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264,240 | When someone nearby is listening to music through earphones or headphones, usually you can only hear the bass (unless it’s really loud, where you can hear the singer and the other instruments too). Why are the bass sounds more audible? The answer probably has to do with the bass sound waves having the highest wavelengths. I was able to gain some insight from this question , however, it focuses on the way sound waves travel across a room and through walls and not on ear/headphones. | You don't. You actually hear the high frequency notes from headphones. The bass really doesn't travel at all well, but the attack noise from the drum or bass guitar is what leaks from headphones. This is why on a bus or train you hear "tsss tsss tsss tsss" and very little else. From @leftaroundabout's answer on the post that valerio92 linked : Normal headphones are basically dipole speakers, and especially for bass frequencies (wavelength much larger than the speakers) this describes their behaviour well. So the amplitide decreases ∝ 1/r 4 . At higher frequencies, they also have some monopole components which decay more slowly, with the familiar inverse-square. So if you're listening from far away, you'll mostly hear those treble frequencies and little or no bass. OTOH, while wearing the headphones there's little difference since you're in the near field where neither frequency range has decayed substantially at all. | {
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264,412 | This question is going to look a lot like a duplicate, but I've read dozens of related posts and they don't touch the subject. Here we go. Why are observables represented by hermitian operators? Because then we'll measure real stuff, since the eigenvalues are real; Because hermitian operators (actually normal operators) have a complete set of eigenvectors, and we'll be sure the accessible states of our experiment will be modelled consistently; Those are the two main reasons for the Physicists of the early XX century to have chosen hermitian operators as a mathematical model of what we can observe: mathematics proves they provide both the needed properties. My question is of philosophical inclination: which one of the above requirements is more fundamental? I'm inclined to think the most important property is to be able to represent the system by the right model, which is II). Real eigenvalues come as a bonus. I've read that having real eigenvalues is more important, but I don't see why they couldn't be complex... Maybe Nature is tricking us some way and we only measure their real part. The problem for me would be describing the system with a model that doesn't mirror our accessible states, and is therefore flawed. Maybe both are fundamental on the same level. Maybe the question is stupid. What do you think? | Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are used not because measurements must use real numbers, but rather because we almost always decide to use real numbers. As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able to use a so-called normal operator to represent the 2-dimensional observable. (Contrary to what Dirac thought, nothing whatsoever goes wrong here.) It should not be too hard then to accept my next claim: You can use whatever measurement scale you want to measure a quantum observable! You can label pointer positions with items of fruit if you want to, and you can still build a perfectly legitimate observable. There is no question that the reals and complexes have enormous advantages over other more arbitrary measurement scales (due to their rich internal structure which we capitalize on in the functional analysis), but the idea that real numbers are somehow endowed with a prestigious metaphysical status is baloney. How to define an observable with any measurement scale you want to Step 1. Set up a bunch of particle detectors Step 2. Attach a label to each detector The set of labels we'll denote by $\Omega$ . Examples include: $$\Omega = \{0,1\},\mathbb{R},\mathbb{C},\{\heartsuit,\clubsuit,\diamondsuit,\spadesuit\}$$ Step 3. Write out the list of all possible events By event , I mean a subset $\Delta$ of $\Omega$ that represents a possible question like "Did a detector in $\Delta$ fire?". We'll label the event structure $\Sigma$ , e.g. $$\Sigma = \{\emptyset,\heartsuit,\clubsuit,\diamondsuit,\spadesuit,\heartsuit\clubsuit,\heartsuit\diamondsuit, \cdots,\heartsuit\clubsuit\diamondsuit\spadesuit\}$$ Step 4. Associate each event in $\Sigma$ with a projection operator This is the hard bit, and there's no recipe for it. But you have to make sure that the family of projectors forms a Boolean algebra that perfectly mirrors the natural algebra of $\Sigma$ . We'll call the association $\sigma$ , so that $\sigma:\Sigma\to\mathscr{P}(\mathscr{H})$ . And that's basically it! The object $\sigma$ (technically, a Projection Valued Measure on $\langle\Omega,\Sigma\rangle$ ) is a quantum observable. It contains all the probabilistic information you need to calculate the probability measure on your chosen measurement scale for any quantum state. For example, suppose the state of the system is $\rho$ and you want the probability that a detector $\Delta\in\Sigma$ fires. The desired probability is just $p=tr[\rho\sigma(\Delta)]$ . What the heck does this have to do with Self-Adjoint operators? Are you ready for the climax? Here it is... IF you choose to use the measurement scale $\langle\mathbb{R},\mathscr{B}(\mathbb{R})\rangle$ , THEN you will be able to build a self-adjoint operator which is precisely equivalent (in terms of the information it stores) to the PVM you constructed. IF you choose a detection screen calibrated by $\langle\mathbb{C},\mathscr{B}(\mathbb{C})\rangle$ , THEN repeat the above sentence replacing 'self-adjoint' with 'normal'. (Neither of the above statements is obvious, by the way. They are famous results in Functional Analysis known as the Spectral Theorems.) IF you choose to be a Fancy Nancy and use $\{\heartsuit,\clubsuit,\diamondsuit,\spadesuit\}$ for a measurement scale (with its power set for the event structure) THEN the fruits of your labors are more modest. In particular, you still get the answers to any questions you care to ask, but you don't get any neat operator to give you computational shortcuts. Instead you will forever be doing calculations like $p=tr[\rho\sigma(\heartsuit\clubsuit)]$ . I haven't even touched eigenvectors yet, but suffice it to say that they also do not have a fundamental status in the theory. There is no doubt that we can learn something by reading the works of the great masters, but taking that work as the state of play can send you back a century. We've learned a lot since Einstein and Dirac. | {
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264,687 | When I wash my hair and go to sleep, my hair is impossible to comb in the morning, stubbornly sticking to the shape it assumed during the night. The only way to get it right is to wet it again and comb it. What's the cause of this memory effect? | Hair, like fingernails and animal horn is made up mostly of a protein called Keratin . The strength and hardness of this polymer is caused by three types of chemical bonds: ionic bonds, hydrogen bonds and disulphide bonds. Water can significantly break the first two types (but not the disulphide ones). Significantly wetting hair thus leads to making it more flexible and softer. But if wet and deformed hair dries it tends to retain the shape it was in while it was wet. The reformed hydrogen and ionic bonds then leads to a 'permanent' deformation (until you wet it again). | {
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264,869 | The picture below is from on top of Mt. Shasta at sunrise. The sun was directly behind me when I took the picture and it appears to be casting a shadow on.... the sky itself? Can anyone explain what the shadow is being cast on? There don't appear to be any clouds or anything. | I found this site Seen from their summits almost all mountain shadows look triangular regardless of the mountain's shape. This is a perspective effect. You are standing at the top edge of a long tunnel of shadowed air and looking along its length. The tunnel's cross section is the shape of the mountain but its "end" is so far away that it looks insignificant. The finite size of the sun also causes the umbral (fully shaded) parts of the shadow to converge and eventually taper away. The tapering sets limits to the umbral length of shadows. That of the Earth is over a million miles. That of a high mountain can be two to three hundred miles. Triangular shadows are not seen from objects much smaller than mountains because their shadows are not long enough. On another page , it says: Mountain shadows at sunrise and sunset are immensely long tunnels of unlit air, crepuscular rays in fact. From the summit, perspective effects nearly always make the shadow triangular regardless of the mountain's profile. You are standing at the top edge of the shadowed tunnel and looking out along its length which can be more than a hundred miles. Only the shadow's end carries much information about the mountain shape and it is so far away and in any event blurred by the 0.5º angular spread of the sun's rays that it is hardly visible To elaborate on "unlit air", as an answer to your "There don't appear to be any clouds or anything," air itself reflects light, as is evident at night when it is dark. It is also seen in the shadow of the earth as the sun rises on a clear sky, a deep blue separated often from the pink/orange of sunrise and sunset. That was the beginning and end of a day in the variable hour ancient calendars. | {
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264,870 | When you add current to a straight piece of wire does it use less electricity than if it was coiled? The power wire on telephone pole's are curved while buried cables are pretty strait in comparison. Does the curve of the wire and proximity to itself make a difference in energy consumption? | I found this site Seen from their summits almost all mountain shadows look triangular regardless of the mountain's shape. This is a perspective effect. You are standing at the top edge of a long tunnel of shadowed air and looking along its length. The tunnel's cross section is the shape of the mountain but its "end" is so far away that it looks insignificant. The finite size of the sun also causes the umbral (fully shaded) parts of the shadow to converge and eventually taper away. The tapering sets limits to the umbral length of shadows. That of the Earth is over a million miles. That of a high mountain can be two to three hundred miles. Triangular shadows are not seen from objects much smaller than mountains because their shadows are not long enough. On another page , it says: Mountain shadows at sunrise and sunset are immensely long tunnels of unlit air, crepuscular rays in fact. From the summit, perspective effects nearly always make the shadow triangular regardless of the mountain's profile. You are standing at the top edge of the shadowed tunnel and looking out along its length which can be more than a hundred miles. Only the shadow's end carries much information about the mountain shape and it is so far away and in any event blurred by the 0.5º angular spread of the sun's rays that it is hardly visible To elaborate on "unlit air", as an answer to your "There don't appear to be any clouds or anything," air itself reflects light, as is evident at night when it is dark. It is also seen in the shadow of the earth as the sun rises on a clear sky, a deep blue separated often from the pink/orange of sunrise and sunset. That was the beginning and end of a day in the variable hour ancient calendars. | {
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264,928 | Say I have my phone on 5% and a large battery pack on 35% and I charge the phone. By the end the phone is on 100% and the pack is on 12%. How can the battery pack charge the phone up to a higher percentage of its current charge? I would expect that the phone would stop charging when the percentage on the power pack was equal to the percentage on the phone, but clearly it doesn't, so what's happening? | Connecting your phone to the battery pack doesn't directly connect the cells in parallel. I assume this is where your guess of an equilibrium with equal voltage -> equal charge percentage comes from. Shorting lithium-ion / lithium-polymer (LiPo) cells together like that would likely cause one or both to literally catch fire from the high currents, or from overcharging / over-discharging. (Circuitry to prevent over-discharging even if the charging cable is shorted out is absolutely essential). There are some links to youtube videos of lithium battery fires on a recent electronics.SE question about designing your own charger. (TL:DR: it's way the hell too dangerous to consider doing for a homebrew design, because Lithium cells need a LOT of protection circuitry to be mostly safe.) So the whole idea of connecting cells together and letting their voltages equalize "naturally" is just completely not viable for modern batteries. Chargers use DC-DC switching power supplies to charge at constant current. They use inductors to efficiently convert to a different voltage (higher or lower). (For example, to produce a lower voltage, see this detailed explanation of a buck converter doesn't use a transformer, just an inductor. Also a discussion of multi-phase buck converters used on computer motherboards .) In the water analogy , where water represents charge and pressure represents voltage: A converter is like a pump that can move charge from a lower reservoir to a higher reservoir. (voltage = pressure = gravitational potential energy (per unit volume / charge).) A small fraction of the energy transferred is lost to inefficiencies in the conversion. (Maybe a couple %, IDK). Since the capacity of the external battery pack is larger than the capacity of the phone's battery pack, it should be obvious that moving charge from the large reservoir to the small reservoir can take the phone battery from 5% to 100% while only dropping the battery pack from 35% to 12%. I don't think this is what the question was really about. Just to make it even more obvious why batteries aren't just connected together to equalize: Some batteries may have multiple cells in series instead of one large cell. This is typically for physical design reasons, more than to get a higher voltage, because DC-DC converters will be used anyway to produce supply voltages in the 1 V to 2 V range to power most electronics. Since Lithium cells are so finicky and dangerous, wiring them in parallel instead of series is unwise. One cell could end up taking most of the current. So instead, they're wired in series with circuitry for each cell to bypass it before it overcharges or undercharges. The power transfer between phone and battery pack happens over a USB cable , which runs at 5 V. (Or, with USB power delivery signalling, the device being charged can signal that it can accept up to 20V, allowing for higher power at the same current to reduce resistive losses and allow faster charging without exceeding safe current limits for the cable / connectors.) | {
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265,071 | Mass (which can here be considered equivalent to energy) curves spacetime, so a body with mass makes the spacetime around it curved. But we live in 3 spatial dimensions, so this curving could only be visualized in a chart with 4 dimensions, and the living being will notice only the effect that this bend causes on matter and light, the gravity, or the curve will expand-contract also the space itself? and can be measured by the change in volume? Let's take an analog on 2 dimensions: The distortion on space is visible only by a 3d being (such as a human.) The 2d creatures can't see the distortion, as they are 2d creatures, but they can see that space is 'bigger' where the distortion occurred, because the total area of the squares (as viewed from the top--i.e., from a different coordinate in a third, orthogonal dimension--increases. In the real world, the axis of distortion is trickier to define. The space needs to be expressed as cubes, not squares, and what changes with distortion is the volume of those cubes. So, if you have two identical boxes, one with a very massive object, and another empty, the one with the massive object fits more things inside it? | In a question like this you need to ask what does the volume change relative to. So it's a little bit ambiguous. However, the answer to your question is "yes" in the following restricted sense. Imagine having a "swarm" of test objects, with mass so small that their effect on the spacetime around them is negligible. Assume that they are in freefall, i.e. all following spacetime geodesics. The swarm has a shape with volume; let's assume it's a sphere, and that their geodesic paths take them through varying curvature spacetime. Let's assume also that our space-bees are signalling their queen so that she can work out her distance to her fellows at all times. Then, in general, the queen will find that the distance to her fellows changes following the geodesic deviation equation (see also Misner, Thorne and Wheeler, "Gravitation" Chapter 1 for a great intuitive explanation). So she will perceive a change in volume of her swarm. Indeed, there is a tensor object whose primary meaning is that of volume change. If one writes the co-ordinates of the queen's swarm in Riemann Normal Co-ordinates (see also Misner Thorne Wheeler section 11.6), whereby something's position is named by naming a direction vector (a tangent vector to spacetime) and a distance along that vector, then the volume element $\mathrm{d} V$, compared to the volume element $\mathrm{d} V_f$ one would calculate assuming flat spacetime, is defined by the Ricci curvature tensor: $$\mathrm{d} V \approx \left(1-\frac{1}{6}R_{j\,k}\,x^j\,x^k+\mathscr{O}(x^3)\right)\mathrm{d} V_f$$ Indeed one can "decompose" the full curvature tensor into expressions involving the Ricci tensor and so-called Weyl tensor. The former measures how the volume of our swarm changes, the latter tells us in detail how the shape of the swarm changes as the swarm freefalls. You may also be interested in Chapter 42 of Volume II, called Curved Space of the "Feynman Lectures on Physics" | {
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265,193 | That may appear as a dumb question, but: Does Hilbert space have curvature, or is it a flat space? How and why? | This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to consider the distance function
$$d_\textrm{proj}(\psi,\phi) =\min_\alpha || \psi - e^{i\alpha} \phi ||$$
Or equivalently, $d_\textrm{proj}(\psi,\phi) = \sqrt{2 -2 |\langle \psi,\phi\rangle|}$. This is for example natural when we are looking at the ground state of a Hamiltonian: this is naturally only defined up to a phase. We might then ask how much our ground state changes when we tune a parameter in our Hamiltonian. The above projective distance function is then the natural quantity to look at (and it becomes singular at a quantum phase transition). Our distance function of course defines a metric, similar to what you might be used to from general relativity. It is in fact the natural metric associated to $\mathbb CP^n$, called the Fubini-Study metric . Moreover this has non-trivial curvature . (In the case of the Bloch sphere, $n = 1$, it coincides with the spherical metric on $\mathbb CP^1 \cong S^2$ but for $n>1$ it is much more non-trivial with e.g. the metric depending on the direction you go.) This metric of course also defines a distance function (defined by taking the shortest geodesic between two states), which turns out to be
$$ d_\textrm{FS}(\psi,\phi) = \arccos \left( \left|\langle \psi \right| \phi \rangle| \right) $$
Note that this is basically the angle between the two states. This is the Fubini-Study distance between $\psi$ and $\phi$. By expanding out the $\arccos$ it is clear that for $\psi$ close to $\phi$ it agrees with $d_\textrm{proj}$ (as it must, since by definition both have the same (local) metric). But they disagree for finite distances, and the fact that $d_\textrm{FS} \neq d_\textrm{proj}$ shows that the metric space defined by $d_\textrm{proj}$ is not an inner metric space/length space. (I.e. its distances are not given by geodesics. Usually in physics we work with inner metric spaces so it might be surprising to see a case pop up where this is not the case. However if we work with $d_\textrm{FS}$ then it is indeed an inner metric space.) The Fubini-Study distance is operationally very meaningful . It arises naturally as follows: First there is the general result in statistics that the natural
``distance'' between two probability distributions
$(p_1,p_2,\cdots)$ and $(q_1,q_2\cdots)$ is given by $\arccos\left( \sum_i \sqrt{p_i q_i} \right)$. This is the so-called Fisher distance. (It is in fact very conceptual as explained here by Alioscia Hamma in his lecture on the quantum informational perspective of quantum phase transitions (at Perimeter Institute 2013) .) Of course we know that any wavefunction $\psi$ defines a probability distribution if we are given an observable $A$. More exactly $p_n = |\langle \psi| A_n\rangle|$ where $A_n$ are the eigenfunctions of $A$. Similarly for $\phi$. So for a given $A$ we then have the natural distance $\arccos\left( \sum_i \sqrt{ \left|\langle \psi| A_i\rangle \langle \phi| A_i\rangle \right|} \right)$. It is then natural to define the operational distance between $\phi$ and $\psi$ by maximizing the previous expression over all possible observables $A$. It is not hard to show that this then gives $d_\textrm{FS}(\psi,\phi)$. The Fubini-Study metric/distance has many nice mathematical properties (e.g. it makes our projective Hilbert space into a so-called Kahler manifold, is related to fun stuff like Hopf fibrations, et cetera). But it is also a natural language for some physical concepts. One example is as I hinted at above: suppose we have a space of parameters $\Lambda$ and for each choice we have a Hamiltonian. I.e. we have a map of $\Lambda$ into the space of Hamiltonians. Then through the ground state this effectively defines a map $\Lambda \to P\mathcal H$ (for each parameter we have a state in our projective Hilbert space). Hence we can pull back our Fubini-Study metric to define a metric on our manifold $\Lambda$. It turns out that then looking at the curvature of this contains the complete physical information of the phase transitions of our model. Another situation where it comes up is concerning entanglement . Given a state $\psi$, we can ask what is the closest separable state nearby. This Fubini-Study distance is then a measure of entanglement (it in fact contains the same information as the largest Schmidt value of our state--in case that means something to you.). More generally it is still quite an open question as to what extent we can relate entanglement and geometry, but it seems like a very promising path. For more information in this direction, here is the freely accessible chapter by Uhlmann and Crell on the geometry of state space from the book Entanglement and Decoherence. | {
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265,285 | I always used to think (I don’t know why!) that the efficiency of human (and animal and plant) cells should be equal to or near the efficiency of a Carnot engine or at least should be the highest efficiency among all practical engines. But I wondered when I saw the answers given to this question . They are talking about an efficiency of $18-26\%$. But you can see here that the efficiency of an Otto engine is between $56\%$ and $61\%$. Is there any explanation for this? Which cycle do human cells work with? Can we compare living cells with heat engines at all? | Can we compare alive cells with heat engines at all? No, not really, because the living being isn't only a heat engine. There are three main points I want to make here. 1. Homeostasis Requires Constant Energy Input This statement is especially true and obvious of homeotherms Mammals (Mammaliaformes, descended from the Therapsid Synapsid Amniotes), and Birds (Avialae / Dinosauria, descended from Dinosauriform Amniotes), which use a great deal of energy simply keeping their body temperature within strict limits, i.e. compensating for (mainly convective) heat loss from their body in cold conditions and actively expelling heat from their bodies in hot conditions. But, more generally, the phenomenon of homeostasis also requires expenditure of energy; a living organism is a highly non-equilibrium thermodynamic system, and excess entropies produced by metabolic processes must be expelled to keep it that way. Thermodynamic equilibrium is only reached when the living creature dies. From this consideration alone, we would expect efficiencies measured when the organism does mechanical work to be considerably less than those of a heat engine. 2. Muscular Action is Not a Heat Engine Muscular action is much more comparable to an electric motor than a heat engine. What I mean by this is an electric motor converts essentially work from one form to another with near to zero entropy change and negligible temperature change; motor proteins convert low-entropy energy stored as ATP to mechanical work through the hydrolysis of ATP with very small temperature change in the reagents as they react. In this case, the most meaningful measure of efficiency is probably expressed in two factors: (1) the ratio of the free energy $\Delta G$ of the ATP hydrolysis reaction to the total enthalpy change $\Delta H$ of the reaction (the difference $T\,\Delta S$ being the work we have to "give up" to expel the excess entropy of the reactants relative to lower entropy reaction products) and (2) the ratio of the mechanical work done to the available $\Delta G$. In a heat engine, we take a quantity of heat from a hot reservoir, reducing the latter's entropy by $Q_i/T_i$ in the process, but find that, if we have a colder reservoir at $T_o<T_i$ we only have to "give back" $Q_o<Q_i$ to the cold reservoir to offset the entropy drop in the hot reservoir, so we get to "keep" energy $Q_i - Q_o > 0$ for doing work with. In biological reactions, the most comparable process to this is that of photosynthesis , where the "working fluid" of light at thermodynamic equilibrium at $6000{\rm K}$ is converted to "stored work" in sugars and, ultimately, ATP, dumping excess heat at ambient temperature $300{\rm K}$ in the process. Thenceforth, all living things use this low-entropy energy store rather like an electric motor converting energy stored in a capacitor, whether it be plants using it for their own life process, or herbivores accessing it through eaten plants or carnivores accessing it through eaten plant eaters. So the plants and their solar energy fixing are the component of the biosphere most comparable to a heat engine in a power station; the plant metabolic processes and animals that eat plants and each other to get access to stored energy in plants are more like the electrical appliances that use the work extracted by the power plant, with very little temperature change. 3. Proteins Denature at Roughly $50{\rm C}$ For any animal process that could be considered to be like a heat engine, the maximum intake temperature can be at most a few or at most a few tens of degrees kelvin above ambient. This is because biological machinery is fatally damaged by temperatures much higher than $40{\rm C}$. Proteins denature and lose their vital life functions at very low temperatures. So if there are any processes in life that can be thought of as reasonably analogous to heat engines, we would foresee their efficiencies to be very low, since the theoretical efficiency is of the order of $3\%$ given this limit. An interesting exception to my point 3 comes up in deep sea life living near hydrothermal vents. John Rennie writes: Re the last point, the efficiency could of course still be 100% if animals had a heat sink at absolute zero. It's the fact there is a very limited temperature difference available that matters, rather than the limited source temperature. Note also that some extremophiles are quite happy living in near boiling water. so we have creatures dwelling in $100{\rm C}$ and over environments and the opportunity to dump heat into the surrounding sea at much lower temperatures. However, my understanding is that these creatures still use the chemical energy from what they can extract from the volcanic vents, rather than working as heat engines taking advantage of the temperature drop. | {
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265,331 | I was reading "The Sixth Extinction: An Unnatural History", by Elizabeth Kolbert, and there she comments that high level of $CO_{2}$ in the atmosphere lowers the pH of oceans (which makes sense) and, consequently, diminishes sound absorption: WHY is ocean acidification so dangerous? The question is tough to answer only because the list of reasons is so long. Depending on how tightly organisms are able to regulate their internal chemistry, acidification may affect such basic processes as metabolism, enzyme activity, and protein function. Because it will change the makeup of microbial communities, it will alter the availability of key nutrients, like iron and nitrogen. For similar reasons, it will change the amount of light that passes through the water, and for somewhat different reasons, it will alter the way sound propagates. (In general, acidification is expected to make the seas noisier.) I do understand that sound absorption depends on the material properties of the medium, but it is not trivial to me that more acidic water should have a different absorption than neutral water. Could anyone explain the physical mechanism behind this? (I'm not exactly sure that this belongs here, but since it deals with material properties, I thought it would fit better here than in Chemistry SE) | Hard though it is to believe, pH does have an effect on sound absorption in water. There are some reactions that are affected by pressure, that is pressure changes their equilibrium. One example is the equilibrium between boric acid and the borate ion: $$ B(OH)_4\,^- + H^+ \rightarrow B(OH)_3 + H_2O $$ Increasing pressure pushes the reaction over to the right, and in doing so it absorbs energy. That energy comes from the energy in the sound wave, so unlikely though it may seem the chemical reaction absorbs sound. When the pressure is released the reaction moves back to the left, but the energy is released as heat not sound. The net result is sound energy gets converted to heat. The problem with lowered pH is that as you lower the pH that moves the equilibrium to the right so there is less borate and more boric acid. That leaves less borate to absorb the sound. So lowered pH means less sound absorption. If you're interested there is an article about this in Scientific American , and a more rigorous scientific publication here (sadly not available online). Lemon helpfully provided this link, Investigation of chemical sound absorption in sea water , and Googling should find you more relevant publications. The effect is very small, not least because boric acid/borate concentrations in seawater are small. However sound propagates for huge distances in water - hundreds of miles. So even a small change in the water chemistry can have a measurable effect. The first of the papers I linked mentions that : Sound attenuation in the low-frequency range is primarily due to boric acid relaxation and is a function of the seawater pH. (my emphasis) Whether that effect is actually big enough to deafen dolphins is a matter of some debate. | {
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265,459 | Wikipedia states that a photon has a spin of value 1 . What I want to know is this: are there two possible spins for photons, i.e. +1 and -1 (like electrons, which can have +1/2 or -1/2 )? If the spin quantum number (the same as "spin"?) has to be essentially positive (to quote Wikipedia : " The conventional definition of the spin quantum number, s, is s = n/2, where n can be any non-negative integer "), then what is the term for possible +/- sign states of that spin-value? Is it called "parity"? Would there be any difference of behavior, e.g. a Circular polarization event, such as through a solution of D-glucose? Since through a solution of such chirally-asymmetric molecules, whole-portion of applied plane-polarized-light, light turns in a single direction (from which the compound is analyzed and given a +/- nomenclature), could I conclude that all the photons present in the applied plane-polarized-light contain only 1 type of spin out of 2 for that particular experiment (optical rotation)? | By definition of spin $S$ it is a positive integer number or zero. Not to confuse with the spin projection possible values $S_z$, which may run from $-S$ to $S$. | {
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265,958 | A common trick to make clear water ice is to boil pure water prior to freezing it. Why does that work and what are the white inclusions in ice that was made from unboiled tap water? | The short answer: Cloudy ice is caused by gases (mainly nitrogen and oxygen) dissolved in the water that come out of solution when the water freezes. The small bubbles trapped in the ice cause the white appearance. Boiling the water removes the air dissolved in it, producing clear ice as a result. Assuming that other impurities don't produce the same cloudy effect. The long answer: Impurities present in water: Gases: Water at 20°C normally contains about 15 ppm dissolved gases, which is the equivalent of 1 volume of air per 50 volumes of water. These are the same gases present in air, but not in the same proportions since some are more soluble than others: it's about 63% nitrogen, 34% oxygen, 1.5% argon and 1.5% carbon dioxide. minerals: Tap water contains dissolved minerals, mainly Ca and Mg. They can be present in the form of bicarbonates: $Ca^{2+}({HCO_3}^-)_2$ and $Mg^{2+}({HCO_3}^-)_2$ (these only exist in solution, not as solid substances), and as calcium and magnesium sulphate. If the water passed through a water softener, the Ca and Mg ions may have been replaced by (twice as many) sodium or potassium ions. The effects of heating the water: removing dissolved gases: higher temperature favors endothermic reactions ( Le Chatelier's principle ). For the gases present in water, dissolution (at room temperature) is an exothermic process, so their solubility decreases when the water is heated. The solubility of gases doesn't reach zero at boiling point, nor does it necessarily decrease over the whole temperature range. For nitrogen in water, the enthalpy of dissolution becomes positive around 75° , and its solubility increases above that temperature.
At 100°C, solubility of air as a whole is $0.93 * 10^{-5}$, about half the solubility at 10°C, $1.82 * 10^{-5}$. removing dissolved minerals: Heating the water promotes the conversion of soluble Ca and Mg bicarbonates to insoluble carbonates ($2 {HCO_3}^-$ → $CO_3^{2-} + H_2O + CO_2$) which will come out of solution (as limescale). The sulphates (sometimes referred to as "permanent hardness"), and the sodium or potassium (bi)carbonates stay in solution. The effect of boiling: Solubility of gas in liquid not only depends on temperature, it is directly proportional to the partial pressure of the gas. When boiling, the gas phase in contact with the water is no longer the air, but the water vapor (in the bubbles and close to the surface). In those bubbles the partial pressure of the gases will be close to zero, so gas molecules will still leave the liquid phase (and the increased surface area and the movement of the water speeds up the process), but hardly any will return. Given sufficient time, the water vapor will remove most of the gas. Boiling is basically the equivalent of degassing by purging: removing a gas (oxygen usually) from a solvent by bubbling an inert gas through it. How do gases make ice "milky/cloudy"? During freezing, the ice layer starts at all sides of the cube and grows inward. Water molecules fit the crystal lattice and will adhere to it, other molecules won't (but if the ice grows faster than the gas molecules can diffuse away, they will get trapped). The concentration of gases (and other impurities) in the remaining liquid rises, the solution becomes supersaturated, microbubbles start forming. All these get trapped in the ice, giving it a milky appearance. | {
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265,966 | An interesting video: youtube video - Railroad tank car vacuum implosion But how is it possible for a metal tank to implode? Even if one can vacuum it, it is subjected to 1 external atmospheric pressure maximum which is around 1 bar or 14.6 psi. How could 14.6 psi crush a metal tank? Please correct me if I am wrong. | Atmospheric pressure is equivalent to supporting a weight of 10 tonnes (about 10 average cars) per metre squared. Put like that, it's not surprising that those metal tanks crumple. However, in the comments you raise the point that you pump your bike tyres to 40 psi (about 3 atm) and yet they don't explode. I think this gets to the crux of your confusion. The crumpling of that tank involves the metal undergoing some fairly insignificant bending . The energy required to bend metal (which, by the way, is naturally pretty soft) is not that great. Whereas the energy required to rip it (or your bike tyres) apart requires significantly more energy. This is because the former case involves simply relocating atomic-scale dislocations while the latter involves breaking atomic bonds. Two very different processes. To illustrate this point, it's easy to compress an empty can of coke. But do you think you could tear a hole in the can? Indeed, if you reversed the situation so that the tank contained 1 atm of pressure, and the outside was instead the vacuum, then it would not explode. | {
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265,974 | I want to calculate the intensity/strength of vibration at a given location. I have measured the acceleration at this location, using an accelerometer. So my measures look for example like: t1 - x:-3.81 y:-10.13 z:5.82
t2 - x:-5.81 y:-9.13 z:3.82
t3 - ... These are measures along the x,y,z axis in m/s^2. My assumption is that the actual strength of the vibration at t1 can be calculated by calculating the distance in 3d space from (0,0,0). I calculate the distance using the Pythagorean theorem for 3 dimensional space. To get the average intensity I just take a lot (e.g. 1.000 .000) measurements and calculate the average of the distances in 3d space of these measures. Is this correct/valid? Are there better ways to do this? Would it be better to use the Jerk?
If I use the Jerk, how would I calculate it? Using the distance in 3D seems to be wrong, since the distance is always positive and is not associated with any direction. So I am not sure if it would be correct. Edit As I was asked in the comments about my usecase:
I want to measure the vibration intensity of a vehicle at different positions of the vehicle. In the end, I want to find the sweet spot of the vehicle, where the least vibration occurs. For example I would want to mount a vibration sensitive device there or add a driver seat to that position to reduce the vibration for the driver. Edit2 This is a plot of approx 1400 samples. I added the 3 axis and the distance in 3d space. The distance is green, z is yellow, y is orange, x is blue. Edit3 Ok so I just created an FFT analysis of the above data. Attached are 2 different plots. One for the 3d distance ( left ) and one for only the x-axis ( right ). Essentially the results tell me that the strongest acceleration occurs in the low frequencies, right? I should definitely get finer grained data. BTW: I used the following function in R plot.frequency.spectrum(fft(accMeasures$x0), xlimits=c(0,1500)) the function itself can be found here . Edit4 This is the normalized FFT Plot of the sample magnitude (left) and the x-axis (right) of my actual samples here: So I tested my accelerometer by hanging it to a rubber band and let it bounce. I sampled with ~100Hz. You can see two spike at 20Hz and 100Hz using the FFT analysis. When I "subtract from every value the calculated average" (I will use the word normalize for that - I hope that is correct). When I compare the non-normalized magnitude (3d distance) to the normalized magnitude, the frequency spikes change from 20 and 100Hz to 40 and 80Hz. This seems to be weird, since the spikes for every axis on its' own are at 20 and 100Hz.
This plot shows the x-axis (the axis the main movement of the rubber band occurred). On the left you can see the non-normalized FFT and on the right you can see the normalized FFT. This looks like what I would expect from normalizing the values. This plot shows the magnitude, left is non-normalized, right is normalized. The change of frequency spikes is weird IMHO. | Atmospheric pressure is equivalent to supporting a weight of 10 tonnes (about 10 average cars) per metre squared. Put like that, it's not surprising that those metal tanks crumple. However, in the comments you raise the point that you pump your bike tyres to 40 psi (about 3 atm) and yet they don't explode. I think this gets to the crux of your confusion. The crumpling of that tank involves the metal undergoing some fairly insignificant bending . The energy required to bend metal (which, by the way, is naturally pretty soft) is not that great. Whereas the energy required to rip it (or your bike tyres) apart requires significantly more energy. This is because the former case involves simply relocating atomic-scale dislocations while the latter involves breaking atomic bonds. Two very different processes. To illustrate this point, it's easy to compress an empty can of coke. But do you think you could tear a hole in the can? Indeed, if you reversed the situation so that the tank contained 1 atm of pressure, and the outside was instead the vacuum, then it would not explode. | {
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265,979 | The work done in thermodynamic process is given by the integral of Pdv and also we can write so assuming a quasi static process between two points. But this work is then non dissipative work or work done by conservative force field , then how can it be path dependent ? | Atmospheric pressure is equivalent to supporting a weight of 10 tonnes (about 10 average cars) per metre squared. Put like that, it's not surprising that those metal tanks crumple. However, in the comments you raise the point that you pump your bike tyres to 40 psi (about 3 atm) and yet they don't explode. I think this gets to the crux of your confusion. The crumpling of that tank involves the metal undergoing some fairly insignificant bending . The energy required to bend metal (which, by the way, is naturally pretty soft) is not that great. Whereas the energy required to rip it (or your bike tyres) apart requires significantly more energy. This is because the former case involves simply relocating atomic-scale dislocations while the latter involves breaking atomic bonds. Two very different processes. To illustrate this point, it's easy to compress an empty can of coke. But do you think you could tear a hole in the can? Indeed, if you reversed the situation so that the tank contained 1 atm of pressure, and the outside was instead the vacuum, then it would not explode. | {
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266,246 | Problems The question I am about to make is either too stupid or hasn't bothered anyone because its obvious because I can't really find the answer anywhere. I am currently studying electricity and magnetism and my book starts by telling that matter consists of atoms that are made of a heavy positively charged nucleus and a cloud of light negatively charged electrons orbiting around it and it bases the rest of the book on these facts. However electricity, positive and negative charges were studied before the atomic theory was confirmed. Actually at that time it wasn't even remotely supported. Benjamin Franklin for example thought that electrical charge was some kind of cloud/gas. When there was abundance of this cloud the charge was positive while otherwise it was negative. Questions Firstly, I suppose that classically we define charge as the physical property that objects have to have in order to interact electrically - at least for now. How did physicists back then know about the existence of two charges, positive and negative? Sure, if you bring two glass rods close after rubbing them with silk they repel each other, while if you bring glass rod and one plastic rod they attract each other, but is that really enough? Moreover, how did they know that opposite charges attract while same charges repel each other? You can't arbitrarily choose to be so, since electrons really do repel each other. Was there a way to tell that a glass rod had abundance of that cloud after the rubbing while the plastic had a deficit? In my book it is said that Franklin, after a series of experiments, determined that there are two kinds of charges without elaborating. I cannot find those experiments anywhere. All I get is about the famous kite experiment. | Get together a collection of charges. As many different ways to generate a charge as you can think of. Go ahead and invite your friends so they can think of some more. (As a practical matter you make static charges just before you use them, but still...) Now, test them pair wise to see if they attract or repel one-another. Keep careful records. Find the largest set that are all mutually attractive and the largest set that are all mutually repulsive. You'll find that the attractive set has exactly two members (though you can make many different examples of this set) and the repulsive set consists of half (either half!) the charges you've created. Ponder that for a while. It also gives you the answer to how like charges respond to one another (though you can get that directly by preparing two similar charges). | {
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266,308 | Recently (in the last week or two), various articles about pear shaped nuclei have appeared, such as this one from Science Alert and this from the BBC The Science Alert article includes the quote We've found these nuclei literally point towards a direction in space. This relates to a direction in time, proving there's a well-defined direction in time and we will always travel from past to present. and further on, So what does all of this have to do with time travel? It's a pretty out-there hypothesis, but Scheck says that this uneven distribition of mass and charge causes Barium-144's nucleus to 'point' in a certain direction in spacetime, and this bias could explain why time seems to only want to go from past to present, and not backwards, even if the laws of physics don't care which way it goes. They've also linked to a copy of the paper on the Arxiv . The paper's way over my head so I couldn't do much more than scan, but no mention of time travel seems to be mentioned there. Do these assertions about time travel make sense? And if so, why do they make sense? What does the direction which the nucleus points in have to do with the "direction of time"? | To be honest, much of this feels like very irresponsible journalism, partly on the part of the BBC and very much so on the part of Science alert. If you're looking for an accessible resource to what the paper does, the cover piece on APS Physics and the phys.org piece are much more sedate and, I think, much more commensurate with what's actually reported in the paper. The paper itself is very moderate in its claims and it restricts itself very well, from what I can tell, to what they found: that certain radium and barium nuclei appear to be pear-shaped. Finding pear-shaped nuclei is not new (a similar paper ( eprint ) made the news in 2013, and was discussed on this site here ), though Bucher et al. seem to have found hints of a discrepancy with theory with regard to just how pear-like these nuclei look like. This is, however, not at the level of statistical significance that would require any rethinking of the theory at this time. It is important to note that pear-shaped nuclei are indeed consistent with the Standard Model of particle physics. Pear-shaped nuclei are a bit of a problem because their shape has a direction , that is, you can draw a vector that starts at the flat end and points toward the pointy end (call this the pear vector $\vec P$). (The alternative, a rugby-ball-shaped nucleus, has an axis, but no preferred direction on this axis.) Because of symmetry considerations, this pear vector $\vec P$ needs to be on the same axis as the nucleus' spin $\vec S$, but these symmetries don't tell you which way they have to point, so you get two different versions of the same nucleus: Mathematica code for this image through Import[" http://goo.gl/NaH6rM "][" http://i.stack.imgur.com/HLcYp.png "], CC BY-SA with attribution to this page. In a theory of nuclear physics that is mirror-symmetric, then both of these nuclei need to be completely equivalent (and, particularly, have the same energy), because they are mirror images of each other. What this paper finds (and what Gaffney et al. found in 2013) is that there are nuclei where this is not true, and these two nuclear states have different energies: the ground state is the "pear" state, and not the "anti-pear" one. Fortunately, this is not a problem: in fact, we've known since 1956 that nuclear physics is not mirror symmetric, i.e. it is not invariant under the parity operator $P$. Fortunately, though, there is a related symmetry that takes up the slack, and it is charge conjugation symmetry $C$, which takes matter to antimatter and vice versa. Much of the Standard Model, including a lot of nuclear physics experiments, is $CP$ symmetric: if you take a mirror version of the experiment, and on top of that you swap out all particles for their antiparticles, then the physics is the same. However, $CP$ violations are still compatible with the Standard Model and have already been observed experimentally. On the other hand, the known $CP$ violations are not really enough to explain the matter-antimatter balance in the universe (a.k.a. the baryogenesis puzzle), which is not explained by the Standard Model, so any beyond-the-SM $CP$ violation is a good place to look for solutions to the baryogenesis problem. There is also a bigger, stronger symmetry, which happens when you combine $CP$ inversion with time inversion $T$ to get what's called the $CPT$ transformation. Because of very basic facts about spacetime , all reasonable physical theories must be $CPT$ symmetric. This is one of the reasons $CP$ violations are so interesting: they point to microscopic violations of time-inversion symmetry. So how does the paper at hand relate to all these generalities? The authors have confirmed the existence of $P$ violations, already observed, and they have found hints that these violations - the peariness of the pear-shaped nucleus are stronger than the existing theory. However, they are not comparing against ab-initio theory (i.e. they compare against approximate theoretical models, so the fault could be in the approximations they made) and, to quote from the paper's discussion, the large uncertainty on the present result does not allow one to elaborate further. So how did we get from there to time travel? That's where you need a large amount of journalistic 'creativity' for the joins to work. The APS Physics piece is clear and to the point, and it makes a good show of understanding the limitations of the paper. On the other hand, the press release , from the University of West Scotland, is already pretty breathless. They quote at length from (presumably a direct interview with) Dr. Scheck, but I think his claim that Further, the protons enrich in the bump of the pear and create a specific charge distribution in the nucleus, which shouldn’t be there according to our currently accepted model of physics. is stretching a fair bit the pretty modest claims of the paper (cf. supra ). Dr. Scheck goes on to claim that We’ve found these nuclei literally ‘point’ towards a direction in space. This relates to a direction in time, proving there’s a well-defined direction in time and we will always travel from past to present. but, same as John Rennie, I'm struggling to see how he connects the (known, possibly stronger than accepted) $P$ violation they found to a confirmation of a $T$ violation, let alone relating that to time travel. Onwards from this, the BBC piece offers very little above the UWS press release, and this is a pretty bad sign - while the BBC does often have great science content, this piece is essentially a redigested press release with the hype turned up one notch, and I can't resist pointing out that things like this have been pointed out as one cause of the bigger problems science has at the moment. The Science Alert piece (also syndicated at RedOrbit and Business Insider), on the other hand, goes a bit further along on the dodgy journalism side. In particular, it misidentifies the Scheck interview as with the BBC instead of a press release, and it quotes from a couple of stories from last year, but makes it sound like they are reaction quotes from experts interviewed about the recent paper. Other than that, it feels like a re-re-digest piece with the hype turned up three notches. If you want the hype, then, go to the press release - that's its job. If you want a sober assessment of the implications of the work, go to the APS Physics piece or the phys.org one , which make it clear that there are few implications yet beyond nuclear physics - if indeed the result stands the test of an accurate measurement. | {
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266,426 | I'm curious as to what the Moon's orbit around the Sun looks like. If there's an answer, what's the intuition for it? Here are some things I'm assuming when trying to tackle this question: The Moon's orbit must be concave toward the Sun. The Moon speeds up as it goes toward the Sun, and it slows down as it moves away. For an observer on the Earth, the Moon appears to orbit the Earth roughly $13$ times a year. The Earth, the Moon, and the Sun remain in the same plane. | Incorrect Path I'm curious as to what does the moon's orbit around the sun looks like? One might think the orbit (in the sun's rest frame) follows the path of an epitrochoid . A (very) over exaggerated view of this motion (for unrealistic parameters, thus, not an accurate representation) can be seen in the following animation: Note that if you change the ratio of the different radii to values that are to scale, then the plot would look more like an epicycloid as in the following example animation. The orbit is more realistic but still exaggerated because it would be impossible to show half an orbit to scale. The correct result is shown in the zoomed-in view of David Hammen's post above. Update/Correction The above animations are flawed because the little red dot orbits as fast as the blue circle rotates as it "rolls" around the large red circle with no slippage. For a realistic Earth-moon system , there should be a lag between the rotation of the blue circle and the red dot, as if the blue circle were "slipping." Or equivalently, one would not use a rigid axis connecting the center of the blue circle and the red dot. This would result in there never being a negative velocity of the moon relative to the sun in the fashion shown by the epicycloid path in the 2nd animation above. Incorrect (still exaggerated) Path The correct path uses two different orbit rates, one for the Earth about the sun (i.e., 1 year) and one for the Moon about the Earth (i.e., ~27 days). In the following example, which is still an incorrect (i.e., exaggerated) orbital motion but much better approximation, I exaggerated the ratio of the astronomical unit to Earth radius by a factor of 100 and increased the moon's orbit by a factor of four to help make the visualization more obvious. Correct Path The following example does not exaggerate the orbital periods relative to each other and is a zoom-in of the above graphic (oddly the GIF, created through the same methods, does not loop on my screen). Here the orbit is always convex with respect to the sun, as David has correctly stated. | {
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266,492 | A rather stupid question, why can we see the moon when it is between the Earth and the Sun? | The sun doesn't just illuminate the moon directly . The moon is also illuminated by sunlight reflected from the earth. This is called earthshine . This makes the parts of the moon that face us visible even when the sun is on the other side. According to NASA , it was Leonardo da Vinci who first explained this. As an example, the brightly lit portion of this photo is illuminated directly by the sun. The rest of the moon, though, is still visible and this is due to earthshine: Image credit: Steve Jurvetson | {
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266,834 | In equations that have quantities with physical dimension. Example: $\mathrm{Force} = (\mathrm{mass})(\mathrm{acceleration})$ or $F=ma$ I know that we use that (mass, force...) to help what we should use in the equation. But, why don't we just use $N = \mathrm{kg}\cdot \mathrm{m}/\mathrm{s}^2$ or $\mathrm{Newton} = (\mathrm{kilograms})(\mathrm{meter})/ \mathrm{seconds}^2$ as the equation? Can any equation be wrong by doing that? I know that some equations are used in specific places/ranges/situations. But even with words instead of units in equations, we need to know the application boundaries of that equation. | A specific parameter might correspond to a specific (SI) unit, but not all units correspond to a specific parameter ! Kinetic energy is $$\begin{align}
K&=\frac{1}{2} mv^2 \\
[\text{Joules}]&=\frac{1}{2}[ \text{kilograms}\times\text{meters}^2/\text{seconds}^2]
\end{align}$$ We also have gravitational potential energy : $$\begin{align}U&=mgh \\
[\text{Joules}] &= [\text{kilograms} \times(\text{meters} / \text{seconds}^2) \times \text{meters}]\\
&= [\text{kilograms} \times \text{meters}^2 / \text{seconds}^2]
\end{align}$$ So, is Joules both $\frac{1}{2} \text{kilograms}\times\text{meters}^2/\text{seconds}^2$ and $ \text{kilograms}\times\text{meters}^2/\text{seconds}^2$ at the same time? If you have a value in Joules and you need to find the number of kilograms, then how would you go backwards? How would you do the algebra? You could start from any of these unit-formulations, and you would get difference answers for the number of kilograms. The answer is not unique seen from the units since the original formula could have contained unit-less parameters. The problem is that there are many kinds of energy with the same unit. In general, parameters have unique units, but units don't belong to unique parameters. You cannot go "backwards" from the unit formulation of a formula. | {
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266,992 | I am a mathematics student with a hobby interest in physics. This means that I've taken graduate courses in quantum dynamics and general relativity without the bulk of undergraduate physics courses and sheer volume of education into the physical tools and mindset that the other students who took the course had, like Noether's theorem, Lagrangian and Hamiltonian mechanics, statistical methods, and so on. The courses themselves went well enough. My mathematical experience more or less made up for a lacking physical understanding. However, I still haven't found an elementary explanation of gauge invariance (if there is such a thing). I am aware of some examples, like how the magnetic potential is unique only up to a (time-)constant gradient. I also came across it in linearised general relativity, where there are several different perturbations to the spacetime metric that give the same observable dynamics. However, to really understand what's going on, I like to have simpler examples. Unfortunately, I haven't been able to find any. I guess, since "gauge invariance" is such a frightening phrase, no one use that word when writing to a high school student. So, my (very simple) question is: In many high school physics calculations, you measure or calculate time, distance, potential energy, temperature, and other quantities. These calculations very often depend only on the difference between two values, not the concrete values themselves. You are therefore free to choose a zero to your liking. Is this an example of gauge invariance in the same sense as the graduate examples above? Or are these two different concepts? | The reason that it's so hard to understand what physicists mean when they talk about "gauge freedom" is that there are at least four inequivalent definitions that I've seen used: Definition 1: A mathematical theory has a gauge freedom if some of the mathematical degrees of freedom are "redundant" in the sense that two different mathematical expressions describe the exact same physical system. Then the redundant (or "gauge dependent") degrees of freedom are "unphysical" in the sense that no possible experiment could uniquely determine their values, even in principle. One famous example is the overall phase of a quantum state - it's completely unmeasurable and two vectors in Hilbert space that differ only by an overall phase describe the exact same state. Another example, as you mentioned, is any kind of potential which must be differentiated to yield a physical quantity - for example, a potential energy function. (Although some of your other examples, like temperature, are not examples of gauge-dependent quantities, because there is a well-defined physical sense of zero temperature.) For physical systems that are described by mathematical structures with a gauge freedom, the best way to mathematically define a specific physical configuration is as an equivalence class of gauge-dependent functions which differ only in their gauge degrees of freedom. For example, in quantum mechanics, a physical state isn't actually described by a single vector in Hilbert space, but rather by an equivalence class of vectors that differ by an overall scalar multiple. Or more simply, by a line of vectors in Hilbert space. (If you want to get fancy, the space of physical states is called a "projective Hilbert space," which is the set of lines in Hilbert space, or more precisely a version of the Hilbert space in which vectors are identified if they are proportional to each other.) I suppose you could also define "physical potential energies" as sets of potential energy functions that differ only by an additive constant, although in practice that's kind of overkill. These equivalence classes remove the gauge freedom by construction, and so are "gauge invariant." Sometimes (though not always) there's a simple mathematical operation that removes all the redundant degrees of freedom while preserving all the physical ones. For example, given a potential energy, one can take the gradient to yield a force field, which is directly measurable. And in the case of classical E&M, there are certain linear combinations of partial derivatives that reduce the potentials to directly measurable ${\bf E}$ and ${\bf B}$ fields without losing any physical information. However, in the case of a vector in a quantum Hilbert space, there's no simple derivative operation that removes the phase freedom without losing anything else. Definition 2: The same as Definition 1, but with the additional requirement that the redundant degrees of freedom be local . What this means is that there exists some kind of mathematical operation that depends on an arbitrary smooth function $\lambda(x)$ on spacetime that leaves the physical degrees of freedom (i.e. the physically measurable quantities) invariant. The canonical example of course is that if you take any smooth function $\lambda(x)$, then adding $\partial_\mu \lambda(x)$ to the electromagnetic four-potential $A_\mu(x)$ leaves the physical quantities (the ${\bf E}$ and ${\bf B}$ fields) unchanged. (In field theory, the requirement that the "physical degrees of freedom" are unchanged is phrased as requiring that the Lagrangian density $\mathcal{L}[\varphi(x)]$ be unchanged, but other formulations are possible.) This definition is clearly much stricter - the examples given above in Definition 1 don't count under this definition - and most of the time when physicists talk about "gauge freedom" this is the definition they mean. In this case, instead of having just a few redundant/unphysical degrees of freedom (like the overall constant for your potential energy), you have a continuously infinite number. (To make matters even more confusing, some people use the phrase "global gauge symmetry" in the sense of Definition 1 to describe things like the global phase freedom of a quantum state, which would clearly be a contradiction in terms in the sense of Definition 2.) It turns out that in order to deal with this in quantum field theory, you need to substantially change your approach to quantization (technically, you need to "gauge fix your path integral") in order to eliminate all the unphysical degrees of freedom. When people talk about "gauge invariant" quantities under this definition, in practice they usually mean the directly physically measurable derivatives, like the electromagnetic tensor $F_{\mu \nu}$, that remain unchanged ("invariant") under any gauge transformation. But technically, there are other gauge-invariant quantities as well, e.g. a uniform quantum superposition of $A_\mu(x) + \partial_\mu \lambda(x)$ over all possible $\lambda(x)$ for some particular $A_\mu(x).$ See Terry Tao's blog post for a great explanation of this second sense of gauge symmetry from a more mathematical perspective. Definition 3: A Lagrangian is sometimes said to posses a "gauge symmetry" if there exists some operation that depends on an arbitrary continuous function on spacetime that leaves it invariant, even if the degrees of freedom being changed are physically measurable. Definition 4: For a "lattice gauge theory" defined on local lattice Hamiltonians, there exists an operator supported on each lattice site that commutes with the Hamiltonian. In some cases, this operator corresponds to a physically measurable quantity. The cases of Definitions 3 and 4 are a bit conceptually subtle so I won't go into them here - I can address them in a follow-up question if anyone's interested. Update: I've written follow-up answers regarding whether there's any sense in which the gauge degrees of freedom can be physically measurable in the Hamiltonian case and the Lagrangian case . | {
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267,035 | The conventional way physicists describe neutrinos is that they have a very small amount of mass which entails they are traveling close to the speed of light. Here's a Wikipedia quote which is also reflected in many textbooks: It was assumed for a long time in the framework of the standard model of particle physics, that neutrinos are massless. Thus they should travel at exactly the speed of light according to special relativity. However, since the discovery of neutrino oscillations it is assumed that they possess some small amount of mass. 1 Thus they should travel slightly slower than the speed of light... -- Wikipedia (Measurements of Neutrino Speed) Taken at face value, this language is very misleading. If a particle has mass (no matter how small), its speed is completely relative, and to say that neutrinos travel close to the speed of light, without qualification, is just as incorrect as saying electrons or billiard balls travel close to the speed of light. So what is the reason everyone repeats this description? Is it because all the neutrinos we detect in practice travel close to the speed of light? If so, then I have this question: Neutrinos come at us from all directions and from all sorts of sources (stars, nuclear reactors, particle accelerators, etc.), and since they have mass, just like electrons, I would have thought we should see them traveling at all sorts of speeds. (Surely some cosmic neutrino sources are traveling away from the earth at very high speeds, for example. Or what about neutrinos emitted from particles in accelerators?) So like I said at the start: Where are all the slow neutrinos? And why do we perpetuate the misleading phrase: 'close to the speed of light' (i.e. without contextual qualification)? | Strictly speaking, it is indeed incorrect that neutrinos travel at "close to the speed of light". As you said, since they have mass they can be treated just like any other massive object, like billiard balls. And as such they are only traveling at nearly the speed of light relative to something. Relative to another co-moving neutrino it would be at rest. However, the statement is still true for almost all practical purposes. And it doesn't even matter in which reference frame you look at a neutrino. The reason is that a non-relativistic neutrino doesn't interact with anything. Or in other words: all the neutrinos you can detect necessarily have to have relativistic speeds. Let me elaborate. Since neutrinos only interact weakly they are already extremely hard to detect, even if they have high energies (> GeV). If you go to ever lower energies the interaction cross-section also decreases more and more. But there is another important point. Most neutrino interaction processes have an energy threshold to occur. For example, the inverse beta decay $$ \bar\nu_e + p^+ \rightarrow n + e^+$$ in which an antineutrino converts a proton into a neutron and a positron, and which is often used as a detection process for neutrinos, has a threshold of 1.8 MeV antineutrino energy. The neutron and the positron are more massive than the antineutrino and the proton, so the antinneutrino must have enough energy to produce the excess mass of the final state (1.8 MeV). Below that energy the (anti)neutrino cannot undergo this reaction any more. A reaction with a particularly low threshold is the elastic scattering off an electron in an atom. This only requires a threshold energy of the order of eV (which is needed to put the electron into a higher atomic energy level). But a neutrino with eV energies would still be relativistic! Assuming that a neutrino has a mass of around 0.1 eV, this would still mean a gamma factor of $\gamma\approx 10$. For a neutrino to be non-relativistic it would have to have a kinetic energy in the milli-eV range and below. This is the expected energy range of Cosmic Background Neutrinos , relics from the earliest times of the universe. They are so to say the neutrino version of the Cosmic Microwave Background. So not only do non-relativistic neutrinos exist (according to mainstream cosmological models), they are also all around us. In fact, their density at Earth is $\approx$50 times larger than neutrinos from the Sun! There is a big debate if they can ever be detected experimentally. There are a few suggestions (and even one prototype experiment ), but there are differing opinions about the practical feasibility of such attempts. The only process left for neutrinos at such small energies is neutrino-induced decay of unstable nuclei . If you have an already radioactive isotope, it's like the neutrino would give it a little "push over the edge". The $\beta$-electron released in the induced decay would then receive a slightly larger energy than the Q-value of the spontaneous decay and the experimental signature would be a tiny peak to the right of the normal $\beta$-spectrum. This will still be an extremely rare process and the big problem is to build an apparatus with a good enough energy resolution so that the peak can be distinguished from the spectrum of normal spontaneous nuclear decay (amidst all the background).
The Katrin experiment is trying to measure the endpoint of $\beta$-spectrum of Tritium in order to determine the neutrino mass. But under very favorable circumstances they even have some chance to detect such a signature of cosmic background neutrinos. TL;DR: In fact there are non-relativistic neutrinos all over the place, but they they interact so tremendously little that they seem to not exist at all. | {
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267,091 | A nucleus is very small and very dense. Its density is approximately $2.3 \times 10^{17}~\mathrm{kg/m^3}.$ So why can't a nucleus itself become a black hole? | Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high. The Schwarzschild radius of a black hole is given by: $$ r_s = \frac{2GM}{c^2} $$ and for a mass of $1.99 \times 10^{-26}$ kg this gives us: $$ r_s = 2.95 \times 10^{-53} \,\text{m} $$ This is far below the Planck length , so it is unlikely matter could be squeezed into a region that small i.e. a single carbon nucleus cannot form a black hole. If we take the Planck length as $r_s$ and calculate the associated black hole mass the result is half the Planck mass $\tfrac{1}{2}\sqrt{hc/2\pi G}$, which is about $11\,\mu\text{g}$ or about $5.5 \times 10^{17}$ times larger than the mass of the carbon nucleus. This is the smallest mass that we expect could form a black hole. | {
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267,196 | I was in doubt, so I went to wikipedia . There it says "the photon has zero rest mass", but on the side description it says the mass is $<1.10^{-18} \:\mathrm{eV}/c^2$. So is the mass of the photon really zero or do we just consider it to be zero because it's negligible? More generally, are there massless particles at all? | Here is a quick & simple answer until the professionals arrive. In the Standard Model , it is zero. This $< 1\cdot 10^{-18} \frac{\mathrm{eV}}{c^2}$ is an experimental upper limit (i.e. if it has a rest mass, because of physics beyond the Standard Model, it must be smaller than this value). This value is very small, compared to the estimated rest mass of the neutrinos (which is of the order of some tenths of an $\mathrm{eV}$). | {
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267,208 | I was soldering a very thin wire today, and when I had one end firmly soldered, I accidentally bumped the wire diagonally with my tweezers. What I'd expect to happen is that the wire oscillates for a little while in one axis, then stops. However, what actually occurred is quite different and much more interesting! I recorded it in real-time; https://youtu.be/O5nFNly7L7s (sorry for the poor macro focus), and recorded it again at 480FPS and imported it into Tracker video analysis; https://youtu.be/9jhDsypkqKQ . As you can see, the rotational motion fully reverses! Here are some still frames from Tracker: The wire begins to rotate clockwise after being excited: The wire begins to oscillate in one axis: And, mindbogglingly, begins to rotate counterclockwise! (clearer views in the videos above) The X and Y axis motion plotted by tracker raises even more questions: As you can see, the X axis motion simply stops, then restarts! What's going on? My first thought was that (because this wire wasn't originally straight) there was some sort of unusual standing wave set up, but this occurs even with a completely straight wire. I'm absolutely sure there's something about two-axis simple harmonic motion that I'm missing, but I just cannot think of what is causing this. I've seen many other "home-experiment" questions on this site, so I thought this would an acceptable question; I hope it's not breaking any rules. EDIT: Okay, I've got some more data! I've set up a little solenoid plunger system that produces no torque or two-axis motion, and it's very repeatable. Here: https://youtu.be/ZAni6VMOVD8 What I've noticed is that I can get almost any wire (even with a 90-degree bend!) to exhibit single-axis motion with this setup, with no spinning or deviation; and if I try enough, the same thing can happen with the tweezers. It seems like if I slide the tweezers slightly when exciting the wire, I can reliably produce this odd motion. I don't know what that indicates. EDIT2: Okay, seems like with the plunger-solenoid I still can get this circular motion even with a straight wire. EDIT3: Okay, so I wanted to test @sammy's suggestion once and for all. I assume that changing the moment of inertia to torsion of the wire would affect his theory, so I soldered a small piece of wire perpendicularly to the end of the main wire: Then I recorded the motion; And then I took off the perpendicular wire, and re-recorded the data: And then I did it again (got noisy data first time): EDIT N:
The final test! Floris's hypothesis requires that the resonant frequency of a wire in each cardinal direction be different. To measure this, I used my solenoid setup that did not cause rotation, as above. I put a straight piece of wire between a light source and a light-dependent resistor and connected it to an oscilloscope; The signal was very faint (42 millivolts), but my scope was able to pull it out of the noise. I have determined this: In the +x direction, the resonant frequency of a just-straightened straight sample wire (unknown cycle frequency) is 51.81hz,+/-1hz; In the +y direction, the resonant frequency of a sample wire is 60.60hz,+/-1hz; So there's definitely a significant difference (~10 percent!) between the cardinal directions. Good enough proof for me. EDIT N+1: Actually, since my light detector above produces two pulses per sine wave, the actual vibration frequency is f/2; so the actual frequencies are 25.5 hz, and 30hz, which agrees roughly with @floris's data. | Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes 1 . You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, is exactly what you are showing. The first signal you show - which seems to "die" then come back to life, is exactly what you expect to see when you have two oscillations of slightly different frequency superposed; in fact, from the time to the first minimum we can estimate the approximate difference in frequency: it takes 19 oscillations to reach a minimum, and since the two waves started out in phase, that means they will be in phase again after about 38 oscillations, for a 2.5% difference in frequency. Update Here is the output of my little simulation. It took me a bit of time to tweak things, but with frequencies of 27 Hz and 27.7 Hz respectively and after adjusting the angle of excitation a little bit, and adding significant damping I was able to generate the following plots: which looks a lot like the output of your tracker. Your wire is describing a Lissajous figure. Very cool experiment - well done capturing so much detail! Here is an animation that I made, using a frequency difference of 0.5 Hz and a small amount of damping, and that shows how the rotation changes from clockwise to counterclockwise: For your reference, here is the Python code I used to generate the first pair of curves. Not the prettiest code... I scale things twice. You can probably figure out how to reduce the number of variables needed to generate the same curve - in the end it's a linear superposition of two oscillations, observed at a certain angle to their principal axes. import numpy as np
import matplotlib.pyplot as plt
from math import pi, sin, cos
f1 = 27.7
f2 = 27
theta = 25*pi/180.
# different amplitudes of excitation
A1 = 2.0
A2 = 1.0
t = np.linspace(0,1,400)
#damping factor
k = 1.6
# raw oscillation along principal axes:
a1 = A1*np.cos(2*pi*f1*t)*np.exp(-k*t)
a2 = A2*np.cos(2*pi*f2*t)*np.exp(-k*t)
# rotate the axes of detection
y1 = cos(theta)*a1 - sin(theta)*a2
y2 = sin(theta)*a1 + cos(theta)*a2
plt.figure()
plt.subplot(2,1,1)
plt.plot(t,-20*y2) # needed additional scale factor
plt.xlabel('t')
plt.ylabel('x')
plt.subplot(2,1,2)
plt.plot(t,-50*y1) # and a second scale factor
plt.xlabel('t')
plt.ylabel('y')
plt.show() 1. The frequency of a rigid beam is proportional to $\sqrt{\frac{EI}{A\rho}}$, where $E$ is Young's modulus, $I$ is the second moment of area, $A$ is the cross sectional area and $\rho$ is the density (see section 4.2 of "The vibration of continuous structures" ). For an elliptical cross section with semimajor axis $a$ and $b$, the second moment of area is proportional to $a^3 b$ (for vibration along axis $a$). The ratio of resonant frequencies along the two directions will be $\sqrt{\frac{a^3b}{ab^3}} = \frac{a}{b}$. From this it follows that a 30 gage wire (0.254 mm) with a 2.5% difference in resonant frequency needs the perpendicular measurements of diameter to be different by just 6 µm to give the effect you observed. Given the cost of a thickness gage with 1 µm resolution, this is really a very (cost) effective way to determine whether a wire is truly round. | {
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267,357 | Zero dimensional points do not take up space, so then wouldn't everything in the universe be literally empty? Or is there something that I'm missing? | Although it's commonly said that fundamental particles are point particles you need to be clear what this means. To measure the size of the particle to within some experimental error $d$ requires the use of a probe with a wavelength of $\lambda=d$ or less i.e. with an energy of greater than around $hc/\lambda$ . When we say particles are pointlike we mean that no matter how high the energy of your probe, or how small its wavelength, you will never measure a particle radius greater than your experimental limit $d$ . That is the particle will always appear pointlike no matter how precise your experiment is. But this does not mean that the particles are actually zero dimensional, infinite density, dots whizzing around. An elementary particle does not have a position in the way we think of a macroscopic object as having a position. It is always delocalised to some extent, i.e., it exists across a region of some non-zero volume. More precisely the probability of finding the particle is non-zero anywhere within that region. So an atom is not empty space. The usual analogy is that it is a fuzzy blob, and actually that's a not a bad metaphor. If we take any small volume $\mathrm dV$ within the atom then the probability of finding the electron in that region is given by: $$ P = \int \psi^*\psi\,\mathrm dV $$ where $\psi$ is the wavefunction describing the electron in the atom. And since this probability is just the charge density that means the charge density varies smoothly throughout the atom. It is important to be clear that this is not just some time average due to the electron whizzing about the atom very fast. It is not the case that the electron has a precise position in the atom and our probability is some time average. The electron genuinely has no position in the usual macroscopic sense. | {
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267,492 | This question is inspired by (or a follow-up to) the threads Where are all the slow neutrinos? and Is it possible that all “spontaneous nuclear decay” is actually “slow neutrino” induced? The cosmic neutrino background (CνB) consists of the "primordial" neutrinos from the time when the universe had cooled/expanded enough for the neutrinos to "decouple" and become free. This of course is similar to the cosmic microwave background except that photons decoupled at a much later time (age $1.2\cdot 10^{13}$ s) than neutrinos (age $1$ s). My question is, what would we expect the CνB to look like today, given that we know that neutrinos possess mass. According to the linked threads, the neutrinos would have become non-relativistic today because of the great expansion of space since the time when the universe was only one second old. So will the relic neutrinos from CνB be captured gravitationally by galaxies and stars and similar objects, in a way that today these neutrinos are orbiting galaxy centers, stars etc.? I imagine this is like how most cold/slow hadronic matter (like dust or isolated hydrogen atoms or whatever) will be gravitationally bound to galaxies, stars etc. Because if that is the case, this neutrino "background" will no longer be have a "background" appearance. So what to expect: Is the neutrino background still quite isotropic and homogenous and "background-like", maybe even with a temperature associated to it? Or are those particles now captured by all sorts of objects, and no longer constituting a "background" of cold/slow particles? If one assumes a particular rest mass for each neutrino, say $0.1$ eV, maybe it is even possible to answer this quantitatively? | As far as theory goes, the Cosmic Neutrino Background (CvB) was created within the first second after the Big Bang, when neutrinos decoupled from other matter. Nevertheless, while the universe was still hot neutrinos stayed in thermal equilibrium with photons. Neutrinos and photons shared a common temperature until the universe cooled down to a point where electrons and positrons annihilated and transfered their temperature to the photons. With the continuing expansion of the universe both the photon background and the neutrino background continued to cool down. From these assumptions one can derive the properties of the Cosmic Neutrino Background today. The calculations are neither particularly lengthy or difficult, but I'll skip them here. As a result of these calculations one expects the CvB to have a temperature of $$ T_\nu = 1.95~\mathrm{K} = 1.7\cdot 10^{-4}~\mathrm{eV},$$ an average momentum of $$ \left< p \right> = 5.314 \cdot 10^{-4}~\mathrm{eV},$$ a root mean squared momentum dispersion of $$ \sqrt{\left< p^2 \right>} = 6.044 \cdot 10^{-3}~\mathrm{eV}$$ and a density of $$ 112~\nu/\mathrm{cm}^3 $$ for each of the three neutrino flavors. This density is many orders of magnitude more abundant in that energy range than neutrinos from any other sources. This number is equally divided into neutrinos and antineutrinos. These are rather hard predictions of Big Bang cosmology. This makes the CvB so important: if we could measure it, any deviation of these numbers cited above would mean that there is a serious and fundamental flaw in our cosmological models. However, one has to keep in mind that all these numbers are averaged over the whole universe. Since neutrinos do have a non-zero rest mass, they are indeed affected by gravity. They cannot cluster like Dark Matter, because even though CvB Neutrinos are "slow", they are still too fast (many hundreds of km/s) to form clusters and therefore no viable Dark Matter candidate. But they may form gigantic weakly bound halos around galaxies that go far beyond the Dark Matter clusters. This may lead to a local enhancement of the CvB density due to the gravitational attraction of the massive neutrinos to large-scale structures in the universe. Unfortunately, this density enhancement cannot be quantified yet, because it depends very much on the absolute neutrino mass, which is still unknown today. For a mass of 0.1$~$eV, which you assumed in your question, there would probably be no relevant density enhancement of CvB neutrinos near our galaxy. The neutrinos would be too fast and simply stream out of the gravitational potential. If the neutrino masses turn out to be larger, on the other hand, the effect of gravity can become significant and density enhancement factors of $\approx$100 might be possible. There would also be a "CvB neutrino wind". Just like the Cosmic Microwave Background, the neutrino background is not co-moving with our reference frame. Rather, our galaxy and the Earth are passing through the gigantic cloud of CvB neutrinos, so the neutrino distribution would not appear completely isotropic to us. It would appear a bit blue-shifted in one direction and a bit red-shifted in the other. I would like to emphasize though, that a possible CvB detection experiment would probably not yield much information about the properties of the CvB. The only feasible method conceived to detect CvB neutrinos uses the neutrino induced $\beta$-decay of unstable nuclei. This process mainly provides us a yes/no-answer about the existence of the Cosmic Background Neutrinos. It does not tell us anything about the temperature of the CvB. In principle it would be possible to determine the density (via the rate) or even the anisotropy (via an annular rate modulation), but I doubt that it we could get anything better than the right order of magnitude. What one can determine from neutrino induced $\beta$-decay is the absolute neutrino mass. But this is not a property specific to the CvB and can be measured in other ways, too. | {
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267,852 | Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower? Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast. | It's the second one: the reason the speed $299792458\ \mathrm{m/s} = c$ is special is because it's the universal speed limit. Light always travels at the speed $c$, whatever that limit may be. The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be there. Or to be more precise: if you took the theoretical description of our universe, and remove light in the most straightforward possible way, it wouldn't affect $c$. There are many other things that depend on the speed $c$. A particularly important one is that it's the "speed of causality": one event happening at a particular time and place can't affect another event unless there's a way to get from the first event to the second without exceeding that speed. (This is sort of another way of saying it has to do with the structure of spacetime.) | {
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268,009 | Sound waves travel with constant speed, but air molecules that transfer action move with different speeds than the ones described by Maxwell distribution. Why does the sound wave not smear out and dissipate quickly? UPDATE . It is not just that should ''smear out'', but that it should do it quickly. If we look at distribution of speed of atoms https://upload.wikimedia.org/wikipedia/commons/0/01/MaxwellBoltzmann-en.svg then we will see that there is no steep peak. There is a lot of atoms that are 30-50 percent quicker or slower that the rest. With length of sound wave measuring in 1-10 meter noticeable smear out should happen in tenths and hundreds of a second = tens of meters. And in ultrasonic frequency a lot quicker. | It does, but the effects are negligible in the regions we think about. If you think about a volume of air as a box of atoms bouncing around, you can apply an oscillating pressure gradient across that box and show that it behaves close enough to an ideal wave propagation medium that you can get away with using such an ideal model. The variations you are looking at "smooth out" on a timescale much shorter than the timescale of the sound wave being transmitted. This is a case where the central limit theorem is quite useful - you can basically show that the variance of the statistical medium you are thinking of is sufficiently negligible when occurring over the timescales we think of when we think of sound waves. That's not to say the effects you are thinking about don't occur, just that they are small enough compared to other effects that we can get away with handwaving them away and still have a useful predictive model left over. The term used for this is "relaxation." The assumption is that the stochastic system you bring up "relaxes" fast enough compared to the behaviors we care about that we don't have to concern ourselves with those details. The random behaviors obscure any information that might have been held in the exact structure of the medium. All that is left is a homogeneous system which, because of the central limit theorem and large number of particles, behaves almost as an ideal wave propagation medium. This assumption is not always valid. There are times where you need to use a more complete model, which includes the statistical model of the air molecules. One particular case where we have to do this is when dealing with objects that approach the speed of sound. As you approach the speed of sound, the assumption that the stochastic effects are on a short enough timescale that we can ignore them starts to fall apart. The timescale of the events we care about start to get closer to the relaxation time of the stochastic system of particles. Now we have to account for the sorts of effects you are looking at, because they have a substantial effect. Now we start seeing behaviors like shock waves which never appeared at lower speeds. We also have to start considering more complete models when dealing with very loud sounds. Once a sound gets above 196dB, you cannot use the nice simple ideal wave propagation formulas because the low-pressure side of the wave is so low that you get a 0atm vacuum. Modeling this correctly requires including effects that were not in the simple model we use every day for normal volume sounds at normal speeds. | {
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268,062 | Earth's orbit is a slight ellipse, so to conserve momentum its speed increases when it is closest to the Sun. If the speed changes there is an acceleration. If there is an acceleration there is a force. Even if the change is small and gradual, wouldn't we experience a force because the Earth is so massive? | We don't feel any acceleration because the Earth and all of us humans on it is in free fall around the Sun. We don't feel the centripetal acceleration any more than the astronauts on the ISS feel the acceleration of the ISS towards the Earth. This happens because of the way general relativity describes motion in gravitational field. The motion of a freely falling object is along a line called a geodesic , which is basically the equivalent of a straight line in curved spacetime. And because the freely falling object is moving in a straight line it experiences no force. To be a bit more precise about this, the trajectory followed by a freely falling object is given by the geodesic equation : $$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$ Explaining what this means is a bit involved, but actually we don't need the details. All we need to know is that the four-acceleration of a body $\mathbf A$ is given by another equation: $$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{2} $$ But if use equation (1) to substitute for $d^2x^\alpha/d\tau^2$ in equation (2) we get: $$ A^\alpha = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu = 0 $$ So for any freely falling body the four acceleration is automatically zero. The acceleration you feel, the "g force", is the size of the four-acceleration - technically the norm of the four-acceleration or the proper acceleration . Nothing in this argument has referred to the shape of the orbit. Whether the orbit is hyperbolic, parabolic, elliptical or circular the same conclusion applies. The orbitting observer experiences no acceleration. You might be interested to read my answer to How can you accelerate without moving? , where I discuss this in a bit more detail. For an even more technical approach see How does "curved space" explain gravitational attraction? . | {
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268,454 | I'm a highschool sophomore, bear this is mind when answering this question, in other words, the answer doesn't need to be in total layman terms, but it should be understandable by an applied highschool student. I have no idea what a standing wave is. Please, explain? Bonus question: How is a standing wave related to the atomic orbit? It is my understanding that the atomic orbit is a mathematical function that describes the probability of an electron being at a certain place, but it is also the image of this function in terms of real space, (i.e. the actual 3 dimensional volume around the nucleus that a particular electron calls "home"). | An animation is worth a million words: | {
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268,456 | I found some theoretical questions in an old test of my school, and would like some help with the ones I cant answer: It is said that the electric field is an intermediary of the electric interaction. What does this mean? Why the participation of this intermediate is necessary and can not establish direct interaction between charges ( as supposed by Coulomb ) ? Electrostatic field lines are open. How can you establish a relation between this and the fact that electric force is conservative? If a charge is placed inside a non-uniform electrostatic field, will it move alongside the field lines? (I think this is true, because the electric field is parallel to the electric force which is what causes the movement state variation, but I'm not sure...) | An animation is worth a million words: | {
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268,568 | I was trying to figure out which piano keys were being played in an audio recording using spectral analysis, and I noticed that the harmonics are not integer multiple of the base note. What is the reason for this? Take a look at the spectrogram of a clean sample from a single piano key. I am using Piano.ff.A4 from here . The following is the same as above, with a superimposed reference grid of $ 440 ~\mathrm{Hz}$. As you can see, the harmonics have increasingly higher frequencies than integer multiples of $440 ~\mathrm{Hz}$. At this point you might think that the actual base frequency is just slightly higher than $440 ~\mathrm{Hz}$. So let us make a different reference grid, which lines up with the harmonic at ~$5060 ~\mathrm{Hz}$. You can now clearly see that they aren't actually integer multiples of a base frequency. Question: What is the explanation for this? I am looking both for simple high-level explanations of what is happening, and more in-depth, instrument specific ones, which could maybe allow me to attempt to calculate the harmonics. My first reaction was that this must be some non-linear effect. But you can see that the harmonics do not change frequency at all as time passes and the sound gets quieter. I would expect a non-linear effect to be pronounced only in the loudest part of the sample. Update – I measured the frequencies using peak detection on the Fourier transform from 0.3 to 0.4 seconds in the sample. This table compares the measured values with integer multiples of 440: meas. int. mult.
440. 440.
880. 880.
1330. 1320.
1780. 1760.
2230. 2200.
2680. 2640.
3140. 3080.
3610. 3520.
4090. 3960.
4570. 4400.
5060. 4840.
5570. 5280. | This effect is known as inharmonicity, and it is important for precision piano tuning. Ideally, waves on a string satisfy the wave equation $$v^2 \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}.$$ The left-hand side is from the tension in the string acting as a restoring force. The solutions are of the form $\sin(kx - \omega t)$ , where $\omega = kv$ . Applying fixed boundary conditions, the allowed values of the wavenumber $k$ are integer multiples of the lowest possible wavenumber, which implies that the allowed frequencies are integer multiplies of the fundamental frequency. This predicts evenly spaced harmonics. However, piano strings are made of thick wire. If you bend a thick wire, there's an extra restoring force in addition to the wire's tension, because the inside of the bend is compressed while the outside is stretched. One can show that this modifies the wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$ Upon taking a Fourier transform, we have the nonlinear dispersion relation $$\omega = kv \sqrt{1 + (A/v^2)k^2}$$ which "stretches" evenly spaced values of $k$ into nonuniformly spaced values of $\omega$ . Higher harmonics are further apart. We can write this equation in terms of the harmonic frequencies $f_n$ as $$f_n \propto n \sqrt{1+Bn^2}$$ which should yield a good fit to your data. Note that the frequencies have no dependence on the amplitude, as you noted, and this is because our modified wave equation is still linear in $y$ . This effect must be taken into account when tuning a piano, since we perceive two notes to be in tune when their harmonics overlap. This results in stretched tuning , where the intervals between the fundamental frequencies of different keys are slightly larger than one would expect. That is, a piano whose fundamental frequencies really were tuned to simple ratios would sound out of tune! | {
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268,631 | In a binary star system, two stars $A$ and $B$ follow circular orbits, of radius $R$ and $r$ respectively, centred on their common centre of mass $O$ . The mass of star $A$ is $M$ , and that of star $B$ is $m$ . I am having trouble with the following problem: Explain why the period of rotation of star $A$ is equal to the period of rotation of star $B$ . By using Kepler's Third Law, we know that $$r^3\propto T^2.$$ In this question, however, we want to show that they are the same. How should I approach this question? I only notice that the two stars are always on the straight line joining them and the centre $O$ . | The centre of mass of the binary system cannot move because there are no external forces acting. The line joining the two stars must always pass through the centre of mass, because by definition the centre of mass lies on the line between the two stars. That means the two stars must orbit with the same period. If their periods weren't the same they could not remain on opposite sides of the COM. | {
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268,644 | In the book , I read some remarks about the criticality: Iterations of the renormalization (group) map generate a sequence of points in the space of couplings, which we call a renormalization group trajectory. Since the correlation length is reduced by a factor $b<1$ at each step, a typical renormalization-group trajectory tends to take the system away from criticality. Because the correlation length is infinite at the critical point, it takes an infinite of iterations to leave that point. In general, a system is critical not only at a given point in the coupling space but on a whole "hypersurface", which we call the critical surface, or sometimes the critical line. I think these remarks are fine. But then the authors give a statement: The critical surface is the set of points in coupling space whose renormalization-group trajectories end up at the fixed point:
$$\lim_{n\rightarrow\infty} T^n(J)=J_c,$$
where $T$ is the renormalization-group map and $J$ represents the couplings in general. The question is, how to properly understand this statement? For this, I particularly have two small questions. Why the critical line need to hit the fixed point? And why the critical line need to end up at the fixed point rather leave the fixed point under the renormalization group transformation? | The centre of mass of the binary system cannot move because there are no external forces acting. The line joining the two stars must always pass through the centre of mass, because by definition the centre of mass lies on the line between the two stars. That means the two stars must orbit with the same period. If their periods weren't the same they could not remain on opposite sides of the COM. | {
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268,733 | Concerning the recent detection of gravitational waves produced by colliding black holes, it has been reported that a significant percentage of the combined mass was lost in the resulting production of the gravitational waves. So evidently in addition to Hawking radiation, black holes can also lose mass in collisions with other black holes. Is there a theoretical limit to how much mass, as a percentage, two black holes can lose in a collision as gravitational waves? Could so much mass be lost that the resulting object would no longer have enough gravity to be a black hole? | Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation
$$
E_t = 2M = \sqrt{2}M + E_{g-wave}
$$
and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy! This is the upper bound for the generation of gravitational radiation from mass. The assumption here is that the total entropy of the two black holes equals the entropy of the final black hole. Physically this happens if all the curvature exterior to the merging black holes does not result in mass-energy falling into the final black hole. There would be back scatter of gravitational radiation, much as one has to be concerned about the near field EM wave near an antenna that can couple back on it. The final entropy of the merged black hole will in fact be larger, but of course not larger than the mass-squared determined area of the two black holes. This means $1.41m~\le~m_{tot}~\le~2m$. To estimate this requires numerical methods. Larry Smarr pioneered a lot of this. So far estimates run around $5\%$ of the total mass of the black holes is converted to gravity waves. in this LIGO paper two black holes of mass $39M_{sol}$ and $32M_{sol}$ is computed to have coalesced into a final black hole of $68M_{sol}$, which radiated $3M_{sol}$ is gravitational radiation and accounts for $4.2\%$ of the initial mass. This is about in line with most numerical studies. Consequently a lot of the spacetime curvature generated by these mergers falls back into the final black hole. In terms of area the initial horizon area is $4066M^2_{sol}$ and the final horizon area is $4624M^2_{sol}$, which is an additional area $558M^2_{sol}$ of horizon area with $S~=~k A/4L_p^2$ as the entropy. | {
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268,737 | Chapter 2 of Kleppner & Kolenkow's An Introduction to Mechanics reads as follows - Newton's Laws describe the behavior of point masses. In the case where the size of the body is small compared with the interaction distance, this offers no problem. For instance, the earth and sun are so small compared with the distance between them that for many purposes their motion can be adequately described by considering the motion of point masses located at the center of each. Consider a system consisting of a block kept at rest on a table. Assume that friction is neglected. When drawing a force diagram for such a block, we assume it to be a point mass. (Also see: Section 2.4 of Kleppner & Kolenkow's book... Just to mention that such a step is followed in the book itself.) However, it doesn't seem to me that this assumption is valid, as the interaction distance between the block and the table is so small as compared to their sizes. Then why is the point mass approximation valid in such a case? Note: This chapter has not generalized Newton's laws to describe rigid bodies yet. So it would be great if you could answer my question without any reference to that (of course, if possible!). | Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation
$$
E_t = 2M = \sqrt{2}M + E_{g-wave}
$$
and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy! This is the upper bound for the generation of gravitational radiation from mass. The assumption here is that the total entropy of the two black holes equals the entropy of the final black hole. Physically this happens if all the curvature exterior to the merging black holes does not result in mass-energy falling into the final black hole. There would be back scatter of gravitational radiation, much as one has to be concerned about the near field EM wave near an antenna that can couple back on it. The final entropy of the merged black hole will in fact be larger, but of course not larger than the mass-squared determined area of the two black holes. This means $1.41m~\le~m_{tot}~\le~2m$. To estimate this requires numerical methods. Larry Smarr pioneered a lot of this. So far estimates run around $5\%$ of the total mass of the black holes is converted to gravity waves. in this LIGO paper two black holes of mass $39M_{sol}$ and $32M_{sol}$ is computed to have coalesced into a final black hole of $68M_{sol}$, which radiated $3M_{sol}$ is gravitational radiation and accounts for $4.2\%$ of the initial mass. This is about in line with most numerical studies. Consequently a lot of the spacetime curvature generated by these mergers falls back into the final black hole. In terms of area the initial horizon area is $4066M^2_{sol}$ and the final horizon area is $4624M^2_{sol}$, which is an additional area $558M^2_{sol}$ of horizon area with $S~=~k A/4L_p^2$ as the entropy. | {
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268,741 | Since the train is slowing down at a constant velocity, shouldn't the acceleration equal zero ? Why is it negative in this case? | Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation
$$
E_t = 2M = \sqrt{2}M + E_{g-wave}
$$
and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy! This is the upper bound for the generation of gravitational radiation from mass. The assumption here is that the total entropy of the two black holes equals the entropy of the final black hole. Physically this happens if all the curvature exterior to the merging black holes does not result in mass-energy falling into the final black hole. There would be back scatter of gravitational radiation, much as one has to be concerned about the near field EM wave near an antenna that can couple back on it. The final entropy of the merged black hole will in fact be larger, but of course not larger than the mass-squared determined area of the two black holes. This means $1.41m~\le~m_{tot}~\le~2m$. To estimate this requires numerical methods. Larry Smarr pioneered a lot of this. So far estimates run around $5\%$ of the total mass of the black holes is converted to gravity waves. in this LIGO paper two black holes of mass $39M_{sol}$ and $32M_{sol}$ is computed to have coalesced into a final black hole of $68M_{sol}$, which radiated $3M_{sol}$ is gravitational radiation and accounts for $4.2\%$ of the initial mass. This is about in line with most numerical studies. Consequently a lot of the spacetime curvature generated by these mergers falls back into the final black hole. In terms of area the initial horizon area is $4066M^2_{sol}$ and the final horizon area is $4624M^2_{sol}$, which is an additional area $558M^2_{sol}$ of horizon area with $S~=~k A/4L_p^2$ as the entropy. | {
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268,818 | How can it be that, if like charges repel, they don't repel themselves? In other words, why don't charges break apart? About the possible duplicate: I want to know about charges in general, not just that of an electron. My response to Lawrence B. Crowell's answer : Thank you very much for the extensive explanation. Unfortunately it's a bit above my level (I'm a first year electrical engineering bachelor student). This is how I understand it: If the charge is the sum of multiple separate charges, there has to be an external force that keeps this charges together. However we most often think of electrons and protons as point charges. In other words, we don't think of them as if they are made up from different (smaller) parts. But the idea of electrons and protons as point charges has its own problems. I didn't know that the electric field has a mass. From the formula, I understand that $m\propto \frac{1}{r}$ . This would mean that $m\rightarrow \infty$ as $r \rightarrow 0$ (or $r = 0$ ), which of course is physically not possible. But there is a way to calculate the radius of an electron based on the mass, the speed of light and the constant of Planck. (Unfortunately I don't know why.) You can get around this by a technique called called renormalization, which causes the integral to converge. Unfortunately, I don't understand the rest of the answer due to my level in physics. Nevertheless I am grateful for you answer. It would be awesome if you could confirm or correct my understanding. | This was one of those big questions in the 19th century. It still causes some consternation. If you have a composite system, such as the nucleus of an atom, some other force is necessary. This force of course is the nuclear interaction. This keeps the protons from flying apart, though for some unstable nuclei there are transitions that eject charged particles, electrons or positrons, due to weak interactions. In the case of the proton it is composed of three quarks and these are bound to each other by the QCD (quantum chromodynamics) interaction. The gauge bosons called gluons interact most strongly at low energy and these keep the quarks, with charges $2/3,2/3,-1/3$ in a bound state. Things are a bit more mysterious with point-like particles, such as the electron and other leptons and quarks. We generally do not regard such particles as composite, though this has not stopped people from proposing constituents called preons or rishons that make them up. There is a problem with defining the mass of the electron or any point-like electrically charged particle. The mass of the electric field is $$
m_\textrm{em}~=~\frac{1}{2}\int E^2~\mathrm d^3r~=~\frac{1}{2}\int_r^\infty\left(\frac{e}{4\pi r^2}\right)^24\pi r^2~\mathrm dr~=~\frac{e^2}{8\pi r}.
$$ if the electron has zero radius this is divergent. There is the classical radius of the electron $r~=~\alpha\lambda_c$ $=~2.8\times10^{-13}~\mathrm{cm}$ for $\lambda_c~=~\hbar/mc$ the Compton wavelength. This raises some questions, for the classical radius suggests "structure," and it also has a relationship to something called Zitterbewegung . A more standard approach to this is renormalization. A screenshot of this is to look at this integral with the variables $p~=~1/r$ so in this integral above $\mathrm dr/r~\rightarrow~-\mathrm dp/p$ . Here we are thinking of momentum and wavelength or position as reciprocally related. This integral is then evaluated for a finite $r$ as equivalent to being evaluated for a finite momentum cut off $\Lambda$ $$
I(\Lambda)~=~\int_0^\Lambda\frac{\mathrm dp}{p}~\simeq~1~+~2^{-1}~+~3^{-1}~\dots
$$ which is equal to $$
\lim_{\Lambda\rightarrow\infty}I(\Lambda)~=~-\zeta(1)
$$ In some ways this is a removal of infinities. Another curious way to look at this is with $p$ -adic number theory. This is a topic that could consume a lot of bandwidth. We have another way to look at this. This comes down to the question of what do we mean by "composite." It also forces us to think about what we mean by the locality of field operators. The Dirac magnetic monopole is a solenoid with an opening to an infinite coil. The condition for the Dirac monopole is that the Aharonov-Bohm phase of a quantum system is zero as it passes the "tube" of the solenoid $\psi~\rightarrow~\exp\left(ie/\hbar\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}\right)\psi$ . This might be compared to "cutting off the tail" on the magnetic monopole charge. The vanishing of this is equivalent to saying $$
2\pi N~=~\frac{e}{\hbar}\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}~=~\frac{e}{\hbar}\iint\nabla\times{\vec A}\cdot{\vec a},
$$ for the integral evaluated over units of area of the opening. This is of course the magnetic field ${\vec B}~=~-\nabla\times{\vec A}$ evaluated in a Gauss' law that gives the magnetic monopole charge $g~=~\displaystyle\iint\nabla\times{\vec A}\cdot{\vec a}$ and we use this expression to see the S-duality relationship between the electric and magnetic monopole charge $$
eg~=~2\pi N\hbar,
$$ sometimes called the Montonen-Olive relationship. This means that if we have an electric charge we can use the renormalization machinery to illustrate how the vacuum around it is polarized with virtual particles according to $\alpha~=~\frac{e^2}{4\pi\epsilon\hbar c}$ . The electric charge is comparatively weak in strength with a modest polarization of the vacuum expanded in orders of $\alpha$ for $N$ internal lines or loops. This S-dual relationship tells us that while this is modest, the magnetic monopole is very strong and the vacuum is a "bee's nest" of lots of particles. This then means the dual of the electric field is a magnetic monopole field that in some ways appears composite. This means in some ways we have questions needed to be asked about the locality of field operators. Something that appears local, point-like and "nice" may be dual to something that appears not so local, more composite-like and not renormalizable. As a result there are still open questions on this, and even Feynman agreed with Dirac that the situation with QED was not perfectly satisfactory. | {
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268,831 | I'm trying to understand entanglement in terms of scarcity and abundance . Given an arbitrary vector $v$ representing a pure quantum state of, say, dimension 4, i.e. $v \in \mathcal{H}^{\otimes 4}$ , Is $v$ more likely to be entangled than non-entangled (separable)? By trying to answer it myself , I can see that the separability test is based on an existential quantifier , namely trying to prove that $\exists v_1, v_2 \in \mathcal{H}^{\otimes 2} $ such that $v_1 \otimes v_2 = v $ . The entanglement test on the other hand is based on a universal quantifier , $$\forall v_1, v_2 \in \mathcal{H}^{\otimes 2}, v_1 \otimes v_2 \neq v.$$ So, this reasoning could suggest that entangled vectors are much more scarce than separable ones because it is easier to find one simple example (existential) that satisfies the condition than to check for every single one (universal). This result would make sense physically since entanglement is a valuable resource so, intuitively, it should be scarce. Does this reasoning make any sense at all, or am I saying nonsense?
Any help would be greatly appreciated. PS: I would assume extending this reasoning to (density) matrices would be obvious. | I'm assuming that you have a finite-dimensional base Hilbert space $\mathcal H_0$ and that you're building your full Hilbert space as $\mathcal H=\mathcal H_0\otimes \mathcal H_0$. In these conditions, the set of separable states has measure zero . (It gets a bit more complicated if you have $\mathcal H_0^{\otimes 4}$ and you're allowed to split it any way you want among those two factors, and the answer is negative if you're allowed to look for any tensor-product structure in your space, as you can always take one factor along your given $|\psi⟩$.) Consider, then, a given basis $\{|n⟩:n=1,\ldots,N\}$ for $\mathcal H_0$, which means that any arbitrary state $|\psi⟩\in\mathcal H$ can be written as
$$
|\psi⟩=\sum_{n,m} \psi_{nm}|n⟩\otimes|m⟩.
$$
If, in particular, $|\psi⟩$ can be written as a tensor product $|\psi⟩=|u⟩\otimes|v⟩$, then you have
$$
|\psi⟩
=\left(\sum_n u_n |n⟩\right)\left(\sum_m v_m |m⟩\right)
=\sum_{n,m} u_nv_m |n⟩\otimes|m⟩;
$$
that is, the coefficient matrix $\psi_{nm}$ has the form $\psi_{nm}=u_n v_m$. This means that this matrix has rank one, which then means that it must have determinant equal to zero. Since the determinant is a continuous polynomial function $\det\colon \mathbb{C}^{N\times N}\to\mathbb C$, its zero set has Borel measure zero inside $\mathbb{C}^{N\times N}$, and therefore correspondingly inside $\mathcal H$. This means, finally, that if you choose a random vector $|\psi⟩\in\mathcal H$ using a probability measure that is absolutely continuous with respect to the canonical Borel measure on $\mathcal H\cong\mathbb C^{N\times N}$, then it is almost certainly entangled. As an added bonus from exactly the same argument, such a vector will actually (almost certainly) have a full Schmidt rank . A bit more intuitively, what this argument is saying is that separable states form a very thin manifold inside the full Hilbert space, and this is caught quite well by the spirit of zeldredge's answer. In particular, to describe an arbitrary separable state, you need $2N-1$ complex parameters ($N$ each for the components of $|u⟩$ and $|v⟩$, minus a shared normalization), so roughly speaking the separable states will form a submanifold of dimension $2N-1$. However, this is embedded inside a much bigger manifold $\mathcal H$ of dimension $N^2$, which requires many more components to describe, so for $N$ bigger than two the separable states are a very thin slice indeed. | {
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269,313 | If a charge travels in a circle it must accelerate, thereby producing EM. However, a wire in a circular loop is analogous to many charges moving in a circle. So, why don't circular currents produce EM? (I have not found any evidence that circular currents can produce EM). A similar question is: Why doesn't alternating current produce light while a vibrating single particle with a charge will . However, my question asks about circular wires and direct current. Thanks. | Circular currents do produce EM, and indeed this is exactly how X-rays are produced by synchotrons such as the (sadly now defunct) synchotron radiation source at Daresbury. In this case the current is flowing in a vacuum not in a wire, but the principle is the same. Current flowing in loops of wire don't produce radiation in everyday life because the acceleration is so small. The electrons are moving at the drift velocity , which is only around a metre per second, so the amount of radiation released is immeasurably small. Synchotrons produce radiation because the electrons are moving at almost the speed of light. | {
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269,332 | Let's say the system is moving at constant velocity. What are the forces applied on B? Is there a static friction force? I find it hard to imagine what is happening to object B. $\Sigma F$ is $0$ on A and on B too since it moves with constant velocity. The only one who can apply force on it is A. If A was to apply force on B, then there would be also an opposite force applied by A (so $\Sigma F$ is $0$ on B) so it is a contradiction thus no force is applied on B. However, lets look at the molecule level. B's molecules are moved by A molecules. So A must apply force on B! Like in this picture: A's surface must move B's surface to the direction of the movement, otherwise B would not move. I'm really confused. | Circular currents do produce EM, and indeed this is exactly how X-rays are produced by synchotrons such as the (sadly now defunct) synchotron radiation source at Daresbury. In this case the current is flowing in a vacuum not in a wire, but the principle is the same. Current flowing in loops of wire don't produce radiation in everyday life because the acceleration is so small. The electrons are moving at the drift velocity , which is only around a metre per second, so the amount of radiation released is immeasurably small. Synchotrons produce radiation because the electrons are moving at almost the speed of light. | {
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269,355 | In Brian Greene's book "The elegant Universe", he talks about the double slit experiment and Feynman's interpretation of Quantum Mechanics. According to the book, Feynman said that one vaild interpretation is that on its way from the emitter to the photoscreen, the photon actually takes every possible path. Greene actually says that some paths include a trip to the Andromeda galaxy and back, as a photon takes (as said before) $\textit{every}$ possible path. Now if this were really the case and we measure the time between the emission of a photon and its impact on the photoscreen, doesn't this mean that some photons would actually travel with a velocity far greater than the speed of light? If a photon actually travelled to Andromeda and back, there is no way it could arrive at the photoscreen in just a fraction of a second as is observed during experiment.... | The paths of the Feynman path integral are not actually taken . The phrase "takes every possible path" is a mangled statement of the mathematical instruction to take the integral of $\exp(-\mathrm{i}S)$ over all possible paths for the action $S$ to get the probability amplitude of something happening. It is a fact of quantum mechanics that this integral computes the correct quantum mechanical amplitude, but the formalism of quantum mechanics never says anything about the particle "taking" these paths, which is in particular absurd because quantum objects are not point particles that have a well-defined path in the first place. So, well, you can say that it "takes" every possible path as long as you don't literally imagine a point particle zipping along each path. Which is what "taking" a path usually means. Which is why this figure of speech does not actually convey any physical insight. The physical insight lies in understanding how the path integral reproduces the correct quantum mechanical amplitude, which cannot be done on the level of such crude heuristic statements based on classical notions of "path" and "particle". There is no path a quantum particle takes unless you continually track it, and then you'll get a perfectly ordinary classical path (see, for instance, the perfectly normal paths in bubble chambers, where the continual interaction with the bubble chamber effectively tracks the particle). | {
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269,563 | Is it possible to "construct" the Hamiltonian of a system if its ground state wave function (or functional) is known? I understand one should not expect this to be generically true since the Hamiltonian contains more information (the full spectrum) than a single state vector. But are there any special cases where it's possible to obtain the Hamiltonian? Some examples would be really helpful. | IF you know that your Hamiltonian is of the form
$$
\hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r)
\tag 1
$$
for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state $\Psi_0(\mathbf r)$ obeys
$$
\hat H\Psi_0(\mathbf r)
=\frac{-\hbar^2}{2m}\nabla^2\Psi_0(\mathbf r)+V(\mathbf r)\Psi_0(\mathbf r)
=E_0 \Psi_0(\mathbf r),
$$
which means that if you know $\Psi_0(\mathbf r)$ then you can calculate its Laplacian to get
$$
\frac{ \nabla^2 \Psi_0(\mathbf r) }{ \Psi_0(\mathbf r) }
=
\frac{2m}{\hbar^2}\left(V(\mathbf r)-E_0\right).
$$
If you know the particle's mass, then you can recover $V(\mathbf r)-E_0$, and this is all you really need (since adding a constant to the Hamiltonian does not change the physics). However, it's important to note that this procedure guarantees that your initial $\Psi_0$ will be an eigenstate of the resulting hamiltonian, but it does not preclude the possibility that $\hat H$ will admit a separate ground state with lower energy. As a very clear example of that, if $\Psi_0$ is a 1D function with a node, then (because 1D ground states have no nodes ) you are guaranteed a unique $V(x)$ such that $\Psi_0$ is an eigenstate, but it will never be the ground state. If you don't know that your Hamiltonian has that structure, there is (in the general case) no information at all that you can extract about the Hamiltonian from just the ground state. As a simple example, without staying too far from our initial Hamiltonian in $(1)$, consider that Hamiltonian in polar coordinates,
$$\hat H=\frac{-\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} + \frac{1}{\hbar ^2r^2}L^2\right)+V(r),$$
where I'm assuming $V(\mathbf r)=V(r)$ is spherically symmetric, and encapsulating the angular dependence into the total angular momentum operator $L^2$. Suppose, then, that I give you its ground state, and that it is an eigenstate of $L^2$ with eigenvalue zero (like e.g. the ground state of the hydrogenic Hamiltonian). How do you tell if the Hamiltonian that created it is $H$ or a similar version,
$$\hat H{}'=\frac{-\hbar^2}{2m}\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} +V(r),$$
with no angular momentum component? Both versions will have $\Psi_0$ as a ground state (though here $\hat H'$ will have a wild degeneracy on every eigenspace, to be fair). Carrying on with this thought, what about
$$\hat H{}''=\frac{-\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} + f(r)L^2\right)+V(r),$$
where I've introduced an arbitrary real function $f(r)$ behind the angular momentum? This won't affect the $\ell=0$ states, but it will take the rest of the spectrum to who knows where. (In fact, you can even tack on an arbitrary function of $L_x$, $L_y$ and $L_z$, while you're at it.) A bit more generally, any self-adjoint operator which vanishes on $\left|\Psi_0\right>$ can be added to the Hamiltonian to get you an operator that has $\left|\Psi_0\right>$ as an eigenstate. As a simple construction, given any self-adjoint operator $\hat A$, the combination
$$\hat H {}'''
= E_0 \left|\Psi_0\right>\left<\Psi_0\right|
+ \left(\mathbf 1 - \left|\Psi_0\right>\left<\Psi_0\right| \right)
\hat A
\left(\mathbf 1 - \left|\Psi_0\right>\left<\Psi_0\right| \right)
$$
(where the factors in brackets are there to modify $\hat A$ into vanishing at $\left|\Psi_0\right>$ and its conjugate) will always have $\left|\Psi_0\right>$ as an eigenstate. Even if you know all the eigenstates, it's still not enough information to reconstruct the Hamiltonian, because they do not allow you to distinguish between, say, $\hat H$ and $\hat{H}{}^2$. On the other hand, if you know all the eigenstates and their eigenvalues, then you can simply use the spectral decomposition to reconstruct the Hamiltonian. In general, if you really insist, there is probably a trade-off between what you know about the Hamiltonian's structure (e.g. "of the form $\nabla^2+V$" versus no information at all) and how many of the eigenstates and eigenvalues you need to fully reconstruct it (a single pair versus the whole thing), particularly if you allow for approximate reconstructions. Depending on where you put one slider, you'll get a different reading on the other one. However, unless you have a specific problem to solve (like reconstructing a Hamiltonian of vaguely known form from a specific set of finite experimental data) then it's definitely not worth it to explore the details of this continuum of trade-offs beyond the knowledge that it exists and the extremes I noted above. | {
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269,669 | As you might already know, frequency of musical notes is arranged in a such a way that if, for example, an A note has frequency of $x$, another A note which is placed one octave higher would produce frequency of $2x$. So here's my childhood toy glockenspiel: This is where my question raised. If wave speed inside the bars were the same, as in strings of a guitar, then we would expect two notes that are one octave apart, like the two C's, would have a 2:1 length ratio. But measurement shows that they are designed in approximately 7:5 ratio or maybe $\sqrt2$. Now I'm wondering what exactly causes this speed variation? Thickness or width doesn't vary meaningfully so it must be about the length but how would length of a bar affect wave speed inside it is what I'm asking here. And why this doesn't happen for strings. I'll measure and report length, width and thickness of the bars if necessary. | The answer to this question has significant overlap with my answer on piano tuning . There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to
$$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$
This case is the other way around: the tension is negligible, so we only have the 'extra' term. The wave equation becomes
$$-A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^2}.$$
Plugging in an ansatz of $\cos(kx-\omega t)$ gives the dispersion relation
$$Ak^4 = \omega^2.$$
That is, $\omega \propto k^2$. Since $k$ is inversely proportional to length,
$$\omega \propto 1/L^2$$
as desired. A bar $\sqrt{2}$ times shorter makes a tone twice as high. As you saw, the wave speed must change for the results to make sense. The phase velocity of a wave is $v_p = \omega / k$, and this is constant only for the simplest dispersion relation, the ideal wave equation $\omega = vk$. In this case, we have $\omega \propto k^2$, which implies $v_p \propto k$. Waves with shorter wavelength, like the ones on the smaller bars, travel faster. But this doesn't mean anything about the smaller bars is different. The phase velocity changes because wave propagation is fundamentally different on bars than strings; it exhibits dispersion . | {
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269,852 | I'm reading Thomson, Modern Particle Physics, and in chapter 16 author says that the decay width of the Z boson is $\Gamma_Z =2.452 \pm 0.0023 \,\mathrm{GeV}$. He also says the total width of the decay is the sum of the partial widths, $$\Gamma_Z=\Gamma_{ee} +\Gamma_{\mu\mu} +\Gamma_{\tau\tau} +\Gamma_\mathrm{hadrons} +\Gamma_{\nu_e \nu_e} +\Gamma_{\nu_\mu \nu_\mu} + \Gamma_{\nu_\tau \nu_\tau},$$ But I still am struggling to understand the language here. My questions are: What do we mean by the decay width ? Why is it given in energy units (instead of a fraction of seconds, say)? I know these questions may sound very naive, but I want to make sure I get the basics of the terminology. | Like @DavidZ I found this a very good question but unlike him I am not a professional Physicist and so will try and answer the question on a simplistic level which may not suit @Martin as I do not know the level that he is working at but given that he is reading Thomson's book it must be quite advanced. I start with the decay law of Rutherford and Soddy which states that the rate of decay $\dot N(t)$ of an unstable particle (or nucleus) at a given time $t$ is proportional to the number of those particles which are present $N$ at that time $t$. $$\dfrac {dN}{dt} = \dot N(t) = \propto N(t) \Rightarrow \dot N = - \lambda N$$ where $\lambda$ is the decay constant. Now it is often the case that there is more than one decay mode and for each of the decay modes there is a corresponding decay constant so now one must write that the rate of decay depends on the sum of all the decay modes. $$ \dot N_{\text{all}} (t) = -\lambda_A N(t) -\lambda_B N(t)-\lambda_C N(t) - . . . . .$$ where $\lambda_A, \lambda_B, \lambda_C$ etc are the decay constants for the various decay modes. From this you get $\dot N_{\text{all}} = - \lambda_{\text{all}} N$ where $\lambda_{\text{all}} = \lambda_A + \lambda_B + \lambda_C+ . . . . $ So that decaying particle has a decay constant which is the sum of the decay constants for all of the possible modes of decay. At the moment of decay the decaying particle chooses one particular mode of decay and the probability of such a decay is expressed as a branching fraction or branching ratio. The average lifetime of an unstable particle $\tau$ is related to the decay constant $\tau = \dfrac 1 \lambda$. As is discussed in Particle lifetimes from the uncertainty principle , "the uncertainty principle in the form $\Delta E \Delta t > \hbar/2$ suggests that for particles with extremely short lifetimes, there will be a significant uncertainty in the measured energy." When measurements are made of the rest mass energy of an unstable particle with no instrument errors a graph of the following type is obtained. The width of such a distribution of mass energies is called the decay width $\Gamma$ and is measured in units of energy. The decay width is related to the uncertianty in energy as follows. $$\Gamma = 2 \Delta E = \dfrac \hbar \tau = \hbar \lambda$$ where $\lambda$ is the decay constant. However the Z-boson has many decay modes so this equation should be written as $\Gamma_{\text{all}} = \hbar \lambda_{\text{all}}$ Remembering that $\lambda_{\text{all}} = \lambda_A + \lambda_B + \lambda_C+ . . . . $ leads to the expression $$\Gamma_{Z,\text{all}}=\Gamma_{ee} +\Gamma_{\mu\mu} +\Gamma_{\tau\tau} +\Gamma_{hadrons} +\Gamma_{v_e v_e} +\Gamma_{v_\mu v_\mu} +\Gamma_{\tau \tau},$$ where the various decay modes of the Z-boson are listed. Multiple experimental are then performed to investigate the decay modes and their probability. All this experimental data is then compared with the theoretical data which is based on the Standard Model and the agreements have been found to be very good to a high degree of precision. The accuracy to which the experimental work is done can be gauged by one of Professor Thomson's slides and this is using the old accelerator at CERN. Entering decay width into this website's search engine is also most illuminating. | {
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269,892 | The Wikipedia article for Venus's alleged moon Neith has this picture: And a tantalizing remark regarding paramecia (vandalism?). I know the grey ball is supposed to be Neith, the white is Venus, the little rods are the halo of light. But why is Venus shaped like a kidney? Is this supposed to be a partially lit Venus? Then why is Neith never behind it? | According to the caption for that picture on the same Wikipedia article, it is Francesco Fontana’s drawing of the supposed satellite(s) of Venus. Woodcuts from Fontana’s work (1646). The fringes of light around Venus are produced by optical effects.$^1$ Fontana lived from around 1580 to around 1656. He was an Italian lawyer at the University of Naples and an astronomer, using a hand-made telescope. He observed the Moon and the planets of the solar system, recording observations and drawing them. Most of these were published in his book Novae coelestium terrestriumq[ue] rerum observationes, et fortasse hactenus non vulgatae in 1646. 1645 was when he made his observation of a Venusian moon. It is interesting to note that according to the book cited by Wikipedia ( The Moon that Was not: The Saga of Venus' Spurious Satellite by Helge Kragh), Fontana's contemporaries thought he was a good telescope maker (his were some of the best of the day) but a poor astronomer. Contemporaries said his observations were ridiculous and fantasies. One even wrote about his published book of observations, "I have the book of foolishnesses observed, or rather dreamed, by Fontana in the heavens. If you want to see insane things...I will send you the book." He was scorned by Galileo and one of Galileo's protégés as well. It also appears that he didn't think the obviously optical effects in his woodcut were optical. Fortana was the first to observe a "moon" of Venus, according to Wikipedia. This he also published in his book of observations, and he was thought ridiculous for this (at least until other, more reputable astronomers sighted something similar). The mention of paramecium is just there saying the depiction of Venus and later depictions of paramecium are similar. Interestingly, Fontana shows two moons in one panel. I found no references to a second theoretical Venusian moon anywhere, but maybe Neith is sometimes in front and sometimes behind, but moon number two is making things confusing. As for why Venus is kidney shaped, that's probably the least confusing part. Venus often appears as a not-complete circle, as it has phases like the moon, as shown below: Hope this helps! Kragh's book can be found here ; the relevant part is chapter two, section 2.1. $^1$See this website | {
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270,030 | According to the Third Newton's law of motion: For every action there is an equal and opposite reaction. So, I understand that if I hit a brick wall with $50\, \mathrm{lbs}$ of force, the brick wall also hits me with $50\, \mathrm{lbs}$ of force (usually painfully). In this instance, Newton's third law makes sense. What I'm confused about is, if I hit a patch of drywall with $50\, \mathrm{lbs}$ of force, it's probably going to break, and due to the lack of pain in my hand, I can tell it did not hit me back with $50\, \mathrm{lbs}$ of force. How does Newton's third law apply to situations when one object or the other is destroyed? It certainly seems like at that point it is incapable of delivering the full force of my blow back to me. What happens with the energy? | You've caught a non-intuitive part of Newton's 3rd law. It's actually applying in the case you mention, but because the objects involved are of dissimilar hardness it's easy to perceive the impact as a violation of the law. Impacts are actually really complicated. Consider this slow motion video of a punch to the gut . We won't be able to cover all of the complexities we see here, but we can layer a few of them together to try to explain why the non-inutitive results you get are actually correct applications of Newton's 3rd law. The key thing which makes impacts so complicated is that we have to pay attention to momentum. When you punch the brick wall or the drywall, your hand has quite a lot of momentum. When you punch the brick wall, that momentum has to be stopped. The only way to do this is through the reactionary force of the wall pushing back on your hand. The more momentum your hand has, the more reactionary force you deal with. In your brick example, that reactionary force is 50lbs, and the corresponding force of your hand on the wall is also 50lbs. In the drywall case, we need to make a few adjustments. The first is to note that your hand goes through the drywall. It does not have to be stopped by the wall. This points out that the reactionary force will be less than it was in the brick wall case, because the brick wall had to stop the fist. Well, almost. I cheated slightly, and that cheat may be a source of non-intuitive behavior. The more correct statement is that the brick wall had to impart more impulse to your hand, because it had to stop your hand. Impulse is force*time, and is a measure of change in momentum. The key detail here is that the distance can change. If you drop a superball on the ground, it rebounds almost back to where you threw it from. The impulse applied to the ball by the ground is very high. Contrast that with a steel ball bearing with the same mass as the superball, which does not rebound as much. The impulse applied to the bearing is lower. However, the superball deforms a great deal on impact, so it has a longer time to apply that impulse over. It is reasonable that superball could be subjected to less force than the ball bearing, and yet bounce higher because that lower force was applied for a longer distance. In the case of the punch, we're lucky that 99% of the deformation in your punch occurs in your hand. Your skin and fat squish out of the way until your bones start to have to move. The shock in theory works its way all the way up your arm. However, we can ignore all of that for now, because we're just doing comparisons. It's the same hand in both the brick wall punch and the drywall punch, so it can be expected to deform in similar ways over similar distances and similar times. This is how we can claim that the brick wall punch must have a higher force. We know the impulse must be higher (because it stopped your hand, and the drywall punch didn't), and the times are the same for the reaction to both punches, so the brick wall punch must have more force. Thus, the truth is that you did not punch the drywall with 50lbs. You actually supplied less force than that. In fact, you supplied just enough force to break the internal bonds that were keeping the drywall solid. Intuitively, we like to measure punches in forces (claiming a 50lb punch), but it's actually not possible to punch that hard unless the thing being punched is capable of providing a corresponding reactionary force! If you layered enough pieces of drywall to have the structural integrity to provide 50lbs of force, you would find that you don't break through, and it hurts almost as much as the bricks did (the first sheet of drywall will deform a little, so it wont hurt as much as the brick) The issue of breaking through the wall is actually a very important thing for martial artists. Those who break boards or bricks in demonstrations all know that it hurts far more if you fail to break the board or brick. That's because the board stopped all of your forward momentum, meaning you had a lot of impulse over a short time, meaning a lot of force. If you break the brick, the reactionary forces don't stop your hand, so they are less. I would wager that the greatest challenge of breaking bricks with a karate chop is not breaking them, but in having conditioned your body and mind such that you can withstand the impulse when you fail to break them. | {
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270,032 | What is the intuition behind the Choi-Jamiolkowski isomorphism ? It says that with every superoperator $\mathbb{E}$ we can associate a state given by a density matrix $$ J(\mathbb{E}) = (\mathbb{E} \otimes 1) (\sigma)$$ where $\sigma = \sum_{ij} | ii \rangle \langle jj |$ is the density matrix of some maximally entangled state $\sum_{i} | ii \rangle$. And then the action of the superoperator is equal to $$\mathbb{E}(\rho) = \operatorname{tr}_2(J(\mathbb{E}) \cdot 1 \otimes \rho^T).$$ What is the point of this? How does one use this in practice? Is it to simulate the action of the channel $\mathbb{E}$ by first preparing a specific state? I really don't understand the intuition behind this concept. | The intuition Let us consider a channel $\mathcal E$ , which we want to apply to a state $\rho$ . (This could equally well be part of a larger system.) Now consider the following protocol for applying $\mathcal E$ to $\rho$ : Denote the system of $\rho$ by $A$ . Add a maximally entangled state $|\omega\rangle=\tfrac{1}{\sqrt{D}}\sum_{i=1}^D|i,i\rangle$ of the same dimension between systems $B$ and $C$ : Now project systems $A$ and $B$ on $|\omega\rangle$ : [This can be understood as a teleportation where we have only consider the "good" outcome, i.e., where we don't have to make a (generalized) Pauli correction on $C$ , see also the discussion.] Our intuition on teleportation (or a simple calculation) tells us that we now have the state $\rho$ in system $C$ : Now we can apply the channel $\mathcal E$ to $C$ , yielding the desired state $\mathcal E(\rho)$ in system $C'$ : However, steps 2 and 3 commute (2 acts on $A$ and $B$ , and 3 acts on $C$ ), so we can interchange the ordering and replace 2+3 by 4+5: Apply $\mathcal E$ to $C$ , which is the right part of $|\omega\rangle$ : This results in a state $\eta=(\mathbb I\otimes \mathcal E)
(|\omega\rangle\langle\omega|)$ , which is nothing but the Choi state of $\mathcal E$ : (This is the original step 3.) We can now carry out the original step 3: Project $A$ and $B$ onto $|\omega\rangle$ : Doing so, we obtain $\mathcal E(\rho)$ in $C'$ : Steps 4 and 5 are exactly the Choi-Jamiolkowski isomorphism: Step 4 tells us how to obtain the Choi state $\eta$ for a channel $\mathcal E$ Step 5 tells us how we can construct the channel from the state Going through the math readily yields the expression for obtaining $\mathcal E$ from $\mathcal \eta$ given in the question: $$
\begin{align*}
\mathcal E(\rho) &= D\langle \omega|_{AB}\rho_A\otimes \eta_{BC}|\omega\rangle_{AB}\\
&= \sum_{i,j} \langle i|\rho_A|j\rangle_{A} \langle i|_B\eta_{BC} |j\rangle_B \\
& = \mathrm{tr}_B[(\rho_B^T\otimes \mathbb I_C) \eta_{BC}]\ .
\end{align*}
$$ Discussion The intuition above is closely linked to teleportation-based quantum computing and measurement based quantum computing. In teleportation-based computing, we first prepare the Choi state $\eta$ of a gate $\mathcal E$ beforehand, and subsequently "teleport through $\eta$ ", as in step 5. The difference is that we cannot postselect on the measurement outcome, so that we have to allow for all outcomes. This is, depending on the outcome $k$ , we have implemented (for qubits) the channel $\mathcal E(\sigma_k \cdot \sigma_k)$ , where $\sigma_k$ is a Pauli matrix, and generally $\mathcal E$ is a unitary. If we choose our gates carefully, they have "nice" commutation relations with Pauli matrices, and we can account for that in the course of the computation, just as in measurement based computing. In fact, measurement based computing can be understood as a way of doing teleportation based computation in a way where in each step, only two outcomes in the teleportation are allowed, and thus only one Pauli correction can occur. Applications In short, the Choi-Jamiolkowski isomorphism allows to map many statements about states to statements about channels and vice versa. E.g., a channel is completely positive exactly if the Choi state is positive, a channel is entanglement breaking exactly if the Choi state is separable, and so further. Clearly, the isomorphism is very straightforward, and thus, one could equally well transfer any proof from channels to states and vice versa; however, often it is much more intuitive to work with one or the other, and to transfer the results later on. | {
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270,289 | One of my friends and I had an argument over this topic. He stressed the fact that in real life many forces exist, whereas in physics we deal only with ideal situations. He put the following arguments:- Newton's First Law is invalid because friction exists in real life. Newton's second law is invalid due to the same reasons. Newton's third law is invalid because in a trampoline, there is excessive reaction. In defence, I put forward the following arguments:- Newton's laws are true but the equations have to be modified to take into account the other forces in real life. For example, if a force $F$ is applied on a body of mass $m$ , and $f_s$ is the force of friction, then, the equation becomes $F - f_s = ma$ . Thus, we have just modified the equation $F = ma$ . So basically I mean to say that we have to adjust the laws to suit our purpose. In the end, there was a stalemate between us. Even now I am confused after this argument. Please clarify my doubt. | Regardless of relativistic effects: Newton's First Law is invalid because friction exists in real life. False, the first law talks about the case when no forces are present, if forces are present go to the second law. Newton's second law is invalid due to the same reasons. False, you add friction to the total force. Newton's third law is invalid because in a trampolin, there is excessive reaction. False, why do you think there is excessive reaction? | {
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270,600 | If I were to mix yellow paint (which reflects yellow light) and blue paint (which reflects blue light), I would get a mixture of paints that I would perceive as green. Is that because the mixture is now reflecting green light (and absorbing blue/yellow)? OR Is it because both blue and yellow light is being reflected simultaneously and my visual cortex is interpreting that as "green?" In other words, are humans capable of perceiving both green frequency light as well as mixtures of colors that our brains interpret as one color/shade? How much of color perception is physics (blue = 450 nm) vs subjective brain interpretation (450 nm + 570 nm = GREEN)? Experimental example of my question: Man A and Man B both describe a laser emitting 525 nm light as "green." Is it possible for Man A to perceive a mixture of two different wavelength lasers to be a different color than Man B? | Color is not a physical phenomenon , it is how light of different frequencies/wavelength are perceived by humans. While the cones are developed by our animal ancestors and we share them with other species, it is entirely possible (but unlikely) that they perceive 700 nm light not a "red" as we do, but as completely different impression. What we do know is
that people cannot perceive some shades at all (they are color-blind, a red-green blind person will never be able to perceive red or green) and some species and extremely rare humans can see more colors than we do, they have one additional receptor ( tetrachromacy ). Back to your question. We are only looking at the visible part of the electromagnetic waves.
If we order this part from the lowest wavelength to the highest wavelength, we see a spectrum. The curves inside the picture displays the sensitivity of the three different receptors we are using. Image taken from en.wikipedia.org by BenRG, Public Domain So light of wavelength 575 nm is perceived by our eye, stimulates both L and M receptors and our brain processes it as "yellow". But we can also use two wavelengths 540 nm and 610 nm , vary their intensity and get the exact same "yellow" impression. Principally you have a vast range of possibilities to display the exact same color. I think I have made clear the difference between physical wavelength and perceived color . One specific wavelength always creates one color, but the same color can be created by many possible combinations of physical wavelengths. For the sake of shortness I now define the part with the shortest wavelength as "blue" light, the part with the longest wavelength as "red" light and the middle part as "green" light. You see from the picture that you cannot define strict boundaries because the sensitivity of the receptors is overlapping. If light is completely missing, we see "black", if every component (blue, green, red) are approximately equal in intensity we call it "white". For luminous objects color creation is easy to understand: They are sending light out and the resulting light mix is interpreted by our eyes. Monitors use light of 476, 530 and 622 nm to approximate each input . But paint and non-luminous objects in general need light to be visible. A monitor can be seen in a dark room, but every other object is black. So the only possibility for non-luminous objects to be perceived as colorful is reflecting back some wavelengths more than others. Let's say our object absorbs "blue" light completely and throws everything else back. I illuminate it with "blue" light, it seems to be black. with "green" light, it looks green. with "red" light, it looks red. with "white" light, the "blue" component is removed, only "red" and "green" remains...it is looking yellow . I have now another paint which absorbs "red" light completely. I again illuminate it with "red" light, it seems to be black. with "green" light, it seems to be green. with "blue" light, it seems to be blue. with "white" light, the "red" component is removed, only "green" and "blue" remains, it is looking like cyan (a greenish blue). A material absorbing "green" light looks purple, for the impression see the overlapping sections in the following picture. The first image shows what happens if you overlay luminous light (additive colors), the second image shows what happens if you mix paint (subtractive colors). First image taken from en.wikipedia.org by SharkD, Public Domain; second image taken from de.wikipedia.org by Quark67 CC BY-SA 2.5 If I mix the pigments of paints, each component will absorb its wavelength component(s). In case of "blue" paint mostly "red" light is absorbed, in case of "yellow" paint mostly "blue" light is absorbed, so the dominant remaining color is green . This is the exact reason plants are looking green because plants are mainly absorbing "red" and "blue" light; plant lights are emitting therefore mainly "red" and "blue" light, the color of a plant light looks purple. If every component is absorbed, mixing yellow, purple and cyan together should give black. In real-life you get a dark brown because the pigments are not mixing perfectly so a color tint is remaining. For that reason we use black ink in our printers for printing grey or black. | {
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270,621 | Let's say I have a bucket or something bigger, like a bathtub full of water. I weigh that bath and get some value, $x$. Then I add a small boat made of wood to that bathtub that doesn't touch the borders (i.e., it is only touching the water). Will this boat add to the total weight of the bathtub, or does the weight of the tub stay the same? | Mass is mass. If you add something which has mass to the bucket, the bucket now has more mass. It doesn't matter if it was more dense or less dense. If you add 1kg to a bucket, you add 1kg. Now there are two areas where this falls apart. One is in the case where the object you are adding is actually less dense than the air (not just less dense than the water, but actually less dense than air, like a helium balloon). In this case, we can forget about the minor detail of whether the object is floating on the water or not, we can focus on the entire bucket which is immersed in air. This object displaces a larger mass of air than its own mass, so it will actually add lift to the bucket. The mass of the bucket will still be bucket+water+object, but the bouyancy forces will make the bucket feel lighter. Indeed, this is precisely the mechanism used by hot air balloons. The other corner case is the case where the bucket is already filled to the brim with water. Adding a single drop of water would cause a drop to have to spill over the brim. Now when we add our toy boat the situation gets a little more complicated. The boat will displace a mass of water equal to the mass of the boat. This displacement will cause the water to rise and spillover. If you tally up all of the masses in this case, you will find that the bucket+water-in-the-bucket+toy boat will have the same mass as the bucket+water did before you added the toy boat. Where did the extra mass go? There's a puddle on the ground outside of the bucket, whose mass is exactly equal to that of the toy boat. | {
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270,835 | You have to break the bonds either way, be it lifting the object or moving it. Why is it that we don’t have to overcome the force required to break the cluster of molecules? Thank you | It does! In most cases the force is too small to be noticable however it is there. I spent a happy few months measuring the friction between carbon whiskers, and the adhesion force between the whiskers was large enough to cause a noticable deformation of the whiskers when I separated them. In fact you had to add the adhesion force to the applied load to get a correct value for the friction coefficient. For large objects the force is usually small compared to the weight of the object however in special cases, e.g. very clean surfaces, it is large enough to cause cold welding . | {
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270,982 | All the youtube videos I have seen on the double slit experiment broadly fall into one of the following three categories: Documentaries and fan made videos heavy on animation which 'admire' the wave-particle duality of light rather than 'explaining' it. University lectures where the professors draw diagrams on the board and go into the mathematics (e.g. how to calculate the distance between two bright bands) but never explain how the actual apparatus is built. Double slit DIY videos on how to do the experiment with sunlight or laser but they seem to leave out the most critical part, how do you make the light source so faint that only one photon is emitted at a time? As far as I understand, the classical wave theory of light fails to explain the interference only when you do the experiment with one photon at a time. My question here is, how do you make the light source fire a single particle of light at a time when you are trying to determine whether or not light is made of particles in the first place? I am more interested in the engineering of the apparatus rather than the mathematical explanation. A follow up question might be, how can you be sure that indeed only one photon is coming out of the source at a time? (a single photon is hitting the screen at a time is not the same as a single photon is coming out of the source at a time, especially given the weird wave-particle duality). | I think there's a bit of confusion here. The double-slit experiment was not performed with "single photons" - it's very hard to even consider what that would mean. At its heart, it is a thought experiment, and it's not really possible to make a real-life device that tests it. The first low-intensity experiment (Taylor 1909) was challenging the EM field interpretation of photons - the idea was that if photons were localised concentrations of the EM field, as you lowered the intensity, there would be no photons to interfere with each other, and the diffraction pattern would disappear. When the experiment was low-intensity enough that Taylor couldn't distinguish between photons emitted and photons absorbed, he noted that the diffraction pattern still exists, so the photons couldn't just be localised concentrations of the EM field. Dirac had a different explanation - he considered that each individual photon was capable of interacting with itself. Later, the experiments were repeated not with light, but rather, electrons. Electrons are a lot more convenient than photons, since they obey the Pauli principle: it makes a lot more sense to say "an electron here, an electron there". And you can emit individual electrons, which was first tested in 1974, and it was found that individual electrons do in fact display the same interference pattern. Later, it was found that the same pattern also appears for atoms and complex molecules (the current "world record" has the experiment done with a molecule with more than 800 atoms, at ~10 000 atomic weights; the experiment gets much harder with bigger "particles", since it requires much more precision). But we'll stick to electrons, since they're quite convenient. Emission of individual electrons is still quite tricky, but they have a few important properties. They carry a charge, and they have mass. Both of these can be measured, and while this does disturb the electron (change its trajectory), it doesn't absorb it. Photons, on he other hand, will be absorbed by any measurement, which makes them tricky to deal with. So you can measure with precision to individual electrons how many electrons were emitted from your emitor, and how many were absorbed on the detection surface. More importantly, you can try experimenting with what happens when you measure the electrons on the way between the emitor and the detector - and that's when real quantum phenomena come in. Which path does the electron take through the double-slit apparatus? If you're already in the modern quantum physics mindset, the question doesn't really make any sense (and by modern, I still mean the first half of the 20th century :). But as we've seen, electrons can be measured on the way, so what if we put detectors on the two slits? Well, as was of little surprise to the people involved, the pattern disappears completely (this was predicted by Feynmann and others long before the actual experiment was done in 1961). What's more interesting (and still predicted in advance), putting the detectors behind the two slits also destroys the pattern. But this is really a matter for another question entirely. At the core, the answer to your question is "the full experiment cannot be done with light, and was never attempted with light". The best that was attempted was to reduce the intensity of the incident light so low that the energy corresponded to one photon of a given energy - by adding distance and barriers in the light's path. But we know of no way to emit individual photons, or to detect them with certainty, or to measure their paths - and it may very well be outright impossible, not just infeasible. Now, this in itself is quite enough to discredit the dual theory of light (unless you go with Dirac's assertion), but the real experiment was done with single electrons , and continued with bigger and bigger quantum particles, up to the 800+-atom monster I mentioned earlier. | {
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271,031 | Why is the potential energy equals to the negative integral of a force? I am really confused with this negative sign. For example, why there is a negative sign in the gravitational potential energy and what does it mean? I read that the negative sign means you do the same force but in the opposite direction. Doesn't that mean the object shouldn't move? | When you do conservative work on an object, the work you do is equal to the negative change in potential energy $W_c = - \Delta U$ . As an example, if you lift an object against Earth's gravity, the work will be $-mgh$ . Gravity is doing work on the object by pulling it towards the Earth, but since you are pushing it in the other direction, the work you do on the box (and therefore the force) is negative. The field does negative work when you increase a particle's potential energy. Mathematically, it is just that $F=\frac{dW}{dx}$ , which means that if the work is conservative, then $F=\frac{-dU}{dx}$ , since $W_c = - \Delta U$ . Then $-dU = Fdx$ , so $U = - \int F dx$ . We can also say that work is negative when the force and displacement are in opposite directions, since $W = \vec F \cdot d\vec x = Fdxcos\phi$ . When $\phi=\pi$ , then $\cos\phi = -1$ . An example of this conceptually is friction. An object sliding down a plane has kinetic friction acting on it. The friction is in the direction (up the ramp) opposite to the object's motion/displacement. So we say that the friction force is doing negative work. | {
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271,419 | I've just started studying quantum mechanics, and I've come across this correlation between Pauli matrices ($\sigma_i$) and quaternions which I can't grasp: namely, that $i\sigma_1$, $i\sigma_2$ and $i\sigma_3$, along with the 2x2 identity matrix $I$, correspond identically to the four 2x2 matrix representation of unit quaternions. My first guess was that this should have something to do with quaternions being useful for representing orientations and rotations of objects in three dimensions and Pauli matrices being related to the three spatial components of spin, but I didn't really know how to put together those two ideas. Google wasn't much help either: the relation is mentioned, for instance, in this Wikipedia article , but no further explanation is given. Although I suspect there is no direct answer to this question, I would appreciate if someone could enlighten me on the subject. In particular, what is the role of the $i$ factor? | At the level of formulas, the three quaternionic units $i_a$ , $a\in~\{1,2,3\}$ , in $\mathbb{H}\cong \mathbb{R}^4$ satisfy $$i_a i_b ~=~ -\delta_{ab} + \sum_{c=1}^3\varepsilon_{abc} i_c, \qquad\qquad a,b~\in~\{1,2,3\}, \tag{1}$$ while the three Pauli matrices $\sigma_a \in {\rm Mat}_{2\times 2}(\mathbb{C})$ , $a\in~\{1,2,3\}$ , $\mathbb{C}=\mathbb{R}+\mathrm{i}\mathbb{R}$ , satisfy $$\sigma_a \sigma_b ~=~ \delta_{ab} {\bf 1}_{2\times 2} + \mathrm{i}\sum_{c=1}^3\varepsilon_{abc} \sigma_c\quad\Leftrightarrow \quad \sigma_{4-a} \sigma_{4-b} ~=~ \delta_{ab} {\bf 1}_{2\times 2} - \mathrm{i}\sum_{c=1}^3\varepsilon_{abc} \sigma_{4-c}, $$ $$ \qquad\qquad a,b~\in~\{1,2,3\},\tag{2}$$ with complex unit $\mathrm{i}\in\mathbb{C}.$ In other words, we evidently have an $\mathbb{R}$ -algebra monomorphism $$\Phi:~~\mathbb{H}~~\longrightarrow ~~{\rm Mat}_{2\times 2}(\mathbb{C}).\tag{3}$$ by extending the definition $$\Phi(1)~=~{\bf 1}_{2\times 2},\qquad \Phi(i_a)~=~\mathrm{i}\sigma_{4-a}, \qquad\qquad a~\in~\{1,2,3\},\tag{4}$$ via $\mathbb{R}$ -linearity. This observation essentially answers OP title question (v2). However OP's question touches upon many beautiful and useful mathematical facts about Lie groups and Lie algebras, some of which we would like to mention. The image of the $\mathbb{R}$ -algebra monomorphism (3) is $$\Phi(\mathbb{H})
~=~ \left\{\left. \begin{pmatrix} \alpha & \beta \cr -\bar{\beta} & \bar{\alpha} \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{C}) \right| \alpha,\beta \in\mathbb{C}\right\}$$ $$~=~ \left\{ M\in {\rm Mat}_{2\times 2}(\mathbb{C}) \left| \overline{M} \sigma_2=\sigma_2 M\right. \right\}.\tag{5}$$ Let us for the rest of this answer identify $\mathrm{i}=i_1$ . Then the $\mathbb{R}$ -algebra monomorphism (3) becomes $$ \mathbb{C}+\mathbb{C}i_2~=~\mathbb{H}~\ni~x=x^0+\sum_{a=1}^3 i_a x^a ~=~\alpha+\beta i_2$$ $$~~\stackrel{\Phi}{\mapsto}~~ \begin{pmatrix} \alpha & \beta \cr -\bar{\beta} & \bar{\alpha} \end{pmatrix} ~=~ x^0{\bf 1}_{2\times 2}+\mathrm{i}\sum_{a=1}^3 x^a \sigma_{4-a}~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}),$$ $$ \alpha~=~x^0+\mathrm{i}x^1~\in~\mathbb{C},\qquad \beta~=~x^2+\mathrm{i}x^3~\in~\mathbb{C},\qquad x^0, x^1, x^2, x^3~\in~\mathbb{R}.\tag{6}$$ One may show that $\Phi$ is a star algebra monomorphism, i.e. the Hermitian conjugated matrix satisfies $$ \Phi(x)^{\dagger}~=~\Phi(\bar{x}), \qquad x~\in~\mathbb{H}. \tag{7}$$ Moreover, the determinant becomes the quaternionic norm square $$\det \Phi(x)~=~ |\alpha|^2+|\beta|^2~=~\sum_{\mu=0}^3 (x^{\mu})^2 ~=~|x|^2, \qquad x~\in~\mathbb{H}.\tag{8}$$ Let us for completeness mention that the transposed matrix satisfies $$\Phi(x)^t~=~\Phi(x|_{x^2\to-x^2})~=~ \Phi(-j\bar{x}j), \qquad x~\in~\mathbb{H}. \tag{9} $$ Consider the Lie group of quaternionic units, which is also the Lie group $$U(1,\mathbb{H})~:=~\{x\in\mathbb{H}\mid |x|=1 \} \tag{10}$$ of unitary $1\times 1$ matrices with quaternionic entries. Eqs. (7) and (8) imply that the restriction $$\Phi_|:~U(1,\mathbb{H})~~\stackrel{\cong}{\longrightarrow}~~ SU(2)~:=~\{g\in {\rm Mat}_{2\times 2}(\mathbb{C})\mid g^{\dagger}g={\bf 1}_{2\times 2},~\det g = 1 \} $$ $$~=~\left\{\left. \begin{pmatrix} \alpha & \beta \cr -\bar{\beta} & \bar{\alpha} \end{pmatrix} \in {\rm Mat}_{2\times 2}(\mathbb{C}) \right| \alpha, \beta\in\mathbb{C}, |\alpha|^2+|\beta|^2=1\right\}\tag{11}$$ of the monomorphism (3) is a Lie group isomorphism. In other words, we have shown that $$ U(1,\mathbb{H})~\cong~SU(2).\tag{12}$$ Consider the corresponding Lie algebra of imaginary quaternionic number $$ {\rm Im}\mathbb{H}~:=~\{x\in\mathbb{H}\mid x^0=0 \}~\cong~\mathbb{R}^3 \tag{13}$$ endowed with the commutator Lie bracket. [This is (twice) the usual 3D vector cross product in disguise.] The corresponding Lie algebra isomorphism is $$\begin{align}\Phi_|:~{\rm Im}\mathbb{H}~~\stackrel{\cong}{\longrightarrow}~~
su(2)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C})\mid m^{\dagger}=-m \}\cr
~=~&\mathrm{i}~{\rm span}_{\mathbb{R}}(\sigma_1,\sigma_2,\sigma_3),\end{align}\tag{14}$$ which brings us back to the Pauli matrices. In other words, we have shown that $$ {\rm Im}\mathbb{H}~\cong~su(2).\tag{15}$$ It is now also easy to make contact to the left and right Weyl spinor representations in 4D spacetime $\mathbb{H}\cong \mathbb{R}^4$ endowed with the quaternionic norm $|\cdot|$ , which has positive definite Euclidean (as opposed to Minkowski) signature, although we shall only be sketchy here. See also e.g. this Phys.SE post. Firstly, $U(1,\mathbb{H})\times U(1,\mathbb{H})$ is (the double cover of) the special orthogonal group $SO(4,\mathbb{R})$ . The group representation $$\rho: U(1,\mathbb{H}) \times U(1,\mathbb{H}) \quad\to\quad SO(\mathbb{H},\mathbb{R})~\cong~ SO(4,\mathbb{R}) \tag{16}$$ is given by $$\rho(q_L,q_R)x~=~q_Lx\bar{q}_R, \qquad q_L,q_R~\in~U(1,\mathbb{H}), \qquad x~\in~\mathbb{H}. \tag{17}$$ The crucial point is that the group action (17) preserves the norm, and hence represents orthogonal transformations. See also this math.SE question. Secondly, $U(1,\mathbb{H})\cong SU(2)$ is (the double cover of) the special orthogonal group $SO({\rm Im}\mathbb{H},\mathbb{R})\cong SO(3,\mathbb{R})$ . This follows via a diagonal restriction $q_L=q_R$ in eq. (17). | {
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271,501 | Can Newton's 1st and 3rd laws be assumed given just $F=ma$. I know that the argument would be, "No, then there would only be 1 law". But I can't think of any situation where 1 and 3 aren't superfluous. If you just told me $F=ma$:
I would assume nothing else causes an acceleration besides a force. So things not experiencing a force don't change velocity, even when velocity is 0. 1 ✔️ And, when two things that exist interact they use only their mass and acceleration to do so so they both must change in opposite ways. 3 ✔️ | The position you are taking seems to depend on hindsight. Put yourself in the position of Newton being the first person to state these laws. The first law was a flat-out statement that Aristotle was wrong when he stated that "nothing moves at all, unless a force which causes it to move is acting on it." Of course everybody now "knows" that Aristotle was wrong about that, so the "shock and awe factor" of Newton building his entire argument from that starting point no longer exists. The second law then gives a definition of how to numerically measure the notion called "force." Of course it is consistent with the first law, since common sense would say that "no force" must have the measured value of $0$. In modern terminology, the third law is a statement of the principle of conservation of momentum. It is independent of the first two laws - and apparently, the many crackpots who are still trying to invent perpetual motion machines and "free energy" devices still don't believe it is true, despite the empirical evidence (not to mention Noether's theorem ). | {
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272,893 | The title question is rather illustrative. I suppose the real question would be: Is heat cumulative? Put back into an example: If I have a lit candle right beneath an iron bar, assuming the candle will remain lit indefinitely, and that the heat-losing rate is below the heat-getting rate. Will the bar eventually reach the needed temperature for it to melt? If the answer is no : Once the iron bar reached the max temp the candle can get it to. Where does all of the energy (heat) go after? EDIT: The question "is heat cumulative?" can be ignored as it is out of place and is misleading. Althrough the answer for it is "yes", it doesn't mean the answer for the general question is also "yes". The point is not if it is actually possible to melt iron with a candle. Iron and candle are mere parts of the illustration, their properties are irrelevant. A better phrasing of the main question would be: Could a heating object contained in a perfectly closed system push the temperature of the system above its own temperature? | I'll try a simple explanation. Assume that there are no phase transitions initially. As you heat a body, its temperature rises, and it radiates energy into the surrounding space according to $$P = A\varepsilon \sigma T^4$$ ($\sigma$ is the Stefan Boltzman constant, $A$ is the surface area, $T$ is the temperature, $\varepsilon$ is the emmissivity). Obviously, $P$ increases as $T$ increases. So, if before the bar reaches its melting point, $P$ becomes equal to the power input (from the flame), there will be no net flow of energy across the surrounding-bar interface (for if there were, $P$ would be greater than the input power, and more energy will be lost than gained, taking it back to the stable equilibrium point). At this point, notice that the temperature is constant, because any energy supplied by the candle is equivalently emmitted by the body (sort of dynamic equilibrium). Of course, this is only possible if the bar doesn't melt before this temperature is reached. | {
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272,918 | This might be a stupid question, but I'm a newbie to physics. An object less dense than water (or any other fluid, but I'm going to use water for this example) floats normally on Earth when placed in water. But if the object was placed in a hypothetical place where there is no gravity and there is air, it would not float on water. So if the object was placed in water on a planet with more gravity than Earth, would it float more or would it float less, or float the same as on Earth? Would it float more because it doesn't float without gravity, but it does float with Earth gravity, therefore it'd float even more with more gravity. Or would it float less because more gravity would pull the object down, so it won't float as much. Or would it'd float the exact same as on Earth because the above two scenarios cancel each other out. EDIT: By "float more," I mean it rises to the surface of the water faster, and it takes more force to push it down. By "float less," I mean it rises to the surface of the water slower, and it takes less force to push it down. | The object would actually float exactly the same for both values of $g$. Let $V$ be the volume of the body, $d$ its relative density, and $V'$ be the volume inside water.
Then for equilibrium of the body, $V \cdot d \cdot g=V' \cdot 1 \cdot g$ So, $V'/V$ is independent of acceleration due to gravity. | {
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272,923 | In chapter 3 of Schroeder's Thermal Physics, the following table is given on pp. 100. For the case of $N = 100$, the table shows values of various dimensionless quantities related to the multiplicity of a two-state non interacting para magnet. I get all but the last two columns. How have these expressions been derived for $T$ and $C$? Note that Schroeder derives the closed form analytical expressions for $\Omega$, $T$ etc. | The object would actually float exactly the same for both values of $g$. Let $V$ be the volume of the body, $d$ its relative density, and $V'$ be the volume inside water.
Then for equilibrium of the body, $V \cdot d \cdot g=V' \cdot 1 \cdot g$ So, $V'/V$ is independent of acceleration due to gravity. | {
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273,032 | Consider the question, "What is a photon?" . The answers say, "an elementary particle" and not much else. They don't actually answer the question. Moreover, the question is flagged as a duplicate of, "What exactly is a quantum of light?" – the answers there don't tell me what a photon is either. Nor do any of the answers to this question mentioned in the comments. When I search on "photon" , I can't find anything useful. Questions such as, "Wave function of a photon" look promising, but bear no fruit. Others say things like, "the photon is an excitation of the photon field." That tells me nothing. Nor does the tag description , which says: The photon is the quantum of the electromagnetic four-potential, and therefore the massless bosonic particle associated with the electromagnetic force, commonly also called the 'particle of light'... I'd say that's less than helpful because it gives the impression that photons are forever popping into existence and flying back and forth exerting force. This same concept is in the photon Wikipedia article too - but it isn't true. As as anna said , "Virtual particles only exist in the mathematics of the model." So, who can tell me what a real photon is, or refer me to some kind of authoritative informative definition that is accepted and trusted by particle physicists? I say all this because I think it's of paramount importance. If we have no clear idea of what a photon actually is, we lack foundation. It's like what kotozna said : Photons seem to be one of the foundation ideas of quantum mechanics, so I am concerned that without a clear definition or set of concrete examples, the basis for understanding quantum experiments is a little fuzzy. I second that, only more so. How can we understand pair production if we don't understand what the photon is? Or the electron? Or the electromagnetic field? Or everything else? It all starts with the photon. I will give a 400-point bounty to the least-worst answer to the question. One answer will get the bounty, even if I don't like it. And the question is this: What exactly is a photon? | The photon is a construct that was introduced to explain the experimental observations that showed that the electromagnetic field is absorbed and radiated in quanta. Many physicists take this construct as an indication that the electromagnetic field consists of dimensionless point particles, however of this particular fact one cannot be absolutely certain. All experimental observations associated with the electromagnetic field necessarily involve the absorption and/or radiation process. So when it comes to a strictly ontological answer to the question "What is a photon?" we need to be honest and say that we don't really know. It is like those old questions about the essence of things; question that could never really be answered in a satisfactory way. The way to a better understanding often requires that one becomes comfortable with uncertainty. | {
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273,724 | What is the shape of a deuterium nucleus? I can think of two obvious extremes. A positive proton end intersecting with a neutral neutron end. Or a cylinder with spherical caps on the ends that is positive on one end and neutral on the other. | This is an extended comment on count_to_10's answer, please upvote that answer not this one. It's tempting to think that because a deuterium nucleus is made up from two different particles, one positive and one neutral, that it must necessarily be asymmetric. However a hydrogen atom is also made up from two different particles, one positive and one negative, and it is spherically symmetric. The reason for the spherical symmetry of the hydrogen atom is that the proton and electon are both delocalised. The same argument applies to the deuterium nucleus except that the strong nuclear force is not a central force but depends on orientation, so we should not expect spherical symmetry. There won't be one end of the nucleus that is the proton and the other end that is the neutron because both particles are delocalised. In fact, as count_to_10 describes, the deuterium nucleus is axially symmetric and centrosymmetric. This means the two ends are identical - there is not a neutron end and a proton end. | {
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273,780 | I'm fairly new to the subject of quantum field theory (QFT), and I'm having trouble intuitively grasping what a n-point correlation function physically describes. For example, consider the 2-point correlation function between a (real) scalar field $\hat{\phi}(x)$ and itself at two different space-time points $x$ and $y$, i.e. $$\langle\hat{\phi}(x)\hat{\phi}(y)\rangle :=\langle 0\rvert T\lbrace\hat{\phi}(x)\hat{\phi}(y)\rbrace\lvert 0\rangle\tag{1}$$ where $T$ time-orders the fields. Does this quantify the correlation between the values of the field at $x=(t,\mathbf{x})$ and $y=(t',\mathbf{y})$ (i.e. how much the values of the field at different space-time points covary, in the sense that, if the field $\hat{\phi}$ is excited at time $t$ at some spatial point $\mathbf{x}$, then this will influence the "behaviour" of the field at later time $t'$ at some spatial point $\mathbf{y}$)? Is this why it is referred to as a correlation function? Furthermore, does one interpret $(1)$ as physically describing the amplitude of propagation of a $\phi$-particle from $x$ to $y$ (in the sense that a correlation of excitations of the field at two points $x$ and $y$ can be interpreted as a "ripple" in the field propagating from $x$ to $y$)? | Yes, in scalar field theory, $\langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is the amplitude for a particle to propagate from $x$ to $y$. There are caveats to this, because not all QFTs admit particle interpretations, but for massive scalar fields with at most moderately strong interactions, it's correct. Applying the operator $\phi({\bf x},t)$ to the vacuum $|0\rangle$ puts the QFT into the state $|\delta_{\bf x},t \rangle$, where there's a single particle whose wave function at time $t$ is the delta-function supported at ${\bf x}$. If $x$ comes later than $y$, the number $\langle 0 | \phi({\bf x},t)\phi({\bf y},t') | 0 \rangle$ is just the inner product of $| \delta_{\bf x},t \rangle$ with $| \delta_{\bf y},t' \rangle$. However, the function $f(x,y) = \langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is not actually a correlation function in the standard statistical sense. It can't be; it's not even real-valued. However, it is a close cousin of an honest-to-goodness correlation function. If make the substitution $t=-i\tau$, you'll turn the action
$$iS = i\int dtd{\bf x} \{\phi(x)\Box\phi(x) - V(\phi(x))\}$$
of scalar field theory on $\mathbb{R}^{d,1}$ into an energy function
$$-E(\phi) = -\int d\tau d{\bf x} \{\phi(x)\Delta\phi(x) + V(\phi(x))\}$$
which is defined on scalar fields living on $\mathbb{R}^{d+1}$. Likewise, the oscillating Feynman integral $\int \mathcal{D}\phi e^{iS(\phi)}$ becomes a Gibbs measure $\int \mathcal{D}\phi e^{-E(\phi)}$. The Gibbs measure is a probability measure on the set of classical scalar fields on $\mathbb{R}^{d+1}$. It has correlation functions $g(({\bf x}, \tau),({\bf y},\tau')) = E[\phi({\bf x}, \tau)\phi({\bf y},\tau')]$. These correlation functions have the property that they may be analytically continued to complex values of $\tau$ having the form $\tau = e^{i\theta}t$ with $\theta \in [0,\pi/2]$. If we take $\tau$ as far as we can, setting it equal to $i t$, we obtain the Minkowski-signature "correlation functions" $f(x,y) = g(({\bf x},it),({\bf y},it'))$. So $f$ isn't really a correlation function, but it's the boundary value of the analytic continuation of a correlation function. But that takes a long time to say, so the terminology gets abused. | {
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273,852 | The Perseid meteor shower is caused by debris left behind Comet Swift-Tutte. My understanding is that the meteor shower occurs once per year because that's when the Earth passes near the orbital path of the comet. My question is why this shower only occurs once per year. I'm imagining that the Earth has an elliptical orbit around the sun, as does the comet. If the comet's orbit intersects the Earth's orbit at some point, since they're both ellipses, I would have thought that there would be (at least) two intersection points with the Earth's orbit, which would mean that we'd see this meteor shower twice or four times in the year as the Earth passes through these intersection points. I'm assuming there's something wrong with my reasoning - can anyone clarify this for me? | Comets do orbit in ellipses, but there is no requirement for their orbital planes to match the Earth's exactly. Short-period comets have planes that are relatively parallel to the Earth's (with a good degree of lee-way), and long-period comets can come in from any direction and with any orbital inclination. Here is a good representation of the orbit of comet 109P/Swift-Tuttle: When the comet's orbit crosses the orbital plane of the Earth again, it is way out past Jupiter, and this means that there is just the single meteor shower from this debris track. Also, it's worthwhile noting that while ellipses in general can intersect four times, keplerian orbits share a focus, which means that they can intersect at most twice (which they do when they're coplanar and the sizes are right). | {
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274,227 | As the main title says. I'm finding myself wondering about helicopters. The tail rotor is a vulnerable and key piece of equipment, especially on military helicopters. I know some helicopters instead use two main rotors (for example the KA-50). Why not use a reaction wheel? The main engine could power the wheel, and it could be placed in an armored area and less vulnerable to fragmentation munition. Is it because any reaction wheel would be prohibitively large? | You're talking about a device (in helicopters the tail fan imparting horizontal thrust) that counteracts the torque imparted on the main rotor (and therefore on the helicopter) by the surrounding air as the main rotor is dragged through the air. You propose instead to impart an opposite torque through a reaction wheel. That would indeed impart an opposite torque for short lengths of time . However, you don't get a torque from spinning a reaction wheel at constant angular velocity but by changing and accelerating that angular velocity. Now the torque imparted on the helicopter by the air through the main rotor is steady - or at least its of roughly constant direction. Therefore, to counter that torque, the reaction wheel would have to accelerated uniformly and indefinitely. Clearly this is impossible from an engineering standpoint. You can also think of this from a conservation of angular momentum, without thinking about the origin of the torques. The air imparts a steady angular impulse to the helicopter. Therefore, the helicopter system's angular momentum must increase steadily (unless there's a countering torque from the tailfan). So either that angular momentum is the spinning of the helicopter's body (which is what we're trying to avoid) or that of the reaction wheel, whose angular momentum must be steadily increasing under the action of the angular impulse to the system. | {
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274,858 | According to this article , a muon decays into one electron and two neutrinos. According to this article , elementary particles or fundamental particles are particles "whose substructure is unknown, thus it is unknown whether it is composed of other particles." I have also seen somewhere that it is a particle that cannot be reduced into other constituent particles. While perhaps not a sure thing, seems like the decay indicates that the muon may be just a composite particle, perhaps consisting of one electron and two neutrinos? Based on this, why does the muon fit with the above definition of an elementary or fundamental particle? I realize there are much more complicated, historical reasons as to why it was included in the Standard Model, but this question is just related so how it fits (or doesn't fit) the stated definition above. It seems to me that we really can only get solid evidence of elementary vs. composite when we smash the particles together and see what comes out and compare that to all the masses, energies and momentum before and after? Until we do that with muons, how can we know with much certainty? And perhaps we'll have a better answer with a Muon collider: https://en.wikipedia.org/wiki/Muon_collider To that point, seems that electrons may not be fundamental after all: https://www.sciencedaily.com/releases/2016/04/160404111559.htm | Addressing misconceptions First, I address some misconceptions in your question. the decay indicates that the muon may be just a composite particle The fact that the muon decays at all is not evidence that it's composite. It's tempting to say that if a particle $A$ can decay into $B$ and $C$ , then it must be "made of" $B$ and $C$ . However, this doesn't work out, because almost all particles have multiple decay channels. For example, hydrogen in the $2s$ state can release a photon to go to the $1s$ state, but it can also rarely do this by releasing two photons. As a more extreme example, parapositronium can completely annihilate, turning into two photons, but it can also turn into four . We think about particle decay in terms of couplings of quantum fields to each other: an excitation in one field can decay into excitations in others. As Feynman put it, those final excitations don't exist "inside" the original one, any more than the word "cat" is bouncing around inside you because you can spend energy to say it. To that point, seems that electrons may not be fundamental after all: https://www.sciencedaily.com/releases/2016/04/160404111559.htm This article is about some of the weird ways that large collections of electrons in solids can behave collectively, but it's not related to whether or not electrons themselves are composite. It's important to keep this in mind when reading news releases, because the people who study what electrons in solids do unfortunately tend to give the resulting phenomena the same names as the particles we search for in colliders, leading to a lot of popular confusion. Answering the question With that in mind, you're still right, in the sense that it's completely natural to think that the muon might be composite. If you were a scientist in the 1950s, for example, the muon would be just one more particle discovered along with a zoo of mesons and hadrons. Today, we know that all of those mesons and hadrons turned out to be composites of quarks. So why not think of the muon as composite as well? Indeed, in the early days, the similarity of the muon and electron was taken as possible evidence that the muon was an excited state of the electron, just like the $2s$ state is an excited state of hydrogen. If this were the case, one would expect the muon to often decay by emitting a photon, $\mu \to e \gamma$ , but this was found not to be the case . Instead, the decays involving neutrinos dominate. Now you might ask, why can't the muon be a composite of the electron bound to some neutrinos? This idea doesn't work out because there's no force we know of that would do the job: even in the 1950s it was known that neutrinos interacted extremely weakly. Getting a neutrino to interact with an electron at all is less likely than winning the lottery, so it seems extremely unlikely that it would be simultaneously possible to bind them together. Another difficulty of any composite muon theory is explaining the muon g-factor , which determines its magnetic moment. Elementary particles are expected to have $g \approx 2$ . The composite proton and neutron violate this by a good margin, $$g_p \approx 5.59, \quad g_n \approx -3.82$$ while the electron and muon have $$g_e \approx 2.002, \quad g_\mu \approx 2.002.$$ That $0.002$ isn't evidence for compositeness either, because it's precisely what you would expect for a perfectly elementary particle, once you include quantum field theoretic effects. In fact, the electron and muon $g$ -factors have been measured to far more decimal places than I've shown, and the results match the Standard Model prediction to great precision. Making the electron and muon composite without upsetting this agreement would seem to require a seriously contrived model, or a miracle. A meta-difficulty These already are big difficulties, but if you imagine being a scientist in the 1950s, the quark model had its own problems (such as the complete nonobservability of individual quarks), but it earned support because of its ability to account for huge numbers of hadrons, and predict new ones. And today, people consider theories where the Higgs boson is composite, because it helps give it an appropriate mass. The meta-difficulty for the muon is that it's only worth trying to make it composite if there's some payoff you expect, such as (1) the completion of a theoretical picture, (2) new predictions, or (3) ways to calculate quantities (such as the muon mass) that we otherwise have to take as inputs. The first reason doesn't apply, because the muon already has a perfectly good place in the Standard Model: it has to be there because of the family structure of the theory, and this structure is rigid enough that without the muon, the Standard Model would be mathematically inconsistent because of gauge anomalies . The second reason doesn't apply, either. It's not like we have a series of weird particles lying around that could be explained as further composites of the electron. And since we've measured properties of the muon to exquisite precision, just about any theory of muon compositeness will make "predictions" that we already know to be wrong! You have to work extremely hard just to avoid that. (Admittedly, the muon $g$ -factor does seem to deviate a bit from the predicted value, and this does receive attention -- it's just that compositeness isn't the kind of thing that would help here.) The third reason could potentially apply. However, explaining the masses of particles like the electron and muon is an infamously hard problem, even if you don't take them as composite. Certainly, heads would turn if you came up with a simple theory that gave the muon-electron mass ratio to many decimal places, but decades of failed attempts have made this seem unlikely. If you just disregarded these reasons, and made a contrived model where the muon was composite, tuning all the constants involved to precisely the values needed to hide all deviations from the Standard Model, then it would "work"... but it would also be scientifically useless. Of course, it's also completely possible that muons might turn out to be non-elementary, because in science it's impossible to ever prove a negative! At the moment, this possibility is not under active investigation. But it's not heresy either. If sufficiently strange experimental results appeared in the future, scientists could be right back to tinkering with composite electrons and muons, trying their best to understand the results, and the universe. | {
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275,089 | I've been told I should defrost my freezer to save energy, wiki , here and here for example, but none of the linked sites is a peer-reviewed paper explaining why (the wiki article doesn't even have references), and I don't find it obvious. I don't understand how the mechanism works, and I ask you for a good paper read on the subject or an explanation. | Refrigerators and freezers work by running a really cold liquid through cooling pipes fitted in the cavity to be cooled. This flow (the compressor) is switched off when the set temperature is reached, the faster the set temperature is reached, the less energy the appliance uses. Cold fluid at $T_c$ runs through the cooling pipes. The cavity to be cooled is at $T_f$. Now let's look at small area $A$ on the surface of a cooling pipe. When the cooling pipe is clean (not frosted over) then Newton's cooling law tells us that the heat flux (amount of heat removed per unit of time) $\dot{q}$ through $A$ is: $$\dot{q}_\textrm{clean}=hA(T_f-T_c)$$ Where $h$ is the heat transfer coefficient . But when the surface is frosted over with porous ice, then: $$\dot{q}_\textrm{frosted}=uA(T_f-T_c)$$ It can be shown that: $$\frac{1}{u}=\frac{1}{h}+\frac{\theta}{k}\implies u=\frac{hk}{k+h\theta}$$ Where $\theta$ is the thickness of the frosty material and $k$ the thermal conductivity of the frosty material. Because the frosty material is a poor conductor of heat ($k$ has a low value): $$h>\frac{hk}{k+h\theta}$$ ( Note that the frosty material isn't pure ice, it's highly porous ice that contains much entrapped air, thereby further lowering the $k$ value of the frost ). And this means that, all other things being equal: $$\dot{q}_\textrm{clean}>\dot{q}_\textrm{frosted}$$ Multiply this of course for the total surface area of the cooling pipes. So clean cooling pipes carry away the heat more quickly, resulting in the compressor running for shorter times to reach the set temperature. This saves energy, Note also how freezers that have been frosted over more (higher frost thickness $\theta$) perform worse. A slightly more detailed approach: $$\dot{q}_\textrm{clean}=u_1A(T_f-T_c)$$
$$\dot{q}_\textrm{frosted}=u_2A(T_f-T_c)$$
Here it can be shown that: $$\frac{1}{u_1}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{1}{h_2}$$
And:
$$\frac{1}{u_2}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{\theta_2}{k_2}+\frac{1}{h_3}$$ But here too, because the frost conducts heat poorly ($k_2$ is small): $$u_1>u_2$$ So that clean pipes carry off the heat more quickly, all other things being equal. Symbols used in this section : $h_1$: convection heat transfer coefficient, cooling fluid to metal . $h_2$: convection heat transfer coefficient, metal to air . $h_3$: convection heat transfer coefficient, frost to air . $k_1$: thermal conductivity, metal . $k_2$: thermal conductivity, frost . $\theta_1$: thickness, metal . $\theta_2$: thickness, frost . | {
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275,091 | ERROR: type should be string, got "https://en.wikipedia.org/wiki/Observable_universe Mass (ordinary matter) 10^53 kg Ordinary (baryonic) matter (4.9%) Diameter 8.8×10^26 m OU=observable universe SR=Schwarzschild radius Mass including dark matter/energy = 10^53 kg / 4.9% = 2E54 kg SR = 2*G*m/c^2 = 3E27 m = 7 times the radius of the OU. Anything wrong so far? I want to write that in wikipedia. Any unaccounted effect of matter expanding thus gravity at a distance not reflecting the position of where matter is now or so? I read the SR of the OU is equal to the radius of the OU, making the OU \"flat\". Normally I would call a mass in a sphere 1/7 of it's SR a black hole. But here I'm assuming that the universe is filled with matter fairly evenly (despite irregularities like galaxies, supermassive black holes, stars or voids on a small scale :), so gravity generally cancels and everything is free to go wherever it wants and cross Schwarzschild radii which are everywhere depending on where you pick the center of the OU. Now, would it be possible to get stuff out of a black hole if it's gravity was cancelled by nearby matter similar to the OU? Maybe a black hole cluster with overlapping event horizons. Would singularities remain (naked?) singularities and stuff could escape the singularities it was originally bound to? Could stuff eventually escape the black hole cluster, perhaps if there were also non singularities like neutron stars and normal stars in the cluster?" | Refrigerators and freezers work by running a really cold liquid through cooling pipes fitted in the cavity to be cooled. This flow (the compressor) is switched off when the set temperature is reached, the faster the set temperature is reached, the less energy the appliance uses. Cold fluid at $T_c$ runs through the cooling pipes. The cavity to be cooled is at $T_f$. Now let's look at small area $A$ on the surface of a cooling pipe. When the cooling pipe is clean (not frosted over) then Newton's cooling law tells us that the heat flux (amount of heat removed per unit of time) $\dot{q}$ through $A$ is: $$\dot{q}_\textrm{clean}=hA(T_f-T_c)$$ Where $h$ is the heat transfer coefficient . But when the surface is frosted over with porous ice, then: $$\dot{q}_\textrm{frosted}=uA(T_f-T_c)$$ It can be shown that: $$\frac{1}{u}=\frac{1}{h}+\frac{\theta}{k}\implies u=\frac{hk}{k+h\theta}$$ Where $\theta$ is the thickness of the frosty material and $k$ the thermal conductivity of the frosty material. Because the frosty material is a poor conductor of heat ($k$ has a low value): $$h>\frac{hk}{k+h\theta}$$ ( Note that the frosty material isn't pure ice, it's highly porous ice that contains much entrapped air, thereby further lowering the $k$ value of the frost ). And this means that, all other things being equal: $$\dot{q}_\textrm{clean}>\dot{q}_\textrm{frosted}$$ Multiply this of course for the total surface area of the cooling pipes. So clean cooling pipes carry away the heat more quickly, resulting in the compressor running for shorter times to reach the set temperature. This saves energy, Note also how freezers that have been frosted over more (higher frost thickness $\theta$) perform worse. A slightly more detailed approach: $$\dot{q}_\textrm{clean}=u_1A(T_f-T_c)$$
$$\dot{q}_\textrm{frosted}=u_2A(T_f-T_c)$$
Here it can be shown that: $$\frac{1}{u_1}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{1}{h_2}$$
And:
$$\frac{1}{u_2}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{\theta_2}{k_2}+\frac{1}{h_3}$$ But here too, because the frost conducts heat poorly ($k_2$ is small): $$u_1>u_2$$ So that clean pipes carry off the heat more quickly, all other things being equal. Symbols used in this section : $h_1$: convection heat transfer coefficient, cooling fluid to metal . $h_2$: convection heat transfer coefficient, metal to air . $h_3$: convection heat transfer coefficient, frost to air . $k_1$: thermal conductivity, metal . $k_2$: thermal conductivity, frost . $\theta_1$: thickness, metal . $\theta_2$: thickness, frost . | {
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275,219 | If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it.
This was solved here . Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it. I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.) $$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$ $m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder. | You should stand at 2/3 of the height of the ladder. If you land with the same kinetic energy as without a ladder, then the ladder should land with the same kinetic energy as without you. Equating the kinetic energy of the ladder with its potential energy at the beginning: $$\frac{1}{2} mgL = \frac{1}{2} I_L \omega^2 = \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2$$ gives:
$$\omega = \sqrt{\frac{3g}{L}}$$ where $L$ is the length, $m$ the mass, $I_L$ the moment of inertia and $\omega$ is the angular velocity of the ladder. For you the same equation holds, but now $\omega$ is known: $$MgH = \frac{1}{2} I_M \omega^2 = \frac{1}{2} (MH^2) \left(\frac{3g}{L}\right)$$
with $M$ your mass, $I_M$ your moment of inertia and $H$ your height. Solving for $H$ gives: $$H=\frac{2}{3}L$$ or of course $H=0\;.$ | {
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275,340 | I'm trying to design a script for a Unity game that can accurately (give or take) simulate heat based on distance from a source. Using the inverse square law seems to be a way I can do this, but setting distance to zero always seems to set intensity to infinity. Am I misunderstanding something about the inverse square law? | This is just an artefact from assuming your heat source is infinitesimal in size. Also, more generally, physics tends to break down in the limit $r\to 0$. So, in many cases, you have the luxury of doing 'something else' in the vicinity of $r=0$. For example, it is common in various simulation settings (n-body, free energy perturbation, etc) to use so-called soft-core potentials. These are potentials that are modified so that the singularities disappear. For a $1/r^n$ potential, it's common to replace it with something of the form
$$ \frac{1}{(1+r^{ns})^{1/s}} $$ Here is a plot for $n=2,s=3$. | {
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275,363 | In classical physics, particles and fields are completely different stuff. However, when a field is quantized, the particles appear as its excitations (e.g. photon appears as a field excitation in the quantization of electromagnetic field). In fact, for all the elementary particles, there is a corresponding field. I am interested whether this is also true for any composite particles. Could we define, for any given composite particle, a field for which, upon quantization, that composite particle appears as its excitation? Is there, for example, anything like "hydrogen atom field"? | It depends on the exact circumstance whether or not such an idea is a good approximation for the physics you want to describe. For the hydrogen atom, you're usually not interesting in it's "scattering behaviour", you're interested in its internal energy states, how it behaves in external electromagnetic fields, etc. Such internal energy states are not well-modelled by QFT. In particular, you'll usually want to consider the proton as "fixed" and the electron as able to jump between its different energy levels. Considering the "hydrogen atom" as an indivisible (or atomic, as it were...) object is not particularly useful. But there are composite particles where associating a field is perfectly sensible, for example the pion , whose effective field theory describes the nuclear force between hadrons - and the hadrons are also composite particles that are treated with a single field here, for instance by means of chiral perturbation theory . There are, besides an interest in scattering behaviour (which you also might legitimately have for the hydrogen atom or other atoms, I'm not implying you should never treat the hydrogen atom this way), other reasons to model certain objects as the particles of a field: Modern many body physics as in condensed matter theory is essentially quantum field theory, too, and it is very frequent there to have fields for composite particles, or even pseudo-particles like phonons . For instance, a simple but powerful model for superconductivity, the Landau model , just treats a conductor as a bunch of charged bosonic particles, thought of as the quanta of a field, coupled to the electromagnetic field, and superconductivity is then another instance of the Higgs mechanism of quantum field theory. | {
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275,375 | From my childhood till now, I have been wondering about the "pink", or sometimes "magenta" color of the evening sky. According to the scattering of light phenomenon, the colors pink/magenta do not directly appear in the spectrum. Moreover, it is said that pink appears as a result of red+white. Do these colors mix somewhere in the sky? Please explain the appearance of sky colors other than blue, red, yellow. | It depends on the exact circumstance whether or not such an idea is a good approximation for the physics you want to describe. For the hydrogen atom, you're usually not interesting in it's "scattering behaviour", you're interested in its internal energy states, how it behaves in external electromagnetic fields, etc. Such internal energy states are not well-modelled by QFT. In particular, you'll usually want to consider the proton as "fixed" and the electron as able to jump between its different energy levels. Considering the "hydrogen atom" as an indivisible (or atomic, as it were...) object is not particularly useful. But there are composite particles where associating a field is perfectly sensible, for example the pion , whose effective field theory describes the nuclear force between hadrons - and the hadrons are also composite particles that are treated with a single field here, for instance by means of chiral perturbation theory . There are, besides an interest in scattering behaviour (which you also might legitimately have for the hydrogen atom or other atoms, I'm not implying you should never treat the hydrogen atom this way), other reasons to model certain objects as the particles of a field: Modern many body physics as in condensed matter theory is essentially quantum field theory, too, and it is very frequent there to have fields for composite particles, or even pseudo-particles like phonons . For instance, a simple but powerful model for superconductivity, the Landau model , just treats a conductor as a bunch of charged bosonic particles, thought of as the quanta of a field, coupled to the electromagnetic field, and superconductivity is then another instance of the Higgs mechanism of quantum field theory. | {
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275,645 | A heatsink can be stuck on your CPU to cool it down. That heatsink feels cold when the system is not turned on. However when the CPU is turned on the heatsink is extremely hot. Isn't that contradictory to a certain extent? I expect a heatsink to always be cold and to have the air blowing system turned on for convection purposes. Could someone explain what is wrong in the way I see things? | "I expect a heatsink to always be cold" The purpose of the heat sink is to transfer heat. The rate of heat transfer depends in a complicated way on (a) the temperature difference between heat sink and air, (b) the exposed surface area, and (c) the air speed. The heat sink is at its most effective if it is at the same temperature as the CPU. This is because this gives the maximum temperature difference between heat sink and air. This is why heat sinks are often made of copper: copper conducts heat well. Let us suppose that the CPU is producing heat $Q$ (Watts) and that the heat sink transfer heat at a rate linear with the temperature difference: $$ Q = K_\text{eff} (T_\text{CPU} - T_\text{air})$$ where $K_\text{eff}$ is the effective heat transfer coefficient. Solving for the CPU temperature: $$ T_\text{CPU} = T_\text{air} + Q / K_\text{eff}$$ So, the CPU temperature is lowest when $K_\text{eff}$ is highest and $K_\text{eff}$ is highest when the heat sink, due it to having high thermal conductivity, is at nearly the same temperature as the CPU. | {
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275,669 | In spite of their different dimensions, the numerical values of $\pi^2$ and $g$ in SI units are surprisingly similar,
$$\frac{\pi^2}{g}\approx 1.00642$$ After some searching, I thought that this fact isn't a coincidence, but an inevitable result of the definition of a metre, which was possibly once based on a pendulum with a one-second period. However, the definition of a metre has changed and is no longer related to a pendulum (which is reasonable as $g$ varies from place to place), but $\pi^2 \approx g$ still holds true after this vital change. This confused me: is $\pi^2 \approx g$ a coincidence? My question isn't about numerology, and I don't think the similarity between the constant $\pi^2$ and $g$ of the planet we live on reflects divine power or anything alike - I consider it the outcome of the definitions of SI units. This question is, as @Jay and @NorbertSchuch pointed out in their comments below, mainly about units and somewhat related to the history of physics. | The differential equation for a pendulum is $$\ddot{\phi}(t) = -\frac{g}{l}\cdot\sin{\phi(t)}$$ If you solve this, you will get
$$\omega = \sqrt{\frac{g}{l}}$$
or
$$T_{1/2}=\pi\sqrt{\frac{l}{g}}$$
$$g=\pi^2\frac{l}{T_{1/2}^2}$$ If you define one metre as the length of a pendulum with $T_{1/2}=1\,\mathrm{s}$ this will lead you inevitably to $g=\pi^2$. This was actually proposed, but the French Academy of Sciences chose to define one metre as one ten-millionth of the length of a quadrant along the Earth's meridian. See Wikipedia’s article about the metre . That these two values are so close to each other is pure coincidence. (Well, if you don't take into account that the French Academy of Sciences could have chosen any fraction of the quadrant and probably took one matching the one second pendulum.) Besides that, $\pi$ has the same value in every unit system, because it is just the ratio between a circle’s diameter and its circumference, while $g$ depends on the chosen units for length and time. | {
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275,672 | I came across the following problem in my textbook: "Two particles A and B are projected in air. A is thrown with a speed of 30m/s and B with a speed of 40m/s as shown in the figure. What is the separation between them after 1 sec?" I approached this problem by resolving both velocity vectors into its components along the x(positive toward right) and y(positive upwards) axes. Then, I found out the velocity of A w.r.t B and solved the problem by using this relative velocity and acceleration. In the solutions, however, they have approached it in a much quicker method as follows: I don't understand how they've been able to find the relative velocity by taking the square root of 30²+40² directly. What point am I missing? Please do help. Thanks in advance :) | The differential equation for a pendulum is $$\ddot{\phi}(t) = -\frac{g}{l}\cdot\sin{\phi(t)}$$ If you solve this, you will get
$$\omega = \sqrt{\frac{g}{l}}$$
or
$$T_{1/2}=\pi\sqrt{\frac{l}{g}}$$
$$g=\pi^2\frac{l}{T_{1/2}^2}$$ If you define one metre as the length of a pendulum with $T_{1/2}=1\,\mathrm{s}$ this will lead you inevitably to $g=\pi^2$. This was actually proposed, but the French Academy of Sciences chose to define one metre as one ten-millionth of the length of a quadrant along the Earth's meridian. See Wikipedia’s article about the metre . That these two values are so close to each other is pure coincidence. (Well, if you don't take into account that the French Academy of Sciences could have chosen any fraction of the quadrant and probably took one matching the one second pendulum.) Besides that, $\pi$ has the same value in every unit system, because it is just the ratio between a circle’s diameter and its circumference, while $g$ depends on the chosen units for length and time. | {
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275,678 | Suppose in an entanglement experiment, angles of measurement are fixed. QM predicts that $50$% pairs will measure same spin. Ignoring other details, just consider QM predicted value is $50$%. Then does QM also predicts the distribution of same spin pairs? If so, is that predicted distribution any different from that of any other event with $50$% probability, like tossing a coin? To clarify further - say probability of getting a head in a coin toss is $50$%. If we toss the coin $100$ times, probability says - expected number of $10$ consecutive heads is = $(100 - 9)*(.5)^{10}$. Would this be also the expected number of $10$ consecutive same spin pairs out of $100$ pairs with same spin pair prediction at $50$%. Please ignore the fact that $100$ is a small number for entanglement. The question is just about the concept. Further clarification - As far as state is concerned, you may consider/assume any state. Only thing is that the number of same spin pairs is predicted at 50%. I said 50 because it can be compared with a coin toss. The real question is - given a prediction of 50%, is expected number of m consecutive same spin pairs, same as the expected number of m consecutive heads in a coin toss. The probability of getting a same spin pair and that of getting a head being same. Total number of entangled pairs measured also same as total number of coin tosses. | The differential equation for a pendulum is $$\ddot{\phi}(t) = -\frac{g}{l}\cdot\sin{\phi(t)}$$ If you solve this, you will get
$$\omega = \sqrt{\frac{g}{l}}$$
or
$$T_{1/2}=\pi\sqrt{\frac{l}{g}}$$
$$g=\pi^2\frac{l}{T_{1/2}^2}$$ If you define one metre as the length of a pendulum with $T_{1/2}=1\,\mathrm{s}$ this will lead you inevitably to $g=\pi^2$. This was actually proposed, but the French Academy of Sciences chose to define one metre as one ten-millionth of the length of a quadrant along the Earth's meridian. See Wikipedia’s article about the metre . That these two values are so close to each other is pure coincidence. (Well, if you don't take into account that the French Academy of Sciences could have chosen any fraction of the quadrant and probably took one matching the one second pendulum.) Besides that, $\pi$ has the same value in every unit system, because it is just the ratio between a circle’s diameter and its circumference, while $g$ depends on the chosen units for length and time. | {
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275,733 | Imagine you're an astronaut on the International Space Station and your fellow astronauts played a prank on you by taking all your clothes and putting you in the center of a module so that you cannot reach anything with either your hands or your feet. What would be the most effective way to escape that situation if you're reluctant to start peeing? | Since you have air around you, you can just take a deep breath and blow it out. There's actually no need to turn you head while doing this, as some other answers suggest. Air has a high Reynolds number at human scales, and so the scallop theorem does not hold: even though the movements of inhaling and exhaling are reciprocal, the airflows they create are not. (You can test this yourself by holding a hand in front of your mouth: you can easily feel the jet of air created by blowing out, even with your hand fully extended, but you can't produce a "reverse jet" no matter how hard you inhale.) In practice, the momentum produced by inhaling is pretty negligible, as air flows in towards your mouth and nose from all sides, and so the only thing that matters is which way you exhale. By blowing air out of your mouth in one direction, you create a net airflow in that direction, and so, by conservation of momentum, propel yourself in the opposite direction. It works for squid and jellyfish (and scallops!) , and it will work for you, too. Maybe not very efficiently, but surely enough to reach a wall in the tight confines of the ISS. Now, if we ever start building space stations with huge air-filled bubbles hundreds of meters across, then this might become a problem, but until then you should be fine. Besides, you may not even need to resort to such huffing and puffing. Any real space station designed for human habitation in microgravity needs to have active air circulation fans anyway, both for heat distribution (important for both humans and equipment, since convection doesn't work in microgravity) and to keep exhaled air from accumulating around your body e.g. when you're sleeping. So in practice, the air around you will be moving slowly anyway, and you just need to wait until this ambient airflow pushes you close to a wall. And of course, on the actual ISS, I doubt there's even any space big enough to properly pull off this prank. The largest open spaces on the ISS, like the Kibo pressurized module , are surrounded by ISPRs that are about 2 meters (6 ½ feet) wide, effectively making the interior cross-section a 2×2 meter square. Even if your crewmates somehow managed to position your body lengthwise along the center axis of an otherwise empty module so that you couldn't just reach out and grab a handhold, you'd just need to twist around like a cat (or, more likely, just flail around semi-randomly) until you managed to turn yourself 90° around, at which point either your toes or your hands should surely be able to reach a wall. | {
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275,921 | I heard that the way light speed is shown in the movie "Star Wars" is not realistic. About 1 month ago in a documentary on TV they said that it would be the opposite of the way it looks in "Star Wars", where lights are moving in front of you.
What would light speed really look like if we could see it with our own eyes? | I think you mean, what does the outside world look like if you could travel at near light speed? Get ready to be disappointed. Apologies to all Star Wars & Star Trek fans and SF readers everywhere. This idea of a streaky starfield may be incorrect, as Hollywood has forgotten about the CMB. Instead, this picture below is apparently correct: Not much is it, after all the effort of getting to nearly the speed of light? Image Source and extracts from Looking out the Window at Near Light Speed A group of physics students at the University of Leicester has discovered, assuming a ship could travel at nearly the speed of light, a crew would see a giant, fuzzy orb in the distance. For their study, the students assumed that the Millennium Falcon (yes, this was the wording used in the study) is traveling at 99.99995 percent the speed of light (c) as it zips past the Earth towards the Sun (at a distance of 1 AU). Obviously, in keeping with the laws established by Albert Einstein, and unlike some sci-fi interpretations of faster-than-light space travel (i.e. "hyperspace"), the students could not assume a value greater than c. The reseach group found that, as you approach light speed, the cosmic background radiation dominates the view A Doppler blueshift effect would be created by the electromagnetic radiation — including visible light — that is rapidly moving towards the crew. This effect, say the researchers, would shorten the wavelength of electromagnetic radiation. From the perspective of the crew, the higher frequency of the light from neighboring stars would transform the previously visible spectrum into the X-ray range — thus making these stars invisible to the human eye.
It was also discovered that, X ray pressure would reduce the velocity of space craft. I want to include 4 possibly salient points made by Rob Jeffries regarding the above image and my answer: What is the angular scale of this picture supposed to be? To blueshift the microwave background so that it looks blue/white takes a redshift of a factor $>2000$, which is indeed what the speed quoted produces. This corresponds to a Lorentz factor of $\gamma=1000$. So wouldn't the "blob" actually be a small spot of angular extent $\sim \gamma^{-1} = 0.06$ degrees. The purported picture has no angular or intensity scale, so its correctness cannot be ascertained. I contend that at this speed, the"blob" shown should actually be a tiny spot a fraction of a degree across, so what is shown is quite misleading (and doesn't feature in the actual papers referred to). Collisions with space dust would destroy your ship at these values of $\gamma$ (unless you can build it to withstand impacts equivalent to tonnes of TNT from dust grains of mass $10^{−7}\,$g. More details at arxiv.org/pdf/1503.05845.pdf . A full answer should, at least, include some mention of relativistic aberration. Based on Relativistic Abberation - Wikipedia This effect is not dependent on the actual distance between the observer and the source of the radiation. Assume we are in the frame of the observers. To them, the source is travelling at an angle $\theta _{s}$ with velocity $v $, relative to a vector drawn from the source (at the time the light was emitted) to the observers in the spacecraft. The equation that then applies to describe the aberration of the light source is
$$\cos \theta _{o} = \frac{\cos\theta _{s} - \frac vc}{1- \frac vc\cos \theta _{s} }.$$ Relativistic beaming occurs: light rays that are emitted from the source towards the observer are tilted towards the direction of the source's motion (relative to the observer), they effectively form a cone, in the direction of motion of the spacecraft. Light received by a moving object (e.g. the view from a very fast spacecraft) also appears concentrated towards its direction of motion. One consequence of this is that a forward observer should normally be expected to intercept a greater proportion of the object's light than a rearward one; this concentration of light in the object's forward direction is referred to as the "searchlight effect. | {
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276,351 | We live in an expanding universe - so I'm told. But how can that be possible? Everything imaginable is attracted by a bigger thing. So, why can't gravitation stop the expansion of the universe? I know the "Big Bang" theory, but is it possible that the expansion of the universe is caused by the attraction of a bigger object? | Suppose you throw a ball up into the air. You could ask how the ball manages to move upwards when gravity is pulling it down, and the answer is that it started with an upwards velocity. Gravity pulls on the ball and slows it down so it will eventually reach a maximum height and fall back, but the ball manages to move upwards against gravity because of its initial velocity. Basically the same is true of the expansion of the universe. A moment after the Big Bang everything in the universe was expanding away from everything else with an extremely high velocity. In fact if we extrapolate back to time zero those velocities become infinite. In the several billion years following the Big Bang gravity was slowing the expansion, in basically the same way gravity slows the ball you threw upwards, but the gravity didn't stop the expansion - it only slowed it. The obvious next question is how did the universe get to start off expanding with such high velocities, and the answer is that we don't know because we have no theory telling us what happened at the Big Bang. There is a slight complication that I'll mention in case anyone is interested: dark energy acts as a sort of anti-gravity and makes the expansion faster not slower. This has only become an important effect in the last few billion years, but as a result of dark energy right now gravity isn't slowing the expansion at all - in fact it's making the expansion faster. | {
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276,372 | If an equiconvex lens is divided equally, we will obtain a planoconvex lens. Let assume focal length of our equiconvex lens is 'f'. I would like to know what will be the focal length of planoconvex lens that we obtained after dividing equiconvex lense. Will it be twice of 'f'? Or will it be 0.5 of 'f'? How to determine it? | Suppose you throw a ball up into the air. You could ask how the ball manages to move upwards when gravity is pulling it down, and the answer is that it started with an upwards velocity. Gravity pulls on the ball and slows it down so it will eventually reach a maximum height and fall back, but the ball manages to move upwards against gravity because of its initial velocity. Basically the same is true of the expansion of the universe. A moment after the Big Bang everything in the universe was expanding away from everything else with an extremely high velocity. In fact if we extrapolate back to time zero those velocities become infinite. In the several billion years following the Big Bang gravity was slowing the expansion, in basically the same way gravity slows the ball you threw upwards, but the gravity didn't stop the expansion - it only slowed it. The obvious next question is how did the universe get to start off expanding with such high velocities, and the answer is that we don't know because we have no theory telling us what happened at the Big Bang. There is a slight complication that I'll mention in case anyone is interested: dark energy acts as a sort of anti-gravity and makes the expansion faster not slower. This has only become an important effect in the last few billion years, but as a result of dark energy right now gravity isn't slowing the expansion at all - in fact it's making the expansion faster. | {
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276,932 | This doesn't make a lot of sense to me, from a physics 101 point of view. I've read a few blog entries on why this is, but none of them explain it well or are convincing. "something-something launch control. something-something computers." Nothing in physics terms or equations. For instance, Car and Driver magazine tested the Porsche Macan GTS. The $x-60$ times are: Rolling start, $5-60\; \mathrm{mph}: 5.4\;\mathrm{ s}$ $0-60\;\mathrm{mph}: 4.4\;\mathrm{s}$ That's a whole second - about $20$% faster from a dead stop than with some momentum - which seems rather huge. edit: here is the article for this particular example. But I've noticed this with many cars that are tested for $0-60$ and $5-60$ times. Here is another example - an SUV. Another example. And finally, interesting, even for the Tesla Model S (EV) where power doesn't depend on engine RPM, $0-60$ is still slightly faster than $5-60.$ | Ok, from the link given by @count_to_10, I think the answer is clear from this response : You can launch from a dead stop at any RPM you want, whereas from 5
MPH it's assumed the car is already in gear at low RPMs. When you start from a standstill, you can rev the engine to any RPM you like before throwing the clutch to engage the axle. Maybe you could match the static friction of the surface to achieve the maximum possible acceleration. When they start at 5 mph, another answer on that site makes it clear that they assume your RPMs are matched to your motion: "What about rolling at 5mph and dropping the clutch like a regular
launch? Wouldn't that help?" Yeah, but that's not how they test 5-60 or any other rolling
acceleration tests. That's the point of them: to test how much passing
power you have while already rolling, in gear without a clutch-drop. So the engine has to move through the entire range of RPMs, which takes more power. | {
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276,958 | Solar System resides in a plane, thanks to Conservation of Angular Momentum . The Milky Way is also a disc, not sphere. What angle does our Solar System's plane (or, normal to plane) make with The Milky Way's plane (or, normal to plane)? Does our Solar System also resides in the same plane of our Galaxy, making the angle zero? If yes, why? If no, what exactly is the angle? Is it not constant? Any function describing the relationship? | The Sun is approximately in the plane of our Galaxy - see this Astronomy SE question . The ecliptic plane (plane of the solar system) and the Galactic plane (the plane of the disc of the Milky Way) are inclined to each other at an angle of 60.2 degrees. This is a point you can confirm yourself by noting that the Milky Way does not follow the signs of the zodiac (which follow the ecliptic plane). There is really no reason that there should be any alignment. Star formation is a turbulent, chaotic process. The evidence so far is that this leaves the angular momentum vectors of individual stars, their discs and ultimately their planetary systems, essentially randomised. The question only asks "What angle does our Solar System's plane (or, normal to plane) make with The Milky Way's plane" -- to which 60 degrees is the answer. To completely specify the relative geometry of the planes we can ask what are the Galactic coordinates of the ecliptic north pole? The ecliptic north pole (the pole of the Earth's orbit and the direction in which a normal to the ecliptic plane points) is currently at around RA $=18$ h, Dec $=+67$ degrees in the constellation of Draco. In Galactic coordinates this is $l=97$ degrees, $b=+30$ degrees, compared with the normal to the Milky Way plane which is at $b=+90$ degrees (where $l=0$ points towards the Galactic centre and $b=0$ is roughly defines the plane of the Milky Way). See also https://astronomy.stackexchange.com/questions/28071/in-which-direction-does-the-ecliptic-plane-make-an-angle-of-63-degrees-with-gala Edit: Note that as pointed out in comments, the ecliptic plane is not quite the same as the plane of the solar system. They differ by about 1.5 degrees. | {
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277,125 | According to this answer on Astronomy.SE , The Sun executes oscillations around its mean orbit in the Galaxy, periodically crossing the Galactic plane. I borrowed this illustration (not to scale!) from http://www.visioninconsciousness.org/Science_B08.htm to show this oscillatory motion.
As the Sun is currently above the plane and moving upwards, and each cycle takes about 70 million years with an amplitude of 100pc ( Matese et al. 1995 ), it will be roughly 30 million years before we cross the plane again. What causes this kind of motion? Does Newtonian Physics have an explanation? If yes, what is it? If no, what General Relativity explanation have we got? | It's good old Newtonian gravity! The plane of the galaxy can be approximated as a disk made up of stars and gas, with a density $\rho(|z|)$, that decreases with absolute distance $|z|$ from the plane. If you were to assume that the Sun was close enough to $z=0$ and that the radial variation in $\rho$ was negligible enough to treat the disk as an infinite plane (this is not bad, the amplitude of the Sun's motion is only about 10% of the radial scale length of the disk density), then you could construct a little cylinder through the plane, with one face at $z=0$, where $g=0$, and use Gauss's law for gravity to estimate the gravitational acceleration at height $z$.
$$ g(z) \simeq -4\pi G \int_0^{z} \rho(z)\ dz$$ Now $\rho(z)$ approximates to an exponentially decaying function with a scale height of maybe 200-300 pc. If we are closer to $z=0$ than that, then the density can be roughly said to be a constant $\rho_0$. Putting this into the equation above, we see that
$$g(z) =-4\pi G\rho_0 z.$$
But this is simple harmonic motion with an angular frequency $\sqrt{4\pi G\rho_0}$. The density of the disk near the Sun is estimated to be 0.076 solar mass per cubic parsec ( Creze et al. 1998 ). Using this value, we get an approximate predicted oscillation period up and down through the disk plane of 95 million years. Note added: The previous paragraph is the reverse of what is actually done - the dynamics of stars in the solar vicinity are used to estimate the density in the plane. However, just counting up stars and estimating the contribution of gas does give a similar result - and in the process, illustrates that the contribution of dark matter to the density of the disk is very small. | {
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277,183 | I come from a continuum mechanics background, and I make numerical simulations of fluids/solids using the Finite Element Method. The basic equation we solve then is Newton's law of motion, written in terms of relevant vectors and tensors. Using a Lagrangian description we have the equilibrium equation: $\rho \underline{\ddot u} = div(\underline{\underline{\sigma}}) + \underline{f}_v$ where $\underline{\ddot u}$ is the acceleration 3-vector (second time derivative of the displacement vector), $\underline{\underline{\sigma}}$ is the second-order Cauchy stress tensor and $\underline{f}_v$ is the 3-vector of external forces. The system of equations is completed by a constitutive equation, linking the stress tensor to a measure of strain (typical solids) or a measure of strain rate (viscoelastic solids and fluids). For the sake of the example let us consider a usual linear elastic relation: $\underline{\underline{\sigma}} = \mathbb{C} : \underline{\underline{\varepsilon}}$ with $\mathbb{C}$ being the 4th-order elasticity tensor and $\underline{\underline{\varepsilon}}$ the linear strain tensor, classically defined in the compatibility equation as the symmetric part of the displacement gradient: $\underline{\underline{\varepsilon}} = \frac{1}{2}(\underline{\underline{\nabla}} u + \underline{\underline{\nabla}}^T u)$. Now, the equations I stated are all I need in order to discretize the system and simulate small deformations of an arbitrary geometry under a system of forces in the context of Newtonian mechanics, using the Finite Element Method. What I want to know is what is the way to adapt these equations so that they satisfy General Relativity. That is, I want to simulate the deformation of a mechanical structure when the velocities involved are close to the speed of light and/or when a very massive object is near.
I am familiar with nonlinear elasticity if needed, but as far as tensors go I'm unfamiliar with co-variant/contra-variant notation and I prefer intrinsic notation, even though I'll take answers expressed in any way. What form do the equilibrium, constitutive and compatibility equations take? Is the simulation of deformation of bodies in a relativistic context something that was properly done already? Does the elasticity tensor need to be redefined in terms of the metric tensor maybe?
I couldn't find any good reference that addresses this issue, even though I feel like this is possible to achieve. I would be very thankful for any help on this matter. | The answer you're looking for seems to be contained in Rezzolla & Zanotti: Relativistic Hydrodynamics (Oxford U.P. 2013) https://books.google.com/books/?id=KU2oAAAAQBAJ but it is not a trivial generalization. Quoting Disconzi's On the well-posedness of relativistic viscous fluids (Nonlinearity 27 (2014) 1915, arXiv:1310.1954 ): we still lack a satisfactory formulation of viscous phenomena within
Einstein's theory of general relativity. [... T]here have been different proposals for what the correct $T_{\alpha\beta}$ should be. [... A]ttempts to formulate a viscous relativistic theory based on a simple covariant generalization of the classical (i.e., non-relativistic) stress-energy tensor for the Navier-Stokes equations have also failed to produce a causal theory. Similar problems exist for elastic materials. You may also take a look at Chapter 15 ("Relativistic continuum mechanics") of Maugin's Continuum Mechanics Through the Twentieth Century (Springer 2013), and its references: https://books.google.com/books?id=-QhAAAAAQBAJ and at Bressan's Relativistic Theories of Materials (Springer 1978): https://books.google.com/books?id=kMTuCAAAQBAJ General-relativistic continuum mechanics unfortunately has not been given a clear mathematical and conceptual framework yet. Newtonian continuum mechanics is easy to summarize: We choose a reference frame (preferably but not necessarily inertial). We have a set of 11 spacetime-dependent fields with clear physical meaning: mass, momentum or deformation, stress, body force, internal energy, heating flux, body heating, temperature, entropy, entropy flux, body entropy supply. Of these, the "body" ones represent external interventions. We have 5 balance equations: mass, force-momentum, torque-rotational momentum, energy, entropy. They are clearly written in terms of the fields above and are valid for any material. We choose a set of independent fields (usually mass, momentum or deformation, temperature). We choose constitutive equations (compatibly with the balance ones) that relate the remaining fields to the independent ones. These equations express the peculiar properties (fluid, solid, elastic, plastic, with/without memory...) of the material under study. And at this point we have a well-defined set of partial differential equations in a number of unknown fields, for which we can set up well-defined initial- & boundary-value problems to be solved analytically or numerically. (An expanded but analogous framework accommodates electromagnetism and continua with internal structure.) This framework and steps are very neat – we clearly know what the fields are, which of them are dependent and which independent; what are the equations valid for all materials, and what are the equations constitutive to each material. I've never seen a clearly defined procedure like the one above for general relativity, although I believe it could be extracted from Rezzolla & Zanotti's or Bressan's books. Moreover, the core of general-relativistic community uses a different jargon and way of thinking. Most general-relativity books tell you that the Einstein equations determine everything, but they are not so clear about which fields in them are independent and which dependent; even Misner et al.'s Gravitation (ch. 21) has a long discussion and explanation about this point. It was only with 3+1 formulations and the work of Arnowitt, Deser, Misner, York, and others around the 1970s that this point got clarified. Then they tell you that we need "special" additional equations for the stress tensor – that is, constitutive equations. Sometimes other conservation equations, like baryonic number (basically rest-mass), are added with no real explanation. This is a sample of books where "constitutive equations" are mentioned explicitly (only once or twice in most of them): Rezzolla & Zanotti above (and they explain what a "constitutive equation" is as though it was an exotic concept) Choquet-Bruhat: General Relativity and Einstein's Equations Anile & Choquet-Bruhat: Relativistic Fluid Dynamics Bertotti et al.: General Relativity and Gravitation Puetzfeld et al.: Equations of Motion in Relativistic Gravity Tonti: The Mathematical Structure of Classical and Relativistic Physics Bini & Ferrarese: Introduction to Relativistic Continuum Mechanics Tolman (obviously): Relativity, Thermodynamics, and Cosmology but they constitute a very small minority in the huge relativistic literature. Yet, the general-relativistic community cannot be criticized for the confused conceptual state and somehow confused language of the subject. Newtonian continuum mechanics can be neatly formulated today because it has been refined over several centuries. General relativity is still very young instead, and its conceptual refinement still in progress. Some of the steps in the Newtonian framework become extremely complicated in general relativity. For example: step 1. (choose an inertial frame) cannot be done so simply. The Einstein equations, evolved from initial conditions, construct a reference frame "along the way", while they determine the dynamics. This gives rise to peculiar fields like "lapse" and "shift", which aren't really physical, and all sorts of redundancy (gauge freedom) in the equations. Another example: the metric becomes a dynamical field variable, and you suddenly realize that it is hidden almost everywhere in the Newtonian framework – divergences, curls, vectors/covectors... So its evolution can't be easily divided among some new balance and constitutive equations (like we can do with electromagnetism instead). Are all of its appearances in the Newtonian framework dynamically significant? or can the metric be eliminated from some places? There's some research today on this "de-metrization" of Newton's equations; see for example Segev's Metric-independent analysis of the stress-energy tensor , J. Math. Phys. 43 (2002) 3220. This line of research has shown that some Newtonian physical objects actually don't need a metric: they be expressed via differential forms and other metric-free differential-geometrical objects (e.g., van Dantzig's On the geometrical representation of elementary physical objects and the relations between geometry and physics , Nieuw Archief voor Wiskunde II (1954) 73; there is a vast literature on this, let me know if you want more references). This is still work in progress – which means that it's obviously not completely clear yet how mass-energy-momentum-stress and metric are coupled. To conclude, I think another good starting point to understand how things work in general-relativistic continuum mechanics is to look in books on numerical formulations of general relativity and matter dynamics. The conceptual framework in them is a bit confused, but you can see how they actually do it . If from the practice of these books you manage to reverse-engineer a framework like the Newtonian one above, please write a pedagogical paper about it! Here are some books and reviews on numerical relativity with continua: Rezzolla & Zanotti above Gourgoulhon: 3+1 Formalism in General Relativity (Springer 2012, arXiv:gr-qc/0703035 ) Baumgarte & Shapiro: Numerical Relativity (Cambridge U.P. 2010) Alcubierre: Introduction to 3+1 Numerical Relativity (Oxford U.P. 2008) Palenzuela-Luque & Bona-Casas: Elements of Numerical Relativity and Relativistic Hydrodynamics (Springer 2009) Lehner: Numerical relativity: a review , Class. Quant. Grav. 18 (2001) R25, arXiv:gr-qc/0106072 Guzmán: Introduction to numerical relativity through examples , Rev. Mex. Fis. S 53 (2007) 78 I'm happy to provide or look for additional references. | {
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277,192 | Does the spin of an atom affect the velocity of it? If so, in layman terms, can someone explain to me why/how the spin of an atom can effect velocity. | The answer you're looking for seems to be contained in Rezzolla & Zanotti: Relativistic Hydrodynamics (Oxford U.P. 2013) https://books.google.com/books/?id=KU2oAAAAQBAJ but it is not a trivial generalization. Quoting Disconzi's On the well-posedness of relativistic viscous fluids (Nonlinearity 27 (2014) 1915, arXiv:1310.1954 ): we still lack a satisfactory formulation of viscous phenomena within
Einstein's theory of general relativity. [... T]here have been different proposals for what the correct $T_{\alpha\beta}$ should be. [... A]ttempts to formulate a viscous relativistic theory based on a simple covariant generalization of the classical (i.e., non-relativistic) stress-energy tensor for the Navier-Stokes equations have also failed to produce a causal theory. Similar problems exist for elastic materials. You may also take a look at Chapter 15 ("Relativistic continuum mechanics") of Maugin's Continuum Mechanics Through the Twentieth Century (Springer 2013), and its references: https://books.google.com/books?id=-QhAAAAAQBAJ and at Bressan's Relativistic Theories of Materials (Springer 1978): https://books.google.com/books?id=kMTuCAAAQBAJ General-relativistic continuum mechanics unfortunately has not been given a clear mathematical and conceptual framework yet. Newtonian continuum mechanics is easy to summarize: We choose a reference frame (preferably but not necessarily inertial). We have a set of 11 spacetime-dependent fields with clear physical meaning: mass, momentum or deformation, stress, body force, internal energy, heating flux, body heating, temperature, entropy, entropy flux, body entropy supply. Of these, the "body" ones represent external interventions. We have 5 balance equations: mass, force-momentum, torque-rotational momentum, energy, entropy. They are clearly written in terms of the fields above and are valid for any material. We choose a set of independent fields (usually mass, momentum or deformation, temperature). We choose constitutive equations (compatibly with the balance ones) that relate the remaining fields to the independent ones. These equations express the peculiar properties (fluid, solid, elastic, plastic, with/without memory...) of the material under study. And at this point we have a well-defined set of partial differential equations in a number of unknown fields, for which we can set up well-defined initial- & boundary-value problems to be solved analytically or numerically. (An expanded but analogous framework accommodates electromagnetism and continua with internal structure.) This framework and steps are very neat – we clearly know what the fields are, which of them are dependent and which independent; what are the equations valid for all materials, and what are the equations constitutive to each material. I've never seen a clearly defined procedure like the one above for general relativity, although I believe it could be extracted from Rezzolla & Zanotti's or Bressan's books. Moreover, the core of general-relativistic community uses a different jargon and way of thinking. Most general-relativity books tell you that the Einstein equations determine everything, but they are not so clear about which fields in them are independent and which dependent; even Misner et al.'s Gravitation (ch. 21) has a long discussion and explanation about this point. It was only with 3+1 formulations and the work of Arnowitt, Deser, Misner, York, and others around the 1970s that this point got clarified. Then they tell you that we need "special" additional equations for the stress tensor – that is, constitutive equations. Sometimes other conservation equations, like baryonic number (basically rest-mass), are added with no real explanation. This is a sample of books where "constitutive equations" are mentioned explicitly (only once or twice in most of them): Rezzolla & Zanotti above (and they explain what a "constitutive equation" is as though it was an exotic concept) Choquet-Bruhat: General Relativity and Einstein's Equations Anile & Choquet-Bruhat: Relativistic Fluid Dynamics Bertotti et al.: General Relativity and Gravitation Puetzfeld et al.: Equations of Motion in Relativistic Gravity Tonti: The Mathematical Structure of Classical and Relativistic Physics Bini & Ferrarese: Introduction to Relativistic Continuum Mechanics Tolman (obviously): Relativity, Thermodynamics, and Cosmology but they constitute a very small minority in the huge relativistic literature. Yet, the general-relativistic community cannot be criticized for the confused conceptual state and somehow confused language of the subject. Newtonian continuum mechanics can be neatly formulated today because it has been refined over several centuries. General relativity is still very young instead, and its conceptual refinement still in progress. Some of the steps in the Newtonian framework become extremely complicated in general relativity. For example: step 1. (choose an inertial frame) cannot be done so simply. The Einstein equations, evolved from initial conditions, construct a reference frame "along the way", while they determine the dynamics. This gives rise to peculiar fields like "lapse" and "shift", which aren't really physical, and all sorts of redundancy (gauge freedom) in the equations. Another example: the metric becomes a dynamical field variable, and you suddenly realize that it is hidden almost everywhere in the Newtonian framework – divergences, curls, vectors/covectors... So its evolution can't be easily divided among some new balance and constitutive equations (like we can do with electromagnetism instead). Are all of its appearances in the Newtonian framework dynamically significant? or can the metric be eliminated from some places? There's some research today on this "de-metrization" of Newton's equations; see for example Segev's Metric-independent analysis of the stress-energy tensor , J. Math. Phys. 43 (2002) 3220. This line of research has shown that some Newtonian physical objects actually don't need a metric: they be expressed via differential forms and other metric-free differential-geometrical objects (e.g., van Dantzig's On the geometrical representation of elementary physical objects and the relations between geometry and physics , Nieuw Archief voor Wiskunde II (1954) 73; there is a vast literature on this, let me know if you want more references). This is still work in progress – which means that it's obviously not completely clear yet how mass-energy-momentum-stress and metric are coupled. To conclude, I think another good starting point to understand how things work in general-relativistic continuum mechanics is to look in books on numerical formulations of general relativity and matter dynamics. The conceptual framework in them is a bit confused, but you can see how they actually do it . If from the practice of these books you manage to reverse-engineer a framework like the Newtonian one above, please write a pedagogical paper about it! Here are some books and reviews on numerical relativity with continua: Rezzolla & Zanotti above Gourgoulhon: 3+1 Formalism in General Relativity (Springer 2012, arXiv:gr-qc/0703035 ) Baumgarte & Shapiro: Numerical Relativity (Cambridge U.P. 2010) Alcubierre: Introduction to 3+1 Numerical Relativity (Oxford U.P. 2008) Palenzuela-Luque & Bona-Casas: Elements of Numerical Relativity and Relativistic Hydrodynamics (Springer 2009) Lehner: Numerical relativity: a review , Class. Quant. Grav. 18 (2001) R25, arXiv:gr-qc/0106072 Guzmán: Introduction to numerical relativity through examples , Rev. Mex. Fis. S 53 (2007) 78 I'm happy to provide or look for additional references. | {
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277,347 | My knowledge of physics comes down to kinematics and Newton's laws. I would like to get answers which do not deviate from this basic knowledge. I am currently questing for exact definitions of work and energy, and what is behind these definitions. I searched in physics sites, and here in Physics in particular, but I have not found a satisfactory answer. Here is a list of the main resources I have reviewed: Over the net: https://en.wikipedia.org/wiki/Energy http://www.energyquest.ca.gov/story/chapter01.html https://www.reference.com/science/work-physics-a310b0a21e7039a5 http://www.edinformatics.com/math_science/work_energy_power.htm Here on Physics: Intuitively Understanding Work and Energy Why does kinetic energy increase quadratically, not linearly, with speed? Understanding relationship between work and energy Relationship between work and energy in an electrostatic system To preface my question, I will explain the process I went through until I got my current understanding, which I want to sharpen. First, here are the main definitions of work and energy I found around the web: Energy Is a property, or state of objects representing the ability to do work. It can be transferred to other objects, and it can be reflected
in many forms, which are convertible. Work is a transfer of energy from one object to another by applying force, and it is equal to $\vec{F_{x}}\cdot \vec{x}$ for a constant force, or to $\int_{x_{1}}^{x_{2}}\vec{F_{x}}\enspace d\vec{x}$ for a changing force. These definitions seem circular because work is a transfer of energy , and energy is the ability to do work . To solve this problem I adopted NeuroFuzzy's approach : Sometimes when you're stuck on things, it's helpful to look at the mathematics of what's being asserted. A basic mathematical analysis of work definition $W=\vec{F_{x}}\cdot \vec{x}$ raises the following two points: When multiplying the force or the displacement by $k$ , the work is also multiplied by $k$ . For example, if I apply a force of $2\vec{F}$ on an object, the work will be twice as big as if I applied a force of $\vec{F}$ . Also, if a given force on an object led to $\frac{1}{2}\vec{x}$ displacement, the work it did is twice as smaller as in a case it led to $\vec{x}$ displacement. A force does a positive work when: The object displaces. The force on the object is in the direction of the displacement. A force does negative work when: The object displaces. The force on the object is in the opposite direction of the displacement. A force does no work when: The object does not displace. No force is applied to the displacement axis. Both of these conditions are met. In other words, when I apply a force on an object in certain circumstances (environment, other forces, time period, etc.) and the object's displacement is different than the displacement in the same conditions, only without the force, I could say that the force has influenced the displacement. It can have a "positive" influence (if the displacement was larger than the displacement under the same conditions without applying the force) or have a "negative" influence (if the displacement was smaller than the displacement under the same conditions, without applying the force). If I applied a force, and the displacement was equal to the displacement under the same conditions, without applying the force, then the force did not influence the displacement. My intuitive conclusion of this analysis is that work is the degree of influence a force has on an object's displacement . If we accept my definition, and we combine it with the definitions above, we could describe work and energy as follows: Energy Is a property, or state of objects representing the ability to apply a force that will influence an object's displacement . It can be transferred to other objects, and it can be reflected
in many forms, which are convertible. Work is the degree of influence a force has on an object's displacement . When a force does work, besides the fact that it moves the object, it also transfers to the object the ability to influence other objects' displacement by applying force. I have a few questions about these definitions I cannot find an answer. I do not know whether the definitions are correct and the questions have answers - and then I will be grateful for answers, or the definitions are incorrect in the first place - and then I will be grateful if you could correct my definitions. Here are the questions: Does the definition of energy mean that all energy is potential energy? Because if energy describes the capability of an object to do work, doesn't it means that we are talking about the potential of the object to do work? I defined energy as a property, or state of an object representing the amount of work it can do. But how could we quantify it? If, for example, a man stores within itself $100J$ of energy, does that mean he is able to apply a force of $10^{100}N$ over $10^{-98}m$ ? Surely a human being is not capable of applying so much force, so why we still say he stores $100J$ ? How potential energy is reflected while being potential? Can we see a difference between a person that stores $x$ energy and a person consuming that $x$ energy? An answer I heard a lot is that the mass of the person is actually its potential energy, and therefore doing physical activities, for example, is consuming energy, and reducing the mass. But if it is correct, wouldn't we measure energy in $kg$ or mass in $Joule$ ? Although my definitions do explain the inner nature of work and energy, I could still mathematically describe this nature in an infinite number of ways. For example, if I describe work as $2\vec{F}_{x}\cdot \vec{x}$ , I would still get to the conclusion that work is the influence a force has on an object's displacement. So why is that the equation? How is consuming energy in other forms than motion work? How are heat energy, or sound waves, for example, work? Thanks. | Don't be surprised that physics has a lot of definitions that are circular. Ultimately, we are just describing the universe. Work and energy have been defined in a certain way in newtonian physics to explain a kinematic model of reality. This is a model , not reality - you will find no such thing in reality. However, in many scenarios, it is close enough to reality to be useful. For example, let's say that a human has a 10% efficiency at converting food to mechanical work. So if you spend 1000 kJ of food energy to press against a wall, are you doing 1000 kJ of work, or 100 kJ of work, or 0 kJ of work? In strict mechanical sense, you did no work whatsoever, and all of the energy you used was wasted as heat. If you instead used this energy to push a locomotive, you would have wasted "only" 900 kJ of the energy as heat, with 100 kJ being work. But the locomotive has its own friction, and it wil stop eventually, wasting all the energy as heat again. And overall, you did expend all those 1000 kJ of food energy that is never coming back . All of those are simplifications. Kinematics is concerned with things that move. Using models is all about understanding the limits of such models. You're trying to explain thermodynamics using kinematics - this is actually quite possible (e.g. the kinematic theory of heat), but not quite as simple as you make it. Let's look at the fire example. You say there is no displacement, and therefore no work. Now, within the usual context kinematics is used, you are entirely correct - all of that energy is wasted, and you should have used it to drive a piston or something to change it to useful work. Make a clear note here: what is useful work is entirely a human concept - it's all 100% relevant only within the context of your goals; if you used that "waste" to heat your house, it would have been useful work. It so happens that if you look closer, you'll see that the heat from the fire does produce movement. Individual molecules forming the wood wiggle more and more, some of them breaking free and reforming, and rising with the hot air away from the fire, while also drawing in colder air from the surroundings to feed the fire further. There's a lot of displacement - individual molecules accelerate and slow down, move and bounce around... But make no mistake, the fact that kinematics can satisfactorily explain a huge part of thermodynamics is just a bonus - nobody claimed that kinematics explains 100% of the universe. It was a model to explain how macroscopic objects move in everyday scenarios. It didn't try to explain fire. For your specific questions, you really shouldn't ask multiple questions in one question. It gets very messy. But to address them quickly: There is no potential energy in the kinematic model. The concept is defined for bound states, which do not really exist as a concept in kinematics. In other models, you might see that there's a difference between, say, potential energy and kinetic energy - no such thing really exists in reality. You need to understand the context of the model. In a perfectly kinematic world, this is 100% correct. However, as noted before, kinematics isn't a 100% accurate description of reality, and there are other considerations that apply, such as the fact that humans have limited work rate, limited ability to apply force, and the materials we are built of aren't infinitely tough, perfectly inflexible and don't exist in perfect isolation from all the outside (and inside) effects. In real world applications of models, these differences are usually eliminated through understanding the limits of given models, and using various "fixup" constants - and if that isn't good enough, picking (or making) a better model. You're mixing up many different models at different levels of abstraction and of different scope so confusion is inevitable. Within the simplified context of kinetics, there is no concept of "potential energy". You simply have energy that can be used to do work, and that's it; it doesn't care about how that energy is used to do work, about the efficiency of doing so etc. In another context, it might be very useful to think of energy and mass as being the same thing - and in yet another, they might be considered interchangeable at a certain ratio, or perhaps in a certain direction, or at a certain rate. It's all about what you're trying to do. How is that equation useful? That's the only thing that matters about both definitions and equations. I can define a million things that are completely useless if I wanted to - but what's the point? Within the original context, those aren't considered at all. Within a more realistic context, both heat and sound are also kinematic. The reason you have so much trouble finding the answer to your questions on physics sites and forums is that the question doesn't make much sense in physics. It's more about the philosophy of science, and the idea of building models of the world that try to describe reality to an approximation that happens to be useful to us. You think that those words have an inherent meaning that is applicable in any possible context - this simply isn't true. From the very inception of the idea of physics, people have known that it isn't (and never will be) an accurate representation of reality; and we've known for a very long time that, for example, different observers may disagree on the energy of one object. You just need to understand where a given model is useful, and pick the right model for the job. Don't try to drive a screw with a garden rake. | {
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277,625 | Gas barbeque manufacturers place metal bars, ceramic plates or lava rocks above the gas burner so that they radiate more heat towards the grill. Cooking directly over a single gas flame just wouldn't work very well. Why don't gas flames radiate much heat directly but a metal/ceramic object heated by the same flame does? For example, you can put your hand near a flame and not feel anything but lava rocks will scorch your skin easily. | The luminous flame itself is hot; the site where the chemical process of combustion takes place. But the product of combustion, typically CO₂, H₂O, and perhaps some CO gas carry off a great quantity of the heat energy created in the reaction. The main purpose of the lava rocks, grill etc. is to capture a good part of that heat flux so it doesn't just blow by the food into the atmosphere. Another reason is to provide a more uniform distribution over the cooking field. The reason the lava rocks 'feel' hotter than the bare flame is that they have the capacity to store a lot of that heat. The flame and hot product gases transfer heat over time and that heat integrates over time to raise the temperature of the rocks. | {
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277,831 | It seems to me that if on Earth soap bubbles pop, it is because the gravity makes the water and soap go to the bottom of it. When there is not enough water at the top of the bubble, it pops. But what if we remove gravity and make a bubble in the ISS? Would it last forever (as long as it never touches a surface)? And what about creating a bubble in space? | Evaporation of its "shell" would be the main source of the eventual POP. Air turbulence may distort it beyond the surface tension forces capacity to hold it together. If the air is very dry, even without touching anything, it will burst at some point. Although this might surprise you, pushing through the bubble wall with a wet nail seems to work. Image Credit: Popping Bubbles In space, i. e. vacuum, it would have a much higher inner pressure, so I think you could guess what would happen. Forgive me for going slightly off topic, but I can personally vouch for the fact that soap bubbles can get this big. The air is pushed in through a mouth opened with two sticks. | {
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278,282 | I'm a layman interested in learning more about nuclear radiation. Heavy elements like plutonium or uranium will eject protons and neutrons as alpha particles, electrons as beta particles, and photons as gamma radiation in an attempt to become more stable. But where do the photons that make up gamma radiation originate? To me it appears as though they just appear from nowhere. How is that possible? | Standing in a lake, I can make a ripple in the water by moving my hand. But this wave wasn't in the water before I did that, and it certainly wasn't inside my hand either. So where did the ripple come from? We can say it was formed by the interaction between my hand and the water. On a slightly deeper level, it was created from the energy in the muscles that moved my hand, which came from the food I ate earlier. The same principle goes for gamma ray emission from an atomic nucleus. The charges in the nucleus are making ripples in the electromagnetic field, but those ripples weren't 'inside' the nucleus beforehand. The energy used to create them was, though. This kind of thinking is necessary to understand more than just gamma ray photons. For example, in beta decay, a neutron turns into a proton, and electron, and a neutrino. None of these three objects were 'inside' the neutron to begin with. (We've smashed particles together to check this.) Instead, one should think of the neutron as a ripple in a quantum field that makes ripples in other quantum fields, in the same way that a hand makes waves in water. | {
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278,285 | Charging by contact requires conductors. When a positively charged glass rod touches a neutral metal body, the body gets positive charge. But how is it possible as glass rod is insulator? | Standing in a lake, I can make a ripple in the water by moving my hand. But this wave wasn't in the water before I did that, and it certainly wasn't inside my hand either. So where did the ripple come from? We can say it was formed by the interaction between my hand and the water. On a slightly deeper level, it was created from the energy in the muscles that moved my hand, which came from the food I ate earlier. The same principle goes for gamma ray emission from an atomic nucleus. The charges in the nucleus are making ripples in the electromagnetic field, but those ripples weren't 'inside' the nucleus beforehand. The energy used to create them was, though. This kind of thinking is necessary to understand more than just gamma ray photons. For example, in beta decay, a neutron turns into a proton, and electron, and a neutrino. None of these three objects were 'inside' the neutron to begin with. (We've smashed particles together to check this.) Instead, one should think of the neutron as a ripple in a quantum field that makes ripples in other quantum fields, in the same way that a hand makes waves in water. | {
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278,402 | Suppose we don't know anything about statistical mechanics, not even the existence of atoms. Why is entropy defined as
$$\delta S=\frac{\delta Q}{T}$$
instead of, say,
$$\delta S=\frac{\delta Q}{T^2}$$
or any other function that will conserve the 2nd Law of thermodynamics? In a nutshell, is our entropy definition unique? | In thermodynamics, the definition of entropy is unique up to redefinitions of temperature . The Zeroth Law of thermodynamics tells us that a temperature scale exists, but it doesn't specify anything more than that. Therefore, we are free to replace temperature $T$ with any monotonic function $f(T)$ , in which case the definition of entropy becomes $$\Delta S = \frac{\Delta Q}{f(T)}.$$ As you've seen, this doesn't upset the Second Law. It does change what "temperature" means, though. The Carnot efficiency, the form of the ideal gas law, etc. all have to be changed. This might sound puzzling, because nobody ever seems to mention this. The reason is that there are many ways of sneaking the choice $f(T) = T$ in. For example, a standard thermodynamics book could begin with the ideal gas law, which defines temperature by $T = pV/nR$ . Then it can be used to derive the efficiency for an ideal gas Carnot engine. Comparing other cycles with this one in turn leads to the Clausius inequality, and hence the usual definition of entropy $\Delta S = \Delta Q / T$ . On the other hand, if one starts with the Carnot engine without specifying the working fluid, then the most one can conclude is that a Carnot engine running between reservoirs of temperature $T_1$ and $T_2$ has an efficiency $\eta$ obeying $$1 - \eta(T_1, T_2) = \frac{g(T_2)}{g(T_1)}$$ as can be shown by considering the composition of two Carnot engines in series. If the book is sloppy, then at some point in this analysis it will implicitly take $g(T) = T$ , thereby fixing a temperature scale. This, of course, agrees with the "ideal gas" temperature $T = pV/nR$ . But we're also free to take any function $g$ , and choosing a nontrivial $g$ is equivalent to choosing a nontrivial $f = g^{-1}$ above. Luckily, all of this discussion is moot, because there really is a sense that $f(T) = T$ is the best choice. That's because in statistical mechanics, we have a more fundamental definition of entropy, $$S = k_B \log \Omega.$$ This definition is unique, and it forces the choice $f(T) = T$ . | {
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278,472 | Suppose I have two identical ropes, one of which I manually tie some knots in. If I use them to hang clothes, which one is more durable? Personally I think that rope with knots will be more durable, but I can't come up with a satisfactory reason. | The fibers in a rope strand are 'layed' stretched out straight, then
twisted. Multiple strands are combined, by twisting, to make the rope.
When a rope is pulled taut, all the fibers have similar stress, and all contribute nearly equally to the tensile strength. When in a knot, however, the fibers on the inside of the curve are not in tension, and those on the outside are stretched as they follow the curve. Strain is uneven. Under tension, the rope is stronger if there are no knots, because the uneven strains caused by the knot shape can snap the most-stressed fibers at tension levels that do not threaten to damage a straight rope. Various knots, and
types of splices, have been tested to see how well they preserve the strength of
the original line; an eye splice with a thimble (steel internal support) is one
of the better designs; the overhand knot (illustrated in the question) is one of the worst. | {
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278,870 | According to special relativity, a photon traveling at the speed of light, does not “experience” time. It is traveling into the future at an infinite rate, right? That would mean that everything that could ever happen (my death, Trump winning the election, death of sun, end of the universe), has already happened (from the frame of a photon). We're just too "slow" to experience this. Is this correct? Because that would bring some weird implications about free will, and the very nature of our existence. If everything has already happened, then we're just following a predetermined path. Edit 1: I'm genuinely fascinated by this topic, however, I am also an idiot with very superficial knowledge, gained mostly from youtube videos and Wikipedia. Bear that in mind :) Edit 2: The nature of my question seems to leave the realm of physics and cross into a philosophical issue. I was seeking physical explanation, to help answer something of philosophical nature, as both these realms seem to coincide for this particular topic. Some have pointed out that thinking about the frame of a photon doesn't make sense. Instead, one could assume a spaceship traveling at $0.9999c$ . | This is an idea known as the block universe . The example of a photon is a bit extreme, since photons have no rest frames, but the idea is the same: to a fast-moving observer, our future might lie in their past, suggesting that our future already exists. However, these are purely philosophical notions. In order to see what physics has to say about it, we need to translate "the future already exists", which is a bunch of words, into an experiment that can actually be performed. One such experiment would be to attempt to receive signals from the future, such as the result of a coin flip. However, relativistic causality forbids this from happening, and also rules out any and all similar experiments. Physics really has nothing to say about this question. You might complain, why would we have to experimentally test this? Isn't the future already existing directly baked into the math of special relativity, which we know is true? It is, but you have to be careful not to mix up features of the mathematics of a theory with the theory's predictions. Showing that the predictions are correct does not show that the mathematical structures used to make the predictions are ontologically real. For example, quantum mechanics has a huge variety of flavors, all of which have different ontologies (wavefunction collapse, many worlds, pilot waves) but identical predictions. Physics can't distinguish between them. Deciding which interpretation of quantum mechanics is right, or whether Lorentz invariance says anything about free will, is a job for the philosophers. | {
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279,783 | If you wrap an active electric cord around your body, do you become an electromagnet? | Okay, accuse me of having too much time on my hands, but here's what I did: If you can't tell from the pic. I wrapped the vacuum cord around a steel bar. I turned on the vacuum and tried to pick up the screw. Absolutely nothing not even a hint of attraction so maybe BowlofRed has a point. In case the comment gets deleted later, here is BowlofRed's comment: The power cord has two conductors in it. The current is moving in opposite directions in each, so the net current flow through the cord is zero. No net current, no bulk magnetic field. – BowlOfRed | {
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279,798 | Determine the maximum ratio $h/b$ for which the homogenous block will
slide without toppling under the action of force F.The coefficient of
static friction between the block and the incline is $\mu_s$. I have a doubt.About which point should the rotational equilibrium be applied?Should it be applied about centre of mass?Or should it be applied about the vertex opposite to the vertex where F is applied?Why? MY ATTEMPT: Translational Equations $F+mg\sin(\theta) \geq \mu N$
and $N=mgcos(\theta)$ Rotational Equations This is where I'm facing a problem.Depending upon which point the equilibrium is applied the required ratio will be obtained. MY VIEWS: Rotational equilibrium should hold at all points if no toppling/rotation happens.However the answer varies depending on the point of application of equilibrium.Strange. I hope this is a conceptual doubt and will not be closed as off-topic or homework.If it needs to be closed please inform me if the post can be improved somehow. | Okay, accuse me of having too much time on my hands, but here's what I did: If you can't tell from the pic. I wrapped the vacuum cord around a steel bar. I turned on the vacuum and tried to pick up the screw. Absolutely nothing not even a hint of attraction so maybe BowlofRed has a point. In case the comment gets deleted later, here is BowlofRed's comment: The power cord has two conductors in it. The current is moving in opposite directions in each, so the net current flow through the cord is zero. No net current, no bulk magnetic field. – BowlOfRed | {
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280,282 | I know that a gas is made of atoms or molecules moving freely in space. When these particles hit the walls of where they're kept in they cause something called pressure. But these particles never pile up on each other and push a surface down by their weight so that we can measure it as weight, not pressure. So why do gases have weight? | Imagine a gas molecule in a closed box bouncing vertically between the top and bottom of the box. Let's suppose the mass of the gas molecule is $m$ and its speed at the top of the box is $v_t$ . When the gas molecule moving upwards hits the top of the box and bounces back the change in momentum is $2mv_t$ . If it does this $N$ times a second then the rate of change of momentum is $2Nmv_t$ , and rate of change of momentum is just force, so the upwards force the molecule exerts is: $$ F_\text{up} = 2Nmv_t $$ And the same argument tells us that if the velocity of the molecule at the bottom of the box is $v_b$ , then the downwards force it exerts on the bottom of the box is: $$ F_\text{down} = 2Nmv_b $$ So the net downwards force is: $$ F_\text{net} = 2Nmv_b - 2Nmv_t = 2Nm(v_b - v_t) \tag{1} $$ But when the molecule leaves the top of the box and starts heading downwards it is accelerated by the gravitational force so when it reaches the bottom it has speeded up i.e. $v_b \gt v_t$ . So that means our net downward force is going to be positive i.e. the molecule has a weight. We can make this quantitative by using one of the SUVAT (see 'Physics For You' by Keith Johnson ) equations: $$ v = u + at $$ Which in this case gives us: $$ v_b - v_t = gt $$ where $t$ is the time the molecule takes to get from the top of the box to the bottom. The number of times per second it makes this round trip is: $$ N = \frac{1}{2t} $$ Substituting these into our equation (1) for the force we get: $$ F_\text{net} = 2 \frac{1}{2t} m(gt) = mg $$ And $mg$ is of course just the weight of the molecule. | {
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280,611 | Why do we use AC (Alternating Current) for long distance transmission of electrical power? I know that AC is such a current that changes polarity (magnitude and direction) and has fixed poles. | The first point to make is: We don't always use AC. There is such a thing as high voltage DC for long-distance power transmission. However its use was rare until the last few decades, when relatively efficient DC-to-AC conversion techniques were developed. The second point is to debunk the common answer given, which is "because DC won't go long distances". Sure it will. In fact DC is sometimes better for long distance (because you don't have capacitive or EM radiation losses). But, yes, AC has been used traditionally. The "why" is because of a series of "a leads to b leads to c leads to...": You want to lose as little power as possible in your transmission lines. And all else being equal, the longer the distance, the more power you'll lose. So the longer the distance, the more important it is to cut the line losses to a minimum. The primary way that power lines lose power is in resistive losses. They are not perfect conductors (their resistance is non-zero), so a little of the power that goes through them is lost to heat - just as in an electric heater, only there, of course, heat is what we want! Now, the more power is being carried, the more is lost. For a given amount of power being transferred, the resistive loss in the transmission line is proportional to the square of the current! (This is because power (in watts) dissipated in a resistance is equal to current in amperes, squared, multiplied by the resistance in ohms. These losses are commonly called "I-squared-R" losses, pronounced "eye-squared-arr", "I" being the usual symbol for current in electrical work.) So you want to keep the current as low as possible. Low current has another advantage: you can use thinner wires. So, if you're keeping the current low, then for the same amount of power delivered, you'll want the voltage high (power in watts = EMF in volts multiplied by current in amps). e.g. to halve the current, you'll need to double your voltage. But this will cut your losses to one fourth of what they were! That's a win. Now high voltage does have its issues. The higher the voltage, the harder it is to protect against accidental contact, short circuits, etc. This means higher towers, wider spacing between conductors, etc. So you can't use the highest possible voltage everywhere; it isn't economical. But in general, the longer the transmission line, the higher the voltage that makes sense. Unfortunately you can't deliver power to the end use point (wall outlets and light sockets) at the high voltages that make sense for the long distance transmission lines. (that could be several hundred thousand volts!) Practical generators can't put out extremely high voltages either (they would arc horrendously). So you need an easy way to convert from one voltage to another. And that's most easily done with AC and transformers. Transformers can be amazingly efficient: power distribution transformers routinely hit 98 or 99 percent efficiency, far higher than any mechanical machine. By contrast, to convert DC voltages you essentially have to convert to AC, use a transformer, and then convert back to DC. The DC-to-AC step, in particular, will have losses. Modern semiconductors have made this a lot better in recent years, but it still generally isn't worth doing until you're talking about very long transmission lines, where the advantages of DC outweigh the conversion losses. Another reason that AC prevailed over Edison's DC was that the AC system scaled better, as it permitted a small number of power plants far from the city, instead of a large number of small plants about a mile apart. Edison didn't just want to sell light bulbs; he (or, rather, his investors) wanted to sell lighting systems to businesses. There was no power distribution network and he didn't want to have to build one before selling light bulbs. At first he was selling lighting systems to commercial buildings, maybe some large apartment buildings; each building would have its own independent generator in the basement, just as you typically have water heaters today. He was initially successful because he (unlike other developers of light bulbs) was selling and installing complete systems, generator and switchgear and wiring and all, not just bulbs. This would have saved a lot of the clutter of overhead wires in cities, but it was clear that this would not work well for small businesses or homes (what homeowner or shopkeeper wants to worry about keeping a generator running?). Westinghouse wanted to build a hydroelectric power generation plant at Niagara Falls - one plant to run all of New York City and beyond. Tesla designed an entire AC distribution system involving AC induction generators, step-up transformers to boost their output as necessary for long distances, then conversion through a series of step-downs to what is called "distribution voltage", and then finally to the lines that are connected to houses and light commercial buildings. This was a far more scalable system than Edison's. And, of course, AC works for light bulbs as well as for motors. Speaking of that... Yet another reason for preferring AC is that AC, and particularly the three-phase AC that Westinghouse's system used (everywhere except at the last drop, from pole distribution transformer to house), was and remains far better for running high-power motors. All practical motors are really AC motors at heart; "DC" motors use commutators to switch the polarity to the coils back and forth as needed, to maintain rotation - essentially they make their own AC internally. But commutators require brushes, which wear out and require maintenance; they make sparks (which interfere with radio), etc. Whereas an AC induction motor needs no commutator nor even slip rings. AC power transmission systems start with three-phase AC generators and maintain three-phase right up to the pole transformer. So they can easily deliver three-phase where it's needed (medium and larger commercial and industrial), but the pole transformer can tap off single-phase for homes and light commercial use. Three-phase AC power distribution has another advantage in not needing a dedicated "return" wire. (Just FYI, the system Tesla originally designed for Westinghouse was two-phase. They changed to three-phase after the work of Mikhail Dolivo-Dobrovolsky in 1888-1891.) During the "war of the currents" Edison made much of the greater danger of AC. It's true that a given level of current, through a given path through the body, is more dangerous at AC than at DC. That's because AC at power line frequencies will cause involuntary muscle contractions - paralysis - and heart fibrillation at far lower current than DC (about a tenth). (See allaboutcircuits.com) However the end-user connectors were designed to minimize risk of contact with live parts, and we keep making them better in that regard. (Aside: I have long held that the electrical transformer should be regarded as one of the basic machines, along with the lever, the inclined plane, the block and tackle, etc. They have the same property of trading off one thing for another. In the case of the mechanical basic machines it's power traded for distance, for an equivalent amount of work done; in the transformer it's voltage for current, at equivalent power. Hydraulic cylinder master-slave pairs should be in the "simple machines" list too. ;) ) | {
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280,627 | I have found that current always is from high voltage end of resistor to the low voltage end.
But in battery sometimes it flows from + end of battery to - and mostly from - to +.
I can find the direction in one loop circuit(with two butteries like this +--+) but its hard in multiloop circuits.
What determines the direction? in this picture 7 volt battery and 6 volt battery have "almost" same condition, but why current is reverse in 6 volt one?
is it because of the limit 10 ohm resistor gives to the circuit? | The first point to make is: We don't always use AC. There is such a thing as high voltage DC for long-distance power transmission. However its use was rare until the last few decades, when relatively efficient DC-to-AC conversion techniques were developed. The second point is to debunk the common answer given, which is "because DC won't go long distances". Sure it will. In fact DC is sometimes better for long distance (because you don't have capacitive or EM radiation losses). But, yes, AC has been used traditionally. The "why" is because of a series of "a leads to b leads to c leads to...": You want to lose as little power as possible in your transmission lines. And all else being equal, the longer the distance, the more power you'll lose. So the longer the distance, the more important it is to cut the line losses to a minimum. The primary way that power lines lose power is in resistive losses. They are not perfect conductors (their resistance is non-zero), so a little of the power that goes through them is lost to heat - just as in an electric heater, only there, of course, heat is what we want! Now, the more power is being carried, the more is lost. For a given amount of power being transferred, the resistive loss in the transmission line is proportional to the square of the current! (This is because power (in watts) dissipated in a resistance is equal to current in amperes, squared, multiplied by the resistance in ohms. These losses are commonly called "I-squared-R" losses, pronounced "eye-squared-arr", "I" being the usual symbol for current in electrical work.) So you want to keep the current as low as possible. Low current has another advantage: you can use thinner wires. So, if you're keeping the current low, then for the same amount of power delivered, you'll want the voltage high (power in watts = EMF in volts multiplied by current in amps). e.g. to halve the current, you'll need to double your voltage. But this will cut your losses to one fourth of what they were! That's a win. Now high voltage does have its issues. The higher the voltage, the harder it is to protect against accidental contact, short circuits, etc. This means higher towers, wider spacing between conductors, etc. So you can't use the highest possible voltage everywhere; it isn't economical. But in general, the longer the transmission line, the higher the voltage that makes sense. Unfortunately you can't deliver power to the end use point (wall outlets and light sockets) at the high voltages that make sense for the long distance transmission lines. (that could be several hundred thousand volts!) Practical generators can't put out extremely high voltages either (they would arc horrendously). So you need an easy way to convert from one voltage to another. And that's most easily done with AC and transformers. Transformers can be amazingly efficient: power distribution transformers routinely hit 98 or 99 percent efficiency, far higher than any mechanical machine. By contrast, to convert DC voltages you essentially have to convert to AC, use a transformer, and then convert back to DC. The DC-to-AC step, in particular, will have losses. Modern semiconductors have made this a lot better in recent years, but it still generally isn't worth doing until you're talking about very long transmission lines, where the advantages of DC outweigh the conversion losses. Another reason that AC prevailed over Edison's DC was that the AC system scaled better, as it permitted a small number of power plants far from the city, instead of a large number of small plants about a mile apart. Edison didn't just want to sell light bulbs; he (or, rather, his investors) wanted to sell lighting systems to businesses. There was no power distribution network and he didn't want to have to build one before selling light bulbs. At first he was selling lighting systems to commercial buildings, maybe some large apartment buildings; each building would have its own independent generator in the basement, just as you typically have water heaters today. He was initially successful because he (unlike other developers of light bulbs) was selling and installing complete systems, generator and switchgear and wiring and all, not just bulbs. This would have saved a lot of the clutter of overhead wires in cities, but it was clear that this would not work well for small businesses or homes (what homeowner or shopkeeper wants to worry about keeping a generator running?). Westinghouse wanted to build a hydroelectric power generation plant at Niagara Falls - one plant to run all of New York City and beyond. Tesla designed an entire AC distribution system involving AC induction generators, step-up transformers to boost their output as necessary for long distances, then conversion through a series of step-downs to what is called "distribution voltage", and then finally to the lines that are connected to houses and light commercial buildings. This was a far more scalable system than Edison's. And, of course, AC works for light bulbs as well as for motors. Speaking of that... Yet another reason for preferring AC is that AC, and particularly the three-phase AC that Westinghouse's system used (everywhere except at the last drop, from pole distribution transformer to house), was and remains far better for running high-power motors. All practical motors are really AC motors at heart; "DC" motors use commutators to switch the polarity to the coils back and forth as needed, to maintain rotation - essentially they make their own AC internally. But commutators require brushes, which wear out and require maintenance; they make sparks (which interfere with radio), etc. Whereas an AC induction motor needs no commutator nor even slip rings. AC power transmission systems start with three-phase AC generators and maintain three-phase right up to the pole transformer. So they can easily deliver three-phase where it's needed (medium and larger commercial and industrial), but the pole transformer can tap off single-phase for homes and light commercial use. Three-phase AC power distribution has another advantage in not needing a dedicated "return" wire. (Just FYI, the system Tesla originally designed for Westinghouse was two-phase. They changed to three-phase after the work of Mikhail Dolivo-Dobrovolsky in 1888-1891.) During the "war of the currents" Edison made much of the greater danger of AC. It's true that a given level of current, through a given path through the body, is more dangerous at AC than at DC. That's because AC at power line frequencies will cause involuntary muscle contractions - paralysis - and heart fibrillation at far lower current than DC (about a tenth). (See allaboutcircuits.com) However the end-user connectors were designed to minimize risk of contact with live parts, and we keep making them better in that regard. (Aside: I have long held that the electrical transformer should be regarded as one of the basic machines, along with the lever, the inclined plane, the block and tackle, etc. They have the same property of trading off one thing for another. In the case of the mechanical basic machines it's power traded for distance, for an equivalent amount of work done; in the transformer it's voltage for current, at equivalent power. Hydraulic cylinder master-slave pairs should be in the "simple machines" list too. ;) ) | {
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281,483 | Is there any substance with segments of its phase change diagram lines going in a negative direction? To explain: Generally, as phase change diagrams go, with heat increasing, and pressure constant, substances tend to evaporate, sublimate or melt; cooling produces the opposite, fusion (freezing), deposit or condensation into liquid from vapor. But the diagrams are rarely straightforward - water, for example, has a multitude of forms of ice, where not all forms are obtainable through transitions "from any direction". There are many obscure factors that define when each transition can occur. Is there any substance, that - without transforming into another substance (say, polymerization) - possesses an area of the phase change diagram (possibly way off "room conditions") where the transition goes in the opposite direction - heating leads to a - not necessarily more dense - but 'more solid' state? Something like thermally hardened glue, but without a chemical transition? | Yes, here's the phase diagram for Helium-3: Notice that around 3 MPa, an increase in temperature causes a transition from liquid to solid. | {
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281,667 | I know this may sound weird. But could a photon be given mass? And if so what would the effects be? | A "photon" propagating in an optical medium can be ascribed a nonzero rest mass. I put "photon" in quotes because light in a medium is not, strictly speaking, pure photons but a quantum superposition of excited electromagnetic field and matter quantum states. In a medium with refractive index $n=1.5$, my calculation here reckons the rest mass of a quantum of this superposition to be: $$m_0 = \frac{E}{c^2}\sqrt{1-\frac{1}{n^2}}$$ For $n=1.5$ (common glasses like window panes or N-BK7 - microscope slide glass) at $\lambda = 500\rm\,nm$, we get, from $E=h\,c/\lambda$, $m_0=3.3\times 10^{-36}{\rm kg}$ or about 3.6 millionths of an electron mass. This take on your question is perhaps a little different from what your original question asks for, and is not to be confused with the assignment of a rest mass to the "pure photon" as discussed in the other answers. The assignment of nonzero rest mass to the photon replaces Maxwell's equations with the Proca equation , whose most "startling" characteristic is screening , i.e. photon fields would dwindle exponentially with distance from their sources and light as we know it could not propagate through the universe. | {
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