the only thing that can unambiguously be said to have changed, is the fine structure constant. This $\alpha$, as any atomic physicist will tell you, is one over the speed of light in atomic units, which is exactly what we're measuring. More specifically,
$$\alpha=\frac{e^2}{\hbar c},$$
and $e^2/\hbar$ is easily seen to be the atomic unit of velocity. This abstract 'atomic unit' is of course a very physical quantity: it is, within a constant, calculable factor, the mean velocity of any electron around its atom. Now, as it happens, this crazy experiment I proposed is indeed being performed. The actual experimental realization does not rely on meter-long rulers or on external clocks, but it relies instead on the natural length and time scales of ytterbium ions, which of course are completely determined by the same length and time scales as all atomic physics. Atoms, it turns out, have natural built-in velocimeters for measuring the speed of light: their kinetic energies $p^2/2m$ change by relativistic corrections on the order of $p^4/8m^3c^2$, and the different electronic and spin currents have magnetic interactions on the order of $p/c$. Different states respond differently, and even in different directions, to these perturbations, so it's possible to monitor $c$ by observing the precise location of the different energy levels. (For more information, see NPL | Physics | PRL | arXiv .) There is, so far, no observed change in $\alpha$. But if there is, we simply won't be able to tell whether a change in $\alpha=e^2/\hbar c$ is because of a change in the speed of light $c$, the size $\hbar$ of the fundamental phase-space cell, or the strength $e^2$ of the electrostatic interaction, or of the finely-tuned joint changes of these constant that would implement the three implementations stated above. Those individual changes have no meaning by themselves. Finally, some useful references for further reading are How fundamental are fundamental constants? M.J. Duff, Contemp. Phys. 56 no. 1, 35-47 (2014) , arXiv:1412.2040 , which supersedes arXiv:hep-th/0208093 (Comment on time-variation of fundamental constants, 2002), and Trialogue on the number of fundamental constants. M.J. Duff, L.B. Okun and G. Veneziano, J. High Energy Phys. 03 (2002) 023 , arXiv:physics/0110060 . | {
"source": [
"https://physics.stackexchange.com/questions/78684",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/8563/"
]
} |
79,331 | The title says it all. If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right? So my son posed me an interesting question today: since light travels as fast as anything can go, what if I shined light when moving in a car? How should I answer his question? | If I was on a bus at 60 km/h, and I started walking on the bus at a
steady pace of 5 km/h, then I'd technically be moving at 65 km/h,
right? Not exactly right. You would be correct if the Galilean transformation correctly described the relationship between moving frames of reference but, it doesn't. Instead, the empirical evidence is that the Lorentz transformation must be used and, by that transformation, your speed with respect to the ground would be slightly less than 65 km/h. According to the Lorentz velocity addition formula , your speed with respect to the ground is given by: $$\dfrac{60 + 5}{1 + \dfrac{60 \cdot 5}{c^2}} = \dfrac{65}{1 + 3.333 \cdot 10^{-15}} \text{km}/\text h \approx 64.9999999999998\ \text{km}/\text h$$ Sure, that's only very slightly less than 65 km/h but this is important to your main question because, when we calculate the speed of the light relative to the ground we get: $$\dfrac{60 + c}{1 + \dfrac{60 \cdot c}{c^2}} = c$$ The speed of light, relative to the ground remains c ! | {
"source": [
"https://physics.stackexchange.com/questions/79331",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/11890/"
]
} |
79,733 | I have took a discrete mathematics course this summer and there we talked about power of groups and functions,and yesterday I though and realize that if we can map all the 3D coordinates with a one on one and surjective function from $(x,y,z)$ to $(a)$, we can prove that our planet is 1D, that's mind blowing. Is that proof is right? | I think that you are referring to space-filling curves and how they can map a line segment to more than one dimension. For example the Hilbert space filling curve can be used to map the interval $[0,1]$ to $[0,1]\times[0,1]$. I am afraid while a continuous bijection is possible one-way, it is not possible to have a homeomorphism between two different Euclidean spaces of different dimensions. A homeomorphism is a mapping that is continuous, bijective and has its inverse continuous. You cannot construct a homeomorphism. Thus the Earth cannot be 1D! See: Topological properties of Real Coordinate Space | {
"source": [
"https://physics.stackexchange.com/questions/79733",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/28172/"
]
} |
79,860 | a while ago I tried to charge a glow-in-the-dark sticker using a simple red laser pointer. It was a large sticker, of the type used to mark emergency exits and fire extinguishers here in Germany. I thought I could draw a picture this way, with glowing strokes. To my surprise, the exact opposite happened - the sticker was already glowing a little bit and the areas I illuminated with my laser turned dark instantly, as if the charge of the sticker was removed by the laserlight. The effect was reversible, the darkened areas would charge and glow as usual if exposed to normal light. So I was able to draw a picture after all - however, it was a negative. Can anyone explain why the sticker was discharged instead of charged by the laserlight? | How interesting. Presumably the "glow-in-the-dark" effect comes from the decay of a meta-stable excited state. It gets charged by sufficiently energetic photons, and decays slowly because some selection rule prevents a direct transition without an external influence. If this is the case, we can guess that the laser is exciting the meta-stable state to a still higher energy state which is not subject to the selection rule so that it promptly falls to the ground state 1 , from which the low energy photons in the laser don't have the energy to push back into the meta-stable state. I think I'll try to make a demo out of that. Thanks. 1 Presumably you could see the light of this decay if you filtered out the intense red from the laser. Try a green or blue filter. Or a spectrograph. | {
"source": [
"https://physics.stackexchange.com/questions/79860",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30667/"
]
} |
80,159 | The Earth is billions of years old, yet its core has not yet cooled down and become solid. Will this happen in the foreseeable future? | I generally take "foreseeable future" to mean "in my lifetime" in which case the answer is No . However, if you are really asking, "What does science say about the solidification 1 of earth's core?" then we can answer this. In the core of the planet, we actually have this pretty picture: The "solid inner core" is pretty much pure iron at a nice pretty temperature of about 5700 K (strangely close to the sun's surface temperature ), which seems to be greater than iron's melting point of about 1800 K, but the Clausius-Clapeyron relation shows that melting points change with an increase in pressure, so all is well there. The reason the heat has been retained in the core is that the only processes to move the heat are convective transport and thermal conduction , the latter being a fairly slow process (aside: it's the convective currents that cause the magnetic field ). Degeun et al (2013) state, ...the dynamical time-scale of the thermal convection in the inner core [is] $\sim$1 My or more. Which, though shorter than the age of the planet, is still pretty long. But the inner core is also growing at a rate of about 0.5 mm/year (possible paywall) and has been doing so for about a billion years. Though unreferenced, this Earth.StackExchange answer suggests that the core has cooled only 250 K since it was formed (a rate of 55 K/Gyr). At that rate, "...it would take something like 91 billion years to cool to 0 K." So no, not in our lifetime for sure! Note also that the mantle is being partially heated by radioactive decays of Uranium-238, Uranium-235, Thorium-232, and Potassium-40, all of which have half-lives of greater than 700 million years (up to about 14 billion years for Thorium). This will protect us for some time from the complete cooling of the core. So, fortunately for our tiny planet, the core will stay nice and toasty warm for a few more billion years (my estimation). Unfortunately for the planet, the sun will go red giant phase and eat it up before the core cools. I'm being liberal with this word. As I state elsewhere, the core is solid. OP intends it to mean a cold lump of iron rather than the toasty lump of iron that we have | {
"source": [
"https://physics.stackexchange.com/questions/80159",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/29849/"
]
} |
80,245 | Recently I've been puzzling over the definitions of first and second order phase transitions. The Wikipedia article starts by explaining that Ehrenfest's original definition was that a first-order transition exhibits a discontinuity in the first derivative of the free energy with respect to some thermodynamic parameter, whereas a second-order transition has a discontinuity in the second derivative. However, it then says Though useful, Ehrenfest's classification has been found to be an inaccurate method of classifying phase transitions, for it does not take into account the case where a derivative of free energy diverges (which is only possible in the thermodynamic limit). After this it lists various characteristics of second-order transitions (in terms of correlation lengths etc.), but it doesn't say how or whether Ehrenfest's definition can be modified to properly characterise them. Other online resources seem to be similar, tending to list examples rather than starting with a definition. Below is my guess about what the modern classification must look like in terms of derivatives of the free energy. Firstly I'd like to know if it's correct. If it is, I have a few questions about it. Finally, I'd like to know where I can read more about this, i.e. I'm looking for a text that focuses on the underlying theory, rather than specific examples. Modern Classification The Boltzmann distribution is given by $p_i = \frac{1}{Z}e^{-
\beta E_i}$ , where $p_i$ is the probability of the system being in state $i$ , $E_i$ is the energy associated the $i$ -th state, $\beta=1/k_B T$ is the inverse temperature, and the normalising factor $Z$ is known as the partition function. Some important parameters of this probability distribution are the expected energy, $\sum_i p_i E_i$ , which I'll denote $E$ , and the "dimensionless free energy" or "free entropy", $\log Z$ , where $Z$ is the partition function. These may be considered functions of $\beta$ . It can be shown that $E = -\frac{d \log Z}{d \beta}$ . The second derivative $\frac{d^2 \log Z}{d \beta^2}$ is equal to the variance of $E_i$ , and may be thought of as a kind of dimensionless heat capacity. (The actual heat capacity is $\beta^2 \frac{d^2 \log Z}{d \beta^2}$ .) We also have that the entropy $S=H(\{p_i\}) = \log Z + \beta E$ , although I won't make use of this below. A first-order phase transition has a discontinuity in the first derivative of $\log Z$ with respect to $\beta$ : Since the energy is related to the slope of this curve ( $E = -d \log Z / d\beta$ ), this leads directly to the classic plot of energy against (inverse) temperature, showing a discontinuity where the vertical line segment is the latent heat: If we tried to plot the second derivative $\frac{d^2 \log Z}{d\beta^2}$ , we would find that it's infinite at the transition temperature but finite everywhere else. With the interpretation of the second derivative in terms of heat capacity, this is again familiar from classical thermodynamics. So far so uncontroversial. The part I'm less sure about is how these plots change in a second-order transition . My guess is that the energy versus $\beta$ plot now looks like this, where the blue dot represents a single point at which the slope of the curve is infinite: The negative slope of this curve must then look like this, which makes sense of the comment on Wikipedia about a [higher] derivative of the free energy "diverging". If this is what second order transitions are like then it would make quite a bit of sense out of the things I've read. In particular it makes it intuitively clear why there would be critical opalescence (apparently a second-order phenomenon) around the critical point of a liquid-gas transition, but not at other points along the phase boundary. This is because second-order transitions seem to be "doubly critical", in that they seem to be in some sense the limit of a first-order transition as the latent heat goes to zero. However, I've never seen it explained that way, and I have also never seen the third of the above plots presented anywhere, so I would like to know if this is correct. Further Questions If it is correct then my next question is why are critical phenomena (diverging correlation lengths etc.) associated only with this type of transition? I realise this is a pretty big question, but none of the resources I've found address it at all, so I'd be very grateful for any insight anyone has. I'm also not quite sure how other concepts such as symmetry breaking and the order parameter fit into this picture. I do understand those terms, but I just don't have a clear idea of how they relate to the story outlined above. I'd also like to know if these are the only types of phase transition that can exist. Are there second-order transitions of the type that Ehrenfest conceived, where the second derivative of $\log Z$ is discontinuous rather than divergent, for example? What about discontinuities and divergences in other thermodynamic quantities and their derivatives? | I'll give a very qualitative answer / overview. The classification 'first-order phase transition vs. second-order phase transition' is an old one, now replaced by the classification 'first-order phase transition vs. continuous phase transition'. The difference is that the latter includes divergences in 2nd derivatives of $F$ and above - so to answer your question, yes there are other orders of phase transitions, in general. Note that there are phase transitions that do not fall into the above framework - for example, there are quantum phase transitions, where the source of the phase transitions is not thermal fluctuations but rather quantum fluctuations. And then there are topological phase transitions such as the Kosterlitz–Thouless transition in the XY model. The framework to understanding the thermal phase transitions is statistical field theory. A very important starting point is Ginzburg theory, and then you upgrade it to Landau-Ginzburg theory. In a nutshell, phases are distinguished by the symmetries they possess. For example, the liquid phase of water is rotationally symmetric and translationally symmetric, but the solid phase (ice) breaks that rotational symmetry because now it only has discrete translational symmetry. So there must be some phase transition between these two phases. Liquid and gas possess the same symmetry and so actually can be identified as the same phase, as evidenced by being able to go from liquid to gas by going around the critical point instead of through the liquid-gas boundary in the phase diagram. LG theory involves writing a statistical field theory of the system respecting symmetries of the system, and then studying how the solution to the field equations respects the symmetry or not against the temperature. Now we don't really deal with first order phase transitions as much as continuous phase transitions. I can give a few reasons: First-order phase transitions aren't very interesting. You can model them by Landau-Ginzburg theory in the mean field approach by adding appropriate terms in the effective action (like $m^3$, $m^4, m^5, m^6$, $m$ being the order parameter [yes, note that odd terms are allowed - they explicitly break the symmetry. Although for reasons of positive-definiteness the largest power must be even.]). First-order phase transitions depend on the microscopic details of the system, so we don't learn much information about such a PT from analyzing one system. Or perhaps, we just don't know how to really deal with first-order phase transitions very well. Continuous phase transitions have a diverging correlation length (first order ones typically do not). This implies a few very important things: a) Microscopic details are washed out because of the diverging correlation length. So we expect continuous phase transitions to be classified into universality classes. By that I mean that near such a critical point, thermodynamic properties diverge with some critical exponents with the order parameter, and these set of critical exponents fall into classes that can be used to classify different PTs. Refer to Peskin and Schroeder pg 450 - we see that the critical point in a binary liquid system has the same set of exponents as that of the $\beta$-brass critical point! And the critical point in the EuO system is the same as the critical point in the Ni system. Interesting, no? b) We can use established techniques such as renormalization to extract information of the critical exponents of the critical points. Try this paper by Kadanoff. Ok, so as I said this is a very qualitative answer, but I hope it points you in some (hopefully right) direction. | {
"source": [
"https://physics.stackexchange.com/questions/80245",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/5477/"
]
} |
80,256 | It is said that iron fusion is endothermic and star can't sustain this kind of fusion (not until it goes supernova). However star is constantly releasing energy from fusion of elements like Hydrogen and Helium. So, can't that energy be used for fusion of Iron nuclei? | The Sun obviously produces far more energy per second than is required to fuse an iron nucleus with some other nucleus. The problem is concentrating all that energy on the iron nucleus. It's not enough to know that it takes the energy from $n$ hydrogen fusions to fuse one iron nucleus, it's getting the energetic products from those $n$ hydrogen fusion events to all collide with the iron nucleus at the same time. Under normal conditions the probability of this is negligible. However, under extreme conditions it can occur. For example in supernovae the pressures and temperatures are so high that iron and heavier nuclei undergo fusion reactions to produce the elements heavier than iron. | {
"source": [
"https://physics.stackexchange.com/questions/80256",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30838/"
]
} |
80,365 | I will try to be as explanatory as possible with my question. Please also note that I have done my share of googling and I am looking for simple language preferable with some example so that I can get some insight in this subject. My question is what is so special about $c$? Why only $c$. Its like chicken and egg puzzle for me. Does Einstein reached to $c$ observing light or does he got to light using some number which turned out equal to $c$. Why is $c$ not relative. If something has zero rest mass like a photon why they only travel at $c$ in vacuum and not with $c+1$ or $c-1$? | Special Relativity is based on the invariance of a quantity called the proper time, $\tau$, which is the time measured by a freely moving (i.e. not accelerated) observer. The proper time is defined by: $$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$ This is similar to Pythagoras' theorem as learned by generations of schoolchildren, except that it includes time (converted to a distance by multiplying by $c$) and it has a mixture of plus and minus signs. The mixture of signs is responsible for all the weird effects like time dilation and length contraction, and because there is a mixture of signs the value of $d\tau^2$ can be positive, negative or zero. If $d\tau^2$ is less than zero then $d\tau$ must be imaginary, and therefore unphysical. A quick bit of maths will show you that $d\tau^2$ can only be negative if you travel faster than light, and therefore that $c$ is the fastest speed anything in the universe can travel. So $c$ is special because it determines a fundamental symmetry of the universe. Footnote: I've said $c$ is special while Kostya has said the opposite, but actually we are both right. Kostya is right that there is nothing special about the speed 299,792,458 m/s (though if you change it by much you'll change physics enough that we may not be here :-). However the speed at which light travels is very special because anything travelling at this speed follows a null geodesic, i.e. $d\tau^2 = 0$. This is the sense in I mean that $c$ is special. | {
"source": [
"https://physics.stackexchange.com/questions/80365",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30838/"
]
} |
80,434 | According to Wikipedia , if a system has $50\%$ chance to be in state $\left|\psi_1\right>$ and $50\%$ to be in state $\left|\psi_2\right>$, then this is a mixed state. Now, consider the state
$$\left|\Psi\right>=\frac{\left|\psi_1\right>+\left|\psi_2\right>}{\sqrt{2}},$$ which is a superposition of the states $\left|\psi_1\right>$ and $\left|\psi_2\right>$. Let $\left|\psi_i\right>$ be eigenstates of the Hamiltonian operator. Then measurements of energy will give $50\%$ chance of it being $E_1$ and $50\%$ of being $E_2$. But this then corresponds to the definition above of mixed state! However, superposition is defined to be a pure state. So, what is the mistake here? What is the real difference between mixed state and superposition of pure states? | The state \begin{equation}
|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right)
\end{equation} is a pure state. Meaning, there's not a 50% chance the system is in the state $|\psi_1\rangle$ and a 50% it is in the state $|\psi_2\rangle$. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state $|\Psi\rangle$. The point is that these statements are all made before I make any measurements. It is true that if I measure the observable corresponding to $\psi$ ($\psi$-gular momentum :)), then there is a 50% chance after collapse the system will end up in the state $|\psi_1\rangle$. However, let's say I choose to measure a different observable. Let's say the observable is called $\phi$, and let's say that $\phi$ and $\psi$ are incompatible observables in the sense that as operators $[\hat{\psi},\hat{\phi}]\neq0$. (I realize I'm using $\psi$ in a sense you didn't originally intend but hopefully you know what I mean). The incompatibliity means that $|\psi_1 \rangle$ is not just proportional to $|\phi_1\rangle$, it is a superposition of $|\phi_1\rangle$ and $|\phi_2\rangle$ (the two operators are not simulatenously diagonalized). Then we want to re-express $|\Psi\rangle$ in the $\phi$ basis. Let's say that we find
\begin{equation}
|\Psi\rangle = |\phi_1\rangle
\end{equation} For example, this would happen if
\begin{equation}
|\psi_1\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle+|\phi_2\rangle)
\end{equation}
\begin{equation}
|\psi_2\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle-|\phi_2\rangle)
\end{equation}
Then I can ask for the probability of measuring $\phi$ and having the system collapse to the state $|\phi_1\rangle$, given that the state is $|\Psi\rangle$, it's 100%. So I have predictions for the two experiments, one measuring $\psi$ and the other $\phi$, given knowledge that the state is $\Psi$. But now let's say that there's a 50% chance that the system is in the pure state $|\psi_1\rangle$, and a 50% chance the system is in the pure state $|\psi_2\rangle$. Not a superposition, a genuine uncertainty as to what the state of the system is. If the state is $|\psi_1 \rangle$, then there is a 50% chance that measuring $\phi$ will collapse the system into the state $|\phi_1\rangle$. Meanwhile, if the state is $|\psi_2\rangle$, I get a 50% chance of finding the system in $|\phi_1\rangle$ after measuring. So the probability of measuring the system in the state $|\phi_1\rangle$ after measuring $\phi$, is (50% being in $\psi_1$)(50% measuring $\phi_1$) + (50% being in $\psi_2$)(50% measuring $\phi_1$)=50% . This is different than the pure state case. So the difference between a 'density matrix' type uncertainty and a 'quantum superposition' of a pure state lies in the ability of quantum amplitudes to interfere, which you can measure by preparing many copies of the same state and then measuring incompatible observables. | {
"source": [
"https://physics.stackexchange.com/questions/80434",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21441/"
]
} |
80,775 | I've seen many applications of topology in Quantum Mechanics (topological insulators, quantum Hall effects, TQFT, etc.) Does any of these phenomena have anything in common? Is there any intuitive explanation of why topology is so important? Is there a similar application of topology in Classical Mechanics? | Topology is of fundamental importance even to systems in classical mechanics. The configuration space (or phase space) of a generic classical mechanical system is a manifold and manifolds are topological spaces with some extra structure (e.g. a smooth structure in the case of smooth manifolds). At the very start of any classical mechanics problem, you need to specify topological information. For example, when we consider a single particle moving in one-dimension, we could consider a particle constrained to move either on a compact manifold (like a circle), or a non-compact manifold (like the entire real line). In each of these cases, the global topology drastically changes the nature of the solutions. If, for example, the particle were otherwise free in each of these cases, then in the compact case (the circle), the particle will always return to the same point after some finite time, while in the non-compact case (the real line), this cannot happen. Addendum. Beyond it's fundamental importance to mechanics as described above, topological properties of classical mechanical systems are important for proving high-powered theorems about dynamical systems. If you, for example, open Foundations of Mechanics by Abraham and Marsden (which is really more a math than physics), then you'll find a chapter on so-called "topological dynamics" where you'll find results like corollary 6.1.9; Let $M$ be a compact, connected, two-dimensional manifold, $X\in\mathscr X(M)$ and $A$ a minimal set of $X$. Then either (i) A is a critical point, (ii) A is a closed orbit, (iii) $A=M$ and $M=S^1\times S^2$. Notice that the statement of this corollary depends on the topological assumptions that $M$ is compact and connected. There are all sorts of theorems like this in dynamical systems. See the Poincare Recurrence Theorem as another example. Update. (2014 - July - 10) More interesting information and discussion in the following physics.SE post: What kind of manifold can be the phase space of a Hamiltonian system? See also the links in the comments to that post, especially to mathoverflow. Update. (2014 - July - 16) Even more interesting related information and discussion in the following post: Is there a physical system whose phase space is the torus? | {
"source": [
"https://physics.stackexchange.com/questions/80775",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21817/"
]
} |
80,807 | I know plus pushes another plus away, but why, really, do they do that? On the other hand, molecules of the same type are attracted to each other. I find that weird. I do know some stuff about four universal forces. But why in general the general "rule" is that opposite charges pull each other? Yes, I do realize this could be connected to very basic stuff that science is still trying to figure out, and can be traced to the Higgs, but still, there must be something to tell. | Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's hard to make this really compelling (i.e., make it seem as inevitable as it is) without the math: Particles and forces are now understood to be the result of fields . Quantum fields to be exact. A field is a mathematical object that takes a value at every point in space and at every moment of time. Quantum fields are fields that carry energy and momentum and obey the rules of quantum mechanics. One consequence of quantum mechanics is that a quantum field carries energy in discrete "lumps". We call these lumps particles . Incidentally this explains why all particles of the same type (e.g. all electrons) are identical: they are all lumps in the same field (e.g. the electron field). The fields take values in different kinds of mathematical spaces that are classified by special relativity. The simplest is a scalar field. A scalar field is a simple number at every point in space and time. Another possibility is a vector field: these assign to every point in space and time a vector (an arrow with a magnitude and direction). There are more exotic possibilities too. The jargon term to classify them all is spin , which comes in units of one half. So you can have fields of spin $0, \frac{1}{2}, 1, \frac{3}{2}, 2, \cdots$ . Spin $0$ are the scalars and spin $1$ are the vectors. It turns out (this is another consequence of relativity) that particles with half integer spin ( $1/2, 3/2, \cdots$ ) obey the Pauli exclusion principle . This means that no two identical particles with spin $1/2$ can occupy the same place. This means that these particles often behave like you expect classical particles to behave. We call these matter particles, and all the basic building blocks of the world (electrons, quarks etc.) are spin $1/2$ . On the other hand, integer spin particles obey Bose-Einstein statistics (again a consequence of relativity). This means that these particles "like to be together," and many of them can get together and build up large wavelike motions more analogous to classical fields than particles. These are the force fields; the corresponding particles are the force carriers. Examples: spin $0$ Higgs, spin $1$ photons, weak force particles $W^\pm, Z$ , and the strong force carriers the gluons, and spin $2$ the graviton, carrier of gravity. (This fact and the previous one are called the spin-statistics theorem .) Now the interaction between two particles with "charges" $q_{1,2}$ goes like $\mp q_1 q_2$ for all the forces ( this is a consequence of quantum mechanics), but the sign is tricky to explain. Because of special relativity, the interaction between a particle and a force carrier has to take a specific form depending on the spin of the force carrier (this has to do with the way space and time are unified into a single thing called spacetime ). For every unit of spin the force carrier has you have to bring in a minus sign (this minus sign comes from a thing called the " metric ", which in relativity tells you how to compute distances in spacetime; in particular it tells you how space and time are different and how they are similar). So for spin $0$ you get a $-$ : like charges attract. For spin $1$ you get a $+$ : like charges repel! And for spin $2$ you get a $-$ again: like charges attract. Now for gravity the "charge" is usually called mass, and all masses are positive. So you see gravity is universally attractive! So ultimately this sign comes from the fact that photons carry one unit of spin and the fact that the interactions between photons and matter particles have to obey the rules of special relativity. Notice the remarkable interplay of relativity and quantum mechanics at work. When put together these two principles are much more constraining than either of them individually! Indeed it's quite remarkable that they get along together at all. A poetic way to say it is the world is a delicate dance between these two partners. Now why do atoms and molecules generally attract? This is actually a more complicated question! ;) (Because many particles are involved.) The force between atoms is the residual electrical force left over after the electrons and protons have nearly cancelled each other out. Here's how to think of it: the electrons in one atom are attracted to the nuclei of both atoms and at the same time repelled by the other electrons. So if the other electrons get pushed away a little bit there will be a slight imbalance of charge in the atom and after all the details are worked out this results in a net attractive force, called a dispersion force . There are various different kinds of dispersion forces (London, van der Waals, etc.) depending on the details of the configuration of the atoms/molecules involved. But they are all basically due to residual electrostatic interactions. Further reading: I recommend Matt Strassler's pedagogical articles about particle physics and field theory. He does a great job at explaining things in an honest way with no or very little mathematics. The argument I went through above is covered in some capacity in just about every textbook on quantum field theory, but a particularly clear exposition along these lines (with the math included) is in Zee's Quantum Field Theory in a Nutshell. This is where I would recommend starting if you want to honestly learn this stuff, maths and all, but this is an advanced physics textbook (despite being written in a wonderful, very accessible style) so you need probably at least two years of an undergraduate physics major and a concerted effort to make headway in it. | {
"source": [
"https://physics.stackexchange.com/questions/80807",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30901/"
]
} |
81,190 | What constitutes protons? When I see pictures, I can't understand. Protons are made of quarks, but some say that they are made of 99% empty space. Also, in this illustration from Wikipedia, what's between the quarks? | Ah, I know this one! What's in a proton? A proton is really made of excitations in quantum fields (kind of like localized waves). Remember that. Any time you hear any other description of the composition of a proton, it's just some approximation of the behavior of quantum fields in terms of something people are likely to be more familiar with. We need to do this because quantum fields behave in very nonintuitive ways, so if you're not working with the full mathematical machinery of QCD (which is hard), you have to make some kind of simplified model to use as an analogy. One of the more confusing things about quantum field excitations is that they react differently depending on how they are observed. More specifically, the only way to measure the properties of an excitation in a quantum field is to make it interact with another excitation and see how the excitations affect each other. Or in particle language, you have to hit the particle with another particle (the "probe") and see what comes out. Depending on the charge, energy, momentum and other properties of the probe, you can get various results. People have been doing this for decades, and they've compiled the results into a few general conclusions. For example, in a slow collision, with very little energy involved, a proton acts like a single point particle. If we give the particles slightly more energy, the proton looks more like a blob with three points in it --- this is part of why it's often said that the proton consists of three quarks. (Incidentally, the reason you see images like the one you found on Wikipedia is that for a long time, people were colliding protons at the intermediate energies where they appear to behave as a group of three quarks.) If we give the colliding particles even more and more energy, the proton will appear to be an ever-more-dense amalgamation of all sorts of particles: quarks, antiquarks, gluons, photons, electrons, and everything else. We call these particle partons (because they're part of the proton). The following diagram shows representative examples of the effective composition of the proton in different kinds of collisions. The vertical axis basically corresponds to collision energy, and the horizontal axis corresponds to the "resolving power" of the incident ("probe") particle. (Resolving power is basically transverse momentum, but I can't explain how that connection works without getting into more detail of quantum mechanics than I think is necessary.) The contents of each circle represents, roughly, a sample "snapshot" of how the proton behaves in a collision at the corresponding energy and resolving power. The exact numbers, locations, and colors of the dots aren't significant (except sort of in the bottom left), just note how they get larger or smaller and more or less numerous as you move around the plot. So for example, if you hit a proton with a beam of high-energy probes (top) that have weak resolving power (left), it behaves like a dense cluster of partons (quarks and gluons etc.), each of which is fairly large. Or if you hit the proton with a beam of low-energy probes (bottom) with high resolving power (right), it behaves like a sparse cluster of partons, each of which is small. If you hit it with a beam of low-energy (bottom), low-resolving-power (left) probes, it behaves like a collection of three particles. Physicists describe this apparently-changing composition using parton distribution functions (PDFs), often denoted $f(x, Q^2)$ . Under certain not-too-crazy assumptions, $f(x, Q^2)$ can be interpreted as the probability density of the probe interacting with a particular type of parton with a particular amount of momentum. Visually, $f(x, Q^2)$ is related to the number of particles in the circle at the corresponding $(x,Q)$ point on the plot (though again, the exact numbers are not chosen to exactly reflect reality). For more information on parton distributions, I would refer you to this answer of mine and the resources named therein, as well as this one . What's the gray region? In the preceding image, I displayed each snapshot of the proton as a set of partons (quarks and gluons etc.) uniformly distributed within a circle, as if the proton has a definite edge and there is nothing outside that edge. But in reality, that's not the case. The quantum fields that make up a proton gradually fade away to zero as you move further away from the center, giving the proton a fuzzy edge. So a (somewhat) more accurate sample snapshot would look something like this: Notice that there are more dots near the center of the proton, and progressively fewer as you move toward the edge; this represents the fact that a probe which hits a proton dead-center is more likely to interact than a probe which hits it near the edge. The ordinary parton distributions that I mentioned above, $f(x, Q^2)$ , are part of a simplified model in which we ignore this fact and pretend that partons are distributed uniformly throughout space. But we can make a more complicated model that does take into account the fact that partons are clumped up toward the center of the proton. In such a model, instead of regular parton distributions, you get more complicated functions, called impact parameter-dependent parton distributions , and denoted $f(x, Q^2, b)$ , where $b$ is the radial distance from the center at which the probe hits - the impact parameter. There have been some theoretical studies showing that these impact parameter-dependent parton distributions trail off gradually as you go to large radii. For example, see figure 5 of this paper ( arXiv ) or figure 7 in this one ( arXiv ): Here $N(y)$ is a quantity related to the parton distributions (specifically, it's the color dipole scattering amplitude), which kind of "condenses" the many different parton distributions into one quantity. (Huge oversimplification, but it's good enough for this.) You can then define the spatial extent of the proton as the region in which $N(y)$ is above, say, 5% of its maximum value. Or 10%. Or 50%. The exact number is somewhat arbitrary, but the point is, whatever number you pick, you'll wind up with a circle that encompasses the region in which the parton distribution function is large, kind of like this: This is roughly what the gray circle in the image from Wikipedia represents. It's a region with a size on the order of $1\text{ fm}$ (that's about $5\text{ GeV}^{-1}$ in natural units ), where the chance of an incident particle (a probe) scattering off the proton is relatively significant. Equivalently, it's the region in which the parton distributions are large, and also the region in which the quantum fields that constitute the proton are much different from zero. As you can guess, all this is pretty imprecise. You can make a more rigorous definition of the size of a proton by using the scattering cross section . You can also get a definition without using scattering, using the charge radius , which can be measured or calculated using various other methods. I won't go into those, as the details would be material for a whole separate question, but the results of all these methods come out to a radius a little less than $1\text{ fm}$ . Incidentally, this claim of a proton being 99% empty space is probably false using any reasonable definition. You might be thinking of atoms , where the volume in which the electron's quantum field has an appreciable value is much larger than the size of the electron itself, whatever it may be. People sometimes simplify that to say that the atom consists of a large fraction of empty space. But you can't really do the same with a proton, given the large number of particles in it and the strength of their interactions. | {
"source": [
"https://physics.stackexchange.com/questions/81190",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/31223/"
]
} |
81,448 | As simple as in the title.. I would like to know also some mathematics about it! | It cannot. This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon. It needs interaction with at least a second charged particle in order to do so (as in Bremsstrahlung). The mathematic involved is that of the energy of a photon $E=\hbar \omega$, energy of a particle $E^2 = m^2 c^4 + p^2 c^2$, momentum of a photon $p = \hbar \omega /c$ and simple trigonometry and basic algebra, very much as in the classical version of Compton scattering. | {
"source": [
"https://physics.stackexchange.com/questions/81448",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/31382/"
]
} |
81,656 | I've been digging a lot into quantum physics in the last few weeks. I didn't care much about the maths, just about what empirically happens to get a conceptual idea about quantum phenomena. The most widely accepted interpretation of quantum mechanics seems to be the Copenhagen one. If I got it right, it's heavily relaying on the two following principles (among others): Superposition : a quantum system is at the same time in all the states it could possibly be in. When it's measured, it instantaneously collapses in a single state. Entanglement (aka "spooky action at a distance"): if two or more quantum systems are entangled, it means that some of their properties are correlated. When measuring a system, all the entangled ones collapse in a state coherent to the measured one. Simultaneously. No matter how far away they are to each other. I'm not able to believe it. It allows some unrealistic paradoxes (e.g. Schrödinger's cat paradox ), and I have the feeling that this interpretation (and its consequences) is what makes quantum mechanics look so weird, mysterious, unnatural and spooky to the public. Besides I've read from a few sources (like this Google Tech Talk ) that this interpretation has proven to be broken: the math says everything is continuous and doesn't hint to anything like collapsing, and even more important, the quantum eraser experiment contradicts the Copenhagen interpretation. The second most popular interpretation, many-worlds sounds a lot more natural to me, although it strongly smells like science fiction. I believe there must be many interpretations that would hold better and would be a lot less weird than the two mentioned ones. What I'm wondering is, then: why does the Copenhagen interpretation (and to a lesser degree the many-worlds one) remain the most accredited one? | Why is the Copenhagen interpretation the most accepted one? I would say the answer is this: it's the oldest more or less "complete" interpretation hence you'll find it in many (all?) early text books, which is basically from where people writing modern text books copy from. the overwhelming majority of physicists doesn't really care about the interpretation, since it (up to now) is only a matter of philosophy. We cannot know what interpretation is correct, because we can't measure differences, hence the interpretation question is a matter of taste rather than scientific knowledge. most standard QM courses at university (at least the ones I know) don't bother with the interpretation. They just introduce the concepts, updates of knowledge, etc. and in that sense, the Copenhagen interpretation is just convenient. This implies that if you ask a lot of physicists, some have never even thought about the matter.
If interpretation is a matter of philosophy, why should we worry about it then? I can think of two points here: a) By thinking also about interpretations of our theory we may come up with new theories that give us "nicer" interpretations of existing results, but they are essentially inequivalent to quantum mechanics. Bohmian mechanics from what little I understand about it is such a candidate, which might turn out to at one point make different predictions than classical quantum mechanics (up till now, it's just a different interpretation). This is of course a very good reason to think about it, because if quantum mechanics can not explain everything and there is a better theory, which can explain more with similarly "simple" assumptions, we want to have it. b) It might help our understanding of "reality". This is only interesting, if you believe that your theory describes reality. If you believe that we only ever create effective models that are limited to a certain domain of our variables, then interpretations become uninteresting. Your model isn't the real deal after all, so why bother with something, you can't measure? It doesn't enhance our knowledge. So, if you don't believe that science should (or even can) provide ontologic theories and if you don't think a better theory than quantum mechanics is maybe just beyond the horizon, then you don't care about interpretations of quantum mechanics. Otherwise, you should. | {
"source": [
"https://physics.stackexchange.com/questions/81656",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/31312/"
]
} |
81,661 | I'm trying to help a child research a science project on refrigeration.
Refreshing my incredibly rusty thermodynamics skills.... The ideal gas law: $PV=nRT$. Let's take air at STP: $P = 101\,kPa$
$V = 1\,L = 0.001\,m^3$
$R = 8.3\,J/mol\cdot K$
$T = 298\,K$ $n=PV/RT = (101000) (.001) / (8.3 \cdot 298) = 0.04\, moles$? If we compress the air to ~7atm adiabatically
$P_2 = 7P$ I would think the volume goes to $\frac{1}{7V}$ $V_2 = \frac{V}{7}$ then I would expect the gas to be hotter. But I'm obviously confused because with that pressure and volume, the temperature is obviously the same. I'm assuming that I'm wrong about what the volume would be for an ideal gas if I compress to $7\,atm?$ $T_2 = \frac{P_2V_2}{nR} = ???$ The specific heat of air: $c_p = 1.006\,kJ/kg\cdot K$ Of course air is not quite ideal. I would also appreciate someone explaining what the non-ideal behavior is due to. Is it related to the mixing? Some kind of chemical interaction between the different components? What I want to know is how to calculate the temperature of a gas given an initial temperature, pressure, heat capacity and final pressure, Adiabatically. | Why is the Copenhagen interpretation the most accepted one? I would say the answer is this: it's the oldest more or less "complete" interpretation hence you'll find it in many (all?) early text books, which is basically from where people writing modern text books copy from. the overwhelming majority of physicists doesn't really care about the interpretation, since it (up to now) is only a matter of philosophy. We cannot know what interpretation is correct, because we can't measure differences, hence the interpretation question is a matter of taste rather than scientific knowledge. most standard QM courses at university (at least the ones I know) don't bother with the interpretation. They just introduce the concepts, updates of knowledge, etc. and in that sense, the Copenhagen interpretation is just convenient. This implies that if you ask a lot of physicists, some have never even thought about the matter.
If interpretation is a matter of philosophy, why should we worry about it then? I can think of two points here: a) By thinking also about interpretations of our theory we may come up with new theories that give us "nicer" interpretations of existing results, but they are essentially inequivalent to quantum mechanics. Bohmian mechanics from what little I understand about it is such a candidate, which might turn out to at one point make different predictions than classical quantum mechanics (up till now, it's just a different interpretation). This is of course a very good reason to think about it, because if quantum mechanics can not explain everything and there is a better theory, which can explain more with similarly "simple" assumptions, we want to have it. b) It might help our understanding of "reality". This is only interesting, if you believe that your theory describes reality. If you believe that we only ever create effective models that are limited to a certain domain of our variables, then interpretations become uninteresting. Your model isn't the real deal after all, so why bother with something, you can't measure? It doesn't enhance our knowledge. So, if you don't believe that science should (or even can) provide ontologic theories and if you don't think a better theory than quantum mechanics is maybe just beyond the horizon, then you don't care about interpretations of quantum mechanics. Otherwise, you should. | {
"source": [
"https://physics.stackexchange.com/questions/81661",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/2653/"
]
} |
81,663 | I need an inexpensive instrument to measure cryogenic temperatures (down to -200C). I can build a thermistor-based thermometer using an Arduino that is accurate to under 1 degree for 0 to 100C. First of all, can ordinary NTC thermistors be used at cryogenic temperatures?
Second, if I were to try to use this approach, I would need ways to calibrate the device. I have found a chipset from Analog Devices that would let me use a thermocouple down to that range. I would take other ideas on how to build one, but then I need some equipment to calibrate whatever I build. Is there anything I can buy for < $100, I don't need huge accuracy, but I need something. | Why is the Copenhagen interpretation the most accepted one? I would say the answer is this: it's the oldest more or less "complete" interpretation hence you'll find it in many (all?) early text books, which is basically from where people writing modern text books copy from. the overwhelming majority of physicists doesn't really care about the interpretation, since it (up to now) is only a matter of philosophy. We cannot know what interpretation is correct, because we can't measure differences, hence the interpretation question is a matter of taste rather than scientific knowledge. most standard QM courses at university (at least the ones I know) don't bother with the interpretation. They just introduce the concepts, updates of knowledge, etc. and in that sense, the Copenhagen interpretation is just convenient. This implies that if you ask a lot of physicists, some have never even thought about the matter.
If interpretation is a matter of philosophy, why should we worry about it then? I can think of two points here: a) By thinking also about interpretations of our theory we may come up with new theories that give us "nicer" interpretations of existing results, but they are essentially inequivalent to quantum mechanics. Bohmian mechanics from what little I understand about it is such a candidate, which might turn out to at one point make different predictions than classical quantum mechanics (up till now, it's just a different interpretation). This is of course a very good reason to think about it, because if quantum mechanics can not explain everything and there is a better theory, which can explain more with similarly "simple" assumptions, we want to have it. b) It might help our understanding of "reality". This is only interesting, if you believe that your theory describes reality. If you believe that we only ever create effective models that are limited to a certain domain of our variables, then interpretations become uninteresting. Your model isn't the real deal after all, so why bother with something, you can't measure? It doesn't enhance our knowledge. So, if you don't believe that science should (or even can) provide ontologic theories and if you don't think a better theory than quantum mechanics is maybe just beyond the horizon, then you don't care about interpretations of quantum mechanics. Otherwise, you should. | {
"source": [
"https://physics.stackexchange.com/questions/81663",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/2653/"
]
} |
81,787 | The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become the normal particle and the rare and unusual positrons become the antiparticle . My question is, when you have made the choice of the electron and positron as particle and anti-particle does this automatically identify every other particle (every other fermion?) as normal or anti? For example the proton is a particle , or rather the quarks inside are. By considering the interactions of an electron with a quark inside a proton can we find something, e.g. a conserved quantity, that naturally identifies that quark as a particle rather than an antiparticle ? Or do we also just have to extend our convention so say that a proton is a particle rather than an antiparticle ? To complete the family I guess the same question would apply to the neutrinos. | Yes, to some extent. Once you choose which of the electron or positron is to be considered the normal particle, then that fixes your choice for the other leptons, because of neutrino mixing. Similarly, choosing one quark to be the normal particle fixes the choice for the other flavors and colors of quarks. But I can't think of a reason within the standard model that requires you to make corresponding choices for leptons and quarks. In particle terms, you can think about it like this: say you start by choosing the electron to be the particle and the positron to be the antiparticle. You can then distinguish electron neutrinos and electron antineutrinos because in weak decay processes, an electron is always produced with an antineutrino and a positron with a normal neutrino. Then, because of neutrino oscillations, you can identify the other two species of neutrinos that oscillate with electron antineutrinos as antineutrinos themselves, and in turn you can identify the muon and tau lepton from production associated with their corresponding antineutrinos. In terms of QFT, the relevant (almost-)conserved quantity is the " charge parity ," the eigenvalue of the combination of operators $\mathcal{CP}$. | {
"source": [
"https://physics.stackexchange.com/questions/81787",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/1325/"
]
} |
82,359 | How did Newton discover his third law ? Was it his original finding or was it a restatement of someone else's, like the first law coming from Galileo? What initiated the concept of what is now known as Newton's 3rd law? | Introduction I restored the original title to show how interesting it is that a non-British student (18 at the time) can be more informed than a British physics graduate. He posted this comment: "The question before this must be whether it was his original or
something like his 1st law, which was a restatement of the
experimental findings of Galileo" – Rijul Gupta Oct 26 '13
He then edited the title and the OP text adding: "Was it his original finding or was it a restatement of someone
else's, like the first law coming from Galileo?" This may seem a trivial detail in an answer but it is very important here: he knows that first law wasn't Newton's own (this is sometime acknowledged, though), but he expresses doubts that also third law might not be his own finding, whereas a U.S. academic is skeptical: "...I don't know if there's any evidence that he knew of them beforehand." – Ben Crowell The historical truth is there, recorded in accessible documents and original texts, if one wants to look for it and is prepared to accept it, even if may be shocking for English eyes. I'll present the original documents, readers can draw their conclusions. The historical facts Christiaan Huygens [wiki (1)] was a good-natured, noble generous man, the son of a diplomat who was an advisor to the House of Orange. He was slow to publish his results and discoveries, in the early days his mentor , mathematicians Frans van Schooten was cautious for the sake of his reputation (1), this had the deplorable consequence that his ideas that he naively communicated to his contemporaries were plagiarized. He was too meek, he complained only to friends (even his patent was violated in Egland, France etc.) and therefore his great scientific merits are to date under-evaluated: he found the real law behind ' the conservation of momentum ', discovered the formula of kinetic energy , the conservation of KE in elastic collisions, suggested the term 'vis viva ' to Leibniz and taught him maths and helped him develop 'calculus', even he never believed in its usefulness. during the years 1650 - 1666 [Enc. (2)] he lived at home, except for three journeys to Paris and London: an allowance supplied by his father enabled him to devote himself completely to the study of nature between the years 1652-54, according to his own statements, he developped the theory of collisions in his work (in Latin): "De motu corporum ex percussione" (English translation: Chicago Journals ), there is no proof of that, although : "... there are numerous indications that Huygens had established all the propositions and their proofs by 1656 at the latest (see the Avertissement in Oeuvres, Vol. XVI, pp. 3-14, for the evidence) (3, p. 574) in 1661 he was already famous: in '55 he had discovered the satellite of Saturn (2), in '56 had invented the pendulum clock and in '57 had written his treatise on probability theory (1) . He went to Paris to meet Pascal as " He had been told of recent work in the field by Fermat, Blaise Pascal and Girard Desargues two years earlier" (1). in May of that year he was in London " "..to observe the planet Mercury transit over the Sun, using the telescope of Richard Reeve in London, together with astronomer Thomas Streete and Reeve himself " (1). He also " ..attended meetings in Gresham College, and met Moray, Wallis , and Oldenburg " (2). he told them about his findings and in particular about the theory of collisions". The scholars at Gresham had recently formed the Royal Society Henry Oldenburg (4) was " ...one of the foremost intelligencers of Europe of the seventeenth century, with a network of correspondents .. At the foundation of the Royal Society he took on the task of foreign correspondence, as the first Secretary" , he was the shady figure (shortly imprisoned as a suspected spy) that recruited scientists all over Europe, trying to entice them with a promise ".. they would be assured undying fame by the preservation of their results in the archives of the RS " and to convince them the could rest assured "that no harm to their discoveries would come about through divulging information in advance of publication" and that at RS each is certain of his due" [(5) p.53, passim] the Royal Society, it is notorious, when Newton was a member "...in 1699 accused Leibniz of plagiarism. The dispute then broke out in full force in 1711 when the Royal Society proclaimed in a study that it was Newton who was the true discoverer and labelled Leibniz a fraud . This study was cast into doubt when it was later found that Newton himself wrote the study's concluding remarks on Leibniz" in (June -September) 1663 " Huygens was made a member of the Society " (2), and was invited to London to illustrate his discoveries, in particular on the theory of collisions in 1668 he was invited by the Society to publish his findings on collisions in the Philosophical Transactions of the Royal Society :"He presented the most important theorems to the Royal Society in 1668, simultaneously with studies by Wren and Wallis" (5 p.543). The Rules of Motion by these two, copied from Huygens' paper, were published while his original work was not . In this dishonest way the Society ensured the primacy of the theory to the English authors and Newton (of course, he can't ignore him altogether) can always cite :"In theoria Wrenni & Hugenii ", "together with the third Law, Sir Christ. Wren, Dr. Wallis, and Mr. Huygens ", ".. Dr. Wallis, indeed, was something more early in the publication; then followed Sir Christopher Wren, and, lastly, Mr. Huygens " Huygens was saddened " ...and publicly voiced his anger at being disadvantaged by not having his results published (in PT) at the same time as those of the opposite party" [5, p. 53] in March 1669, having had no satisfaction, he published his paper in French in the Journal des sçavans immediately after, his original Latin paper was published in the PT of the RS in 1670, the following year, Huygens had already forgotten his anger and "... seriously ill, chose Francis Vernon to carry out a donation of his papers to the Royal Society in London, should he die. " (1). If these historical reports are true, by 1671 (the RS and) Newton was in possess of the complete demonstrations concerning the theory of collisions (by Huygens and Mariotte): in 1669 Newton had already been appointed Lucasian Professor. in 1670 Edme Mariotte had announced his intention to compose a major work on the impact of bodies. Completed and read to the Academy in 1671, it was published in 1673 as Traité de la percussion ou choc des corps. The first comprehensive treatment of the laws of inelastic and elastic impact and of their application to various physical problems". In order to verify his suppositions, he used " an experimental apparatus consisting of two simple pendulums of equal length, the replaceable bobs (the impacting bodies) of which meet at dead center" . Here we found the real inventor of "Newton's 'cradle' . Newton cites Wrenn's experiments and Mariotte's book: ".*.veritas comprobata est a Wrenno ...quod etiam Clarissimus Mariottus libro integro exponere mox dignatus est**" (p. 37), but never Huygens'. Newton affirms that Mariotte had just divulged the findings of the British architect: ".. Wren confirmed the truth of the thing before the Royal Society by the experiment of pendulums, which Mr. Mariotte soon after thought fit to explain in a treatise entirely upon that subject." (p.90) Huygens was certainly saddened by the fact that Mariotte did not cite him as his source, but did not respond ( the silence of the lambs ) his nature was so meek that only " seventeen years later, in 1690, when Mariotte was dead, Huygens responded to this slight (see below) by accusing Mariotte of plagiarism. “Mariotte took everything from me,” he protested in a sketch of an introduction to a treatise on impact never completed" (ibidem, you can read about the 'slight'): Clearly he knew of the work of Wallis, Wren, and Huygens published in the Philosophical Transactions of the Royal Society in 1668; and there are enough striking similarities between Mariotte’s treatise and Huygens’ then unpublished paper on impact (De motu corporum ex percussione, in Oeuvres, XVI) to suggest that he knew the content of the latter, perhaps verbally from Huygens himself * Certainly his colleagues in the Academy recognized Mariotte’s debt to others while they praised the clarity of his presentation. And yet Galileo’s name alone appears in the treatise; Huygens’ in particular is conspicuously absent. Since the first meeting in 1661 the RS group had tried to understand the profound meaning of 'conservation of momentum' and even after Huygens sent in his paper in 1668 with the complete list of the rules were not able to interpret them. " In a letter of Feb. 4, 1669, (one month before the delayed publication in RPRS of his submitted paper) Oldenburg "earnestly intreated" Huygens to publish his
theory, but the reply ignored the point. Huygens himself explains his refusal to publish. " (3, p. 575) The reason was not, as one might imagine, that he did not want to reveal the theory, but because he wanted to get to the bottom ( penitus ), to the metaphysical level: "he was concerned to determine the essence and true cause of motion and forces" (ibidem). He mentioned the true principle once, but never wrote it down. from 1669 to 1687 Newton tried for 18 more years to figure out the true essence/law of motion, but lacked Huygens' metaphysical insight He tried to be original (as he had done with the 1st law ) and produced a 3rd law which is partially in contrast with his own laws. (this will be discussed in a another answer, to separate facts from opinions) conclusions This is how Newton found the third law. Is "there enough evidence that he knew of them beforehand" ?, I'll leave the answer to the readers, as this post is unpopular just as it is. Certainly if one should decide that 3rd law is Huygens ', considering that 1st law is Galileo's and 2nd is Leibniz/Coriolis ' gravity's inverse square law is Hooke 's (universally acknowledged, even Newton's agiographers, p.152 affirm Hooke's paternity), and that calculus is probably Leibniz ', one of the greatest urban myth would get a serious blow. Is physics prepared to accept the historical truth? | {
"source": [
"https://physics.stackexchange.com/questions/82359",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21108/"
]
} |
82,613 | Why can't we measure imaginary numbers? I mean, we can take the projection of a complex wave to be the "viewable" part, so why are imaginary numbers given this immeasurable descriptor? Namely with operators in quantum mechanics, why must measurable quantities be Hermitian, and consequently real? | I) Well, one can identify a complex-valued observable with a normal operator $$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$ A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator. Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from. II) But notice that a normal operator $$\tag{2} A~=~B+iC$$ can uniquely$^2$ be written as a sum of two commuting self-adjoint operators $$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$ ($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1). Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$. We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics . For more on real-valued observables, see e.g. this Phys.SE post and links therein. -- $^1$ We will ignore subtleties with unbounded operators , domains, selfadjoint extensions , etc., in this answer. $^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$. | {
"source": [
"https://physics.stackexchange.com/questions/82613",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/-1/"
]
} |
82,625 | If an ferromagnetic object is heated and reaches Tc the magnetization gradually drops as we get closer to Tc or it's a instant drop?
Can I assume as I heat the object, the magnetization is weakening gradually? Likewas as it cools? | I) Well, one can identify a complex-valued observable with a normal operator $$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$ A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator. Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from. II) But notice that a normal operator $$\tag{2} A~=~B+iC$$ can uniquely$^2$ be written as a sum of two commuting self-adjoint operators $$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$ ($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1). Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$. We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics . For more on real-valued observables, see e.g. this Phys.SE post and links therein. -- $^1$ We will ignore subtleties with unbounded operators , domains, selfadjoint extensions , etc., in this answer. $^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$. | {
"source": [
"https://physics.stackexchange.com/questions/82625",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21741/"
]
} |
82,678 | This question was prompted by Can matter really fall through an event horizon? . Notoriously, if you calculate the Schwarzschild coordinate time for anything, matter or light, to reach the event horizon the result is infinite. This implies that the universe ages by an infinite time before someone falling into the black hole reaches the event horizon, so could that person see the universe age by an infinite time? To be more precise, suppose the observer starts falling from rest at time $t = 0$ and some initial distance $r > r_s$ . If we wait for some time $T$ then shine a light ray at the falling observer. Will the light ray always reach the falling observer before they cross the event horizon? If not, what is the formula for the longest time $T$ that we can wait and still be sure the ray will catch the observer? If $T$ is not bounded it implies that observer could indeed see the end of the universe. I can think of a qualitative argument for an upper limit on $T$ , but I'm not sure how sound my argument is. The proper time for the observer to fall to the event horizon is finite - call this $\tau$ . The proper time for the light ray to reach the horizon is zero, therefore the light ray will reach the observer before they cross the event horizon only if $T < \tau$ . Hence $T$ is bounded and the observer won't see the end of the universe. I think a more rigorous approach would be to determine the equations of motion (in the Schwarzschild coordinates) for the falling observer and the light ray, and then find the condition for the light to reach the falling observer at some distance $\epsilon$ from the event horizon. Then take the limit as $\epsilon \rightarrow 0$ . In principle this seems straightforward, but in practice the algebra rapidly defeated me. Even for a light ray the radial distance:time equation isn't closed form (Wolfram claims it needs the $W$ function) and for the falling observer the calculation is even harder. | I would recommend steering clear of Schwarzschild coordinates for these kind of questions. All the classical (i.e. firewall paradox aside) infinities having to do with the event horizon are due to poor coordinate choices. You want to use a coordinate system that is regular at the horizon, like Kruskal-Szekeres . Indeed, have a look at the Kruskal-Szekeres diagram: (source: Wikipedia) This is the maximally extended Schwarschild geometry, not a physical black hole forming from stellar collapse, but the differences shouldn't bother us for this question. Region I and III are asymptotically flat regions, II is the interior of the black hole and IV is a white hole. The bold hyperbolae in regions II and IV are the singularities. The diagonals through the origin are the event horizons. The origin (really a 2-sphere with angular coordinates suppressed) is the throat of a non-traversable wormhole joining the separate "universes" I and III. Radial light rays remain 45 degree diagonal lines on the Kruskal-Szekeres diagram. The dashed hyperbolae are lines of constant Schwarzschild $r$ coordinate, and the dashed radial rays are lines of constant $t$. You can see how the event horizon becomes a coordinate singularity where $r$ and $t$ switch roles. Now if you draw a worldline from region I going into region II it becomes obvious that it crosses the horizon in finite proper time and, more importantly, the past light-cone of the event where it hits the singularity cannot possibly contain the whole spacetime. So the short answer to your question is no , someone falling into a black hole does not see the end of the universe. I don't know the formula you ask for for $T$, but in principle you can read it off from light rays on the diagram and just convert to whatever coordinate/proper time you want to use. | {
"source": [
"https://physics.stackexchange.com/questions/82678",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/1325/"
]
} |
83,303 | Why do small and big holes (the aperture) in a lens change the depth of field in the image? (if you have big hole you have smaller depth of field). I wondered if I should ask this on photo.stackexchange.com , but then realize that I want optic based answer which I think will get here. | The effect, aperture give to the depth of field is caused by the "used part of the lens". As the a system of lenses can only make a certain point being focused, there is the need of a trick to gain a high depth of field. This is (not only but also) done by the small aperture. The reduction of the used part of the lens leads to less aberrations for the not perfect focused light paths. You see the Parts (1) and (3) in the image being mapped better focused on the "photo plate" (5). The light paths from (2) are unaffected, because these are perfect focused. As you use a smaller part of the lens, your image will become more dark as is illustrated in the darker background of the "photo plate" on the right. Source of image: https://en.wikipedia.org/wiki/File:Depth_of_field_illustration.svg | {
"source": [
"https://physics.stackexchange.com/questions/83303",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/12649/"
]
} |
83,307 | According to the Dirac equation, antimatter is the negative energy solution to the following relation: $$E^2 = p^2 c^2 + m^2 c^4.$$ And according to general relativity, the Einstein tensor (which roughly represents the curvature of spacetime) is linearly dependent on (and I assume would then have the same mathematical sign as) the stress-energy tensor: $$G_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}.$$ For antimatter, the sign of the stress-energy tensor would change, as the sign of the energy changes. Would this change the sign of the Einstein tensor, causing spacetime to be curved in the opposite direction as it would be curved if normal matter with positive energy were in its place? Or does adding in the cosmological constant change things here? | Antimatter has the same mass as normal matter, and its interaction with gravity should be the same according to GR and QM. That said, antimatter has only been created in tiny amounts so far and only few experiments have been performed to confirm there is no new physics involved. The gravitational interaction of antimatter with matter or antimatter has not been conclusively observed by physicists. While the overwhelming consensus among physicists is that antimatter will attract both matter and antimatter at the same rate that matter attracts matter, there is a strong desire to confirm this experimentally, since the hypothesis is still open to falsification. https://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter | {
"source": [
"https://physics.stackexchange.com/questions/83307",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/25627/"
]
} |
83,488 | If an observer starts moving at relativistic speeds will he observe the temperature of objects to change as compared to their rest temperatures?
Suppose the rest temperature measured is $T$ and the observer starts moving with speed $v$. What will be the new temperature observed by him? | This is a very good question. Einstein himself, in a 1907 review (available in translation as Am. J. Phys. 45 , 512 (1977) , e.g. here ), and Planck, one year later, assumed the first and second law of thermodynamics to be covariant, and derived from that the following transformation rule for the temperature:
$$
T' = T/\gamma, \quad \gamma = \sqrt{1/(1-v^2/c^2)}.
$$
So, an observer would see a system in relativistic motion "cooler" than if he were in its rest frame. However, in 1963 Ott ( Z. Phys. 175 no. 1 (1963) 70 ) proposed as the appropriate transformation
$$
T' = \gamma T
$$
suggesting that a moving body appears "relatively" warmer. Later on Landsberg ( Nature 213 (1966) 571 and 214 (1967) 903 ) argued that the thermodynamic quantities that are statistical in nature, such as temperature, entropy and internal energy, should not be expected to change for an observer who sees the center of mass of the system moving uniformly.
This approach, leads to the conclusion that some thermodynamic relationships such as the second law are not covariant and results in the transformation rule:
$$
T' = T
$$ So far it seems there isn't a general consensus on which is the appropriate transformation, but I may be not aware of some "breakthrough" experiment on the topic. Main reference: M.Khaleghy, F.Qassemi. Relativistic Temperature Transformation Revisited, One hundred years after Relativity Theory (2005). arXiv:physics/0506214 . | {
"source": [
"https://physics.stackexchange.com/questions/83488",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/26033/"
]
} |
83,531 | Einstein's big revelation was that time and space are inseparable components of the same fabric. Physical observation tells us that distant galaxies are moving away from us at an accelerated rate, and because of the difficulty (impossibility?) of defining a coordinate system where things have well defined coordinates while also moving away from each other without changing the metric on the space, we interpret this to mean that space itself is expanding. Because space and time are so directly intertwined is it possible that time too is expanding? Or perhaps it could be contracting? | The simple answer is that no, time is not expanding or contracting. The complicated answer is that when we're describing the universe we start with the assumption that time isn't expanding or contracting. That is, we choose our coordinate system to make the time dimension non-changing. You don't say whether you're at school or college or whatever, but I'm guessing you've heard of Pythagoras' theorem for calculating the distance, $s$, between two points $(0, 0, 0)$ and $(x, y, z)$: $$ s^2 = x^2 + y^2 + z^2 $$ Well in special relativity we have to include time in the equation to get a spacetime distance: $$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$ and in general relativity the equation becomes even more complicated because we have to multiply the $dt^2$, $dx^2$, etc by factors determined by a quantity called the metric, and usually denoted by $g$: $$ ds^2 = g_{00}dt^2 + g_{11}dx^2 + g_{22}dy^2 + ... etc $$ where the $... etc$ can include cross terms like $g_{01}dtdx$, so it can all get very hairy. To be able to do the calculations we normally look for ways to simplify the expression, and in the particular case of the expanding universe we assume that the equation has the form: $$ ds^2 = -dt^2 + a(t)^2 d\Sigma^2 $$ where the $d\Sigma$ includes all the spatial terms. The function $a(t)$ is a scale factor i.e. it scales up or down the contribution from the $dx$, $dy$ and $dz$, and it's a function of time so the scale factor changes with time. And this is where we get the expanding universe. It's because when you solve the Einstein equations for a homogenous isotropic universe you can calculate $a(t)$ and you find it increases with time, and that's what we mean by the expansion. However the $dt$ term is not scaled, so time is not expanding (or contracting). | {
"source": [
"https://physics.stackexchange.com/questions/83531",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/17173/"
]
} |
83,743 | This question is about a mathematical operation (the tensor product) but thinking about the motivation that comes from physics. Algebraists motivate the tensor product like that: "given $k$ vector spaces $V_1,\dots,V_k$ over the same field $\Bbb K$ we want to find a new space $S$ and a universal multi-linear map $T$ such that for every vector space $W$ and multi-linear mapping $g : V_1\times\cdots\times V_k\to W$ we have a linear map $f : S\to W$ such that $g = f\circ T$ ". Then, they prove this thing exists by constructing it. They take the free vector space $\mathcal{M}=F(V_1\times\cdots\times V_k)$ and consider the subspace $\mathcal{M}_0$ spanned by all elements of the form $$(v_1,\dots,v_i'+av_i'',\dots,v_k)-(v_1,\dots,v_i',\dots,v_k)-a(v_1,\dots,v_i'',\dots,v_k),$$ and define $S=\mathcal{M}/\mathcal{M_0}$ denoting $S=V_1\otimes\cdots\otimes V_k$ and define $T(v_1,\dots,v_k)=(v_1,\dots,v_k)+\mathcal{M}_0$ and denote this by $T(v_1,\dots,v_k)=v_1\otimes\cdots\otimes v_k$ . That's fine, but tensors appear a lot in physics. In General relativity, in electrodynamics, in classical mechanics, in quantum mechanics, etc. So, if someone asked me: "what's the motivation for that definition of tensor product" and I wished to motivate it through physics, what should be the motivation? How would I convince myself that the tensor product as defined like that is useful in physics? I know that one can defined tensors as multi-linear maps, and that is far more intuitive, however I'm interest to see how one would motivate this definition. | It is essentially impossible to answer the general question of "how does multilinearity come up naturally in physics?" because of the myriad of possible examples that make up the total answer. Instead, let me describe a situation that very loudly cries out for the use of tensor products of two vectors. Consider the problem of conservation of momentum for a continuous distribution of electric charge and current, which interacts with an electromagnetic field, under the action of no other external force. I will describe it more or less along the lines of Jackson ( Classical Electrodynamics , 3 rd edition, §6.7) but depart from it towards the end. This will get very electromagneticky for a while, so if you want to skip to the tensors, you can go straight to equation (1) . The rate of change of the total mechanical momentum of the system is the total Lorentz force, given by
$$
\frac{ d\mathbf{P}_\rm{mech}}{dt}
=\int_V(\rho\mathbf{E}+\mathbf{J}\times \mathbf{B})d\mathbf{x}.
$$
To simplify this, one can take $\rho$ and $\mathbf{J}$ from Maxwell's equations:
$$
\rho=\epsilon_0\nabla\cdot\mathbf{E}
\ \ \ \text{ and }\ \ \
\mathbf{J}=\frac1{\mu_0}\nabla\times \mathbf{B}-\epsilon_0\frac{\partial \mathbf{E}}{\partial t}.$$
(In particular, this means that what follows is only valid "on shell": momentum is only conserved if the equations of motion are obeyed. Of course!) One can then put these expressions back, to a nice vector calculus work-out, and come up with the following relation:
$$
\begin{align}{}
\frac{ d\mathbf{P}_\rm{mech}}{dt}
+&\frac{d}{dt}\int_V\epsilon_0\mathbf{E}\times \mathbf{B}d\mathbf{x}
\\ &=
\epsilon_0\int_V \left[
\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E}) + c^2 \mathbf{B} (\nabla \cdot \mathbf{B})- c^2 \mathbf{B} \times (\nabla \times \mathbf{B})
\right]d\mathbf{x}.
\end{align}
$$ The integral on the left-hand side can be identified as the total electromagnetic momentum, and differs from the integral of the Poynting vector by a factor of $1/c^2$. To get this in the proper form for a conservation law, though, such as the one for energy in this setting,
$$
\frac{dE_\rm{mech}}{dt}
+\frac{d}{dt}\frac{\epsilon_0}{2}\int_V(\mathbf{E}^2
+c^2\mathbf{B}^2)d\mathbf{x}
=
-\oint_S \mathbf{S}\cdot d\mathbf{a},
$$
we need to reduce the huge, ugly volume integral into a surface integral. The way to do this, is, of course, the divergence theorem. However, that theorem is for scalars, and what we have so far is a vector equation. To work further then, we need to (at least temporarily) work in some specific basis $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$, and write $\mathbf{E}=\sum_i E_i \mathbf{e}_i$. Let's work with the electric field term first; after that the results also apply to the magnetic term. Thus, to start with,
$$
\begin{align}{}
\int_V \left[
\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})
\right]d\mathbf{x}
=
\sum_i \mathbf{e}_i
\int_V \left[
E_i(\nabla\cdot \mathbf{E})-\mathbf{e}_i\cdot\left(\mathbf{E} \times(\nabla \times \mathbf{E})\right)
\right]d\mathbf{x}.
\end{align}
$$
These terms should be simplified using the vector calculus identities
$$
E_i(\nabla\cdot \mathbf{E})
=
\nabla\cdot\left(E_i \mathbf{E}\right) - \mathbf{E}\cdot \nabla E_1
$$
and
$$
\mathbf{E} \times(\nabla \times \mathbf{E})
=
\frac12\nabla(\mathbf{E}\cdot\mathbf{E})-(\mathbf{E}\cdot\nabla)\mathbf{E},
$$
which mean that the whole combination can be simplified as
$$
\begin{align}{}
\int_V \left[
\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})
\right]d\mathbf{x}
=
\sum_i \mathbf{e}_i
\int_V \left[
\nabla\cdot\left(E_i \mathbf{E}\right)
-
\mathbf{e}_i\cdot\left(
\frac12\nabla(\mathbf{E}\cdot\mathbf{E})
\right)
\right]d\mathbf{x},
\end{align}
$$
since the terms in $\mathbf{E}\cdot \nabla E_i$ and $\mathbf{e}_i\cdot\left( (\mathbf{E}\cdot\nabla)\mathbf{E}\right)$ cancel. This means we can write the whole integrand as the divergence of some vector field, and use the divergence theorem:
$$
\begin{align}{}
\int_V \left[
\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})
\right]d\mathbf{x}
&=
\sum_i \mathbf{e}_i
\int_V \nabla\cdot\left[
E_i \mathbf{E}
-
\frac12 \mathbf{e}_i E^2
\right]d\mathbf{x}
\\ & =
\sum_i \mathbf{e}_i
\oint_S\left[
E_i \mathbf{E}
-
\frac12 \mathbf{e}_i E^2
\right]\cdot d\mathbf{a}. \tag 1
\end{align}
$$ In terms of conservation law structure, we're essentially done, as we've reduced the rate of change of momentum to a surface term. However, it is crying out for some simplification. In particular, this expression is basis-dependent, but it is so close to being basis independent that it's worth a closer look. The first term, for instance, is simply crying out for a simplification that would look something like
$$
\sum_i \mathbf{e}_i
\oint_S
E_i \mathbf{E}\cdot d\mathbf{a}
=
\oint_S
\mathbf{E}\, \mathbf{E}\cdot d\mathbf{a}
$$
if we could only make sense of an object like $\mathbf{E}\, \mathbf{E}$. Even better, if we could make sense of such a combination, then it turns out that the seemingly basis-dependent combination that would come up in the second term, $\sum_i \mathbf{e}_i\,\mathbf{e}_i$, turns out to be basis independent: one can prove that for any two orthonormal bases $\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ and $\{\mathbf{e}_1', \mathbf{e}_2', \mathbf{e}_3'\}$, those combinations are the same:
$$
\sum_i \mathbf{e}_i\,\mathbf{e}_i = \sum_i \mathbf{e}_i'\,\mathbf{e}_i'
$$
as long as the product $\mathbf{u}\,\mathbf{v}$ of two vectors, whatever it ends up being, is linear on each component, which is definitely a reasonable assumption. So what, then, should this new vector multiplication be? One key to realizing what we really need is noticing the fact that we haven't yet assigned any real physical meaning to the combination $\mathbf{E}\,\mathbf{E}$; instead, we're only ever interacting with it by dotting "one of the vectors of the product" with the surface area element $d\mathbf{a}$, and that leaves a vector $\mathbf{E}\,\mathbf{E}\cdot d\mathbf{a}$ which we can integrate to get a vector, and that requires no new structure. Let's then write a list of how we want this new product to behave. To keep things clear, let's give it some fancy new symbol like $\otimes$, mostly to avoid unseemly combinations like $\mathbf{u}\,\mathbf{v}$. We want then, a function $\otimes:V\times V\to W$, which takes euclidean vectors in $V=\mathbb R^3$ into some vector space $W$ in which we'll keep our fancy new objects. Combinations of the form $\mathbf{u}\otimes \mathbf{v}$ should be linear in both $\mathbf{u}$ and $\mathbf{v}$. For all vectors $w$ in $V$, and all combinations $(\mathbf{u},\mathbf{v})\in V\times V$, we want the combination $(\mathbf{u}\otimes \mathbf{v})\cdot\mathbf{w}$ to be a vector in $V$. Even more, we want that to be the vector $(\mathbf{v}\cdot\mathbf{w})\mathbf{u}\in V$. That last one looks actually pretty strong, but there's evidently room for improvement. For one, it depends on the euclidean structure, which is not actually necessary: we can make an equivalent statement that uses the vector space's dual. For all $(\mathbf{u},\mathbf{v})\in V\times V$ and all $f\in V^\ast$, we want $f_\to(\mathbf{u}\otimes \mathbf{v})=f(\mathbf{v})\mathbf{u}\in V$ to hold, where $f_\to$ simply means that $f$ acts on the factor on the right. Finally, if we're doing stuff with the dual, we can reformulate that in a slightly prettier way. Since two vectors $\mathbf{u},\mathbf{v}\in V$ are equal if and only if $f(\mathbf{u})=f(\mathbf{v})$ for all $f\in V^\ast$, we can give another equivalent statement of the same statement: For all $(\mathbf{u},\mathbf{v})\in V\times V$ and all $f,g\in V^\ast$, we want $g_\leftarrow f_\to(\mathbf{u}\otimes \mathbf{v})=g(\mathbf{u})f(\mathbf{v})\in V$. [Note, here, that this last rephrasing isn't really that fancy. Essentially, it is saying that the vector equation (1) is really to be interpreted as a component-by-component equality, and that's not really off the mark of how we actually do things.] I could keep going, but it's clear that this requirement can be rephrased into the universal property of the tensor product , and that rephrasing is a job for the mathematicians. Thus, you can see the story like this: Upon hitting equation (1), we give to the mathematicians this list of requirements. They go off, think for a bit, and come back telling us that such a structure does exist (i.e. there exist rigorous constructions that obey those requirements) and that it is essentially unique, in the sense that multiple such constructions are possible, but they are canonically isomorphic. For a physicist, what that means is that it's OK to write down objects like $\mathbf{u}\otimes \mathbf{v}$ as long as one does keep within the rules of the game. As far as electromagnetism goes, this means that we can write our conservation law in the form
$$
\frac{ d\mathbf{P}_\rm{mech}}{dt}
+\frac{d}{dt}\int_V\epsilon_0\mathbf{E}\times \mathbf{B}d\mathbf{x}
=
\oint_A \mathcal T\cdot d\mathbf{a}
$$
where
$$
\mathcal T
=
\epsilon_0\left[
\mathbf{E}\otimes\mathbf{E}+c^2\mathbf{B}\otimes\mathbf{B}
-\frac12\sum_i\mathbf{e}_i\otimes\mathbf{e}_i\left(E^2+c^2 B^2\right)
\right]
$$
is, of course, the Maxwell stress tensor. I could go on and on about this, but I think this really captures the essence of how and where it happens in physics that a situation is really begging the use of a tensor product. There are other such situations, of course, but this is the clearest one I know. | {
"source": [
"https://physics.stackexchange.com/questions/83743",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/21146/"
]
} |
83,821 | Betavoltaic batteries are devices which creates electricity from beta radiation of a radioactive material. Alphavoltaics operate similarly, using alpha radiation. The concept was invented roughly 50 years ago and they are safe enough to be used, for example, in pacemakers. However, the Wikipedia article on them states that they were "phased out as cheaper lithium-ion batteries were developed." I feel, though, that lithium ion batteries are hardly up to the task that consumers would want them to perform: for instance, iPhones hold their charge for about a day and notebooks can sometimes manage no more than four hours. Betavoltaics, on the other hand, can hold their charge for years. Why, then, are they not used in commercial applications? What are their relative advantages and disadvantages with respect to the current solutions, and in particular to lithium ion batteries? UPDATE Amount of electricity is tied with half-life. For example, if Ni-63 has half-life of 100 years this means, that mole of Nickel will produce Avogadro/2 electrons during that 100 years. This means 10^21 electrons per year and 10^14 electrons per second. This means up to 0.1 mA or electric current. The energy of electrons from Nickel is 67 keV. This means that each electron has 67 kilovolts of electric tension. So, the power of electricity from one mole of Nickel-63 is 67000*0.0001 = up to 6 watts. Other way to calculate. If Nickel-64 produces 10^14 electrons per second, each of 67 keV of energy, then the power is 7 * 10^4 * 10^14 ev/s = 7 * 10^4 * 10^14 * 10^(-19) = 0.7 Watts. So, the numbers are consistent to the order of magnitude. Approximating, one mole of Nickel-63 provides 1 watt of electricity approx. This looks sufficient for many cases including iPhone power consumption. 1 mole of Nickel-63 weights 63 grams. iPhones accumulator can weight more than 100. So atomic batteries can supersede conventional batteries, and serve for years. So why we don't use them? | There are many reasons for this situation. Power produced is non-adjustable . The battery produces power at nearly constant rate (slowly decaying with time). It cannot be increased and if not consumed (or stored) the power is lost. (Mentioned by DumpsterDoofus) low power density . ${}^{63}\text{Ni}$ for instance produces ~5 W/kg (and kg here is just mass of radioactive material, the actual battery would be at least order of magnitude heavier). There are, of course isotopes with power densities much higher but they encounter other problems. Semiconductor damage . If we try to increase power by using isotopes with higher decay energies we find that the high energy electrons damage semiconductors, reducing service life of batteries to times much shorter than isotope halflife. Alpha particles, especially, damage the p-n junctions, so even though (for instance) ${}^{238}\text{Pu}$ produces 0.55 W/g of alpha radiation, it is mainly used in the thermoelectric schemes rather than in direct energy converters. Gamma radiation . Many isotopes has gamma emission as a secondary mode of decay. Since this type of radiation is difficult to shield, this means that the selection of isotopes usable for batteries is limited only to pure beta emitters . Bremsstrahlung . Electrons braking produces this type of radiation, that had to be shielded. Again, this limits our selection of isotopes to those with relatively low decay energies. Low volume of production / Economics . Many isotopes cost too much to be practical in wide array of applications. This is partly explained by low volume of production and partly by production process which will be costly at all volumes because it requires energy consuming isotope separation and special facilities for working with radioactive materials. For instance, tritium (one of the materials for betavoltaics) costs about $30 000 per gram and its world annual production is 400 g ( from wikipedia ). All this means, that nuclear batteries are limited to a selection of niche applications, typically those with low power / long autonomous lifetime requirements. That is not to say that there can't be innovations expanding their use or reducing costs. [1] Tsvetkov, L. A., et al. "Possible Way To Industrial Production of Nickel-63 and the Prospects of Its Use." (2005). online version Update . Your updated calculations on power output from ${}^{63}\text{Ni}$ is essentially correct with one crucial distinction: 67 keV is total decay energy and approximately maximum energy of electron. But, since the decay also produces neutrino the mean energy of electron is much smaller: 17 keV (look at this NUDAT reference , or this java applet for electron spectrum). So the usable power from 1 mole of ${}^{63}\text{Ni}$ is:
$$
W= {}^{63}\text{Ni specific activity} \times 17\,\text{keV} \times 63\,\text{g} = 0.36\,\text{W},
$$
where specific activity could be, for instance, taken from Wolfram Alpha . This is not sufficient to provide iPhone peak power consumption, which is about 1.5 W (see my reason 1). Incidentally, we come to one more reason (though not, strictly speaking, related to physics): Safety / Regulations / Perception : 63 grams of ${}^{63}\text{Ni}$ constitute more than 3500 curie of radioactivity, which would definitely require regulations for handling and probably would not be allowed inside a single unit for unrestricted civilian use. We know that when properly used betavoltaics are safe. But what about im -proper use / improper disposal / potential for abuse? At any rate, current perception of nuclear power by general public is not that good, so marketing nuclear batteries will present certain challenge. | {
"source": [
"https://physics.stackexchange.com/questions/83821",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/7574/"
]
} |
86,141 | Let us say that I am sitting in a room with all the drapes open. Bright sunlight is coming through the window. The whole room is brilliantly lighted. I will not be able to see the dust particles suspended in air. Now, if I draw the drapes close, keeping a small slit open, allowing only a beam of sunlight to come in, I will readily see the suspended dust particles in that beam. The same thing will happen in a dark night with the beam of light from a handheld battery torch. What will be the scientific explanation for this? I can not see the dust particles
when I have more light. But when I actually reduce the light and there is only one narrow beam present, I can see those minuscule particles. How does a narrow beam of light enable me to see those fine elements? | Your inability to see the dust until you narrow the slit has nothing to do with the narrowness of the beam but instead the dynamic range of light that your eye can see at one time. A bit of searching turns up reports of a contrast ratio for you eye at one time as between 100:1 and 1000:1. This means if you're in a room with a range of brightness greater than about 100 to 1 the brightest things will all be washed out as white and the darkest things will all be essentially black. This is obvious in photos that are "backlit" like this one: These horses aren't black but because the ratio of the bright light to the dark horses exceeds the dynamic range of the camera the sky is washed out white and the horses are in silhouette. Your eye can adjust over time to a huge range but it can't utilize the whole range all at once. In the case of dust reflecting light, if you allow a lot of light into the room the relative brightness between the small amount of light the dust is reflecting and the rest of the illuminated room prevent you from seeing the dust. This is fundamental to signal processing. Why can't you hear a whisper in a noisy room? The noise of the crowd obscures the whisper. The difference between the signal you're trying to pick up and the background noise is called the signal-to-noise ratio . In the case of dust, the light let into the room is scattered and reflected in the room and causes the room to be illuminated. This is the noise that obscures the signal from light reflected off of the dust. | {
"source": [
"https://physics.stackexchange.com/questions/86141",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/33512/"
]
} |
86,343 | When I look at electric or magnetic fields, each of them has a constant that defines how a field affects or is affected by a medium. For example, electric fields in vacuum have a permittivity constant $ϵ_0$ embedded in the electric field expression of a point charge: $E = q/4π ϵ_0r^2$. However, if I put this point charge in some dielectric that has a different permittivity constant $ϵ$, the value of the electric field changes. On a similar note, magnetic fields behave very similar but have the permeability constant $μ_0$ instead. From my understanding, I believe that this is not the case for gravitational fields since the universal gravitational constant $G$ is consider to be a fundamental constant. So I am assuming that even though gravitational fields do operate in different types of mediums, this somehow doesn’t affect the gravitational field value. My question is why is this the case, that is, why isn’t there a permittivity-type constant for gravitation? | Permittivity $\varepsilon$ is what characterizes the amount of polarization $\mathbf{P}$ which occurs when an external electric field $\mathbf{E}$ is applied to a certain dielectric medium. The relation of the three quantities is given by $$\mathbf{P}=\varepsilon\mathbf{E},$$ where permittivity can also be a (rank-two) tensor: this is the case in an anisotropic material. But what does it mean for a medium to be polarized? It means that there are electric dipoles, that units of both negative and positive charge exist. But this already gives us an answer to the original question: There are no opposite charges in gravitation, there is only one kind, namely mass, which can only be positive. Therefore there are no dipoles and no concept of polarizability. Thus, there is also no permittivity in gravitation. | {
"source": [
"https://physics.stackexchange.com/questions/86343",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/32210/"
]
} |
86,534 | Bohr's second postulate in Bohr model of hydrogen atom deals with quantisation of angular momentum. I was wondering, though: why did he quantise angular momentum instead of some other quantity? | Bohr postulated that electrons orbit the nucleus in discrete energy levels, and electrons can gain and lose energy by jumping between energy levels, giving off radiation of frequency $\nu$ according to the formula $$\Delta E = E_2 - E_1 = h\nu$$ where $\nu = \frac{1}{T}$, where $T$ is the period of orbit, as in classical mechanics. Now during the transition, let $r$ be the average radius and $v$ be the average velocity of the particle. Making such a simplification allows us to calculate the period of orbit: $$T = \frac{2\pi r}{v}$$ Therefore, $$\Delta E = h\nu = \frac{hv}{2\pi r} \tag 1$$ Also, we know the kinetic energy at a particular energy level is given by \begin{align}
\text{K.E.} & = \frac{mv^2}{2} = \frac{Lv}{2r}, \quad \text{so therefore} \\
-U & = 2KE = \frac{Lv}{r}
\end{align} Again, taking $r$ and $v$ to be the average radius and velocity during the transition, we get $$\Delta E = \frac{(L_2 - L_1)v}{r}. \tag 2$$ Equating $(1)$ and $(2)$ gives $$\frac{(L_2 - L_1)v}{r} = \frac{hv}{2\pi r}.$$ Therefore, $$L_2 - L_1 = \frac{h}{2\pi} = \hbar$$ Therefore, each energy level differs from the next by an angular momentum of $\hbar$. It is therefore reasonable to postulate that if the lowest energy level has no angular momentum, then each energy level from then on has an angular momentum of $n\hbar$ where $n$ is an integer. Below is the modern de-Broglie method: From the definition of angular momentum, $L = rp$, where L is angular momentum, r is radius of orbit and p is momentum. We also know that momentum is related to wavelength of a particle from the de-broglie relation: $$p = \frac{h}{\lambda}.$$ Combining these gives $$L = \frac{rh}{\lambda}.$$ Ok, now let us consider an electron orbiting a nucleus. The circumference of the orbit is $2\pi r$, and because we want the electron to form a standing wave orbit, we require that $\frac{2\pi r}{\lambda}$ be an integer, in order for the wave not to interfere with itself. That is, $$\frac{2\pi r}{\lambda} = n,$$ where $n$ is some integer. Now we can substitute our definition of $L$ from above into this equation to give: $$\frac{2\pi L}{h} = n$$ and re-arranging gives, $$L = \frac{nh}{2\pi} = n\hbar$$ Therefore, quantising angular momentum allows for the electron wave to not interfere with itself during orbit. | {
"source": [
"https://physics.stackexchange.com/questions/86534",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/23742/"
]
} |
86,774 | I noticed that when I had a helium filled, latex balloon inside of my car, it moved forward in the cabin as I accelerated forward. The faster I accelerated forward, the faster the balloon went from the back of the car to the front of the car. The balloon didn't have a string. This became a game with my 4 year old as we drove home. We figured out where the balloon would go based on how fast I accelerated, turned corners etc. I expected that it would act a lot like the water in a cup does, but it was the total opposite it seemed. What forces caused this behavior? I assumed it has something to do with the fluid dynamics in the closed cabin, but I can't figure it out. | It travels forwards instead of backwards in an accelerating car for the same reason that a helium balloon travels upwards instead of downwards under the influence of gravity. Why is that? In an accelerating car, for all intents and purposes the acceleration can be considered a change in the amount and direction of gravity, from pointing straight down to pointing downwards and backwards. The balloon doesn't know and doesn't care if the acceleration is from gravity or from the acceleration of the car; it just tries to move in the direction it naturally moves, namely, against the direction of the acceleration. Thus, it moves forwards when you accelerate. Hopefully you find this explanation intuitively satisfying. Another more rigorous way to view the problem is through Lagrangian minimization. The balloon can be considered a low-density object embedded in a higher-density fluid constrained within the confines of the car. Under the influence of gravity pointing sideways, the total system potential energy decreases the farther forward the balloon is situated. Since the force is the gradient of the potential, the balloon will try to move forward. | {
"source": [
"https://physics.stackexchange.com/questions/86774",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/33986/"
]
} |
87,107 | It is my understanding that metals are a crystal lattice of ions, held together by delocalized electrons, which move freely through the lattice (and conduct electricity, heat, etc.). If two pieces of the same metal are touched together, why don't they bond? It seems to me the delocalized electrons would move from one metal to the other, and extend the bond, holding the two pieces together. If the electrons don't move freely from one piece to the other, why would this not happen when a current is applied (through the two pieces)? | I think that mere touching does not bring the surfaces close enough. The surface of a metal is not perfect usually. Maybe it has an oxide layer that resists any kind of reaction. If the metal is extremely pure and if you bring two pieces of it extremely close together, then they will join together. It's also called cold welding. For more information: What prevents two pieces of metal from bonding? Cold Welding | {
"source": [
"https://physics.stackexchange.com/questions/87107",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30968/"
]
} |
87,342 | I have seen numerous 'derivations' of the Maxwell Lagrangian, $$\mathcal{L} ~=~ -\frac{1}{4}F_{\mu \nu}F^{\mu \nu},$$ but every one has sneakily inserted a factor of $-1/4$ without explaining why. The Euler-Lagrange equations are the same no matter what constant we put in front of the contraction of the field strength tensors, so why the factor of $-1/4$? | Comments to the question: First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like. As Frederic Brünner mentions a normalization of the $J^{\mu}A_{\mu}$ source term with a normalization constant $\pm N$ goes hand in hand with a $-\frac{N}{4}$ normalization of the $F_{\mu\nu}F^{\mu\nu}$ term. Here the signature of the Minkowski metric is $(\mp,\pm,\pm,\pm)$. Recall that the fundamental variables of the Lagrangian formulation are the $4$-gauge potential $A_{\mu}$. Here $A_{0}$ is a non-dynamical Lagrange multiplier. The dynamical variables of the theory are $A_1$, $A_2$, and $A_3$. The $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}_i\dot{A}_i}_{\text{kinetic term}}+\ldots$$ is just the standard $+\frac{1}{2}$ normalization of a kinetic term in field theory. In particular note that the kinetic term is positive definite in order not to break unitarity . | {
"source": [
"https://physics.stackexchange.com/questions/87342",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/34382/"
]
} |
87,731 | I don't know if anyone else has noticed this but in most buildings and most rooms, radiators are predominantly placed under a window. Now, in my eyes, that is the worst place to put them; hot air rises, reaches the window (which no matter how well insulated it's still letting out heat, in loose terms) and the thermal energy of the air disperses around the window area, thus not doing much to warm up the room. Am I wrong to think this? I mean, I can hold my hand close to my window and feel that it is colder there than at the other end of my room, but then again, my room does warm up when the radiator is on. | The reason is because the heat loss occurs mostly in the windows and the fenestration. The idea is that you would like the incoming air to be heated up. Also, it creates an air curtain that prevents more heat from being lost through these exposed areas. Finally, it makes the temperature of the room more or less uniform. If the heaters were placed at the center of the room, you would create a large temperature gradient, resulting in drafts and discomfort for the occupant. | {
"source": [
"https://physics.stackexchange.com/questions/87731",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/32287/"
]
} |
87,751 | It is a common belief that low frequencies travel longer distances. Indeed, the bass is really what you hear when the neighbor plays his HiFi loud (Woom Woom). Try asking people around, a lot of them believe that low sounds carry longer distances. But my experience isn't as straightforward. In particular: When I stand near someone who's listening loud music in headphones, it is the high pitched sounds that I hear (tchts tchts), not the bass. When I sit next to an unamplified turntable (the disc is spinning but the volume is turned off), I hear high pitched sounds (tchts tchts), not the bass. So with very weak sounds, high frequencies seem to travel further? This makes me think that perhaps low frequencies do not carry longer distances, but the very high amplitude of the bass in my neighbor's speakers compensates for that. Perhaps also the low frequencies resonate with the walls of the building? Probably also the medium the sound travels through makes a difference? Or perhaps high frequencies are reflected more by walls than low frequencies? I found this rather cute high school experiment online, which seems to conclude that low and high frequencies travel as far, but aren't there laws that physicist wrote centuries ago about this? | Do low frequencies carry farther than high frequencies? Yes. The reason has to do with what's stopping the sound. If it weren't for attenuation (absorption) sound would follow an inverse square law. Remember, sound is a pressure wave vibration of molecules. Whenever you give molecules a "push" you're going to lose some energy to heat. Because of this, sound is lost to heating of the medium it is propagating through. The attenuation of sound waves is frequency dependent in most materials. See Wikipedia for the technical details and formulas of acoustic attenuation. Here is a graph of the attenuation of sound at difference frequencies (accounting for atmospheric pressure and humidity): As you can see, low frequencies are not absorbed as well. This means low frequencies will travel farther. That graph comes from this extremely detailed article on outdoor sound propagation . Another effect that affects sound propagation, especially through walls, headphones, and other relative hard surfaces is reflection. Reflection is also frequency dependent. High frequencies are better reflected whereas low frequencies are able to pass through the barrier: This is and frequency-based attenuation are why low-frequency sounds are much easier to hear through walls than high frequency ones. Frequency Loudness in Headphones: The above description apply to sounds that travel either through long distances or are otherwise highly attenuated. Headphones start off at such low intensities already they don't travel long enough distances for attenuation to be a dominate factor. Instead, the frequency response curve of the human ear plays a big role in perceived loudness. The curves that show human hearing frequency response are called Fletcher–Munson curves : The red lines are the modern ISO 226:2003 data. All the sound along a curve is of "equal loudness" but as you can see, low frequencies must be much more intense to sound equally as loud as higher frequency sounds. Even if the low frequencies are reaching your ear, it's harder for you to hear them. Headphone sound is doubly compounded by the difficulty of making headphones with good low-frequency response. With loudspeakers you can split the job of producing frequencies among a subwoofer, a midrange speaker, and a tweeter. For low frequencies subwoofers are large and have a resonating chamber which simply isn't an option with headphones that must produce a large range of sound frequencies in a small space. Even a good pair of headphones like Sennheiser HD-650 struggle with lower frequencies : So if it sounds like high frequencies travel farther with headphones, it's because headphones are poor at producing low frequencies and your ear is poor at picking them up. | {
"source": [
"https://physics.stackexchange.com/questions/87751",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/34618/"
]
} |
87,817 | It is a standard exercise in relativistic electrodynamics to show that the electromagnetic field tensor $F_{\mu\nu}$ , whose components equal the electric $E^i=cF^{i0}$ and magnetic $B_i=-\frac12\epsilon_{ijk}F^{jk}$ fields in the taken frame of reference, has two Lorentz invariant quantities, $$\frac12F^{\mu\nu}F_{\mu\nu}=\mathbf{B}^2-\mathbf{E}^2$$ and $$\frac14F_{\mu\nu} {}^\ast F^{\mu\nu}=\frac14\epsilon^{\mu\nu\alpha\beta }F_{\mu\nu}F_{\alpha\beta}=\mathbf{B}\cdot\mathbf{E}.$$ There is, however, a further Wikipedia article which states that these two quantities are fundamental , in the sense that any other invariant of this tensor must be a function of these two. While I find this plausible, I have never seen a proof of this fact, and it is absent from e.g. Jackson. Is there a simple proof of this fact? I'm particularly interested in higher-order invariants, but I would also like answers to include a proof that these are the only two bilinears. To be more precise, I would like to see a proof that Any function $I:F_{\mu\nu}\mapsto I(F)\in\mathbb{R}$ that takes electromagnetic field tensors to real scalars and which is Lorentz invariant (i.e. $I(\Lambda_{\mu}^\alpha \Lambda_{\nu}^\beta F_{\alpha\beta})= I(F_{\mu\nu})$ for all Lorentz transformations) must be a function $I(F)=I'(F^{\mu\nu}F_{\mu\nu},F_{\mu\nu}\ {}^\ast F^{\mu\nu})$ of the two fundamental invariants described above. If there are multiple ways to arrive at this result, I would also appreciate comments on how they relate to each other. | Here is the proof taken from Landau & Lifshitz' "Classical Theory of Fields": Take the complex (3)-vector:
$$
\mathbf{F} = \mathbf{E}+i\, \mathbf{B}.
$$
Now consider the behavior of this vector under Lorentz transformations. It is easy to show that Lorentz boosts correspond to rotations through the imaginary angles, for example boost in $(x,t)$ plane:
\begin{gather}
F_x=F'_x,\\
F_y = F'_y \cosh \psi - i F'_z \sinh \psi = F'_y \cos i \psi - F'_z \sin i \psi. \\
F_z = F'_z \cos i \psi + F'_y \sin i \psi,
\end{gather}
where $\tanh \psi = \frac{v}c$, correspond to rotation of $\mathbf{F}$ through imaginary angle $i \psi$ in the $(y,z)$ plane. Overall, the set of all Lorentz transformations (including
also the purely spacial rotations) is equivalent to the set of all possible
rotations through complex angles in three-dimensional space (where the six angles of
rotation in four-space correspond to the three complex angles of rotation of the three-dimensional system). The only invariant of a vector with respect to rotation is its square: $\mathbf{F}^2 = E^2 - B^2 + 2 i (\mathbf{E}\cdot \mathbf{B})$ thus the real quantities $E^2-B^2$ and $(\mathbf{E}\cdot \mathbf{B})$ are the only two independent
invariants of the tensor $F_{\mu\nu}$. So in essence, we reduce the problem of invariant of $F_{\mu\nu}$ under Lorentz tranform to invariants of a 3-vector under rotations which is square of a vector (and only it). So any invariant $I(F)$ has to be the function of $\Re(F^2)$ and $\Im(F^2)$. | {
"source": [
"https://physics.stackexchange.com/questions/87817",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/8563/"
]
} |
87,986 | It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances? | The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from that star, you are nevertheless getting bombarded by 1 million photons per square centimeter in each second. $$\frac{10^{45}\ \text{photons}/\mathrm s}{4\pi (10 \ \text{lightyears})^2} \approx 10^6\ \text{photons}/(\mathrm{cm^2\ s)}$$ | {
"source": [
"https://physics.stackexchange.com/questions/87986",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/31340/"
]
} |
88,145 | I'm curious why rockets are so big in their size. Since both the gravitational potential one need to overcome in order to put thing into orbit, and the chemical energy burned from the fuel, are proportional to the mass, so if we shrink the rocket size, it would seem to be fine to launch satellites. So why not build small rocket say the size of human? I can imagine small rocket would be easier to manufacture in large quantities and easier to transport. And maybe someone can make a business out of small rocket, carrying one's own satellite. | The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially . This relation , specifically, is
$$
\Delta v=v_e\ln\left(\frac{m_i}{m_f}\right)
$$
where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass. The above can be rearranged to get
$$
m_f=m_ie^{-\Delta v/v_e}\qquad m_i=m_fe^{\Delta v/v_e}
$$
or by taking the difference between the two,
$$
M_f=1-\frac{m_f}{m_i}=1-e^{-\Delta v/v_e}
$$
where $M_f$ is the exhaust mass fraction. If we assume we are starting from rest to reach 11.2 km/s (i.e., Earth's escape velocity ) with a constant $v_e=4$ km/s (typical velocity for NASA rockets), we'd need
$$
M_f=1-e^{-11.2/4}=0.939
$$
which means almost 94% of the mass at launch needs to be fuel! If we have a 2000 kg craft (about the size of a car), we would need nearly 31,000 kg of fuel in a craft that size. The liquid propellant has a density similar to water (so 1000 kg/m$^3$), so you'd need an object with a volume of 31.0 m$^3$ to hold it. Our car sized object's interior would be around 3 m$^3$, a factor of 10 too small! This means we need a bigger craft which means more fuel! And explains why this mass-speed relation has been dubbed " the tyranny of the rocket problem ". This also explains the fact that modern rockets are multi-staged . In an attempt to alleviate the required fuel, once a stage uses all of its fuel, it is released from the rocket and the next stage is ignited (doing this over land is dangerous for obvious reasons, hence NASA launching rockets over water ), and the mass of the craft is lowered by the mass of the (empty) stage. More on this can be found at these two Physics.SE posts: Why do rockets have multiple stages? Why do rockets jettison fuel tanks? | {
"source": [
"https://physics.stackexchange.com/questions/88145",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/15934/"
]
} |
88,322 | We know that if an imaginary astronaut is in the intergalactic (no external forces) and has an initial velocity zero, then he has is no way to change the position of his center of mass. The law of momentum conservation says:
$$ 0=\overrightarrow{F}_{ext}=\frac{d\overrightarrow{p}}{dt}=m\frac{d\overrightarrow{v}_{c.m.}}{dt}$$ But I don't see an immediate proof, that the astronaut can't change his orientation in the space. The proof is immediate for a rigid body (from the law of conservation of angular momentum). But the astronaut is not a rigid body. The question is: can the astronaut after a certain sequence of motions come back to the initial position but be oriented differently (change "his angle")? If yes, then how? | The astronaut can change his or her orientation in the same way that a cat does so whilst falling through the air. After the transformation, the astronaut is still and angular momentum is conserved. There is a rather beautiful way of understanding this rotation as an anholonomy i.e. a nontrivial transformation wrought by the parallel transport of the cat's (or astronaut's) state around a closed loop in cat configuration space. I'll write a bit more about this when I have some more time, but for now, one can give a simple explanation with an idealized "robot cat" (or astronaut) which I made up for the thought experiment: Above I have drawn a simplified cat. I am a very aural person, so this is good enough for me so long as I can imagine it mewing! Now our "cat" comprises two cylindrical sections: the "forecat" ( F ), "hinder-cat" ( H ) and two legs ( L ) which can be drawn in so that they are flush with the hinder-cat's surface. With the legs drawn in, the forecat on one hand and hinder-cat + legs assembly on the other have the same mass moment of inertia about the axis of the body. Here is how the cat rotates: Deploy legs symmetrically, i.e. spread them out as shown in the drawing. Now the hinder cat + legs has a bigger mass moment of inertia than the fore cat. Note that, if the legs are diametrically opposite and identical and are opened out symmetrically, the cat undergoes no motion; With an internal motor, the forecat and hinder cat exert equal and opposite torques on one another to accelerate, then stop. Owing to the differences between the moments of inertia, the forecat undergoes a bigger angular displacement than the hind cat; Pull the legs. Again this begets no motion if done symmetrically; Use the internal motor again with an acceleration / deceleration sequence to bring the forecat and hinder cat back to their beginning alignment (i.e. with the line along the cylinders aligned). Now the two halves have the same mass moment of inertia, so when the cat is aligned again, the rotation angles are equal and opposite. Since the rotation angles are different in step 2, but the same in step 5, our robot cat's angular orientation has shifted. If you want to know more about the "Berry phase" explanation and the anholonomy of the cat configuration space before I get around to expanding on this, see Mathematics of the Berry Phase by Peadar Coyle . This is not peer reviewed, but looks sound and is in keeping with similar treatments along these lines that I have seen. | {
"source": [
"https://physics.stackexchange.com/questions/88322",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/34934/"
]
} |
88,935 | I've encountered in some books (and even completed an exercise from the Goldstein by using it), a strange notation that seems to work exactly like a gradient, I have tried to look for an explanation but found none yet and I think I can reduce the time spent looking for my answer by posting this question, even if you give me some literature to check I will be greatly thankful: Why $$\frac{\partial}{\partial\mathbf{r}}=\nabla$$ As I understand it the partial derivative with respect to a vector is like aplying the gradient. I don't know why it seem so odd to me the notion of differentiating something with respect to a vector. | It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows:
\begin{align}
\frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right)
\end{align}
so, by definition, $\partial f/\partial \mathbf r$ is a vector of functions that precisely equals $\nabla f$. You may also run into the notation $\nabla_{\mathbf r}f$ which means precisely the same thing. An advantage of the notations $\partial f/\partial\mathbf r$ and $\nabla_\mathbf r$ is that they make explicit the symbol used to label the argument of the function, and this can sometimes stave off confusion. Addendum. You may also come across the following notation. Let $\mathbf f(\mathbf r) = \mathbf f(x^1, \dots, x^n)$ be a real, $m$-component vector-valued function of $n$ real variables, then define its derivative with respect to $\mathbf r$ as the following matrix, often referred to as the Jacobian matrix:
\begin{align}
\frac{\partial\mathbf f}{\partial \mathbf r}(\mathbf r)
= \begin{pmatrix}
\frac{\partial f^1}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^1}{\partial x^n}(\mathbf r) \\
\vdots & \ddots & \vdots \\
\frac{\partial f^m}{\partial x^1}(\mathbf r) & \cdots & \frac{\partial f^m}{\partial x^n}(\mathbf r) \\
\end{pmatrix}
\end{align}
Consider, for example, the function $\mathbf g(\mathbf r) = \mathbf r$. In this case, you can convince yourself that its derivative $\partial/\partial\mathbf r$ is the identity matrix;
\begin{align}
\frac{\partial\mathbf g}{\partial \mathbf r}(\mathbf r)
= \begin{pmatrix}
1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & 1 \\
\end{pmatrix}
\end{align} | {
"source": [
"https://physics.stackexchange.com/questions/88935",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/30844/"
]
} |
89,002 | I am new to QFT, so I may have some of the terminology incorrect. Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field:
$$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$
The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT in this book by Robert D. Klauber, "Student Friendly Quantum Field Theory". ) What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$. Is it simply a shortcut to obtaining the equations of motion
$$(\square +m^2)\phi=0\\
(\square + m^2)\phi^*=0~?$$ I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, as is done here ; perhaps this addresses my question in a way that I don't understand. | TL;DR: Yes, it is just a short-cut. The main point is that the complexified map $$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix}
~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix}
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} $$ is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2 $. Notation in this answer: In this answer, let $\phi,\phi^{*}\in \mathbb{C}$ denote two independent complex fields. Let $\overline{\phi}$ denote the complex conjugate of $\phi$. I) Let us start with the beginning. Imagine that we consider a field theory of a complex scalar field $\phi$. We are given a Lagrangian density $$\tag{B} {\cal L}~=~{\cal L}(\phi,\overline{\phi},\partial\phi, \partial\overline{\phi})$$ that is a polynomial in $\phi$, $\overline{\phi}$, and spacetime derivatives thereof. We can always decompose a complex field in real and imaginary parts $$\tag{C} \phi~\equiv~\phi_1+ i \phi_2 ,$$ where $\phi_1,\phi_2 \in \mathbb{R}$. Hence we can rewrite the Lagrangian density (B) as a theory of two real fields $$\tag{D}{\cal L}~=~{\cal L}(\phi_1,\phi_2,\partial\phi_1, \partial\phi_2).$$ II) We can continue in at least three ways: Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$. Originally $\phi_1,\phi_2 \in \mathbb{R}$ are of course two real fields. But we can complexify them, vary the action wrt. the two independent complex variables $\phi_1,\phi_2 \in \mathbb{C}$, if we at the end of the calculation impose the two real conditions
$$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$ Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density (B) with an independent new complex variable $\phi^{*}$, i.e. treat $\phi$ and $\phi^{*}$ as two independent complex variables, vary the action wrt. the two independent complex variables $\phi,\phi^{*} \in \mathbb{C}$, if we at the end of the calculation impose the complex condition
$$\tag{F} \phi^{*} ~=~ \overline{\phi}. $$ III) The Euler-Lagrange equations that we derive via the two methods (1) and (2) will obviously be exactly the same. The Euler-Lagrange equations that we derive via the two methods (2) and (3) will be just linear combinations of each other with coefficients given by the constant matrix from eq. (A). IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition (E), or equivalently, condition (F)] is typically not unitary , and therefore ill-defined as a QFT. Recall for starter that we usually demand that the Lagrangian density is real. References: Sidney Coleman, QFT notes ; p. 56-57. | {
"source": [
"https://physics.stackexchange.com/questions/89002",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/29216/"
]
} |
89,035 | I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy. Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations? I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking. My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that? | There are several reasons for using the Hamiltonian formalism: Statistical physics. The standard thermal states weight of pure states is given according to $$\text{Prob}(\text{state}) \propto e^{-H(\text{state})/k_BT}$$ So you need to understand Hamiltonians to do stat mech in real generality. Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to flowing along a vector field on phase space. This gives a nice geometrical picture of how time evolution works in such systems. People use this framework a lot in dynamical systems, where they study questions like 'is the time evolution chaotic?'. The generalization to quantum physics. The basic formalism of quantum mechanics (states and observables) is an obvious generalization of the Hamiltonian formalism. It's less obvious how it's connected to the Lagrangian formalism, and way less obvious how it's connected to the Newtonian formalism. [Edit in response to a comment:] This might be too brief, but the basic story goes as follows: In Hamiltonian mechanics, observables are elements of a commutative algebra which carries a Poisson bracket $\{\cdot,\cdot\}$ . The algebra of observables has a distinguished element, the Hamiltonian, which defines the time evolution via $d\mathcal{O}/dt = \{\mathcal{O},H\}$ . Thermal states are simply linear functions on this algebra. (The observables are realized as functions on the phase space, and the bracket comes from the symplectic structure there. But the algebra of observables is what matters: You can recover the phase space from the algebra of functions.) On the other hand, in quantum physics, we have an algebra of observables which is not commutative. But it still has a bracket $\{\cdot,\cdot\} = -\frac{i}{\hbar}[\cdot,\cdot]$ (the commutator), and it still gets its time evolution from a distinguished element $H$ , via $d\mathcal{O}/dt = \{\mathcal{O},H\}$ . Likewise, thermal states are still linear functionals on the algebra. | {
"source": [
"https://physics.stackexchange.com/questions/89035",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/5788/"
]
} |
89,039 | See the pictures below. A pair of sunglasses I recently purchased has the polarization axis in one lens offset about 20 degrees (by eyeball estimation) from the other. I don't have much experience with other polarized sunglasses, but this seems very obviously wrong to me. And it's very noticeable while wearing the glasses. When facing into glare from certain directions, one eye filters our considerably more glare than the other, which is very distracting and uncomfortable. I can see arguments for making the polarization axis angle just about anything (though horizontal or vertical seem most logical), but what I cannot see is any argument for having it different in each lens on the same pair of glasses. Thoughts? | There are several reasons for using the Hamiltonian formalism: Statistical physics. The standard thermal states weight of pure states is given according to $$\text{Prob}(\text{state}) \propto e^{-H(\text{state})/k_BT}$$ So you need to understand Hamiltonians to do stat mech in real generality. Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to flowing along a vector field on phase space. This gives a nice geometrical picture of how time evolution works in such systems. People use this framework a lot in dynamical systems, where they study questions like 'is the time evolution chaotic?'. The generalization to quantum physics. The basic formalism of quantum mechanics (states and observables) is an obvious generalization of the Hamiltonian formalism. It's less obvious how it's connected to the Lagrangian formalism, and way less obvious how it's connected to the Newtonian formalism. [Edit in response to a comment:] This might be too brief, but the basic story goes as follows: In Hamiltonian mechanics, observables are elements of a commutative algebra which carries a Poisson bracket $\{\cdot,\cdot\}$ . The algebra of observables has a distinguished element, the Hamiltonian, which defines the time evolution via $d\mathcal{O}/dt = \{\mathcal{O},H\}$ . Thermal states are simply linear functions on this algebra. (The observables are realized as functions on the phase space, and the bracket comes from the symplectic structure there. But the algebra of observables is what matters: You can recover the phase space from the algebra of functions.) On the other hand, in quantum physics, we have an algebra of observables which is not commutative. But it still has a bracket $\{\cdot,\cdot\} = -\frac{i}{\hbar}[\cdot,\cdot]$ (the commutator), and it still gets its time evolution from a distinguished element $H$ , via $d\mathcal{O}/dt = \{\mathcal{O},H\}$ . Likewise, thermal states are still linear functionals on the algebra. | {
"source": [
"https://physics.stackexchange.com/questions/89039",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/35256/"
]
} |
89,375 | The second postulate of special relativity deals with constancy of light in inertial reference frames. But, how did Einstein come to this conclusion? Did he know about the Michelson-Morley experiment ? | Did he knew about the Michelson-Morley experiment? He just knew the name of the experiment not any details. The experiment didn't play any role in the formulation of STR by Albert Einstein. The context is taken from the book: Special Theory of Relativity
by V. A.; Atanov, Yuri (Trans.) Ugarov (Author) Art: Was Michelson's experiment "decisive" for the creation οΙ
the special theory οΙ relativity? An article by R. Shankland,
published in 1963, the following excerpt from his interview with
Einstein dating back to 1950: "When Ι asked him how he had learned of the Michelson Morley experiment, he told me that he had become aware of it through
writings of Η. Α. Lοrentz, but only after 1905 had it come to his
attention! "Otherwise" he said, "I would have mentioned it in my
paper!" indeed, Einstein's 1905 paper contains no mention of Μichelson's experiment or references to Lorentz's papers." A letter written by Albert Einstein: "Ιn my own development Michelson's result had not had a considerable influence. Ι even do not remember if Ι knew of it at all when I wrote my first paper on the subject (1905). Τhe explanation is that Ι was, for general reasons, firmly convinced how this could be reconciled with our knowledge οf electro-dynamics. One can therefore understand why in my personal struggle Michelson's
experiment played no role or at least no decisive role..." | {
"source": [
"https://physics.stackexchange.com/questions/89375",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/23742/"
]
} |
89,518 | "The question you are asking appears to be subjective, and is likely to be closed." Challenge . . . ACCEPTED. Okay, here it is. A friend prone to uplifting aphorisms posted on Facebook: "You Are An Infinite Possibility." I thought about that and it seems to me that, given the Bekenstein bound, one can calculate the maximal number of possible states, in principle, of the human brain (given an average volume). Lo and behold, Wikipedia revealed that someone's actually done the calculation and gives the number as less than approximately $10^{7.8\times 10^{41}}$ states. On the other hand, if you model the brain as a quantum system and calculate its evolution over time with the Schrödinger equation --well, the wavefunction is continuous, so doesn't it in principle describe an infinite number of possible states? | This is a frequently-encountered 'boobytrap' in information theory, but it turns out that having a continuous degree of freedom does not entail free access to an infinite amount of information. First of all, while the wavefunction is continuous, you can discretize it quite easily. We know every nucleus and electron in your brain is confined to within a box, say, 20cm a side, so we can describe it in that basis. You still have an infinite number of eigenstates, but it is now discrete. This shift in perspective from uncountable to countable infinity is due to the fact that physically accessible wavefunctions must be continuous and smooth, and there's not actually that many of those. So far, we still have an infinite number of eigenstates in our description. However, we know that the energy content of the human brain is bounded, which means that after a given point, all the energy eigenstates must have negligible contribution. Indeed, if a sizable fraction of the electrons in our brain had energies above, say, 1 GeV, we would instantly come apart in a blaze of gamma rays and positrons and whatnot. Informationally, this means that you can provide an approximation to the wavefunction that's good for all practical purposes using only a finite number of states. An experiment that would distinguish states at that level of approximation would need so much energy it would incinerate your brain. This 'paradox' is also present in classical information theory, and it comes about when you ask what the information capabilities of analog computing are. Here the state of a system is encoded in, say, a voltage, and in principle you have infinite information there because you can in principle measure as many digits as you want. However, it turns out that the scalings tend to be unfavourable, and noise kills you quickly; the upshot is that in analog computing you need to be very careful about what precision / tolerance you demand of your noise and measuring apparatus when counting information capabilities. As it happens, high precision tends to be harder to achieve than simply having more, coupled systems, with simply one bit per system. | {
"source": [
"https://physics.stackexchange.com/questions/89518",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/27649/"
]
} |
89,590 | I'm preparing for my exam, but I have difficulties in perceiving why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$? | It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :) | {
"source": [
"https://physics.stackexchange.com/questions/89590",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/35535/"
]
} |
89,659 | The hamiltonian for the hydrogen atom,
$$
H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}
$$
is spherically symmetric and it therefore commutes with the angular momentum $\mathbf{L}$; this causes all its eigenfunctions with equal angular momentum number $l$ but arbitrary magnetic quantum number $m$ to be degenerate in energy. The hydrogen atom also has a further degeneracy, in that given any angular momentum there are usually other $l$s with the same energy. This degeneracy is due to the existence of a second constant of motion, usually called the Laplace-Runge-Lenz vector,
$$
\mathbf{A} = \frac{1}{2m} ( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r},
$$
which is the generator of an even bigger symmetry, which is isomorphic for bound states to the group $\rm{SO}(4)$ of rotations in four dimensions, of the Kepler problem. The Runge-Lenz vector also has a rich geometrical interpretation. For a classical elliptical orbit, it points from the focus to the periapsis and its magnitude is proportional to the orbit's eccentricity. For circular orbits, it vanishes. Image source: Wikipedia The hydrogen atom is usually described in the common eigenbasis of the hamiltonian and the angular momentum, with the well-known and well-loved quantum numbers $|nlm\rangle$. However, the Runge-Lenz vector $\mathbf{A}$ is also a constant of the motion. What do its eigenfunctions look like? More concretely, I'm looking for the spatial structure of the common eigenfunctions of $H$ and at least one component of $\mathbf A$, and possibly also of $A^2$ (which, in analogy with the common eigenfunctions of $H$, $L^2$ and $L_z$, is the most one could expect), and if that's not possible then an explanation of why, and a description of suitable third quantum numbers to complete a CSCO . I would like to know what their corresponding eigenvalues are, and what the uncertainty of the other components is, whether one can assign a classical eccentricity to the orbital, and more generally in the relation to the corresponding classical geometry. | 1. Definitions Let's consider the nondimensionalized Hamiltonian $$\hat H=\frac{\hat{p}^2}2-\frac1r.\tag1$$ Its standard eigenfunctions diagonalize operators $\hat H$ , $\hat L_z$ and $\hat L^2$ . Laplace-Runge-Lenz operator can be defined as $$\hat{\vec A}=\frac{\vec r}r-\frac12\left(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p}\right).\tag2$$ Its $z$ -component $\hat A_z$ commutes with $\hat H$ , $\hat L_z$ , but doesn't commute with $\hat L^2$ nor with $\hat A^2$ , although $\hat A^2$ does commute with $\hat H$ , $\hat L_z$ and $\hat L^2$ . 2. Eigenfunctions Since $\hat A^2$ commutes with the operators giving quantum numbers $n,l,m$ to the standard hydrogenic eigenstates, these standard eigenstates are also eigenstates of $\hat A^2$ . So no new functions here. For these, we should look at $\hat A_z$ . 2.1. Semi-numerical approach to finding exact eigenstates of $\hat A_z$ Since operators for $H$ and $L_z$ commute with $\hat A_z$ , any eigenstate of $\hat A_z$ is a superposition of states with fixed $n$ and $m$ and different $l$ quantum numbers. This lets us find eigenstates of $\hat A_z$ for given $n,m$ without actually solving the eigenfunction PDE. We can e.g. numerically minimize variance of a sample of values of the expression for the eigenvalue $$A_z=\frac{\hat A_z\sum\limits_l\alpha_l\psi_{n,l,m}}{\sum\limits_l\alpha_l\psi_{n,l,m}},\tag3$$ where $\psi_{n,l,m}$ are the standard simultaneous eigenfunctions of $\hat H$ , $\hat L^2$ and $\hat L_z$ , getting a set of weights $\alpha_l$ . Then we can use the approximations we got to guess the exact values of these weights (and substitute into $(3)$ , simplifying, to confirm the guess). So, for $n=1$ we have the only eigenfunction, and it's obviously an eigenfunction of $\hat A_z$ . For $n>1$ we have more basis functions, which lets us actually form more "interesting" eigenstates of $\hat A_z$ . Here're some tables for $n=2,3,4$ (I calculated them using the above mentioned minimization procedure): $$n=2\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{2,1,1} &0 &1 \\
\psi_{2,1,-1} &0 &-1 \\
\frac1{\sqrt2}(\psi_{2,0,0}+\psi_{2,1,0}) &-\frac12 & 0\\
\frac1{\sqrt2}(\psi_{2,0,0}-\psi_{2,1,0}) &\frac12 & 0\\
\hline
\end{array}$$ $$n=3\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{3,2,-2} & 0 & -2\\
\psi_{3,2,2} & 0 & 2\\
\frac1{\sqrt2}(\psi_{3,1,1}+\psi_{3,2,1}) & -\frac13 & 1\\
\frac1{\sqrt2}(\psi_{3,1,1}-\psi_{3,2,1}) & \frac13 & 1\\
\frac1{\sqrt2}(\psi_{3,1,-1}+\psi_{3,2,-1}) & -\frac13 & -1\\
\frac1{\sqrt2}(\psi_{3,1,-1}-\psi_{3,2,-1}) & \frac13 & -1\\
\sqrt{\frac13}\psi_{3,0,0}-\sqrt{\frac23}\psi_{3,2,0} & 0 & 0\\
\frac1{\sqrt3}\psi_{3,0,0}+\frac1{\sqrt2}\psi_{3,1,0}+\frac1{\sqrt6}\psi_{3,2,0} & -\frac23 & 0\\
\frac1{\sqrt3}\psi_{3,0,0}-\frac1{\sqrt2}\psi_{3,1,0}+\frac1{\sqrt6}\psi_{3,2,0} & \frac23 & 0\\
\hline
\end{array}$$ $$n=4\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{4,3,3} & 0 & 3\\
\psi_{4,3,-3} & 0 & -3\\
\frac1{\sqrt2}(\psi_{4,2,2}+\psi_{4,3,2}) & -\frac14 & 2\\
\frac1{\sqrt2}(\psi_{4,2,2}-\psi_{4,3,2}) & \frac14 & 2\\
\frac1{\sqrt2}(\psi_{4,2,-2}+\psi_{4,3,-2}) & -\frac14 & -2\\
\frac1{\sqrt2}(\psi_{4,2,-2}-\psi_{4,3,-2}) & \frac14 & -2\\
%
\sqrt{\frac3{10}}\psi_{4,1,1}-\sqrt{\frac12}\psi_{4,2,1}+\sqrt{\frac15}\psi_{4,3,1} & \frac12 & 1\\
%
\sqrt{\frac3{10}}\psi_{4,1,1}+\sqrt{\frac12}\psi_{4,2,1}+\sqrt{\frac15}\psi_{4,3,1} & -\frac12 & 1\\
%
\sqrt{\frac3{10}}\psi_{4,1,-1}-\sqrt{\frac12}\psi_{4,2,-1}+\sqrt{\frac15}\psi_{4,3,-1} & \frac12 & -1\\
%
\sqrt{\frac3{10}}\psi_{4,1,-1}+\sqrt{\frac12}\psi_{4,2,-1}+\sqrt{\frac15}\psi_{4,3,-1} & -\frac12 & -1\\
%
\sqrt{\frac25}\psi_{4,1,1}-\sqrt{\frac35}\psi_{4,3,1} & 0 & 1\\
%
\sqrt{\frac25}\psi_{4,1,-1}-\sqrt{\frac35}\psi_{4,3,-1} & 0 & -1\\
%
\frac12\psi_{4,0,0}-\sqrt{\frac1{20}}\psi_{4,1,0}-\frac12\psi_{4,2,0}+\sqrt{\frac9{20}}\psi_{4,3,0} & \frac14 & 0\\
%
\frac12\psi_{4,0,0}+\sqrt{\frac1{20}}\psi_{4,1,0}-\frac12\psi_{4,2,0}-\sqrt{\frac9{20}}\psi_{4,3,0} & -\frac14 & 0\\
%
\frac12\psi_{4,0,0}+\sqrt{\frac9{20}}\psi_{4,1,0}+\frac12\psi_{4,2,0}+\sqrt{\frac1{20}}\psi_{4,3,0} & -\frac34 & 0\\
%
\frac12\psi_{4,0,0}-\sqrt{\frac9{20}}\psi_{4,1,0}+\frac12\psi_{4,2,0}-\sqrt{\frac1{20}}\psi_{4,3,0} & \frac34 & 0\\
\hline
\end{array}$$ The values of $m$ have been added to the table to make it more obvious that they, together with the eigenvalues $A_z$ and $n$ , actually complete the CSCO. We can even plot configurations of the possible states using these quantum numbers, and see a pattern: Looking at these plots, we can guess what the eigenvalues of $\hat A_z$ will be for the higher $n$ . And then we won't even need the procedure of minimization of variance of the sample of $(3)$ to find the weights $\alpha_l$ : we can simply solve this equation with a random sample of points in the domain, setting one of the weights to $1$ , and those corresponding to the $m$ we're not interested in, to $0$ (number of points should be chosen to make number of equations equal to number of weights remaining unknown). Now, $k$ th value for $A_z$ appears to follow this formula: $$A_z(n,m,k)=\frac{|m|-n-1+2k}n,\tag4$$ where $$k=1,2,\dots,n-|m|.\tag5$$ 2.2. Analytical approach to general solution The general solution to eigenproblem of $\hat A_z$ can be found, if the Schrödinger's equation is expressed in the parabolic coordinates . Then the natural eigenfunctions there will be characterized by a different set of quantum numbers than usual: parabolic ones $n_1$ and $n_2$ , and the usual magnetic quantum number $m$ . The complete expression in terms of confluent hypergeometric function ${}_1F_1$ for the eigenfunctions and its derivation can be found e.g. in Landau and Lifshitz, "Quantum Mechanics. Non-relativistic theory" $\S37$ "Motion in a Coulomb field (parabolic coordinates)" . The eigenvalues of $\hat A_z$ can be expressed in terms of the parabolic and magnetic quantum numbers as $$A_z=\frac{n_1-n_2}n,\tag6$$ where $$n=n_1+n_2+|m|+1\tag7$$ is the principal quantum number, and parabolic quantum numbers can have the values $$n_{1,2}=0,1,...,n-|m|-1.\tag8$$ In terms of $n$ , $m$ and $n_1$ , $(6)$ can be rewritten as $$A_z=\frac{|m|-n+2n_1+1}n,\tag9$$ which is consistent with $(4)$ - $(5)$ . 3. What the eigenfunctions of $\hat A_z$ look like Eigenfunctions of $\hat A_z$ with high absolute values of eigenvalues look like bells in shape, with probability density symmetric along the $z$ axis. As $L_z$ is conserved, real and imaginary parts of the eigenfunctions oscillate when we go around the $z$ axis with $m\ne0$ . Here're plots of some of the eigenfunctions, with 3D density plot on the LHS and cross-sections in $y=0$ plane on the RHS: $n=4,\,A_z=\frac34,\,m=0:$ $n=4,\,A_z=-\frac14,\,m=0:$ $n=4,\,A_z=-\frac14,\,m=2$ , real part: $n=3,\,A_z=\frac13,\,m=1$ , real part: 4. Relation to classical orbits and their eccentricity 4.1. Eigenfunctions of $\hat A_z$ Unfortunately, the eigenstates of $\hat A_z$ don't appear to be nicely related to classical eccentric orbits. In classical orbits, $\vec A$ is always in the plane of rotation. Classical orbital motion corresponds in quantum regime to the case of high expected values of $L_z$ . But $\hat A_z$ commutes with $\hat L_z$ , not with e.g. $\hat L_x$ , so the "rotating" (in the sense of $e^{im\phi}$ ) eigenstates rotate roughly perpendicularly to direction of the orbital eccentricity. We can interpret these as eccentric standing waves in the $\theta$ direction, which is of course far from classical regime. It's interesting to note that although with the high values of $A_z$ the $z$ component of $\vec A$ indeed dominates, giving the cross-section of the orbital somewhat elliptic shape, it's much less so for lower values. This is because of uncertainty in $A_x$ and $A_z$ . See e.g. the following eigenfunctions. Ellipses show the supposed classical orbits with semi-major axis $a=n^2$ and LRL vectors $\vec A=A_z \vec e_z+s \sqrt{\langle A_x^2\rangle}\vec e_x$ , where $s=-2,-1,0,1,2$ . $n=20,\, A_z=\frac{19}{20},\, m=0,\,\langle A_x^2\rangle=\frac{19}{800}\colon$ $n=20,\, A_z=\frac{9}{20},\, m=0,\,\langle A_x^2\rangle=\frac{159}{800}\colon$ $n=20,\, A_z=\frac{1}{20},\, m=0,\,\langle A_x^2\rangle=\frac{199}{800}\colon$ 4.2. Eigenfunctions of $\hat A^2$ We may have better luck if we consider instead eigenstates of $\hat A^2$ (which, as mentioned above, doesn't commute with $\hat A_z$ ). These are also eigenstates of $\hat L^2$ and $\hat L_z$ , so they are the familiar functions. Eigenvalues of $\hat A^2$ are $$A_{n,l}^2=1-\frac{l(l+1)+1}{n^2}.$$ As we can see, consistently with classical intuition, eccentricity $e=\sqrt{A^2}$ decreases with increasing angular momentum. One might hope to see that at least these states will allow assignment of classical eccentricity. But, despite they do, there's a problem: the eigenstates of $\hat A^2$ are all almost symmetric with respect to rotations around $z$ axis — modulo the $\exp(im\phi)$ oscillations. So we never get anything resembling eccentric ellipses even in eigenstates of $\hat A^2$ . Instead we get the following cross-sections in $xy$ plane of the real parts of wavefunctions (here $n=21$ , $m=l$ , $l$ changes from $0$ to $20$ with the step of $4$ ): Since these states don't actually have any orientation of the LRL vector (aside from avoiding $z$ direction in case of high $m$ values), a better interpretation and drawing of classical eccentricity for them would be like this: where the ellipses show some of the possible orbits one may get if e.g. one were to form a localized wave packet from similar states with this state being dominant. 5. A note on the method of semi-numerical calculations To make the numerical procedure more understandable, I'll show an example of how one can get an eigenfunction of $\hat A_z$ and associated eigenvalue with $n=4$ , $m=1$ . The code in this section is in Wolfram Language. I did all the calculations in Mathematica 11.2, but the code is compatible with versions as old as Mathematica 9. First, some definitions for the operators and functions we'll use here. (* Components of momentum operator *)
px = -I D[#,x] &;
py = -I D[#,y] &;
pz = -I D[#,z] &;
(* z component of LRL vector operator *)
Az = Simplify[
z/Sqrt[x^2+y^2+z^2] # -
1/2 (z px@px@# + px[z px@#] - px[x pz@#] + z py@py@# +
py[z py@#] - py[y pz@#] - x pz@px@# - y pz@py@#)] &;
(* Hydrogenic wavefunction in spherical coordinates *)
ψ[n_,l_,m_,r_,θ_,ϕ_] = Sqrt[(n-l-1)!/(n+l)!] E^(-r/n) (2 r/n)^l 2/n^2 *
LaguerreL[n-l-1, 2 l+1, (2 r)/n] *
SphericalHarmonicY[l,m,θ,ϕ];
(* The same wavefunction converted to Cartesian coordinates *)
Ψ[n_,l_,m_,x_,y_,z_] = ψ[n, l, m, Sqrt[x^2+y^2+z^2],
ArcCos[z/Sqrt[x^2+y^2+z^2]],
ArcTan[x,y]]; Now the example test function for $n=4$ , $m=1$ . (* Test function with parameters α, β and γ. Restricting arguments to
numeric to avoid attempts at symbolic evaluation, which can seriously slow
things down. Simplifying it to speedup calculations and reduce roundoff
errors. *)
test[x_Real,y_Real,z_Real,α_?NumericQ,β_?NumericQ,γ_?NumericQ] =
FullSimplify[Az@#/# &[α Ψ[4,1,1,x,y,z] +
β Ψ[4,2,1,x,y,z] +
γ Ψ[4,3,1,x,y,z]],
(x|y|z) ∈ Reals]; And finally minimization of its variance. Note that we don't need to actually calculate an integral as in variational methods: we only need a "close enough" approximation of the parameters, the rest can be left to Rationalize . So we use a coarse mesh of points to evaluate the function on. Note also that here we let Mathematica go to complex domain, although the parameters should be real-valued. This lets it avoid singularities in the function by simply going around them, and thus gives much faster convergence. (* Table is generated not on integers to avoid problems like
division by zero on evaluation *)
With[{
var = Total[Abs[#-Mean@#]^2]& @ Flatten @
Table[test[x,y,z,α,βR + I βI,γR + I γI],
{x,-10.123,10,4},
{y,-10.541,10,5},
{z,-10.07,10,5}
]
},
{minVal,minim} = NMinimize[{var,Total[#^2] == 1 &[{α,βR,βI,γR,γI}] && α>0.1},
{α,βR,βI,γR,γI}]
] {1.29281973036898*10^-9, {α -> 0.547724712837901, βR -> -0.707106153522197, βI -> 1.84829862406368*10^-7, γR -> 0.447211968838118, γI -> -1.8807768960726*10^-7}} OK, so we see that indeed the imaginary parts are close to zero, so let's guess the form of actual parameters assuming that what we got are square roots of some rationals. (* Tolerance of rationalization is chosen so at to
1) ignore numerical errors of minimization,
2) still give a good enough room to guess the correct number *)
Sqrt[Rationalize[#^2, 10^-4]]Sign[#]&[{α,βR,γR} /. minim] {Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]} This is what we got. Let's check whether this is a correct guess. FullSimplify[Az@#/# &[Sqrt[3/10] Ψ[4,1,1,x,y,z] -
Sqrt[1/2] Ψ[4,2,1,x,y,z] +
Sqrt[1/5] Ψ[4,3,1,x,y,z]],
(x|y|z) ∈ Reals] 1/2 Now we not only confirmed that our test function with the guessed values of parameters is an eigenfunction (since we got constant here), but also obtained the associated eigenvalue, $A_z=1/2$ . This is entry #7 in the table above for $n=4$ . To find another set of parameters we can go the following way. First, we can guess that changing some signs in $\alpha$ , $\beta$ and $\gamma$ might give us some more eigenfunctions. Indeed, it does, so using $+\sqrt{1/2}$ instead of $-\sqrt{1/2}$ for $\beta$ does result in an eigenfunction (entry #8 in the table, with eigenvalue $A_z=-1/2$ ). Another approach at finding other eigenfunctions (useful when there are more parameters, e.g. for $n=4$ , $m=0$ there are $4$ ) is using Orthogonalize to find a basis in the orthogonal subspace of parameters to the one we've already identified. Then we can use that basis to form our new set of parameters for NMinimize to work on. In our example case the situation is trivial, since the whole set of $n=4$ , $m=1$ eigenfunctions consists of 3 elements, and we've already identified two of them, so no need in further NMinimize . So Orthogonalize[{{Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]},
{Sqrt[3/10], 1/ Sqrt[2], 1/Sqrt[5]},
{1, 1, 1}}] // FullSimplify {{Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]}, {Sqrt[3/10], 1/Sqrt[2], 1/Sqrt[5]}, {-Sqrt[(2/5)], 0, Sqrt[3/5]}} The third element of the output list is the third eigenfunction (in the $|n,l,m\rangle$ basis). We can find that associated eigenvalue is $A_z=0$ . | {
"source": [
"https://physics.stackexchange.com/questions/89659",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/8563/"
]
} |
89,661 | While studying thermal physics at school, I have been taught that solids simply have more potential energy than the liquids and gases. Note that it was said that this potential energy is due to the intermolecular bonds between the atoms. However, my intuition makes me doubt this, why would there be more potential energy in a solid??? Lets see: the internal energy is the sum of the kinetic and potential energies of a body. Well liquids and gases have more kinetic - pretty clear. As for, our original point, the potential: well, if the body has potential to do work, via lets say, chemical reactions, then its solid form will have a higher activation energy, as the intermolecular bonds have to be broken first - so the energy then released from combustion or what not, is less than that of its gaseous equivalent! Then how is it, that solids are somehow said to have more "potential" energy than the more energetic states. I, upon questioning this, have been told that this is due to the forces of the molecular bonds, but really I think, these forces signalize a lack of potential energy! | 1. Definitions Let's consider the nondimensionalized Hamiltonian $$\hat H=\frac{\hat{p}^2}2-\frac1r.\tag1$$ Its standard eigenfunctions diagonalize operators $\hat H$ , $\hat L_z$ and $\hat L^2$ . Laplace-Runge-Lenz operator can be defined as $$\hat{\vec A}=\frac{\vec r}r-\frac12\left(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p}\right).\tag2$$ Its $z$ -component $\hat A_z$ commutes with $\hat H$ , $\hat L_z$ , but doesn't commute with $\hat L^2$ nor with $\hat A^2$ , although $\hat A^2$ does commute with $\hat H$ , $\hat L_z$ and $\hat L^2$ . 2. Eigenfunctions Since $\hat A^2$ commutes with the operators giving quantum numbers $n,l,m$ to the standard hydrogenic eigenstates, these standard eigenstates are also eigenstates of $\hat A^2$ . So no new functions here. For these, we should look at $\hat A_z$ . 2.1. Semi-numerical approach to finding exact eigenstates of $\hat A_z$ Since operators for $H$ and $L_z$ commute with $\hat A_z$ , any eigenstate of $\hat A_z$ is a superposition of states with fixed $n$ and $m$ and different $l$ quantum numbers. This lets us find eigenstates of $\hat A_z$ for given $n,m$ without actually solving the eigenfunction PDE. We can e.g. numerically minimize variance of a sample of values of the expression for the eigenvalue $$A_z=\frac{\hat A_z\sum\limits_l\alpha_l\psi_{n,l,m}}{\sum\limits_l\alpha_l\psi_{n,l,m}},\tag3$$ where $\psi_{n,l,m}$ are the standard simultaneous eigenfunctions of $\hat H$ , $\hat L^2$ and $\hat L_z$ , getting a set of weights $\alpha_l$ . Then we can use the approximations we got to guess the exact values of these weights (and substitute into $(3)$ , simplifying, to confirm the guess). So, for $n=1$ we have the only eigenfunction, and it's obviously an eigenfunction of $\hat A_z$ . For $n>1$ we have more basis functions, which lets us actually form more "interesting" eigenstates of $\hat A_z$ . Here're some tables for $n=2,3,4$ (I calculated them using the above mentioned minimization procedure): $$n=2\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{2,1,1} &0 &1 \\
\psi_{2,1,-1} &0 &-1 \\
\frac1{\sqrt2}(\psi_{2,0,0}+\psi_{2,1,0}) &-\frac12 & 0\\
\frac1{\sqrt2}(\psi_{2,0,0}-\psi_{2,1,0}) &\frac12 & 0\\
\hline
\end{array}$$ $$n=3\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{3,2,-2} & 0 & -2\\
\psi_{3,2,2} & 0 & 2\\
\frac1{\sqrt2}(\psi_{3,1,1}+\psi_{3,2,1}) & -\frac13 & 1\\
\frac1{\sqrt2}(\psi_{3,1,1}-\psi_{3,2,1}) & \frac13 & 1\\
\frac1{\sqrt2}(\psi_{3,1,-1}+\psi_{3,2,-1}) & -\frac13 & -1\\
\frac1{\sqrt2}(\psi_{3,1,-1}-\psi_{3,2,-1}) & \frac13 & -1\\
\sqrt{\frac13}\psi_{3,0,0}-\sqrt{\frac23}\psi_{3,2,0} & 0 & 0\\
\frac1{\sqrt3}\psi_{3,0,0}+\frac1{\sqrt2}\psi_{3,1,0}+\frac1{\sqrt6}\psi_{3,2,0} & -\frac23 & 0\\
\frac1{\sqrt3}\psi_{3,0,0}-\frac1{\sqrt2}\psi_{3,1,0}+\frac1{\sqrt6}\psi_{3,2,0} & \frac23 & 0\\
\hline
\end{array}$$ $$n=4\\
\begin{array}{|c|c|c|}
\hline
\text{Eigenfunction of }\hat A_z&A_z&m\\
\hline
\psi_{4,3,3} & 0 & 3\\
\psi_{4,3,-3} & 0 & -3\\
\frac1{\sqrt2}(\psi_{4,2,2}+\psi_{4,3,2}) & -\frac14 & 2\\
\frac1{\sqrt2}(\psi_{4,2,2}-\psi_{4,3,2}) & \frac14 & 2\\
\frac1{\sqrt2}(\psi_{4,2,-2}+\psi_{4,3,-2}) & -\frac14 & -2\\
\frac1{\sqrt2}(\psi_{4,2,-2}-\psi_{4,3,-2}) & \frac14 & -2\\
%
\sqrt{\frac3{10}}\psi_{4,1,1}-\sqrt{\frac12}\psi_{4,2,1}+\sqrt{\frac15}\psi_{4,3,1} & \frac12 & 1\\
%
\sqrt{\frac3{10}}\psi_{4,1,1}+\sqrt{\frac12}\psi_{4,2,1}+\sqrt{\frac15}\psi_{4,3,1} & -\frac12 & 1\\
%
\sqrt{\frac3{10}}\psi_{4,1,-1}-\sqrt{\frac12}\psi_{4,2,-1}+\sqrt{\frac15}\psi_{4,3,-1} & \frac12 & -1\\
%
\sqrt{\frac3{10}}\psi_{4,1,-1}+\sqrt{\frac12}\psi_{4,2,-1}+\sqrt{\frac15}\psi_{4,3,-1} & -\frac12 & -1\\
%
\sqrt{\frac25}\psi_{4,1,1}-\sqrt{\frac35}\psi_{4,3,1} & 0 & 1\\
%
\sqrt{\frac25}\psi_{4,1,-1}-\sqrt{\frac35}\psi_{4,3,-1} & 0 & -1\\
%
\frac12\psi_{4,0,0}-\sqrt{\frac1{20}}\psi_{4,1,0}-\frac12\psi_{4,2,0}+\sqrt{\frac9{20}}\psi_{4,3,0} & \frac14 & 0\\
%
\frac12\psi_{4,0,0}+\sqrt{\frac1{20}}\psi_{4,1,0}-\frac12\psi_{4,2,0}-\sqrt{\frac9{20}}\psi_{4,3,0} & -\frac14 & 0\\
%
\frac12\psi_{4,0,0}+\sqrt{\frac9{20}}\psi_{4,1,0}+\frac12\psi_{4,2,0}+\sqrt{\frac1{20}}\psi_{4,3,0} & -\frac34 & 0\\
%
\frac12\psi_{4,0,0}-\sqrt{\frac9{20}}\psi_{4,1,0}+\frac12\psi_{4,2,0}-\sqrt{\frac1{20}}\psi_{4,3,0} & \frac34 & 0\\
\hline
\end{array}$$ The values of $m$ have been added to the table to make it more obvious that they, together with the eigenvalues $A_z$ and $n$ , actually complete the CSCO. We can even plot configurations of the possible states using these quantum numbers, and see a pattern: Looking at these plots, we can guess what the eigenvalues of $\hat A_z$ will be for the higher $n$ . And then we won't even need the procedure of minimization of variance of the sample of $(3)$ to find the weights $\alpha_l$ : we can simply solve this equation with a random sample of points in the domain, setting one of the weights to $1$ , and those corresponding to the $m$ we're not interested in, to $0$ (number of points should be chosen to make number of equations equal to number of weights remaining unknown). Now, $k$ th value for $A_z$ appears to follow this formula: $$A_z(n,m,k)=\frac{|m|-n-1+2k}n,\tag4$$ where $$k=1,2,\dots,n-|m|.\tag5$$ 2.2. Analytical approach to general solution The general solution to eigenproblem of $\hat A_z$ can be found, if the Schrödinger's equation is expressed in the parabolic coordinates . Then the natural eigenfunctions there will be characterized by a different set of quantum numbers than usual: parabolic ones $n_1$ and $n_2$ , and the usual magnetic quantum number $m$ . The complete expression in terms of confluent hypergeometric function ${}_1F_1$ for the eigenfunctions and its derivation can be found e.g. in Landau and Lifshitz, "Quantum Mechanics. Non-relativistic theory" $\S37$ "Motion in a Coulomb field (parabolic coordinates)" . The eigenvalues of $\hat A_z$ can be expressed in terms of the parabolic and magnetic quantum numbers as $$A_z=\frac{n_1-n_2}n,\tag6$$ where $$n=n_1+n_2+|m|+1\tag7$$ is the principal quantum number, and parabolic quantum numbers can have the values $$n_{1,2}=0,1,...,n-|m|-1.\tag8$$ In terms of $n$ , $m$ and $n_1$ , $(6)$ can be rewritten as $$A_z=\frac{|m|-n+2n_1+1}n,\tag9$$ which is consistent with $(4)$ - $(5)$ . 3. What the eigenfunctions of $\hat A_z$ look like Eigenfunctions of $\hat A_z$ with high absolute values of eigenvalues look like bells in shape, with probability density symmetric along the $z$ axis. As $L_z$ is conserved, real and imaginary parts of the eigenfunctions oscillate when we go around the $z$ axis with $m\ne0$ . Here're plots of some of the eigenfunctions, with 3D density plot on the LHS and cross-sections in $y=0$ plane on the RHS: $n=4,\,A_z=\frac34,\,m=0:$ $n=4,\,A_z=-\frac14,\,m=0:$ $n=4,\,A_z=-\frac14,\,m=2$ , real part: $n=3,\,A_z=\frac13,\,m=1$ , real part: 4. Relation to classical orbits and their eccentricity 4.1. Eigenfunctions of $\hat A_z$ Unfortunately, the eigenstates of $\hat A_z$ don't appear to be nicely related to classical eccentric orbits. In classical orbits, $\vec A$ is always in the plane of rotation. Classical orbital motion corresponds in quantum regime to the case of high expected values of $L_z$ . But $\hat A_z$ commutes with $\hat L_z$ , not with e.g. $\hat L_x$ , so the "rotating" (in the sense of $e^{im\phi}$ ) eigenstates rotate roughly perpendicularly to direction of the orbital eccentricity. We can interpret these as eccentric standing waves in the $\theta$ direction, which is of course far from classical regime. It's interesting to note that although with the high values of $A_z$ the $z$ component of $\vec A$ indeed dominates, giving the cross-section of the orbital somewhat elliptic shape, it's much less so for lower values. This is because of uncertainty in $A_x$ and $A_z$ . See e.g. the following eigenfunctions. Ellipses show the supposed classical orbits with semi-major axis $a=n^2$ and LRL vectors $\vec A=A_z \vec e_z+s \sqrt{\langle A_x^2\rangle}\vec e_x$ , where $s=-2,-1,0,1,2$ . $n=20,\, A_z=\frac{19}{20},\, m=0,\,\langle A_x^2\rangle=\frac{19}{800}\colon$ $n=20,\, A_z=\frac{9}{20},\, m=0,\,\langle A_x^2\rangle=\frac{159}{800}\colon$ $n=20,\, A_z=\frac{1}{20},\, m=0,\,\langle A_x^2\rangle=\frac{199}{800}\colon$ 4.2. Eigenfunctions of $\hat A^2$ We may have better luck if we consider instead eigenstates of $\hat A^2$ (which, as mentioned above, doesn't commute with $\hat A_z$ ). These are also eigenstates of $\hat L^2$ and $\hat L_z$ , so they are the familiar functions. Eigenvalues of $\hat A^2$ are $$A_{n,l}^2=1-\frac{l(l+1)+1}{n^2}.$$ As we can see, consistently with classical intuition, eccentricity $e=\sqrt{A^2}$ decreases with increasing angular momentum. One might hope to see that at least these states will allow assignment of classical eccentricity. But, despite they do, there's a problem: the eigenstates of $\hat A^2$ are all almost symmetric with respect to rotations around $z$ axis — modulo the $\exp(im\phi)$ oscillations. So we never get anything resembling eccentric ellipses even in eigenstates of $\hat A^2$ . Instead we get the following cross-sections in $xy$ plane of the real parts of wavefunctions (here $n=21$ , $m=l$ , $l$ changes from $0$ to $20$ with the step of $4$ ): Since these states don't actually have any orientation of the LRL vector (aside from avoiding $z$ direction in case of high $m$ values), a better interpretation and drawing of classical eccentricity for them would be like this: where the ellipses show some of the possible orbits one may get if e.g. one were to form a localized wave packet from similar states with this state being dominant. 5. A note on the method of semi-numerical calculations To make the numerical procedure more understandable, I'll show an example of how one can get an eigenfunction of $\hat A_z$ and associated eigenvalue with $n=4$ , $m=1$ . The code in this section is in Wolfram Language. I did all the calculations in Mathematica 11.2, but the code is compatible with versions as old as Mathematica 9. First, some definitions for the operators and functions we'll use here. (* Components of momentum operator *)
px = -I D[#,x] &;
py = -I D[#,y] &;
pz = -I D[#,z] &;
(* z component of LRL vector operator *)
Az = Simplify[
z/Sqrt[x^2+y^2+z^2] # -
1/2 (z px@px@# + px[z px@#] - px[x pz@#] + z py@py@# +
py[z py@#] - py[y pz@#] - x pz@px@# - y pz@py@#)] &;
(* Hydrogenic wavefunction in spherical coordinates *)
ψ[n_,l_,m_,r_,θ_,ϕ_] = Sqrt[(n-l-1)!/(n+l)!] E^(-r/n) (2 r/n)^l 2/n^2 *
LaguerreL[n-l-1, 2 l+1, (2 r)/n] *
SphericalHarmonicY[l,m,θ,ϕ];
(* The same wavefunction converted to Cartesian coordinates *)
Ψ[n_,l_,m_,x_,y_,z_] = ψ[n, l, m, Sqrt[x^2+y^2+z^2],
ArcCos[z/Sqrt[x^2+y^2+z^2]],
ArcTan[x,y]]; Now the example test function for $n=4$ , $m=1$ . (* Test function with parameters α, β and γ. Restricting arguments to
numeric to avoid attempts at symbolic evaluation, which can seriously slow
things down. Simplifying it to speedup calculations and reduce roundoff
errors. *)
test[x_Real,y_Real,z_Real,α_?NumericQ,β_?NumericQ,γ_?NumericQ] =
FullSimplify[Az@#/# &[α Ψ[4,1,1,x,y,z] +
β Ψ[4,2,1,x,y,z] +
γ Ψ[4,3,1,x,y,z]],
(x|y|z) ∈ Reals]; And finally minimization of its variance. Note that we don't need to actually calculate an integral as in variational methods: we only need a "close enough" approximation of the parameters, the rest can be left to Rationalize . So we use a coarse mesh of points to evaluate the function on. Note also that here we let Mathematica go to complex domain, although the parameters should be real-valued. This lets it avoid singularities in the function by simply going around them, and thus gives much faster convergence. (* Table is generated not on integers to avoid problems like
division by zero on evaluation *)
With[{
var = Total[Abs[#-Mean@#]^2]& @ Flatten @
Table[test[x,y,z,α,βR + I βI,γR + I γI],
{x,-10.123,10,4},
{y,-10.541,10,5},
{z,-10.07,10,5}
]
},
{minVal,minim} = NMinimize[{var,Total[#^2] == 1 &[{α,βR,βI,γR,γI}] && α>0.1},
{α,βR,βI,γR,γI}]
] {1.29281973036898*10^-9, {α -> 0.547724712837901, βR -> -0.707106153522197, βI -> 1.84829862406368*10^-7, γR -> 0.447211968838118, γI -> -1.8807768960726*10^-7}} OK, so we see that indeed the imaginary parts are close to zero, so let's guess the form of actual parameters assuming that what we got are square roots of some rationals. (* Tolerance of rationalization is chosen so at to
1) ignore numerical errors of minimization,
2) still give a good enough room to guess the correct number *)
Sqrt[Rationalize[#^2, 10^-4]]Sign[#]&[{α,βR,γR} /. minim] {Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]} This is what we got. Let's check whether this is a correct guess. FullSimplify[Az@#/# &[Sqrt[3/10] Ψ[4,1,1,x,y,z] -
Sqrt[1/2] Ψ[4,2,1,x,y,z] +
Sqrt[1/5] Ψ[4,3,1,x,y,z]],
(x|y|z) ∈ Reals] 1/2 Now we not only confirmed that our test function with the guessed values of parameters is an eigenfunction (since we got constant here), but also obtained the associated eigenvalue, $A_z=1/2$ . This is entry #7 in the table above for $n=4$ . To find another set of parameters we can go the following way. First, we can guess that changing some signs in $\alpha$ , $\beta$ and $\gamma$ might give us some more eigenfunctions. Indeed, it does, so using $+\sqrt{1/2}$ instead of $-\sqrt{1/2}$ for $\beta$ does result in an eigenfunction (entry #8 in the table, with eigenvalue $A_z=-1/2$ ). Another approach at finding other eigenfunctions (useful when there are more parameters, e.g. for $n=4$ , $m=0$ there are $4$ ) is using Orthogonalize to find a basis in the orthogonal subspace of parameters to the one we've already identified. Then we can use that basis to form our new set of parameters for NMinimize to work on. In our example case the situation is trivial, since the whole set of $n=4$ , $m=1$ eigenfunctions consists of 3 elements, and we've already identified two of them, so no need in further NMinimize . So Orthogonalize[{{Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]},
{Sqrt[3/10], 1/ Sqrt[2], 1/Sqrt[5]},
{1, 1, 1}}] // FullSimplify {{Sqrt[3/10], -(1/Sqrt[2]), 1/Sqrt[5]}, {Sqrt[3/10], 1/Sqrt[2], 1/Sqrt[5]}, {-Sqrt[(2/5)], 0, Sqrt[3/5]}} The third element of the output list is the third eigenfunction (in the $|n,l,m\rangle$ basis). We can find that associated eigenvalue is $A_z=0$ . | {
"source": [
"https://physics.stackexchange.com/questions/89661",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/33768/"
]
} |
90,024 | After a very long time will we see more stars (due to the fact that more light is get to us) or less stars (as the universe expends and light have to pass larger distance)?
In general, can stellar objects go outside of the scope of the observable universe or is it only growing with time? Calculations are always better than just discussion.
Does it have any connection to dark matter? | This is a rather lengthy answer as I tried to go a bit in depth; there is a short summary at the end. Will we see more or fewer stars with time? The short answer to this is: We see less stars with time, due to the fact that cosmic expansion is accelerating. Although what we really see at the relevant distances are galaxies; single stars are far too far away to be resolved. The first thing to realise to understand this is that the speed limit of special relativity doesn't apply to cosmic expansion. If you imagine the galaxies as raisins in an enormous, rising dough, what SR tells you is that nothing can move through the dough faster than light. But if expansion is the same everywhere, and the dough is large enough, then an arbitrarily small speed at which the dough would rise would make all raisins beyond a certain distance recede from your local raisin faster than light. This "certain distance", in the Universe, is called the Hubble Distance . Galaxies farther away from the Hubble Distance are receding from us faster than light. This, however, is not the limit beyond which light cannot reach us. If a galaxy is receding from us faster than light, a photon emitted from said galaxy towards us will initially be receding from us, but it will crawl to regions of space - at local light speed - which are receding from us a bit slower than the one from which it was emitted, and so it will recede a bit slower from us than before, etc., until it finally enters a region of space that is receding from us slower than light, from which point it is trivial that it will eventually reach us. This is not always the case, though. There is, in certain cosmologies (including our own with an accelerating expansion), a distance,from beyond which a photon emitted will never reach us. This happens if the Universe expands faster than the photons can traverse it. This distance is called the cosmic Event Horizon , and while it expands in terms of absolute distances, it is shrinking in terms of co-moving distance. The co-moving distance is the distance measured in a coordinate system that expands along with the Universe - such a system will have galaxies more or less sitting in the same coordinates throughout the history of the Universe, and so the Event Horizon having a shrinking co-moving radius from us directly tells us that fewer objects will be inside this radius in the future. The way we can in theory observe these distant galaxies leave our event horizon - and the reason why it has its name - will be very much like we would observe an object falling into the event horizon of a Black Hole: relativistic effects will redshift the light emitted from such an object indefinitely, which also slows down the time as observed by us. So the object will appear to slow down asymptotically to a complete freeze, while the light will be redshifted towards infinity and become undetectable by us, although you could argue that it will never, in theory, disappear completely. The last we will ever see of such a galaxy is a point in its history (that is not special to the galaxy itself), after which we will never receive more recent information about the galaxy - this is why it is called an event horizon. Can stellar objects go outside of the scope of the observable universe or is it only growing with time? This is not easy to answer with a clear "yes" or "no", because the term "the observable Universe" is not clearly defined.
The event horizon that I mention above is defined as: Event Horizon: The distance, from beyond which a photon emitted towards us now will never reach us. And we just established that this distance is shrinking in co-moving coordinates, so yes, galaxies are escaping our cosmic event horizon as we speak. But we can see that the Event Horizon has the property that is is a cut-off in time, as well as Space: For any galaxy, it is the last event in its history that we will be able to observe. But we could also ask a different question: What is the farthest distance, at which we can see the first event in an object's history? Or in other word, how far can a photon emitted at the time of the Big Bang, have travelled today ? At which distance are we observing the Big Bang? We call this distance the Particle Horizon : Particle Horizon: The distance from which a photon emitted during the Big Bang will reach us now. We normally use the Cosmic Microwave Background as a proxy for this; it currently has a redshift of ~1100 and a proper distance of 46 billion light years from us (remember, that is the current size of that space, but it has expanded continuously while the photon was travelling, so the photon will still have travelled only the 13.8 billion light years we would expect from the age of the Universe). While the event horizon is growing in absolute, but shrinking in co-moving coordinates, the Particle Horizon is growing in both coordinate systems, so that horizon is expanding forever. Summary To answer this question properly, we have to think in terms of both time and space. On one hand ; for all galaxies, there is a point in their history that is the last we will ever observe, so in that sense, galaxies (and thus stars) are receding out of the "observable Universe" as we speak. On the other hand , the distance at which we can see the first events in the Universe is expanding and always will be. But the time limit on how much of the history of these new regions we can study before they slow down to a freeze and redshift to infinity will get ever shorter. PS: The math behind There is an article that is relatively easy to read for someone with a background in physics, written by someone way smarter than me, on ArXiv.org , explaining this whole thing in better theoretical detail. | {
"source": [
"https://physics.stackexchange.com/questions/90024",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/11804/"
]
} |
90,036 | A 9.5 kg block (A) is traveling in the positive x-direction with a speed of 3.0 m/s. At some point, it explodes and breaks into two pieces. After the explosion, block C, which is 6.0 kg, moves off in the negative y-direction, and block B, which is 3.5 kg, moves off at an angle 35° with respect to the x axis. The two parts of this question that I am stuck on are: a) find the momentum of the 2-block combination after the explosion and write it in vector notation Here I know that momentum is conserved, so Pa = Pbc = 28.5 kg*m/s I am confused about how to separate the x and y components of the momentum after the explosion: can I use trigonometry with the mass of block B as the hypotenuse? b) Find the speeds of blocks B and C after the explosion Since momentum is conserved, and Mass(A) = Mass(B)+ Mass(C), I know that V(A)=V(B)+V(C), but I don't understand how to split them (I know it's not in half...) | This is a rather lengthy answer as I tried to go a bit in depth; there is a short summary at the end. Will we see more or fewer stars with time? The short answer to this is: We see less stars with time, due to the fact that cosmic expansion is accelerating. Although what we really see at the relevant distances are galaxies; single stars are far too far away to be resolved. The first thing to realise to understand this is that the speed limit of special relativity doesn't apply to cosmic expansion. If you imagine the galaxies as raisins in an enormous, rising dough, what SR tells you is that nothing can move through the dough faster than light. But if expansion is the same everywhere, and the dough is large enough, then an arbitrarily small speed at which the dough would rise would make all raisins beyond a certain distance recede from your local raisin faster than light. This "certain distance", in the Universe, is called the Hubble Distance . Galaxies farther away from the Hubble Distance are receding from us faster than light. This, however, is not the limit beyond which light cannot reach us. If a galaxy is receding from us faster than light, a photon emitted from said galaxy towards us will initially be receding from us, but it will crawl to regions of space - at local light speed - which are receding from us a bit slower than the one from which it was emitted, and so it will recede a bit slower from us than before, etc., until it finally enters a region of space that is receding from us slower than light, from which point it is trivial that it will eventually reach us. This is not always the case, though. There is, in certain cosmologies (including our own with an accelerating expansion), a distance,from beyond which a photon emitted will never reach us. This happens if the Universe expands faster than the photons can traverse it. This distance is called the cosmic Event Horizon , and while it expands in terms of absolute distances, it is shrinking in terms of co-moving distance. The co-moving distance is the distance measured in a coordinate system that expands along with the Universe - such a system will have galaxies more or less sitting in the same coordinates throughout the history of the Universe, and so the Event Horizon having a shrinking co-moving radius from us directly tells us that fewer objects will be inside this radius in the future. The way we can in theory observe these distant galaxies leave our event horizon - and the reason why it has its name - will be very much like we would observe an object falling into the event horizon of a Black Hole: relativistic effects will redshift the light emitted from such an object indefinitely, which also slows down the time as observed by us. So the object will appear to slow down asymptotically to a complete freeze, while the light will be redshifted towards infinity and become undetectable by us, although you could argue that it will never, in theory, disappear completely. The last we will ever see of such a galaxy is a point in its history (that is not special to the galaxy itself), after which we will never receive more recent information about the galaxy - this is why it is called an event horizon. Can stellar objects go outside of the scope of the observable universe or is it only growing with time? This is not easy to answer with a clear "yes" or "no", because the term "the observable Universe" is not clearly defined.
The event horizon that I mention above is defined as: Event Horizon: The distance, from beyond which a photon emitted towards us now will never reach us. And we just established that this distance is shrinking in co-moving coordinates, so yes, galaxies are escaping our cosmic event horizon as we speak. But we can see that the Event Horizon has the property that is is a cut-off in time, as well as Space: For any galaxy, it is the last event in its history that we will be able to observe. But we could also ask a different question: What is the farthest distance, at which we can see the first event in an object's history? Or in other word, how far can a photon emitted at the time of the Big Bang, have travelled today ? At which distance are we observing the Big Bang? We call this distance the Particle Horizon : Particle Horizon: The distance from which a photon emitted during the Big Bang will reach us now. We normally use the Cosmic Microwave Background as a proxy for this; it currently has a redshift of ~1100 and a proper distance of 46 billion light years from us (remember, that is the current size of that space, but it has expanded continuously while the photon was travelling, so the photon will still have travelled only the 13.8 billion light years we would expect from the age of the Universe). While the event horizon is growing in absolute, but shrinking in co-moving coordinates, the Particle Horizon is growing in both coordinate systems, so that horizon is expanding forever. Summary To answer this question properly, we have to think in terms of both time and space. On one hand ; for all galaxies, there is a point in their history that is the last we will ever observe, so in that sense, galaxies (and thus stars) are receding out of the "observable Universe" as we speak. On the other hand , the distance at which we can see the first events in the Universe is expanding and always will be. But the time limit on how much of the history of these new regions we can study before they slow down to a freeze and redshift to infinity will get ever shorter. PS: The math behind There is an article that is relatively easy to read for someone with a background in physics, written by someone way smarter than me, on ArXiv.org , explaining this whole thing in better theoretical detail. | {
"source": [
"https://physics.stackexchange.com/questions/90036",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/35726/"
]
} |
90,048 | Background I studied Rindler's book on Relativity. Relevant information from this book is available on-line here: http://www.scholarpedia.org/article/Special_relativity:_mechanics Given that we work with four-vectors $\mathbf{R} = \left(\vec{r},ct\right)$ where the four-velocity is defined as $$
\tag{1}
\mathbf{U}=\frac{d\mathbf{R}}{d\tau}=\gamma\frac{d\mathbf{R}}{dt}=\gamma\left(u\right)\left(\vec{u},c\right),
$$ $\vec{u}$ is the spacial "three-velocity" and the four-acceleration is defined as $$
\tag{2}
\mathbf{A}=\gamma\frac{d\mathbf{U}}{d\tau}=\gamma\frac{d}{dt}\left(\gamma\vec{u},\gamma c\right)=\gamma\left(\dot{\gamma}\vec{u}+\gamma\vec{a},\dot{\gamma}c\right),
$$ we can define the four-force as $$
\tag{3}
\mathbf{F}=\frac{d}{ {d\tau } } \mathbf{P}=\frac{d}{ {d\tau}}(m_{0}\mathbf{U})=m_{0}\mathbf{A}+\frac{ {dm}_{0} }{ {d\tau} }\mathbf{U}.
$$
The four-force $\mathbf{F}$ is said to be mass-preserving if $dm_0/d\tau$ equals zero, i.e. the force leaves the resting mass of the particle unchanged. Moreover $$
\tag{4}
\mathbf{F}=\frac{d}{d\tau}\mathbf{P}=\gamma(u)\frac{d}{ {dt} }( \vec{p},mc)=\gamma
(u)\left(\vec{f},\frac{1}{c}\frac{dE}{dt}\right).
$$ We note that $\mathbf{A}\cdot\mathbf{U}=0$ and therefore for a mass preserving four-force $\mathbf{F}\cdot\mathbf{U}=0$ from Eq. (3) or from Eq. (4) $$
\tag{5}
\mathbf{F}\cdot\mathbf{U}=\gamma^2(u)\left(\frac{dE}{dt}-\vec{f}\cdot\vec{u}\right)=0
$$ which means that $$
\frac{dE}{dt}=\vec{f}\cdot\vec{u}
$$ and $$
\tag{6}
\mathbf{F}=\gamma(u)\left(\vec{f},\vec{f}\cdot\vec{u}/c\right).
$$ We also have the relationship $$
\tag{7}
\gamma(u)\vec{f}=m_0\frac{d^2\vec{r}}{d\tau^2}.
$$ Moreover we have that the proper acceleration $\alpha$ is defined as $$
\mathbf{A}\cdot\mathbf{A}=\mathbf{A}^2=-\alpha^2.
$$ I studied the questions formulated in Why proper acceleration is $du/dt$ and not $du/d\tau$? Relativistic factor between coordinate acceleration and proper acceleration and realized that the spatial vectors can be broken down into a parallel component and an orthogonal component, i.e. $\vec{v}=\vec{v}_\parallel+\vec{v}_\perp$ or $\vec{a}=\vec{a}_\parallel+\vec{a}_\perp$ where one makes a distinction between the component of the velocity that is parallel with the acceleration and vice versa. Question Now it can be shown that for a force $\vec{f}$ acting upon a particle traveling with a velocity vector $\vec{u}$ where there is an angle $\theta$ between $\vec{f}$ and $\vec{u}$ the relationship between the proper acceleration of the particle and the force $\vec{f}$ is $$
\tag{8}
\vec{f}^2=m_0^2\frac{\gamma_\parallel^2}{\gamma^2}\alpha^2
$$ where $$
\tag{9}
\gamma_\parallel=\frac{1}{\sqrt{1-u_\parallel^2/c^2}}\;\;\;\text{and}\;\;\;\gamma=\frac{1}{\sqrt{1-u^2/c^2}}.
$$ Now how can I derive or prove equation (8)? I have tried every feasible combination of Eq. (1) to (7) without success. I've also tried the more basic variants of the Lorentz transformation https://en.wikipedia.org/wiki/Lorentz_transformation but it gets very confusing when there are two components to consider when assessing the relativistic effects. Any ideas? | This is a rather lengthy answer as I tried to go a bit in depth; there is a short summary at the end. Will we see more or fewer stars with time? The short answer to this is: We see less stars with time, due to the fact that cosmic expansion is accelerating. Although what we really see at the relevant distances are galaxies; single stars are far too far away to be resolved. The first thing to realise to understand this is that the speed limit of special relativity doesn't apply to cosmic expansion. If you imagine the galaxies as raisins in an enormous, rising dough, what SR tells you is that nothing can move through the dough faster than light. But if expansion is the same everywhere, and the dough is large enough, then an arbitrarily small speed at which the dough would rise would make all raisins beyond a certain distance recede from your local raisin faster than light. This "certain distance", in the Universe, is called the Hubble Distance . Galaxies farther away from the Hubble Distance are receding from us faster than light. This, however, is not the limit beyond which light cannot reach us. If a galaxy is receding from us faster than light, a photon emitted from said galaxy towards us will initially be receding from us, but it will crawl to regions of space - at local light speed - which are receding from us a bit slower than the one from which it was emitted, and so it will recede a bit slower from us than before, etc., until it finally enters a region of space that is receding from us slower than light, from which point it is trivial that it will eventually reach us. This is not always the case, though. There is, in certain cosmologies (including our own with an accelerating expansion), a distance,from beyond which a photon emitted will never reach us. This happens if the Universe expands faster than the photons can traverse it. This distance is called the cosmic Event Horizon , and while it expands in terms of absolute distances, it is shrinking in terms of co-moving distance. The co-moving distance is the distance measured in a coordinate system that expands along with the Universe - such a system will have galaxies more or less sitting in the same coordinates throughout the history of the Universe, and so the Event Horizon having a shrinking co-moving radius from us directly tells us that fewer objects will be inside this radius in the future. The way we can in theory observe these distant galaxies leave our event horizon - and the reason why it has its name - will be very much like we would observe an object falling into the event horizon of a Black Hole: relativistic effects will redshift the light emitted from such an object indefinitely, which also slows down the time as observed by us. So the object will appear to slow down asymptotically to a complete freeze, while the light will be redshifted towards infinity and become undetectable by us, although you could argue that it will never, in theory, disappear completely. The last we will ever see of such a galaxy is a point in its history (that is not special to the galaxy itself), after which we will never receive more recent information about the galaxy - this is why it is called an event horizon. Can stellar objects go outside of the scope of the observable universe or is it only growing with time? This is not easy to answer with a clear "yes" or "no", because the term "the observable Universe" is not clearly defined.
The event horizon that I mention above is defined as: Event Horizon: The distance, from beyond which a photon emitted towards us now will never reach us. And we just established that this distance is shrinking in co-moving coordinates, so yes, galaxies are escaping our cosmic event horizon as we speak. But we can see that the Event Horizon has the property that is is a cut-off in time, as well as Space: For any galaxy, it is the last event in its history that we will be able to observe. But we could also ask a different question: What is the farthest distance, at which we can see the first event in an object's history? Or in other word, how far can a photon emitted at the time of the Big Bang, have travelled today ? At which distance are we observing the Big Bang? We call this distance the Particle Horizon : Particle Horizon: The distance from which a photon emitted during the Big Bang will reach us now. We normally use the Cosmic Microwave Background as a proxy for this; it currently has a redshift of ~1100 and a proper distance of 46 billion light years from us (remember, that is the current size of that space, but it has expanded continuously while the photon was travelling, so the photon will still have travelled only the 13.8 billion light years we would expect from the age of the Universe). While the event horizon is growing in absolute, but shrinking in co-moving coordinates, the Particle Horizon is growing in both coordinate systems, so that horizon is expanding forever. Summary To answer this question properly, we have to think in terms of both time and space. On one hand ; for all galaxies, there is a point in their history that is the last we will ever observe, so in that sense, galaxies (and thus stars) are receding out of the "observable Universe" as we speak. On the other hand , the distance at which we can see the first events in the Universe is expanding and always will be. But the time limit on how much of the history of these new regions we can study before they slow down to a freeze and redshift to infinity will get ever shorter. PS: The math behind There is an article that is relatively easy to read for someone with a background in physics, written by someone way smarter than me, on ArXiv.org , explaining this whole thing in better theoretical detail. | {
"source": [
"https://physics.stackexchange.com/questions/90048",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/35730/"
]
} |
90,533 | What causes time dilation? Acceleration or velocity? I've seen multiple comments on this forum that assert velocity is the cause, but that doesn't seem right to me. You can't have velocity without acceleration. It's the inertial force with acceleration that breaks the symmetry. My understanding is that the asymmetry is where the inertial frame changes. Measuring time between two objects with different inertial frames is where you have time dilation. When the acceleration ends, the object is effectively at rest in a new inertial frame and has velocity relative to another object in the original inertial frame. In other words, acceleration (changing reference frames) is the cause...velocity and time dilation is the effect. Am I right about this? If there are flaws in my logic I'd like to find and correct them. | We need to untangle this a bit but first: the cause of time dilation is the geometry of spacetime which is such that there is an invariant speed c . Now, remember that velocity or speed is not a property of an object ; there is no absolute rest. Further, consider the case of three objects in uniform relative motion with respect to each other. If I choose one of those objects and then ask you "what is the relative velocity of this object?", the only proper response you could give is " velocity relative to which of the other objects? " So, we can't speak of the relative motion of an object but rather the relative motion of a pair of objects. What we can say is that, for two objects in relative uniform motion with respect to each other, the other object's clock runs slow according to each object's own clock. This is called relative velocity time dilation . It is important to realize that in the case of relative time dilation, the two relatively and uniformly moving clocks are spatially separated except at one event. Comparing the readings of the two clocks when spatially separated requires additional spatially separated clocks synchronized and stationary in their respective object's frame of reference But, we find that clocks synchronized in one object's frame are not synchronized in the other relatively moving object's frame. Thus, the relative velocity time dilation is symmetric without contradiction. We can't say that one or the other clock is absolutely running slower. Now, within the context of Special Relativity, acceleration is absolute , i.e., an object's accelerometer either reads 0 or it doesn't. And, a fundamental result in SR is that a clock along an accelerated worldline through two events in spacetime records less elapsed time between those events than a clock along an unaccelerated world line through the same two events. Since, in this case, an accelerated clock and an unaccelerated clock are co-located at two different events, the two clocks can be directly compared and, in this case, the time dilation is absolute - the accelerated clock absolutely shows less elapsed time than the unaccelerated clock. | {
"source": [
"https://physics.stackexchange.com/questions/90533",
"https://physics.stackexchange.com",
"https://physics.stackexchange.com/users/35968/"
]
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.