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161,453 | I understand that time is relative for all but as I understand it, time flows at a slower rate for objects that are either moving faster or objects that are near larger masses than for those that are slower or further from mass. So, the illustrative example I always see is that if I were to leave earth and fly around at near light speed for a while or go into orbit around a black hole the time I experience would be substantially shorter than for those I left behind at home on earth and I'd come back to find that I've only aged however long I felt I was gone by my own clock but that on Earth substantially more time would have elapsed. Following through on this model, the stars in orbit around the black hole at the center of the Milky Way are aging much much slower (relative to us), right? So does it not follow that the center of the galaxy is some appreciable (I have no idea how to go about putting this in an equation so will avoid guessing at the difference) amount "younger" than the stuff further away from the center? If this is not true, could someone please explain why not, and if it is true, can someone please point me to where I can calculate the age of the center of the galaxy :-) And to be clear... what I'm asking is... if there was an atomic clock that appeared at the center of the galaxy when the center was first formed, and we brought it through a worm hole to earth today - how much time would have elapsed on that clock vs. the age we recon the galaxy currently is? (13.2 billion years) | The gravitational potential of the disk of the Milky Way can be approximated as: $$ \Phi = -\frac{GM}{\sqrt{r^2 + (a + \sqrt{b^2 + z^2})^2}} \tag{1} $$ where $r$ is the radial distance and $z$ is the height above the disk. I got this equation from this paper , and they give $a$ = 6.5 kpc and $b$ = 0.26 kpc. In the weak field approximation the time dilation is related to the gravitational potential by: $$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} \tag{2} $$ At the centre of the galaxy $r = z = 0$ and equation (1) simplified to: $$ \Phi = -\frac{GM}{a + b} \tag{3} $$ No-one really knows the mass of the Milky Way because we don't know how much dark matter it contains, but lets guesstimate it at $10^{12}$ Solar masses. With this value for $M$ and using $a$ + $b$ = 6.76 kpc equation (3) gives us: $$ \Phi = 6.4 \times 10^{11} \text{J/kg} $$ Feeding this into equation (2) gives: $$ \frac{\Delta t_r}{\Delta t_\infty} = 0.999993 $$ So over the 13.7 billion year age of the universe the centre of the Milky Way will have aged about 100,000 years less than the outskirts. | {
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161,650 | Could someone intuitively explain to me Ohm's law ? I understand what voltage is and how it is the electric potential energy and that it is the integral of the electric field strength etc. I also understand that current is the rate at which charge flows at a specific point in the circuit, and I get that resistivity is the opposite of conductivity and that it's analogous to friction in some ways, but I cannot at all get the whole picture and connect the 3 together. | In addition to the other answers, here is something for the intuition: $$V=RI$$ More "pressure" $V$ (more correctly: higher "pressure" difference from one side to the other) is required to keep the flow $I$ of charges constant when the flow is resisted by $R$. A thin wire has higher resistance than a thick wire, $R=\rho L/A$, analogous to a "bottleneck" in a traffic jam. | {
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161,975 | There are many documentaries, forums, blogs and more dedicated to Dark Matter . I have been frantically searching for an answer to my question however none of my sources have clarity to the matter of hand. I would really love a clear explanation to: What exactly is Dark Matter? Please help me to have a clear understanding. | We don't know. Though there are several ideas what dark matter could be, e.g. the humorously abbreviated WIMPs , all we know about dark matter is that is it massive (by light deflection, etc., etc.) and that it does not interact electromagnetically, and probably also not with the strong force. Other than that, there is no sufficently tested theory of dark matter to pronounce with confidence what it is. We only know what it is not (i.e. not EM charged, not strongly charged, and there are probably a few other constraints from observation). Also, though highly unlikely, it could be that it is our theory of gravity , i.e. GR, that needs to be modified. In that case, it could be that there is no additional unknown matter, just different gravitational interactions from what we currently think. | {
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162,020 | According to Carnot efficiency formula $\eta=1-T_C/T_H$, can we say that the engines of cars are more efficient on cold days where $T_C$ (the temperature of the surroundings) is less than on hot days? | There are a lot factors involved here, but as internal combustion engines follow the Otto or Diesel cycle, those cycles (which are less efficient than the Carnot cycle) are the right model to follow, rather than the Carnot cycle itself. Let's say we reduce the inlet air temperature from 300 Kelvin to 270 Kelvin (90%) while keeping compression ratio and RPM the same. Now, the engine ingests the same volume of air, but greater mass. This means that a greater mass of fuel will be needed and greater power will be produced. The compression step remains the same, that is, if before it doubled the temperature from 300K to 600K, it will now double the temperature from 270K to 540K. The combustion chamber pressure before ignition will also be the same. The burning of the fuel will now increase the temperature by a fixed amount , so in the first case it will reach 600+2000=2600K and in the second case it will reach 540+2000=2540K. Expansion will now half the temperature (assuming reversibility and neglecting the minor change in composition of the gas during combustion) so the exhaust temperatures are 1300K and 1270K. What we can see from this theoretical internal combustion engine cycle is that the efficiency of both compression and expansion is determined only by the compression ratio. But also note that a significant proportion of the energy generated in the expansion is needed to drive the compression. The proportion is marginally less at lower temperature (due to the fixed temperature increase provided by combustion) so the engine will theoretically run slightly more efficiently at lower temperature. For identical efficiency, we would need the fuel to increase the temperature by 2000K in the 300K ambient case and 1800K in the 270K ambient case. But as the temperature increase from combustion is independent of the ambient temperature, the lower ambient temperature gives higher efficiency, at least in theory. There are many practical factors that may affect this analysis. The first is that my exhaust temperatures are extremely high. I can assure you that a temperature increase of 2000K is typical for complete combustion of hydrocarbon in air (I am a boiler engineer, but here's a table for those who like to check http://en.wikipedia.org/wiki/Adiabatic_flame_temperature .) Reciprocating internal combustion engines do run at close to complete combustion of all the air ingested, so it would appear that there are significant losses to the cooling system during the combustion process. My own experience is more with industrial gas turbines, which have proportionally much less loss to cooling systems, and an exhaust temperature arount 800K. Gas turbines never burn sufficient fuel to use all the oxygen in the inlet air, because the expansion turbine would melt. If suitable materials were available they would use all the oxygen, which would make them a lot more efficient. The next thing to consider is the valve timing. It will be apparent that if we have the same ratio on the compression and expansion stroke, we have effectively the same mass of gas entering the engine at 270-300K and leaving it at a higher temperature. The higher temperature means there is a higher pressure at the end of the expansion stroke, and if the exhaust valve opens too early, about 2-3 bar absolute (1-2 bar gauge) of pressure will be wasted. Although there are from time to time suggestions of ways to improve this situation, including closing the inlet valve late (to reduce compression) and opening the exhaust valve late (to increase the expansion) this is not a common pattern of operation. One reason is that when there is only 1-2 bar gauge pressure left, the amount of friction means that not much shaft power can be usefully extracted. However modern engines are often able to vary their valve timing and this complicates real-world analysis considerably. Thirdly, as per the above analysis an engine will produce more power in colder conditions due to the greater mass of air ingested. If more power is produced and friction stays the same, we would expect a small gain in engine efficiency, too. Finally, let us consider turbocharged engines. Here the pressure in the exhaust gas which would otherwise be wasted is used to drive a turbine, which compresses the air at the engine inlet. The primary motive for this is to increase power by increasing the mass flow into the engine. The work of compression heats the inlet air, so an intercooler is installed between the turbocharger and the engine. The purpose of this is to cool the air back down to ambient to further increase the mass flow of air into the engine. In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors. A number of people continue to believe colder temperatures lead to worse mpg . This is generally true for a variety of reasons: Cold temperatures lead to lower air pressure in tires, wet roads have increased resistance , cold temperatures lead to thicker lubricants etc. The engine is operating more efficiently (unless we choose to disbelieve 1-Tc/Th) but is working against a host of conditions, that lead to reduced mileage, that occur during the winter. Try picking a spot of road you drive every day. Put your car in neutral at a given speed at a given point. You will coast farther during the summer, with your engine disengaged. | {
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162,184 | Given that space is not a perfect vacuum, what is the speed of sound therein? Google was not very helpful in this regard, as the only answer I found was $300\,{\rm km}\,{\rm s}^{-1}$, from Astronomy Cafe , which is not a source I'd be willing to cite. | By popular demand (considering two to be popular — thanks @Rod Vance and @Love Learning), I'll expand a bit on my comment to @Kieran Hunt's answer: Thermal equilibrium As I said in the comment, the notion of sound in space plays a very significant role in cosmology: When the Universe was very young, dark matter, normal ("baryonic") matter, and light (photons) was in thermal equilibrium, i.e. they shared the same (average) energy per particle, or temperature. This temperature was so high, that neutral atoms couldn't form; any electron caught by a proton would soon be knocked off by a photon (or another particle). The photons themselves couldn't travel very far, before hitting a free electron. Speed of sound in the primordial soup Everything was very smooth, no galaxies or anything like that had formed. Stuff was still slightly clumpy, though, and the clumps grew in size due to gravity. But as a clump grows, pressure from baryons and photons increase, counteracting the collapse, and pushing baryons and photons outwards, while the dark matter tends to stay at the center of the overdensity, since it doesn't care about pressure. This creates oscillations, or sound waves with tremendously long wavelengths. For a photon gas, the speed of sound is $$
\begin{array}{rcl}
c_\mathrm{s} & = & \sqrt{p/\rho} \\
& = & \sqrt{c^2/3} \\
& \simeq & 0.58c,
\end{array}
$$ where $c$ is the speed of light, and $p$ and $\rho$ are the pressure and density of the gas. In other words, the speed of sound at that time was more than half the speed of light (for high temperatures there is a small correction to this of order $10^{-5}$ ; Partovi 1994 ). In a non-relativistic medium, the speed of sound is $c_\mathrm{s} = \sqrt{\partial p / \partial \rho}$ , which for an ideal gas reduces to the formula given by @Kieran Hunt. Although in outer space both $p$ and $\rho$ are extremely small, there $are$ particles and hence it odes make sense to talk about speed of sound in space. Depending on the environment, it typically evaluates to many kilometers per second (i.e. much higher than on Earth, but much, much smaller than in the early Universe). Recombination and decoupling As the Universe expanded, it gradually cooled down. At an age of roughly 200,000 years it had reached a temperature of ~4000 K, and protons and electrons started being able to combine to form neutral atoms without immediately being ionized again. This is called the "Epoch of Recombination", though they hadn't previously been combined. At ~380,000 years, when the temperature was ~3000 K, most of the Universe was neutral. With the free electrons gone, photons could now stream freely, diffusing away and relieving the overdensity of its pressure. The photons are said to decouple from the baryons. Cosmic microwave background The radiation that decoupled has ever since redshifted due to the expansion of the Universe, and since the Universe has now expanded ~1100 times, we see the light (called the cosmic microwave background, or CMB) not with a temperature of 3000 K (which was the temperature of the Universe at the time of decoupling), but a temperature of (3000 K)/1100 = 2.73 K, which is the temperature that @Kieran Hunt refers to in his answer. Baryon acoustic oscillations These overdensities, or baryon acoustic oscillations (BAOs), exist on much larger scales than galaxies, but galaxies tend to clump on these scales, which has ever since expanded and now has a characteristic scale of ~100 $h^{-1}$ Mpc, or 465 million lightyears. Measuring how the inter-clump distance change with time provides a way of understanding the expansion history, and acceleration, of the Universe, independent of other methods such as supernovae and the CMB. And beautifully, the methods all agree . | {
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162,704 | I am reading Why we do quantum mechanics on Hilbert spaces by Armin Scrinzi. He says on page 13: What is new in quantum mechanics is non-commutativity. For handling this, the Hilbert space representation turned out to be a convenient — by many considered the best — mathmatical environment. For classical mechanics, working in the Hilbert space would be an overkill: we just need functions on the phase space. I understand the definition of a Hilbert space, but I do not understand why non-commutativity compels us to use Hilbert spaces. Can someone explain? | I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation:
$$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle x+x'|\rho|x-x'\rangle e^{-2ipx'/\hbar}\,\mathrm{d}x'\text{,}$$
where for pure states $\rho = |\psi\rangle\langle \psi|$ as usual, and Hermitian operators correspond to functions through the inverse Weyl transformation. Note that $W(x,p)$ is a real function that's just like a joint probability distribution over the phase space, except that it's allowed to be negative. The uncertainty principle requires us to give up something, but it doesn't actually force Hilbert spaces on us. However , what Scrinzi is saying is two things: (a) Hilbert spaces are very convenient to us in quantum mechanics, and (b) Hilbert spaces could be used in classical mechanics, but because non-commutativity doesn't exist in classical mechanics, it's "overkill" there, whereas it's "just the right amount of kill" in quantum mechanics. Both claims are correct. The reason we can have used Hilbert spaces in classical mechanics is because they can represent very general algebras of observables, while the classical algebra of observables, being commutative, is actually simpler. (Cf. the Gel'fand–Naimark theorem for $C^*$-algebras in particular.) The Hilbert space formulation of classical mechanics was done by Koopman and von Neumann in 1931-1932. But what their formulation actually does is generalize classical mechanics unless one imposes an artificial restriction that you're only ever allowed to measure observables in some mutually commutative set; only then is the classical (in the 19th century sense) mechanics is recovered exactly. It is that artificial restriction that quantum mechanics lifts. Physically, non-commutativity observables corresponds to an uncertainty principle between them. | {
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162,883 | I mean, why is $F=ma$? Why not $m^{0.123}$, $a^{1.43}$ or some random non-integers or irrational? I hope you understand that my question isn't limited just to force, energy, velocity, etc.; it also extends to the area of a square, circle, etc. and all other formulas. I think the whole thing starts with direct proportionality. Most of them tell about the area of a circle, $A = πr^2$, where π is 3.14159..... an irrational number! It's not about the constant. I am talking about the power of a physical quantity. I know why it has pi. It's because we chose the constant for area square to be one.
If we chose it to be 1 for the circle then the area of a square will have a constant, 1/pi. I've edited the question to 'rational exponents' since all are giving me examples of decimal non-integers. | There is a non-subjective and quite mathematical approach to this question. First, we have the simple linear proportionalities that aren't really physical laws but just definitions of physical quantities. Why are different sensible measurable quantities usually in linear or power-law proportions will be further clarified later.
An example is $F=ma$ (just defines what force is - it's convenient to define it like that) and all unit-conversion formulas (essentially, there is only one unit - time and space can be made equal by $x=ct$, energy and momentum as well, then you have $E=\hbar \omega$ from quantum mechanics and so on). Linear relationship is not just a mathematical thing. Linearity means the principle of superposition holds: that a sum of causes creates a sum of effects. It's almost universal that when the effect is small, perturbation theory is valid and you have the first correction as a linear term. Imagine a Taylor expansion: it's a power series, no fractional exponents. This also means that a lot of the linear relations like that are approximation for weak perturbation. There's Ohm's law, heat conduction, hooke's law and so on. Even if you expand it further, it's still a power law. However, this may just be an approximation of some general result with some nonlinear function (could be exponential or something worse). But some of these relations are exact: in electrodynamics/vacuum optics, superposition principle is fundamental. But this brings us to the next point: Natural laws are local (ok, they can be expressed variationally, but that's another discussion). Local means that relationships between quantities obey differential equations. And differential equations are linear and when operated on power laws, they just shift the exponent by one. They are also usually linear (superposition), because nonlinearity most likely has a physical interpretation of a system acting on itself by changing its environment. Linearity in differential equations does not necessarily yield power laws: all exponential and oscillatory phenomena are results of linear differential equations. Here, nonlinearity means something different: dependence of the phenomena on the amplitude. A linear differential law means that twice the cause has twice the effect. Nonlinear means that twice the cause can have a completely unrecognizable effect. For instance, a pendulum at small amplitudes has a constant frequency. But when amplitudes are too big, the nonlinearity kicks in and you can have quite interesting behaviour. Fundamental laws are usually linear (Maxwell equations, for instance), and while there is an inherent question why the universe is so beautiful and elegant , the fact is, that if there are conserved quantities in a system, the relationship between them will be something simple. With differential equations, we again see not only laws but also plain definitions... velocity as derivative of position, acceleration as derivative of velocity, that's all just our decision what to measure. There's also $dE=F\,dx$ to get the work (energy contribution) caused by the force, which leads to all quadratic energy laws (of course: if forces are linear, at least in approximation, then integration brings you to quadratic). One very interesting point is, that fundamental laws of nature don't involve time derivatives higher than two (acceleration). That's somewhat related to the conservation of energy (Lagrangian functional) and tells you "how far a phenomenon can see" -- how much about history influences the now. But even with higher derivatives, we would still just the exponents by 1. So all in all, you can't really define a sensible differential law that would give you constant, but noninteger exponents. You may get rational exponents if you express quantities which have different powers to each other (from $a^3=b^2$ you will get $a=\sqrt[3]{b^2}$), but that's just an algebraic development. You do see weird exponents in empirical relationship: if there is no theoretical physical law behind it, but you measured some dependency and made up a function to draw a curve through the measurements, then a power function is something simple enough for people to try if it works. This is again an approximation and probably hides some more general theoretical result that is not a weird power law but a transcendental function or something that's just too complicated to write out algebraically. This is very common in material science: dependence of heat capacity, conductivity,... on temperature or current, are very strange functions. Transmission spectra are even worse. When things get complicated, a bunch of linear and nonlinear processes together yield a complex behaviour that's best just measured or at least simulated on a computer. However, the base laws and defining formulas of our chosen fundamental quantities, are mostly linear, or at least, something manageable. Superposition and proportionality are the most natural phenomena and even outside physics (economics, general statistics), this is just how things are. | {
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163,082 | How and why does the Huygens principle really work? I mean, does it always work? The Huygens principle: Every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets. For the Huygens principle to be true, light must emit light. Can light emit light? | The key phrase is "may be considered". That is, Huygens principle, as a mathematical procedure, gives (approximately) the right answer. The principle does not say that light is actually being generated at each point on the wavefront. Another way to look at it: the light at a phase front already exists there. There is no need to generate it all over again. It's interesting to note that if the vacuum were a "perfect" polarizable material, then the incident light would polarize every point in space, and each point would scatter light into spherical waves. I don't think that there's any physical justification for the picture, and it might lead to dangerously wrong conclusions, but it might help give you a grasp of how the mathematics works. | {
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163,372 | If a beam of light was shone horizontally, and simultaneously a stone was dropped from the same height, would they both hit the ground a the same time? Of course on Earth they would not, but let's imagine a land mass large enough for the light not to fire off into space and away from the pull of the land's gravity. | Given that the stone is in free fall - that is, in a frame following a geodesic - it would have to observe the light emitted in its frame as behaving normally. This means that the stone would not observe the light bend away from a path radially outwards from it, which is a complex way of saying that it must hit the ground at the same time as the stone. In the comments above, Floris mentioned how the deflection of light predicted by GR is twice what one expects from Newtonian calculations. This is true, however it is the measured deflection from the frame of a non-local observer that it refers to. So an observer standing on the ground would notice the discrepancy in the deflection of light, but the stone would not notice anything, since it is a local observer. If the stone were to see the photon hit the ground at a different time from it, that would violate the weak equivalence principle. | {
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163,691 | I never learnt QFT and I apologize for my (probably) elementary question. Somebody told me that in QFT a particle is viewed as an irregularity in the field. On the other hand, in an article in Wikipedia I see the sentence " A QFT treats particles as excited states of an underlying physical field, so these are called field quanta. " Which one of the true is a better description? The 1st description hints that the particle is a localized phenomenon inside a field that maybe occupies a big region in space. The 2nd description speaks of an " underlying " field. So, is there a field and in addition there is a particle? If it is, then what is the occupation number of that " underlying " field? None of these approaches is clear to me, I know the approach in QM, and none of them resembles the QM. The motivation behind my question is a certain similarity that I find between the above descriptions and the Bohm interpretation of QM, (i.e. the background field - in Bohm's interpretation there is a background quantum potential - and a particle floating in it.) In all, is a particle treated in QFT as a localized phenomenon inside a field occupying a wider volume? I would appreciate a simple and direct answer. | Somewhat surprisingly, the "generic" particle of QFT is in fact totally delocalized. More precisely, particles are thought to come from the mode expansion of free fields . Since every free relativistic field $\phi$ fulfills the Klein-Gordon equation $(\partial^\mu\partial_\mu - m^2)\phi = 0$, a Fourier transform shows that it can be expanded as $$ \phi(x) = \int \frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2p^0}}(a(\vec p)\mathrm{e}^{\mathrm{i}px} + a^\dagger(\vec p)\mathrm{e}^{-\mathrm{i}px})$$ where Lorentz invariance is not manifest, but can nevertheless be shown. A quantum field is operator-valued , and the operator valued objects $a(\vec p),a^\dagger(\vec p)$ fulfill exactly the correct commutation relations to be interpreted as creation and annihilation operators. The $n$-particle state of particles that are associated with the field $\phi$ is now defined as $$ \lvert n;p_1,\dots,p_n \rangle := a^\dagger(p_1)\dots a^\dagger(p_n)\lvert \Omega \rangle$$ where $\lvert \Omega \rangle$ is the (mostly) unique vacuum state. In this way, you first create all particle states that are sharply localized in momentum space (and hence completely delocalized in position space) and you can build localized particle states by the usual building of "wavepackets" with fuzzy momentum out of the sharp momentum states: A QM wavepacket of width $\sigma_x$ localized at $x_0$ is constructed out of the pure momentum states $\lvert \vec p \rangle$ as something like $$\lvert x_0,\sigma_x\rangle = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\mathrm{e}^{\mathrm{2i\sigma_x^2(x - x_0)^2}}\lvert p \rangle$$It works exactly the same for localized QFT particles, except that one should multiply the measure with $\frac{1}{\sqrt{2p^0}}$ to have a Lorentz invariant integration, and, of course, $\lvert p \rangle = a^\dagger(p)\lvert \Omega \rangle$. The idea that "particles are local excitations of the fields" comes from the observation that this mode expansion is almost completely analogous to a classical field fulfilling a wave equation like the Klein-Gordon equation, where the $a(\vec p),a^\dagger(\vec p)$ would directly represent an excitation of the field of wavenumber $\vec p$. It cannot be made precise in the context of QFT because the quantum field is operator-valued and has no definite values, so it is wholly unclear what rigorous sense could be given to it being "excited". It is a nice picture, but nothing you should take too literally. Also, take note that this is for the free field . The true interacting field of a QFT cannot be mode expanded in this way, and particle states are (through the LSZ formalism) only obtained in the asymptotic past and future (when they were far enough apart for interactions to be effectively non-existent) of the theory - the Hilbert space (and hence any states you could or could not identify as particles) of interacting QFTs is essentially unkown . Furthermore, more mathematical methods of constructing QFTs often first construct the $a,a^\dagger$ and the Fock space of particle states, and then define the field out of it - then, the roles of particle and field as "fundamental" and "derived" are somewhat reversed. | {
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164,126 | I know my subject sounds improbable and that's why I posted it. I helped my daughter with a science project for school. We tested five different materials to determine which would keep hot water hot, glass, ceramic, plastic, pewter and foam. To may amazement, the pewter did the best, though only slighter better than foam. I was so surprised I re-did the tests for the foam and pewter and got the same results. We placed boiling water in the cup and measured the temperature over a five minute period. The temp in the foam cup dropped about 24 degrees and in the pewter it dropped about 23 over the five minute period. All the others were a degree or two over that. I can't explain this and I am a chemical engineer. Can't find much about it on the Internet either. | The metal mug will equilibrate with the water much faster than the foam mug will— but after that the heat has no place to go$^\dagger$ except to be transferred away through radiation or be lost as steam via natural convection (which does not depend on the material of the mug). It turns out that radiation plays a very small role as shown below: Say each mug is a cylinder with diameter $10\ cm$ and height $10\ cm$. Inputting the temperature of boiling water ($373\ K$) and the surface area of the mug ($150 \pi\ cm^2$) into the Stefan-Boltzmann Law gives an approximation of the heat radiated away by the cup of 15 Watts. 5 minutes at 15 Watts = 4,500 J of energy radiated away, which we will call $Q$. Assuming you filled the mugs to the top, there is $250 \pi\ cm^3$ of water or a mass $m$ of $250 \pi\ g$. Since water's heat capacity $C$ is $4.18\ J/g\cdot K$ you can calculate the temperature drop $\Delta T$ due to radiation using the definition of heat capacity: $$\Delta T= \frac{Q}{mC} =\ \frac{4,500\ J}{\left(250 \pi\ g\right)\left(4.18\ J/g\cdot K\right)}=1.3\ K$$ That is much less than the 23 degrees temperature drop you report, so the majority of the heat loss comes from evaporative cooling as suggested by Martin Beckett. That means the cooling rate is relatively unaffected by the material of the mug and instead depends mostly on the area of the water surface exposed to air. A great teaching opportunity about the different mechanisms of heat transfer! $^\dagger$This is assuming no additional conduction to the table. If the table was made of metal you might see the effect you were expecting, since the heat would be able to conduct quickly through the pewter mug into the table. | {
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164,511 | A news story is going viral on social media networks claiming that two physicists have found a way to eliminate the Big Bang singularity, or in layman's terms (as claimed by many sensationalist news articles): The Big Bang never happened at all. The paper was published in Phys. Lett. B, and it seems to me that the basic idea of the paper is that it takes the Raychaudhuri equation and replaces the geodesics with Bohmian trajectories (from Bohm's quantum analogue of the Hamilton-Jacobi equation). I do not hold any expertise in this area, but from whatever little I have understood (and as explained by a friend of mine, who may very well be wrong), if we do a few substitutions on this quantum Raychaudhuri equation, we would get the Friedmann equations with a few extra terms, one of which looks like a cosmological constant and another of which eliminates the Big Bang singularity. Now my question is: How legitimate are the claims of the authors of this paper? And if it holds any merit, does it really eliminate the Big Bang singularity? | I'm not a quantum cosmologist, but I am an early-universe cosmologist, so I can give you my opinion after having read this paper. The article claims that Bohmian trajectories is a valid replacement for geodesics. This was claimed in the very beginning of the paper and not much is offered in the way of defense for this assumption. That's not to say that it's an invalid assumption, it's just that hands have been waved and they said "Geodesics are not the trajectories you're looking for. You can use Bohmian trajectories. Move along." Next what they do is refer us to another few papers where slightly different problems have been approached using similar but slightly different methods and they say (not an actual quote) "using a similar result to these works, and making arbitrary but not illegal substitutions, we recover the Friedman equations with higher order corrections" That's all well and good, but did you actually do the work or did you pull the result from the similar yet different problems. And the substitutions you made make no sense. From here, the interpretations they offer are fairly evident. Using Bohmian trajectories has necessitated that there is no initial singularity. They also find the correct magnitude for a cosmological constant in one of the corrections because of how they approach it. There's a complicated term that involves unspecified real-valued functions and they replace this with things corresponding to the graviton mass that would be expected to give the current value of the cosmological constant . Then, they find though some fancy footwork and more "convenient" substitutions that the term would result in a cosmological constant that is related to $c^2/H_0^2$, which results in the correct order of magnitude. They also show that with the Bohmian trajectories, the age of the universe tends toward infinity. What confused me here is that they used the standard method of calculating the age of the universe; the method that was designed for use with the Big Bang model. Essentially, they try to find the time since the singularity, but they automatically removed the singularity by using Bohmian trajectories. So it's obvious they should find infinity. Nonetheless, all of the science presented in the paper is valid. The problem is that it relies so heavily on science not presented in the paper. If what they've done is correct, then the interpretations is also correct. However, there's altogether too much handwaving and unintuitive assumptions made for me to accept it without further study. And as said before, a lot of it seems to rely on similar but different approaches that were used to address similar but different problems by other physicists. This is probably why the media has been able to go nuts with it. There's not enough in this paper to say it's wrong or to explicitly discredit it. There's also not enough to accept it or to use it to discredit other theories. The media sees it has the possibility of being right and that it has interpretations which don't include a Big Bang and they swarmed on it. But this paper doesn't say there wasn't a Big Bang; it says the Big Bang wasn't necessary. At best, this paper makes the possibility of no Big Bang equally likely as having a Big Bang. But I don't want to sound overly pessimistic. My stance remains that if the assumptions they made are correct and if the work they did outside of their writings is correct, then the interpretations they present could be correct as well. | {
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164,893 | What is the difference between states $$ \frac1{\sqrt{2}} |11\rangle+\frac1{\sqrt{2}} |00\rangle $$ and $$ \frac1{\sqrt{2}} |11\rangle- \frac1{\sqrt{2}} |00\rangle~? $$ They will all eventually mean probabilities. So what difference does it make? | This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to measure in other bases . In particular, there exists a measurement in the basis $\{(\lvert 00 \rangle\ + \lvert 11 \rangle)/\sqrt{2},(\lvert 00 \rangle\ - \lvert 11 \rangle)/\sqrt{2}\}$ which can perfectly distinguish the two states. Note that this issue has nothing to do with entangled or two-particle states in particular. Exactly the same could be said of the states $$\lvert \pm\rangle = \frac{1}{\sqrt{2}} ( \lvert 0\rangle \pm \lvert 1 \rangle ).$$
These states are indistinguishable when measured in the $\{ \lvert 0\rangle,\lvert 1 \rangle\}$ basis, but perfectly distinguishable in the $\{ \lvert +\rangle,\lvert - \rangle\}$ basis. | {
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165,381 | Take a video camera and crank up the frames per second rate. Disregarding current technological advancements, could a camera's FPS go so fast that any two captured images be identical ? Would accomplishing this defy "time"? | In addition to the good answer of MSalters it is worth pointing out that individual photons arrive on the detector of the camera at different times. The way that cameras work is that they convert photons arriving into a signal that can be read out. No matter how high the fps rate there will be some point in time where one frame is finished and the new one begins. For the two pictures to be identical the object being imaged must remain stationary on the timescale of fps speed - meaning that the object cannot move more than a distance the camera can resolve from one frame to the next... ....and as MSalters point out the arrival of individual photons at the detector can lead to another point that as the fps increases less photons are detected per frame and the image can change from frame to frame simply due to the statistics of their only being a few photons detected per frame. Taking this to a logical extreme if fps gets so high that less than one photon is absorbed in each frame then you could have pairs of images that are identical because they have no signal at all...., but I do not think this is the answer you wanted. | {
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165,526 | If gravity is zero at the center of the earth, why is there a core of heavy elements, such as iron? Alternate question for the opposite hypothesis: If gravity is greatest at the center of the earth, as classical education tells us, why is the core not dominated by the heaviest elements (elements heavier than iron)? I am a person reasonably familiar with technical terms, but I am not a physicist so I will appreciate answers that don't rely on equations. I am 70 years old and I want to explain it to my mother who is equally curious. | Forget about force. Force is a bit much irrelevant here. The answer to this question lies in energy, thermodynamics, pressure, temperature, chemistry, and stellar physics. Potential energy and force go hand in hand. The gravitational force at some point inside the Earth is the rate at which gravitational potential energy changes with respect to distance. Force is the gradient of energy. Gravitational potential energy is at it's lowest at the center of the Earth. This is where thermodynamics comes into play. The principle of minimum total potential energy is a consequence of the second law of thermodynamics . If a system is not in its minimum potential energy state and there's a pathway to that state, the system will try to follow that pathway. A planet with iron and nickel (and other dense elements) equally mixed with lighter elements is not the minimum potential energy condition. To minimize total potential energy, the iron, nickel, and other dense elements should be at the center of a planet, with lighter elements outside the core. A pathway has to exist to that minimum potential energy state, and this is where pressure, temperature, and chemistry come into play. These are what create the conditions that allow the second law of thermodynamics to differentiate a planet. As a counterexample, uranium is rather dense, but yet uranium is depleted in the Earth's core, slightly depleted in the Earth's mantle, and strongly enhanced in the Earth's crust. Chemistry is important! Uranium is fairly reactive chemically. It has a strong affinity to combine with other elements. Uranium is a lithophile ("rock-loving") element per the Goldschmidt classification of elements. In fact, uranium is an "incompatible element" , which explains the relative abundance of uranium in the Earth's crust. Nickel, cobalt, manganese, and molybdenum, along with the most extremely rare and precious metals such as gold, iridium, osmium, palladium, platinum, rhenium, rhodium and ruthenium, are rather inert chemically, but they do dissolve readily in molten iron. These (along with iron itself) are the siderophile (iron-loving) elements. In fact, iron is not near as siderophilic as the precious metals. It rusts (making iron is a bit lithophilic) and it readily combines with sulfur (making iron a bit chalcophilic). This is where pressure and temperature come into play. Pressure and temperature are extremely high inside the Earth. High pressure and high temperature force iron to relinquish its bonds with other compounds. So now we have pure iron and nickel, plus trace amounts of precious metals, and thermodynamics wants very much to have those dense elements settle towards the center. The conditions are now right for that to happen, and that's exactly what happened shortly after the Earth formed. Finally, there's stellar physics. The Earth would have a tiny little core of rare but dense elements if iron and nickel were as rare as gold and platinum. That's not the case. Iron and nickel are surprisingly abundant elements in the universe. There's a general tendency for heavier elements to be less abundant. Iron (and to a lesser extent, nickel) are two exceptions to this rule; see the graph below. Iron and nickel are where the alpha process in stellar physics stops. Everything heavier than iron requires exotic processes such as the s-process or those that occur in a supernova to create them. Moreover, supernova, particularly type Ia supernovae, are prolific producers of iron. Despite their relatively heavy masses, iron and nickel are quite abundant elements in our aging universe. (source: virginia.edu ) | {
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166,095 | I am trying to understand whether or not tachyons travel faster than light.
The linked Wikipedia page shows some seemingly contradictory statements, and they are confusing. For instance, the first sentence states that tachyons "always travel faster than the speed of light" whereas, in a later section, it is claimed that they are actually propagating subluminally.
Is it true that tachyons represent faster-than-light particles, or not? | A tachyon is a particle with an imaginary rest mass. This however does not mean it "travels" faster than light, nor that there's any conflict between their existence and the special theory of relativity. The main idea here is that the typical intuition we have about particles -- them being billiard ball-like objects -- utterly fails in the quantum world. It turns out that the correct classical limit for quantum fields in many situations is classical fields rather than point particles, and so you must solve the field equations for a field with imaginary mass and see what happens rather than just naively assume the velocity will turn out to be faster than light. The mathematical details are a bit technical so I'll just refer to Baez's excellent page if you're interested ( http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html ), but the conclusion can be stated very simply. There's two types of "disturbances" you can make in a tachyon field: 1) Nonlocal disturbances which can be poetically termed "faster than light" but which do not really represent faster than light propagation since they are nonlocal in the first place. In other words, you can't make a nonlocal disturbance in a finite sized laboratory, send it off to your friend in the andromeda galaxy and have them read the message in less time than it would take for light to get there. No, you could at best make a nonlocal disturbance that is as big as your laboratory, and to set that up you need to send a bunch of slower than light signals first. It's akin to telling all your friends all over the solar system to jump at exactly 12:00 am tomorrow: you'll see a nonlocal "disturbance" which cannot be used to send any information because you had to set it up beforehand. 2) Localized disturbances which travel slower than light. These are the only types of disturbances that could be used to send a message using the tachyon field, and they respect special relativity. In particle physics the term "tachyon" is used to talk about unstable vacuum states. If you find a tachyon in the spectrum of your theory it means you're not sitting on the true vacuum, and that the theory is trying to "roll off" to a state of lower energy. This actual physical process is termed tachyon condensation and likely happened in the early universe when the electroweak theory was trying to find its ground state before the Higgs field acquired its present day value. A good way to think about tachyons is to imagine hanging several pendulums on a clothesline, one after the other. If you disturb one of them, some amount of force will be transmitted from one pendulum to the next and you'll see a traveling disturbance on the clothesline. You'll be able to identify a "speed of light" for this system (which will really be the speed of sound in the string). Now you can make a "tachyon" in this system by flipping all the pendulums upside down: they'll be in a very unstable position, but that's precisely what a tachyon represents. Nevertheless, there's absolutely no way that you could send a signal down the clothesline faster than the "speed of light" in the system, even with this instability. tl;dr: Careful consideration of tachyons makes them considerably different from science fiction expectations. EDIT: As per jdlugosz's suggestion, I've included the link to Lenny Susskind's explanation. http://youtu.be/gCyImLu0HSI?t=58m51s | {
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166,367 | Given the inverse square law force of gravity shouldn't two particles that are infinitely close to each other be infinitely attracted to one another? For example, suppose the hands of some super deity grabbed hold of two neutrinos and put them infinitely close to one another or suppose that some physicists in a high energy particle physics laboratory shot two neutrinos together at super high speeds. Why should any forces be able to pull the neutrinos apart again after they collide? Why don't we find any tiny super dense lumps of matter around us that are the results of high energy collisions and that we are unable to pull apart? | There is no established quantum theory of gravity. Hence, at the microscopic level of particle, we don't know what is going on gravitationally between particle, but it isn't going to be the "inverse square law" we know, just like electromagnetism between two charged particles is, on quantum scales, not just an "inverse square law", but a rich variety of interactions that can be thought of as being mediated by virtual photons, and that gives the inverse square law of electrostatics only in the macroscopic and non-relativistic limit of the simplest of these interactions. Essentially, particles don't clump together as you imagine because they are, on the smallest scales, not particles at all. They are quantum states that are smeared out over an area like an electron orbital in an atom , and they do not behave as our classical intuition leads us to believe. In essence, because we don't have a good description of gravity at these scales, and because quantum objects are not little dots of mass in space, your question might as well be countered with: "Why should they?" | {
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166,568 | According to a textbook I have begun to read, there are seven base SI units: Length Mass Time Temperature Amount of a substance Electric current Luminous intensity What I do not understand is, why have these been chosen as the fundamental units? It seems to me that mass, amount, current and luminous intensity could all be expressed with energy. Instead energy is for some reason a derived unit. Why is this? | Temperature Amount of a substance Luminous intensity are pretty much bogus fundamental units. The unit temperature is just an expression of the Boltzmann constant (or you could say the converse, that the Boltzmann constant is not fundamental as it is merely an expression of the anthropocentric and arbitrary unit temperature). The unit energy will be whatever is the unit of force times the unit of length. A Joule is the same as a Newton - Meter , which are already defined in the SI system. You should read the NIST page on units to get the low-down on it. In my opinion, electric charge is a more fundamental physical quantity than electric current , but NIST (or more accurately, BIPM ) defined the unit current first and then, using the unit current and unit time , they defined the unit charge . I would have sorta defined charge first and then current. Just like the unit charge (or current) is just another way to express the vacuum permittivity or, alternatively the Coulomb constant and the unit temperature is just another way to express the Boltzmann constant, the unit time, unit length, and unit mass, all three taken together could be just another way to express the speed of light , the Planck constant , and the gravitational constant . But because $G$ is not easy to measure (given independent units of measure) and can never be measured as accurately as we can measure the frequency of "radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom" , we will never have $G$ as a defined constant as we do for $c$ and as we will soon for $\hbar$ and perhaps for $\epsilon_0$ and $k_\text{B}$. But once we define length, time, and mass independently, we cannot define energy independently. The Joule is a "derived unit". EDIT: so i will try to explain why the candela is bogus. (i had already for the mol .) so there is a sorta arbitrary specification of frequency, then what is the difference between 1 Candela and $\frac{4 \pi}{683} \approx$ 0.0184 watts? bogus base unit. | {
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166,966 | Say you start off floating in space, in a fixed position and orientation,
with zero linear and angular velocity, with no external forces.
So you are a closed mechanical system. By twisting your body around, you can't change your linear momentum. you can't change your position (center of mass). you can't change your angular momentum. you can change your orientation (i.e. rotation)! The fact that you can change your orientation comes
as a surprise to me-- why isn't it conserved
like the other three quantities?
It's a familiar fact-- cats do it all the time in order to land on their feet,
and you can find videos of astronauts doing it
on the international space station. See the videos linked from https://space.stackexchange.com/questions/2954/how-do-astronauts-turn-in-space .
But it still seems counterintuitive to me that they can do this
while not being able to change the other three quantities.
Is there some intuitively clear explanation as to why? | It is because the moment of inertia is not a conserved quantity. The statement that an isolated body can't change its position is more precisely the statement that an isolated body cannot change the position of its centre of mass. The position of the centre of mass, ${\bf R}$, is given by: $$ {\bf R} = \frac{1}{M}\sum m_i {\bf r}_i $$ where $M$ is the total mass and the $m_i$ are the masses of the individual elements of our system. Mass is a conserved quantity, so all the masses in our equation are constants and if we differentiate with respect to time we get: $$ \dot{\bf R} = \frac{1}{M}\sum m_i \dot{\bf r}_i = \frac{\bf P}{M} $$ where ${\bf P}$ is the total momentum. Since momentum is conserved the total momentum must be a constant and if we differentiate again we get $\ddot{\bf R} = 0$, so the acceleration of the centre of mass must always be zero. Now let's try and apply the same argument to the angular equivalent of the centre of mass. By analogy with the centre of mass we can define a centre of angle as: $$ \Theta = \frac{1}{I}\sum I_i \theta_i $$ The next step is to try and differentiate $\Theta$ twice with respect to time in the hope of obtaining $\ddot{\Theta} = 0$. The problem is that neither the total moment of inertia nor the moments of the individual elements are constants, but instead can be functions of time. In general our result will be: $$ \ddot{\Theta} \ne 0 $$ which means that $\Theta$ is not a constant. | {
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167,077 | In the movie Frozen, the following dialogue takes place: Anna: "It's a hundred-foot drop." Kristoff: "It's two hundred." Anna: "Okay, what if we fall?" Kristoff: "There's 20 feet of fresh powder down there. It will be like landing on a pillow... Hopefully. Then they fall all the way to the bottom and survive. My question is this: would this be actually possible? My instinct tells me no, but I'm too awful at physics to back it. | As a very rude guess, fresh snow (see page vi) can have a density of $0.3 \ \mathrm{g/cm^3}$ and be compressed all the way to about the density of ice, $0.9\ \mathrm{ g/cm^3}$ . Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters. Going from $30\ \mathrm{m/s}$ to $0\ \mathrm{m/s}$ (as @Sean suggested in comments), you'd have $(\frac{4\ \mathrm m}{12.5\ \mathrm{m/s}})$ = 0.32 seconds to decelerate. The acceleration is $\frac{30\ \mathrm{m/s}}{0.32\ \mathrm{s}}$ = $93.75\ \mathrm{m/s^2}$ . That's about: 9.5G's of acceleration Wikipedia lists 25g's as the point where serious injury/death can occur, and 215g's as the maximum a human has ever survived. So it seems plausible. But it should be noted that since the snow at the bottom is under a lot of pressure from the weight of the snow above, it's likely the density would not be $0.3\ \mathrm{g/cm^3}$ throughout. It would help that the force lasts only a fraction of a second. Edit as pointed out in the comments, the force that the snow will exert could vary with its density. So initially, the force would be rather weak, and as you approach $0.9\ \mathrm{\frac{g}{cm^3}}$ that force would increase, probably exponentially. So the above answer is really a "best case scenario" when it comes to snow compressibility | {
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167,250 | Astronomers find ancient black hole 12 billion times the size of the Sun. According to the article above, we observe this supermassive black hole as it was 900 million years after the formation of the universe, and scientists find its extreme specifications mysterious because of the relatively young age of the Universe at that time. Why would the 12 billion Solar Masses mass value be mysterious, unless there was a limit of sorts to the rate of mass consumption by a black hole? (naive point: Why would 900 million years not suffice for this much accumulation, keeping in mind that most supermassive stars which form black holes have life-spans of a few tens of millions of years at most?) | The accretion of matter onto a compact object cannot take place at an unlimited rate. There is a negative feedback caused by radiation pressure. If a source has a luminosity $L$, then there is a maximum luminosity - the Eddington luminosity - which is where the radiation pressure balances the inward gravitational forces. The size of the Eddington luminosity depends on the opacity of the material. For pure ionised hydrogen and Thomson scattering
$$ L_{Edd} = 1.3 \times 10^{31} \frac{M}{M_{\odot}}\ W$$ Suppose that material fell onto a black hole from infinity and was spherically symmetric. If the gravitational potential energy was converted entirely into radiation just before it fell beneath the event horizon, the "accretion luminosity" would be
$$L_{acc} = \frac{G M_{BH}}{R}\frac{dM}{dt},$$
where $M_{BH}$ is the black hole mass, $R$ is the radius from which the radiation is emitted (must be greater than the Schwarzschild radius) and $dM/dt$ is the accretion rate. If we say that $L_{acc} \leq L_{Edd}$ then
$$ \frac{dM}{dt} \leq 1.3 \times10^{31} \frac{M_{BH}}{M_{\odot}} \frac{R}{GM_{BH}} \simeq 10^{11}\ R\ kg/s \sim 10^{-3} \frac{R}{R_{\odot}}\ M_{\odot}/yr$$ Now, not all the GPE gets radiated, some of it could fall into the black hole. Also, whilst the radiation does not have to come from near the event horizon, the radius used in the equation above cannot be too much larger than the event horizon. However, the fact is that material cannot just accrete directly into a black hole without radiating; because it has angular momentum, an accretion disc will be formed and will radiate away lots of energy - this is why we see quasars and AGN -, thus both of these effects must be small numerical factors and there is some maximum accretion rate. To get some numerical results we can absorb our uncertainty as to the efficiency of the process and the radius at which the luminosity is emitted into a general ignorance parameter called $\eta$, such that
$$L_{acc} = \eta c^2 \frac{dM}{dt}$$
i.e what fraction of the rest mass energy is turned into radiation.
Then, equating this to the Eddington luminosity we have
$$\frac{dM}{dt} = (1-\eta) \frac{1.3\times10^{31}}{\eta c^2} \frac{M}{M_{\odot}}$$
which gives
$$ M = M_{0} \exp[t/\tau],$$
where $\tau = 4\times10^{8} \eta/(1-\eta)$ years (often termed the Salpeter (1964) growth timescale ). The problem is that $\eta$ needs to be pretty big in order to explain the luminosities of quasars, but this also implies that they cannot grow extremely rapidly. I am not fully aware of the arguments that surround the work you quote, but depending on what you assume for the "seed" of the supermassive black hole, you may only have a few to perhaps 10 e-folding timescales to get you up to $10^{10}$ solar masses. I guess this is where the problem lies. $\eta$ needs to be very low to achieve growth rates from massive stellar black holes to supermassive black holes, but this can only be achieved in slow-spinning black holes, which are not thought to exist! A nice summary of the problem is given in the introduction of Volonteri, Silk & Dubus (2014) . These authors also review some of the solutions that might allow Super-Eddington accretion and shorter growth timescales - there are a number of good ideas, but none has emerged as a front-runner yet. | {
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167,469 | I can't seem to connect these two stories. Can you please help? I have heard that you have to turn electron by 720 degrees in order to get the same spin state. Has this been seen experimentally? How do you turn an electron? I know that Bloch sphere is used to explain spin states. I sort of understand how it works and how matrices can be constructed. However there you turn the vector by 360 degrees in order to get the same state. Why so? (Is that in reality you have to turn twice as much?) | You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert space $\mathcal{H}$. The crucial thing is that not all elements of this space represent physically different states - if we have two elements $\phi$ and $\psi$ and they are related in such a way that one can be obtained from the other by multiplying it with any complex number $c$, i.e. $\phi = c\psi$, then they are the same state. This is analogous to two arrows with different length pointing in the same direction describing the same direction. Only the direction of the Hilbert space element has immediate physical meaning, not the length (though it is not completely irrelevant, "phases" play a role, but this is not relevant here). Now, it turns out that there are two different ways how such elements of a Hilbert space can behave under a full rotation by $2\pi$ - they either stay the same, $\psi \overset{2\pi}{\mapsto} \psi$, or they change their sign, $\psi \overset{2\pi}{\mapsto} -\psi$. But $-1$ is just a complex number, so $\psi$ and $-\psi$ are the same state, and a rotation by $2\pi$ does not change any state at all . Objects whose states stay the same are called bosons and have "integer spin", objects whose states change sign are called fermions and have "half-integer spin". The Bloch sphere you refer to is not the Hilbert space of a system, but the projective Hilbert space . The projective Hilbert space is obtained by just identifying all vectors in the Hilbert space that lie on the same ray ( = have the same direction = are complex multiples of each other). Thus, $\psi$ and $-\psi$ are the same point in a projective space, hence in particular on the Bloch sphere, and a $2\pi$ rotation does nothing on a projective space either way - as it should, since each point of the projective space is a physically distinct state. | {
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167,888 | The way thrust is created by a rocket is discussed here: How does fire create thrust in rocket? If you look at a typical rocket, say V2, it has two primary tanks — one which stores the fuel and the other which stores oxygen. Is it that the lighter flame doesn't use its own oxygen, the reason that it doesn't generate thrust? Or is it that the thrust is so small that I can't feel it? Or is it something more complicated like the shape of the nozzle? | For the average disposable lighter, when you press the fuel lever a pressurised liquefied gas is released which will create a very small thrust. The combustion, however, will not generate thrust because, unlike a rocket engine, it is not occurring within a chamber. | {
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168,066 | In one dimension, the acceleration of a particle can be written as: $$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$ Does this equation imply that if: $$v = 0$$ Then, $$\Rightarrow a = 0$$ I can think of several situations where a particle has a non-zero acceleration despite being at instantaneous rest. What's going on here? | The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$
x=-\frac{gt^2}{2}+v_0t
$$ $$
v=v_0-gt
$$ and $$
a=-g
$$ and, clearly, "a" can never be zero, but "v" can be zero... so what gives? Well... solving for t(x) gives $$
t(x)=\frac{1}{g}(v_0\pm\sqrt{v_0^2-2gx})
$$ and $$
v(x)\equiv v(t(x))=\pm\sqrt{v_0^2-2gx}
$$ and so $$
\frac{dv}{dx}=\frac{\pm g}{\sqrt{v_0^2-2gx}}
$$ ...which is conveniently infinite whenever v is zero. Furthermore, I think that a more natural way to think about this issue can be found by considering what we really mean by $$
v(x)
$$ and how we go about taking the derivative w.r.t. x. What we really mean is that, given some functional form for "v" as a function of "t" called "v(t)", and given some functional form for "x" as a function of "t" called "x(t)", and given that "x(t)" can be inverted to find "t(x)", then, as mentioned above $$
v(x)=v(t(x))\;,
$$ which is silly physicist notation. It is clearly silly notation because the "v()" on the left-hand side cannot actually have the same form as the "v()" on the right-hand side. So really let's call it $\tilde v$ . I.e., $$
\tilde v(x)=v(t(x))
$$ This function $\tilde v$ is a function of x and the derivative with respect to x is $$
\frac{d\tilde v}{dx}(x)=\frac{dv}{dt}(t(x))\frac{dt}{dx}=\frac{\frac{dv}{dt}(t(x))}{\frac{dx}{dt}}=\frac{a(t(x))}{v(t(x))}
$$ I.e., (switching back to silly notation and not writing arguments of functions) $$
\frac{dv}{dx}=\frac{a}{v}\;,
$$ so, clearly, for constant a, dv/dx is infinite whenever v=0. | {
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168,218 | I have tried to explore the information but still not very clear on the exact difference between diffusion , convection and advection . Can anyone help me out to clear my concept? | Convection is the collective motion of particles in a fluid and actually encompasses both diffusion and advection. Advection is the motion of particles along the bulk flow Diffusion is the net movement of particles from high concentration to low concentration We typically describe the above two using the partial differential equations:
\begin{align}
\frac{\partial\psi}{\partial t}+\nabla\cdot\left(\mathbf u \psi\right)&=0\tag{advection}\\
\frac{\partial\psi}{\partial t}&=\nabla\cdot\left(D\nabla\psi\right)\tag{diffusion}
\end{align}
where $\psi$ is the quantity in consideration, $\mathbf u$ is the fluid velocity and $D$ the diffusion coefficient (sometimes called the diffusivity). There are some nuances to the combined effect for convection (e.g., forced, natural, gravitational mechanisms), but the general definition for it is the total motion. | {
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168,684 | The answer I usually get (and I'm paraphrasing here) is that they disappear and are instead absorbed as heat energy. But I find it hard to believe that the photon simply "disappears." Common sense tells me it must turn into something or other, not just simply poof out of existence; then again, common sense has betrayed me before. Forgive me if this is obvious; high school physics student here who's just learned about light and is greatly confused by all this. | Well, the answer you usually get is half right. They do disappear (more on this in a second). I'd hesitate to say they turn into "heat energy," both because we don't use the term "heat" that way in a technical sense and because most of the time we like to talk about atoms absorbing photons. In this case the energy of the photon becomes potential energy of the electron that made the transition, and there's no need to talk about heat. Now, can the photon disappear? The short answer is yes. When you talk about things "not simply poofing out of existence" what you're really describing is like a conservation law. For instance, we say that energy is neither created nor destroyed. Your intuition that things aren't just "poofed" out of existence is probably due to your everyday experience that objects generally can be broken into parts, but not usually destroyed. This isn't true in the particle physics sense, usually. The energy carried by that photon has to be accounted for, as does its momentum and angular momentum. But "photon number" is not a conserved quantity the way that energy or (for instance) electric charge are. A photon really is just a way of looking at disturbances/excitations in the electric field, and so its "destruction" just represents that energy that was present in the field has been moved into some other mode. | {
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168,713 | As per my knowledge, atoms are small beyond our imaginations. But there is an image on Wikipedia that shows silicon atoms observed at the surface of silicon carbide crystals. The image: How can we see these distinct atoms if they are so small? | This entirely depends on what you mean by "see". Let me start of by noting: As per my knowledge, atoms are small beyond our imaginations No. Atoms are quite big compared to certain other things we play around with, like its constituents (protons, electrons) in particle accelerators. The size of atoms is of the order 0.1 nanometres (of course, there is a variation in size , but I'm not going to bother for now). A nanometre is $10^{-9}$ metres. Protons for example are very much smaller and atoms are in a sense so big that we know for over a hundred years now that they are not indivisible, because we have seen in experiments that they are not. Now, can we "see" atoms? This depends, as I already hinted at, what you mean by "see". If you mean "make a picture in visible light", then you can't do that. In microscopy, there is a rule of thumb that the smallest things you can distinguish with a perfectly engineered microscope have to have a size about half the wavelength of the light you're shining at it. The more exact version of this is known as the Abbé difraction limit . Visible light has wavelength of about 400-700 nanometres. This is of course about 4000-7000 times as much as the diameter of the atom, so there is indeed no way we can see an atom with a (diffraction) microscope using light. [As suggested in the comments, there are a number of methods to get around Abbé's diffraction limit using, in parts, very different techniques to usual microscopy. It seems, however, that a resolution of atoms is not achieved yet.] But there are other things besides light we can use. We could, for instance, use electrons instead of light. Quantum mechanics tells us that electrons, just like light and everything else, have a wavelengths . Of course, such a microscope looks a bit different than a light microscope, because we humans have no good detection mechanism for electrons. This means, in order to make an image from the refracted and difracted electrons, we need to use electronic sensors and then recreate the image. This type of microscope that I just described is more or less a transmission electron microscope (TEM) and they have been around for a long time. Today, such types of microscopes have a resolution of about 0.05 nanometres (usual TEMS are sometimes cited to have a resolution of about 1000 times better than the resolution of light microscopes, but using some correction techniques one can achieve the resolutions of 0.05 nm and maybe below ). This is just about enough to see an atom (see here for an early picture, the other answer contains better and more recent pictures), but it's probably not enough to see the picture you linked to have a slightly better resolution. [Note: a few years ago, you definitely needed the microscope I describe in the next section for such a picture, today you might be able to achieve it via TEMs also. In other words: Today you might be able to "see" atoms with electrons.] So how did we get this: But there is a wikipedia image which shows silicon atoms observed at the surface of silicon carbide crystals. We have to use a different type of electronic microscope, a scanning tunneling microscope (STM) . While the TEM works basically the same as a light microscope, the STM uses different concepts. Therefore, it is even more removed from what you would ordinarily call "seeing". I'm not going to describe how this works in detail, but the microscope consists of a little tip with a voltage applied and it measures the tunneling of electrons into the probe, thereby measuring the distance to the probe. The peak then wanders over the surface of your material and measures the distance of the material to the tip at many points, then constructing a topographic image of the probe. So it measures the electron density around the atom and thereby, as we understand it, the size of the atom. With this, any reasonable STM can get a resolution of about 0.1 nm and good STMs are much better. And this, finally, is how we can see atoms. | {
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168,716 | The internal kinetic energy of an ideal fermi gas at temperature 0K is given by $$U=\frac{3}{5}NE_f$$ What conclusion can we draw from this statement. | This entirely depends on what you mean by "see". Let me start of by noting: As per my knowledge, atoms are small beyond our imaginations No. Atoms are quite big compared to certain other things we play around with, like its constituents (protons, electrons) in particle accelerators. The size of atoms is of the order 0.1 nanometres (of course, there is a variation in size , but I'm not going to bother for now). A nanometre is $10^{-9}$ metres. Protons for example are very much smaller and atoms are in a sense so big that we know for over a hundred years now that they are not indivisible, because we have seen in experiments that they are not. Now, can we "see" atoms? This depends, as I already hinted at, what you mean by "see". If you mean "make a picture in visible light", then you can't do that. In microscopy, there is a rule of thumb that the smallest things you can distinguish with a perfectly engineered microscope have to have a size about half the wavelength of the light you're shining at it. The more exact version of this is known as the Abbé difraction limit . Visible light has wavelength of about 400-700 nanometres. This is of course about 4000-7000 times as much as the diameter of the atom, so there is indeed no way we can see an atom with a (diffraction) microscope using light. [As suggested in the comments, there are a number of methods to get around Abbé's diffraction limit using, in parts, very different techniques to usual microscopy. It seems, however, that a resolution of atoms is not achieved yet.] But there are other things besides light we can use. We could, for instance, use electrons instead of light. Quantum mechanics tells us that electrons, just like light and everything else, have a wavelengths . Of course, such a microscope looks a bit different than a light microscope, because we humans have no good detection mechanism for electrons. This means, in order to make an image from the refracted and difracted electrons, we need to use electronic sensors and then recreate the image. This type of microscope that I just described is more or less a transmission electron microscope (TEM) and they have been around for a long time. Today, such types of microscopes have a resolution of about 0.05 nanometres (usual TEMS are sometimes cited to have a resolution of about 1000 times better than the resolution of light microscopes, but using some correction techniques one can achieve the resolutions of 0.05 nm and maybe below ). This is just about enough to see an atom (see here for an early picture, the other answer contains better and more recent pictures), but it's probably not enough to see the picture you linked to have a slightly better resolution. [Note: a few years ago, you definitely needed the microscope I describe in the next section for such a picture, today you might be able to achieve it via TEMs also. In other words: Today you might be able to "see" atoms with electrons.] So how did we get this: But there is a wikipedia image which shows silicon atoms observed at the surface of silicon carbide crystals. We have to use a different type of electronic microscope, a scanning tunneling microscope (STM) . While the TEM works basically the same as a light microscope, the STM uses different concepts. Therefore, it is even more removed from what you would ordinarily call "seeing". I'm not going to describe how this works in detail, but the microscope consists of a little tip with a voltage applied and it measures the tunneling of electrons into the probe, thereby measuring the distance to the probe. The peak then wanders over the surface of your material and measures the distance of the material to the tip at many points, then constructing a topographic image of the probe. So it measures the electron density around the atom and thereby, as we understand it, the size of the atom. With this, any reasonable STM can get a resolution of about 0.1 nm and good STMs are much better. And this, finally, is how we can see atoms. | {
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168,943 | I was reading about the equipartition theorem and I got the following quotations from my books: A diatomic molecule like oxygen can rotate about two different axes. But rotation about the axis down the length of the molecule doesn't count. - Daniel V. Schröder's Thermal Physics. A diatomic molecule can rotate like a top only about axes perpendicular to the line connecting the atoms but not about that line itself. - Resnick, Halliday, Walker s' Fundamentals of Physics. Why is it so? Doesn't the rotation take place that way? | The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$ B = \frac{\hbar^2}{2I} $$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This means the moment of inertia around an axis along the bond is going to be about $10^{10}$ smaller than the moment of inertia around an axis normal to the bond. Therefore the energy level spacings will be around $10^{10}$ times bigger along the bond than normal to it. In principle we can still excite rotations about the axis along the bond, but you'd need huge energies to do it. | {
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168,962 | Is there significant energy to capture in daily day/night temperature change? For instance, I see from posts such as the following that there are materials with varying coefficients of linear thermal expansion. Would not some have enough power to be useful? We know that water, upon freezing, can lift sidewalks (but I presume that the magic range from 4C to 0C, wherein I see it gets this magic power, is an outlier). Thanks very much for entertaining this layman's question, and I hope this doesn't "generate discussion" because I know how antithetical that is to the Stack philosophy. What kinds of materials contract the most in cold temperatures? | The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$ B = \frac{\hbar^2}{2I} $$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This means the moment of inertia around an axis along the bond is going to be about $10^{10}$ smaller than the moment of inertia around an axis normal to the bond. Therefore the energy level spacings will be around $10^{10}$ times bigger along the bond than normal to it. In principle we can still excite rotations about the axis along the bond, but you'd need huge energies to do it. | {
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168,964 | So recently I have became interested in quantum physics. However when I read up on quantum physics Feynman diagrams appear everywhere, and I do not understand them! Here is a Feynman diagram that is annoying me: So, How do I know if that electron and anti-neutrino were there all along, or one created the other? I also apologize for my immense niaveness. | The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$ B = \frac{\hbar^2}{2I} $$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This means the moment of inertia around an axis along the bond is going to be about $10^{10}$ smaller than the moment of inertia around an axis normal to the bond. Therefore the energy level spacings will be around $10^{10}$ times bigger along the bond than normal to it. In principle we can still excite rotations about the axis along the bond, but you'd need huge energies to do it. | {
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169,003 | If there are two spheres (hollow and solid) with equal mass and radius and we want to find the hollow sphere without using any equipment. What's the best way(s) to recognize the hollow sphere and solid sphere? | Let both roll down an inclined plane. The hollow sphere will accelerate slower than the solid one (due to their different moments of inertia). For a solid sphere, the moment of inertia is $$I = \frac{2}{5}mr^2$$ with mass $m$ and radius $r$. For a hollow sphere it is $$I = \frac{2}{3}mr^2$$ The hollow sphere therefore has a greater moment of inertia and will accelerate slower under the same torque: $$M = I \frac{d\omega}{dt}$$ with angular velocity $\omega$ and torque $M$. | {
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169,009 | I'm not educated in physics, but I learned that a bell struck in a small vacuum chamber will not be heard by people around it (in a school science lab). If it had been surrounded by air, there would have been energy propagating outward from the bell in a sound wave. Because this can't happen, does the bell vibrate harder from the impact? Will it break sooner? | When a bell vibrates in air, it pushes air molecules out of the way which will make the vibrations "decay". If you strike a bell in vacuum, this loss mechanism will not be there so the bell will "ring" for longer (but nobody can hear it). This doesn't mean the initial amplitude is significantly greater - just that it persists longer. Obviously if you rang the bell repeatedly there is a small chance that the moments of impact would amplify the motion of the bell (resonance) but that's not really very likely. I would not expect the bell to break sooner. On the other hand, for an electrical bell the coils used in the mechanism will heat up, and they would normally be air cooled (not terribly efficient but still - it's a cooling mechanism). It is conceivable that an electrical bell operated continuously in a vacuum would burn out because the cooling is less than in air - but that's pure speculation. Back to the question of damped vibration, a sketch of what this might look like (arbitrary units of time along the horizontal, and amplitude along the vertical axis): I am assuming there are "some" loss mechanisms for both - but at least one fewer for the bell in vacuum. The initial amplitude after being struck ought to be (very nearly) the same. | {
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169,209 | The electromagnetic spectrum is a continuum of wavelengths of light, and we have labels for some ranges of these and numerical measurements for many. Question: Is the EM spectrum continuous such that between two given wavelengths (e.g. 200nm and 201nm) there is an infinite number of distincts wavelengths of light? Or is there some cut-off of precision with which light might exist (e.g. can light only have wavelengths of whole number when measured in nanometers, etc.)? | Yes, there are an uncountable infinity of possible wavelengths of light. In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2). Two things to consider in practice: There are situations in which the only relevant frequencies are discrete (such as the modes in a cavity). For any given experimental measurement you will always have a finite precision or bandwidth with which you can measure, and so although light at 200nm and 200.01nm is in principle different, you might not be able to tell in practice. | {
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169,213 | I have always heard that the inconsistency in explaining atomic models with classical mechanics was that the study of electrical charges had shown that whenever a charge is accelerated, it emits light (and thus there is no way to sustain circular movement of electrons around a nucleus). Since then it seemed interesting to me the possibility that perhaps a very natural/commonplace phenomenon in which we can see light emitted from an accelerated charge could be lightning. But then it also occurred to me that perhaps the light of lightning is caused by the chemical reactions that happen in the path of the ray (oxygen is converted to ozone if I am not mistaken, there may be others). I did try to search the internet a bit, but at simple glance at least the answer seemed not to be obvious. Does anybody here know which case is it? what other common or interesting cases of light emission due to charge acceleration are there? NOTE: I am tagging this as quantum mechanics because, like I said before, the most likely place I think this information might come up other than when studying electrodynamics is as a motivation on the development of the quantum theory. Please mods feel free to modify it if you disagree or think other tags should apply and apologies for any inconvenience. | Yes, there are an uncountable infinity of possible wavelengths of light. In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2). Two things to consider in practice: There are situations in which the only relevant frequencies are discrete (such as the modes in a cavity). For any given experimental measurement you will always have a finite precision or bandwidth with which you can measure, and so although light at 200nm and 200.01nm is in principle different, you might not be able to tell in practice. | {
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169,275 | On a few occasions either in bed or sitting around a fire, with my eyes closed, I rarely but sometimes see a very quick fast flash of white and then, with my eyes still closed, the flash disappears immediately. It happens so fast that I sit up and rethink if it was even real. But I know it is real because I have had it happen to me many times in my life. I have also asked other people if it happens to them and 4/5 replied back saying that they had experienced the flash before. Is it possible for a neutrino to pass the brain and in response produce the white flash? After all the brain is made of 73% water and neutrino detectors are predominantly water. I tried submitting this question on biology.stackexchange and I was told that questions like these belonged on the physics.stackexhange site. | The cross-section for neutrino interactions is energy dependent. For solar neutrinos at $\sim 0.4$ MeV, which would likely dominate any neutrinos likely to interact with a brain (the cosmic background neutrinos have way low energies), the cross-sections are $\sigma \sim 10^{-48}$ m $^2$ , for both leptonic processes (elastic scattering from electrons) and neutrino-nucleon interactions. The mean free path of a neutrino will be given by $l \sim (n\sigma)^{-1}$ , where $n$ is number of interacting target particles per cubic metre and $\sigma$ is the cross-section. If your head is basically water with a density of 1000 kg/m $^3$ , then there are $n_e = 3.3\times10^{29}\ m^{-3}$ of electrons and about $6 \times 10^{29} m^{-3}$ of nucleons. Including both nucleonic and leptonic processes, the mean free path is $\sim 10^{18}\ m$ . So unless your head is 100 light years wide, there is little chance of any individual neutrino interacting with it. This is only one part of the calculation though - we need to know how many neutrinos are passing through your head per second. The neutrino flux from the Sun is about $7\times 10^{14}$ m $^{-2}$ s $^{-1}$ . If your head has an area of about 400 cm $^2$ , then there are $3\times 10^{13}$ neutrinos zipping through your brain every second. Thus is we take $x=20$ cm as the path length through your head, there is a chance $\sim x/l$ of any neutrino interacting, where $l$ was the mean free path calculated earlier. This probability multiplied by the neutrino flux through your head indicates there are $6\times 10^{-6}$ s $^{-1}$ neutrino interactions in your head, or roughly one every two days. Whether that would produce any perceptible effect in your brain needs to be shunted back to Biology SE. If we require it (or rather scattered electrons) to produce Cherenkov radiation in the eyeball, then this needs $>5$ MeV neutrinos and so the rate would reduce to 1 per 100 days or even lower due to the smaller number of neutrinos at these energies and the smaller volume of water in the eyeball. EDIT: In fact my original answer may be over-optimistic by an order of magnitude since water only acts as a good detector (via Cherenkov radiation) for neutrinos above energies of 5 MeV. Solar neutrinos are predominantly lower energy than this. My calculation ignored atmospheric neutrinos which are produced in far fewer numbers (but at higher energies $\sim 0.1-10$ GeV). The cross-section for these is 4-6 orders of magnitude higher, but I think they are produced in so much lower numbers that they don't contribute. Conclusion It doesn't have anything to do with neutrinos. The rate would be too low, even if they could be perceived. | {
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169,279 | Say I have a theoretical mountain with a river starting exactly at the peak. Gravity would act on the water, accelerating it down the mountain. There would also be a force limiting the water's velocity. My question is: "Would collecting the energy from the water in intervals, allowing the water to reach its 'terminal velocity' before the next collection point, be more optimal than a single hydro-electric generator at the bottom? I think it would be, but I am not a physicist or engineer (yet) so there may be something integral I am missing. Here is a picture of a similar situation I am referring to to help understand: http://www.directionalenergy.com/wp-content/uploads/2011/04/Run-Of-River-1024x593.jpg | The cross-section for neutrino interactions is energy dependent. For solar neutrinos at $\sim 0.4$ MeV, which would likely dominate any neutrinos likely to interact with a brain (the cosmic background neutrinos have way low energies), the cross-sections are $\sigma \sim 10^{-48}$ m $^2$ , for both leptonic processes (elastic scattering from electrons) and neutrino-nucleon interactions. The mean free path of a neutrino will be given by $l \sim (n\sigma)^{-1}$ , where $n$ is number of interacting target particles per cubic metre and $\sigma$ is the cross-section. If your head is basically water with a density of 1000 kg/m $^3$ , then there are $n_e = 3.3\times10^{29}\ m^{-3}$ of electrons and about $6 \times 10^{29} m^{-3}$ of nucleons. Including both nucleonic and leptonic processes, the mean free path is $\sim 10^{18}\ m$ . So unless your head is 100 light years wide, there is little chance of any individual neutrino interacting with it. This is only one part of the calculation though - we need to know how many neutrinos are passing through your head per second. The neutrino flux from the Sun is about $7\times 10^{14}$ m $^{-2}$ s $^{-1}$ . If your head has an area of about 400 cm $^2$ , then there are $3\times 10^{13}$ neutrinos zipping through your brain every second. Thus is we take $x=20$ cm as the path length through your head, there is a chance $\sim x/l$ of any neutrino interacting, where $l$ was the mean free path calculated earlier. This probability multiplied by the neutrino flux through your head indicates there are $6\times 10^{-6}$ s $^{-1}$ neutrino interactions in your head, or roughly one every two days. Whether that would produce any perceptible effect in your brain needs to be shunted back to Biology SE. If we require it (or rather scattered electrons) to produce Cherenkov radiation in the eyeball, then this needs $>5$ MeV neutrinos and so the rate would reduce to 1 per 100 days or even lower due to the smaller number of neutrinos at these energies and the smaller volume of water in the eyeball. EDIT: In fact my original answer may be over-optimistic by an order of magnitude since water only acts as a good detector (via Cherenkov radiation) for neutrinos above energies of 5 MeV. Solar neutrinos are predominantly lower energy than this. My calculation ignored atmospheric neutrinos which are produced in far fewer numbers (but at higher energies $\sim 0.1-10$ GeV). The cross-section for these is 4-6 orders of magnitude higher, but I think they are produced in so much lower numbers that they don't contribute. Conclusion It doesn't have anything to do with neutrinos. The rate would be too low, even if they could be perceived. | {
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169,294 | I am not sure if this is the most appropriate place to post this but here goes nothing: Assume we were trying to come up with system of numbers $S$ to model our intuition of length. We want $S$ to have these properties intuitively at least: S is an abelian group under some operation (say +). Abelian because we want it count stuff. It should be a $\mathbb{Q}$ - vector space. This corresponds to our notion of divisibility of length, given a ruler, we can imagine $1/k$ th of that ruler. It should be archimedean. No observation contradicts it. We should have a notion of limits in it through something like the intuition behind Zeno's paradoxes. Essentially, we are asking for completeness. I think these are all our intuitions in mathematical language. Of course, these are also enough to force our system to be the familiar reals. I am not sure if anyone has done something similar to this before and I have a few questions. Questions: Does the other stuff we model by the reals (temperature, probabilities, entropy etc) also follow the same/similar intuitions? Is there any reason that all our measurements have these properties? If not, do we measure other properties in physics by other systems? The only one I can think of is complex numbers in Quantum Mechanics but I don't know anything about that. Finally, is it coincidence that we have a uniqueness theorem for exactly those properties that model our intuitions about the world? I am sorry that my questions are vague/philosophical. This seemed like an interesting enough phenomenon to post anyway. x-posted from math.stackexchange. I have also seen a few other questions similar to this one but none of them carried out the analysis to an axiomatic level as far as I am aware. | The cross-section for neutrino interactions is energy dependent. For solar neutrinos at $\sim 0.4$ MeV, which would likely dominate any neutrinos likely to interact with a brain (the cosmic background neutrinos have way low energies), the cross-sections are $\sigma \sim 10^{-48}$ m $^2$ , for both leptonic processes (elastic scattering from electrons) and neutrino-nucleon interactions. The mean free path of a neutrino will be given by $l \sim (n\sigma)^{-1}$ , where $n$ is number of interacting target particles per cubic metre and $\sigma$ is the cross-section. If your head is basically water with a density of 1000 kg/m $^3$ , then there are $n_e = 3.3\times10^{29}\ m^{-3}$ of electrons and about $6 \times 10^{29} m^{-3}$ of nucleons. Including both nucleonic and leptonic processes, the mean free path is $\sim 10^{18}\ m$ . So unless your head is 100 light years wide, there is little chance of any individual neutrino interacting with it. This is only one part of the calculation though - we need to know how many neutrinos are passing through your head per second. The neutrino flux from the Sun is about $7\times 10^{14}$ m $^{-2}$ s $^{-1}$ . If your head has an area of about 400 cm $^2$ , then there are $3\times 10^{13}$ neutrinos zipping through your brain every second. Thus is we take $x=20$ cm as the path length through your head, there is a chance $\sim x/l$ of any neutrino interacting, where $l$ was the mean free path calculated earlier. This probability multiplied by the neutrino flux through your head indicates there are $6\times 10^{-6}$ s $^{-1}$ neutrino interactions in your head, or roughly one every two days. Whether that would produce any perceptible effect in your brain needs to be shunted back to Biology SE. If we require it (or rather scattered electrons) to produce Cherenkov radiation in the eyeball, then this needs $>5$ MeV neutrinos and so the rate would reduce to 1 per 100 days or even lower due to the smaller number of neutrinos at these energies and the smaller volume of water in the eyeball. EDIT: In fact my original answer may be over-optimistic by an order of magnitude since water only acts as a good detector (via Cherenkov radiation) for neutrinos above energies of 5 MeV. Solar neutrinos are predominantly lower energy than this. My calculation ignored atmospheric neutrinos which are produced in far fewer numbers (but at higher energies $\sim 0.1-10$ GeV). The cross-section for these is 4-6 orders of magnitude higher, but I think they are produced in so much lower numbers that they don't contribute. Conclusion It doesn't have anything to do with neutrinos. The rate would be too low, even if they could be perceived. | {
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169,486 | The faucet design depicted below is driving me crazy. The water falling from the tap appears to follow a spiral path. No one seems to agree whether it is physically possible for the water to spin in such a way. Is it possible? Please explain your answer. More info about the tap is available here . | You are right that, without a force acting on it, water falling from a tap could not follow a spiralled path. The tap, however, creates an illusion - the water appears to be spiralling, but it isn't - it's falling straight down. The illusion is created by the "turbine" inside the nozzle, which rotates the ring of spouts that the water falls through. The effect is that a corkscrew shape of water falls downwards. A falling corkscrew, however, is difficult to distinguish from a rotating corkscrew, hence the illusion, Let me also add a caveat, pointed out by @Adam Davis. The pictures of the tap are all computer rendered. This fancy tap is just an idea - the designer has not built a working, proof of concept prototype. There might be serious difficulties in realizing such a design, for example, turbulence from the spinning nozzle might destroy the corkscrew shape of the falling water. | {
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169,489 | Let's assume that I am on an airplane that is at about 4,000 altitude and now let's also assume that I am standing on one of the wings with a light torch, if I point the light torch below to the surface, the photons will be attracted by gravity and hit the ground at their normal speed but if I point a light torch upwards, will the speed of light be slightly slowed (even by the slightest margin) due to the force of gravity? | You are right that, without a force acting on it, water falling from a tap could not follow a spiralled path. The tap, however, creates an illusion - the water appears to be spiralling, but it isn't - it's falling straight down. The illusion is created by the "turbine" inside the nozzle, which rotates the ring of spouts that the water falls through. The effect is that a corkscrew shape of water falls downwards. A falling corkscrew, however, is difficult to distinguish from a rotating corkscrew, hence the illusion, Let me also add a caveat, pointed out by @Adam Davis. The pictures of the tap are all computer rendered. This fancy tap is just an idea - the designer has not built a working, proof of concept prototype. There might be serious difficulties in realizing such a design, for example, turbulence from the spinning nozzle might destroy the corkscrew shape of the falling water. | {
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169,494 | As far as I know, the movement of air molecules can result in what we call wind, sound, or heat (and maybe some other things I'm not thinking of), depending on the assumptions about its motion. Wind generally occurs when air molecules all move in a group in the same direction. Sound occurs when air molecules (or little packets of air), are locally compressed. And I believe heat moves when air molecules vibrate faster than their neighbors do, and some how that energy is moved. This figure from "Sound and heat revolutions in phonics" in Nature helps setup what I'm curious about: So where do we draw the dividing lines between wind, sound, and heat? How does the air "know what to do", given a certain stimulus? If I move a mass, and its position is given by a unit step function, what happens? I understand that by Fourier Analysis, I should be exciting the air around the mass at infinite frequencies, and if this is that case, would this create wind, sound, and heat? And if so, where are the boundaries between these phenomena? Are they just related to the frequency of the motion, or something more complex like coupling or impedance matching? Would the wind, sound, and heat travel out at different speeds? Thank you! | You are right that, without a force acting on it, water falling from a tap could not follow a spiralled path. The tap, however, creates an illusion - the water appears to be spiralling, but it isn't - it's falling straight down. The illusion is created by the "turbine" inside the nozzle, which rotates the ring of spouts that the water falls through. The effect is that a corkscrew shape of water falls downwards. A falling corkscrew, however, is difficult to distinguish from a rotating corkscrew, hence the illusion, Let me also add a caveat, pointed out by @Adam Davis. The pictures of the tap are all computer rendered. This fancy tap is just an idea - the designer has not built a working, proof of concept prototype. There might be serious difficulties in realizing such a design, for example, turbulence from the spinning nozzle might destroy the corkscrew shape of the falling water. | {
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169,553 | Now, I don't really mean to say that Maxwell's equations are wrong. I know Maxwell's equations are very accurate when it comes to predicting physical phenomena, but going through high school and now in college, Maxwell's equations are seen as the equations of electricity and magnetism. Now, it's common knowledge among students that, while Newton's laws are generally accurate when applied to everyday experiences, they are also replaced at high velocities by special relativity (and general relativity for very large gravitational fields). But this is less so the case with Maxwell's equations. I have read that Maxwell's equations are replaced with quantum electrodynamics (which to me has all the effect of mere buzzwords, since I don't know what quantum electrodynamics is ) as a more accurate way of describing electromagnetic waves, but what are the limitations of Maxwell's equations? Or let me phrase this differently. I'm currently an electrical engineering major. I know NASA scientists and engineers can still get away with using Newtonian physics for their calculations, because it's that accurate. I also know, however, that relativity does have to come into play with GPS. So, in what situation as an electrical engineer would Maxwell's equations ever fail me? When (assuming I'm working on such a sufficiently advanced project) would I have to resort to more accurate ways to describe electromagnetic waves? | Maxwellian electrodynamics fails when quantum mechanical phenomena are involved, in the same way that Newtonian mechanics needs to be replaced in that regime by quantum mechanics. Maxwell's equations don't really "fail", as there is still an equivalent version in QM, it's just the mechanics itself that changes. Let me elaborate on that one for a bit. In Newtonian mechanics, you had a time-dependent position and momentum, $x(t)$ and $p(t)$ for your particle. In quantum mechanics, the dynamical state is transferred to the quantum state $\psi$, whose closest classical analogue is a probability density in phase space in Liouvillian mechanics . There are two different "pictures" in quantum mechanics, which are exactly equivalent. In the Schrödinger picture, the dynamical evolution is encoded in the quantum state $\psi$, which evolves in time according to the Schrödinger equation. The position and momentum are replaced by static operators $\hat x$ and $\hat p$ which act on $\psi$; this action can be used to find the expected value, and other statistics, of any measurement of position or momentum. In the Heisenberg picture, the quantum state is fixed, and it is the operators of all the dynamical variables, including position and momentum, that evolve in time, via the Heisenberg equation. In the simplest version of quantum electrodynamics, and in particular when no relativistic phenomena are involved, Maxwell's equations continue to hold: they are exactly the Heisenberg equations on the electric and magnetic fields, which are now operators on the system's state $\psi$. Thus, you're formally still "using" the Maxwell equations, but this is rather misleading as the mechanics around it is completely different. (Also, you tend to work on the Schrödinger picture, but that's beside the point.) This regime is used to describe experiments that require the field itself to be quantized, such as Hong-Ou-Mandel interferometry or experiments where the field is measurably entangled with matter. There is also a large gray area of experiments which are usefully described with this formalism but do not actually require a quantized EM field, such as the examples mentioned by Anna. (Thus, for example, black-body radiation can be explained equally well with discrete energy levels on the emitters rather than the radiation.) This regime was, until recently, pretty much confined to optical physics, so it wasn't really something an electrical engineer would need to worry about. That has begun to change with the introduction of circuit QED , which is the study of superconducting circuits which exhibit quantum behaviour. This is an exciting new research field and it's one of our best bets for building a quantum computer (or, depending on who you ask, the model used by the one quantum computer that's already built. ish.), so it's something to look at if you're looking at career options ;). The really crazy stuff comes in when you push electrodynamics into regimes which are both quantum and relativistic (where "relativistic" means that the frequency $\nu$ of the EM radiation is bigger than $c^2/h$ times the mass of all relevant material particles). Here quantum mechanics also changes, and becomes what's known as quantum field theory , and this introduces a number of different phenomena. In particular, the number of particles may change over time, so you can put a photon in a box and come back to find an electron and a positron (which wouldn't happen in classical EM). Again, here the problem is not EM itself, but rather the mechanics around it. QFT is built around a concept called the action , which completely determines the dynamics. You can also build classical mechanics around the action, and the action for quantum electrodynamics is formally identical to that of classical electrodynamics. This regime includes pair creation and annihilation phenomena, and also things like photon-photon scattering, which do seem at odds with classical EM. For example, you can produce two gamma-ray beams and make them intersect, and they will scatter off each other slightly. This is inconsistent with the superposition principle of classical EM, as it breaks linearity, so you could say that the Maxwell equations have failed - but, as I pointed out, it's a bit more subtle than that. | {
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169,626 | I've recently been reading Path Integrals and Quantum Processes by Mark Swanson; it's an excellent and pedagogical introduction to the Path Integral formulation.
He derives the path integral and shows it to be: $$\int_{q_a}^{q_b} \mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$$ This is clear to me. He then likens it to a discrete sum
$$\sum_\limits{\text{paths}}\exp\left(\frac{iS}{\hbar}\right)$$ where $S$ is the action functional of a particular path. Now, this is where I get confused. He claims that, because some of these paths are discontinuous or non-differentiable and that these "un-mathematical" 1 paths cannot be disregarded, the sum is not mathematically rigorous, and, thus, that the transition amplitude described by the path integral is not rigorous either. Please correct me if I am incorrect here. Furthermore, he claims that this can be alleviated through the development of a suitable measure. There are two things that I don't understand about this. First, why isn't the integral rigorous? Though some of the paths might be difficult to handle mathematically, they aren't explicitly mentioned at all in the integral. Why isn't the answer that it spits out rigorous? And, second, why would a measure fix this problem? 1 Note: this is not the term he uses | There are several points: The first is that for usual self-adjoint Hamiltonians of the form $H=-\Delta +V(x)$, with a common densely defined domain (and I am being very pedantic here mathematically, you may just ignore that remark) the limit process is well defined and it gives a meaning to the formal expression $\int_{q_a}^{q_b} \mathcal{D}p\mathcal{D}q\exp\{\frac{i}{\hbar}\int_{t_a}^{t_b} \mathcal{L}(p, q)\}$ by means of trotter product formula and the corresponding limit of discrete sums. So the object has most of the time meaning, as long as we see it as a limit. Nevertheless, it would be suitable to give a more direct mathematical interpretation as a true integral on paths. This would allow for generalizations and flexibility in its utilization. It turns out that a suitable notion of measure on the space of paths can be given, using stochastic processes such as brownian motion (there is a whole branch of probability theory that deals with such stochastic integration, called Itô integral). To relate this notion with our situation at hand there is however a necessary modification to make: the factor $-it$ in the quantum evolution has to be replaced by $-\tau$ (i.e. it is necessary to pass to "imaginary time") . This enables to single out the correct gaussian factors that come now from the free part of the Hamiltonian, and to recognize the correct Wiener measure on the space of paths. On a mathematical standpoint, the rotation back to real time is possible only in few special situations, nevertheless this procedure gives a satisfying way to mathematically define euclidean time path integrals of quantum mechanics and field theory (at least the free ones, and also in some interacting case). There are recent works of very renowned mathematicians on this context, most notably the work of the fields medal Martin Hairer (see e.g this paper and this one , or the recent work by A. Jaffe that gives an interesting overview; a more physical approach is given by Lorinczi, Gubinelli and Hiroshima among others). The precise mathematical formulation of path integral in QM is called Feynman-Kac formula, and the precise statement is the following: Let $V$ be a real-valued function in $L^2(\mathbb{R}^3)+L^\infty(\mathbb{R}^3)$, $H=H_0+V$ where $H_0=-\Delta$ (the Laplacian). Then for any $f\in L^2(\mathbb{R}^3)$, for any $t\geq 0$:
$$(e^{-tH}f)(x)=\int_\Omega f(\omega(t))e^{-\int_0^t V(\omega(s))ds}d\mu_x(\omega)\; ;$$
where $\Omega$ is the set of paths (with suitable endpoints, I don't want to give a rigorous definition), and $\mu_x$ is the corresponding Wiener measure w.r.t. $x\in\mathbb{R}^3$. | {
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169,631 | Suppose we have two events $(x_1,y_1,z_1,t_1)$ and $(x_2,y_2,z_2,t_2)$ . Then we can define $$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2,$$ which is called the spacetime interval. The first event occurs at the point with coordinates $(x_1,y_1,z_1)$ and the second at the point with coordinates $(x_2,y_2,z_2)$ which implies that the quantity $$r^2 = \Delta x^2+\Delta y^2+\Delta z^2$$ is the square of the separation between the points where the events occur. In that case the spacetime interval becomes $\Delta s^2 = r^2 - c^2\Delta t^2$ . The first event occurs at time $t_1$ and the second at time $t_2$ so that $c\Delta t$ is the distance light travels on that interval of time. In that case, $\Delta s^2$ seems to be comparing the distance light travels between the occurrence of the events with their spatial separation. We now have the following definitions: If $\Delta s^2 <0$ , then $r^2 < c^2\Delta t^2$ and the spatial separation is less than the distance light travels and the interval is called timelike. If $\Delta s^2 = 0$ , then $r^2 = c^2\Delta t^2$ and the spatial separation is equal to the distance light travels and the interval is called lightlike. If $\Delta s^2 >0$ , then $r^2 > c^2\Delta t^2$ and the spatial separation is greater than the distance light travels and the interval is called spacelike. These are just mathematical definitions. What, however, is the physical intuition behind them? What does an interval being timelike, lightlike or spacelike mean? | Let's suppress some dimensions to simplify: $$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 $$ This quantity $$\Delta s^2$$ is preserved by changes of reference frame, just as in Galilean physics the quantity $$\Delta r^2 = \Delta x^2 + \Delta y^2 $$ is preserved by rotations. Notice it is also the equation of a hyperbola. Thus, the effect of a frame shift is to slide events around on hyperbolae of constant $\Delta s^2$ . Here's a helpful image from Wikipedia (attribution below): Ignore the vectors and just look at the hyperbolae. Events on a given hyperbola must, under a given frame boost, remain on that hyperbola . Now you might notice those hyperbolae seem to come in two classes, those on the top and those on the bottom. The "v=c" hyperbolae - the straight lines - divide the two. Events on those are said to be "lightlike (or null) separated from the origin". Notice that for these, $\Delta s^2$ is just zero . The hyperbolae in the purple regions are said to be timelike separated from the origin. This is because no matter how much they slide around on their hyperbolae, their ordering compared to the origin never changes . Any events in the purple regions which occur before (after) the origin will occur before (after) the origin to all observers . Thus, this set of events - plus the null events - is said to be causally connected with the origin . The fact that the ordering of these events with the origin in time is fixed motivates the term. The hyperbolae in the white regions do not have this property. Some observers think they happened before O, while some think they happened after. It had therefore better be true that nothing about O depend logically on happening after (or before) these events! Otherwise we could break logic by running really fast. However, notice that it is not possible to slide the white-region events from one side of the origin to the other. This makes the separation more like our normal ideal of "distance", so we say the events are spacelike separated . Image attribution: "Minkowski lightcone lorentztransform" by Maschen - Own work. Licensed under Public Domain via Commons | {
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169,730 | I would like to really understand how the uncertainty principle in QM works, from a practical point of view. So this is my narrative of how an experiment goes, and I'm quickly in trouble: we prepare a set of many particles in the same state $\psi$ as best we can, then we start measuring 2 observables A and B that don't commute on... each of the particles (?). When we measure A, the wave function collapses to an eigenstate of A with some probability. By accumulating measurements with A, we obtain statistics, and in particular $\langle\psi|A|\psi\rangle$, the expected value of $A$ w.r.t to state $\psi$. But how do I get $\langle\psi|B|\psi\rangle$? Can I measure A and B "simultaneously" on one particle, even if $\psi$ has collapsed to an eigenstate of A, which is not an eigenstate of B, and A and B don't commute... What happens? How do I measure B? Do I need to pull in another particle, on which I'll measure B, but not A this time? | There are many steps: Step 1, select a state $\Psi$. Step 2, prepare many systems in same state $\Psi$ Step 3, select two operators A and B Step 4a, for some of the systems prepared in state $\Psi$, measure A Step 4b, for some of the systems prepared in state $\Psi$, measure B Now if you analyze the results, assuming strong (not weak) measurements then every time you measured A, you got an eigenvalue of A, and every time you measure B you got an eigenvalue of B. Each eigenvalue had a probability (which is equal to the ratio of the squared norm of the projection onto the eigenspace divided by the squared norm before you projected onto the eigenspace). So your eigenvalues of A come from a probability distribution that often has a mean $\langle A\rangle=\langle \Psi|A|\Psi\rangle $ and a standard deviation $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$. And your eigenvalues of B come from a probability distribution that often has a mean $\langle B\rangle=\langle \Psi|B|\Psi\rangle $ and a standard deviation $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$. You never get those from a measurement, or even from a whole bunch, but from steps 4a and 4b you do get a sample mean and a sample standard deviation, and for a large sample these are likely to be very close to the theoretical mean and the theoretical standard deviation. The uncertainty principle says that way back in step 1 (when you selected $\Psi$) you could select a $\Psi$ that gives a small $\Delta A$, or a $\Psi$ that gives a small $\Delta B$ (in fact if $\Psi$ is an eigenstate of A then $\Delta A=0$, same for $B$). However, $$\Delta A \Delta B \geq \left|\frac{\langle AB-BA\rangle}{2i}\right|=\left|\frac{\langle\Psi| AB-BA |\Psi\rangle}{2i}\right|,$$ So in particular noncommuting operators often (i.e. if the expectation value of their commutator does not vanish) have a tradeoff, if the state in question has really low standard deviation for one operator, then the state in question must have a higher standard deviation for the other. If the operators commute, not only is there no joint limit to how low the standard deviations can go, but measuring the other variable keeps you in the same eigenspace of the other operator. However that is a completely different fact since the uncertainty principle is about the standard deviations of two probability distributions for two observables applied to one and the same state, and thus approximately applies to the sample standard deviations generated from identically prepared states. If you have a system prepared in state $\Psi$ and you measure A on it then you generally have to use a different system also prepared in $\Psi$ to measure B. That's because when you measure A on a system it projects the state onto an eigenspace of A, which generally changes the state. And since the probability distribution for B is based on the state, now that you have a different state you will have a different probability distribution for B. You can't find out $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$ if you don't have $\Psi$ and only have $\Psi$ projected onto an eigenspace of A. | {
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169,745 | I have not been able to find a clear or decisive answer about this. On Earth, a ship sails close hauled when the ship is, to some degree or another, facing into the wind. This is made possible with sails that are rigged along the length of the ship rather than perpendicular to the length. Is a similar thing practically or at least theoretically possible with a solar sailing vessel, to sail towards a star in this manner? Why or why not? | A sailboat can make headway against the wind because of the sum of force vectors due to the wind interacting with the sail and, due to the keel interacting with the water . A sailboat without a keel can not make headway into the wind. There is no "water" out there into which a solar sailer could dip its keel. http://newt.phys.unsw.edu.au/~jw/sailing.html http://web.mit.edu/2.972/www/reports/sail_boat/sail_boat.html | {
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169,758 | Quaternions are commonly used to model 4 dimensional systems where the quaternion consists of a real 3 dimensional vector and an imaginary scalar. So on the surface Quaternions seem well suited to model space time if time can be considered imaginary. Do the operations of quaternion math further provide a suitable framework, or are there problems? | There are some problems with using quaternions to describe spacetime. Quaternions have two important properties: (1) they form a four-dimensional vector space; (2) you can multiply quaternions together. [1] The first property is obviously very suggestive, but it's no different from the usual four-vectors that we already use in special relativity. To specifically make use of quaternions, we'd also have to use the second property. Recall that -- at least in ordinary discussions of special relativity -- you'll never multiply four-vectors together and get another four-vector; you only "contract" them (take their dot product). And the standard quaternion multiplication doesn't actually achieve this dot product. The scalar term in the result is something like the usual spacetime interval , but then there are those extra vector components. You can get a pure scalar multiplying a quaternion by itself, but only by conjugating one copy, which gives you a positive-definite result rather than the interval. So it's not quite useful for special relativity. And as far as vectors go, the usual approach is good enough. Even though they have four degrees of freedom, quaternions really do "live in" a physical space of three dimensions. It turns out that quaternions shouldn't be thought of as a scalar plus a vector. Instead, they should be thought of as a scalar plus a bi vector . [2] More specifically, quaternions are actually the natural "spinors" of three-dimensional space. So instead of being like vectors, they act on vectors. For example, probably the most common use for quaternions is to describe rotations of vectors. This bears repeating: Quaternions shouldn't be thought of as vectors in 4-d; they should be thought of as operators acting on vectors in 3-d. So to answer your question, yes you could use quaternions to model spacetime -- though there'd be a lot of useless baggage floating around. But extrapolating the motivation for your question, I might rephrase it as "Can we use the special properties of quaternions to get any computational advantage or philosophical insight in relativity?" To that, the answer is no; they don't have anything useful to say about space time since they're really just about space. But there's good news! There is a natural generalization of quaternions to four-dimensional spacetime, and it really does give us computational advantages and philosophical insights. The cool thing about thinking of quaternions as scalar + bivector is that the idea now generalizes very easily to arbitrary dimensions -- and in particular to Minkowski spacetime . This is a field of study called "Geometric algebra" , or GA for short. [3] Spinors in 4 dimensions [4] turn out to act a lot like spinors in 3 dimensions (quaternions). For example, they can be used to rotate 4-dimensional vectors really nicely. But they can also boost 4-d vectors just as easily -- a boost is sort of a generalized rotation . It turns out that a lot of the usual things we do in special relativity are way easier using spinors. And you can keep going to other dimensions. For example, going back to just two dimensions, you find that complex numbers are the spinors of 2-d! You even start to understand complex algebra better using GA. In fact, I've taught GA to biologists by starting in 2 dimensions. Once you understand this simple example, it's almost trivial to extend GA to arbitrary dimensions. [5] If you want to learn more, there's a fantastic book on this called Geometric Algebra for Physicists . It's actually my favorite physics book, full stop. There are lots of good online references too, if you google around for it. And I have to plug the Geometric Algebra module for sympy , which gives us a nice (open-source) program for doing the math symbolically. Footnotes: Taken together, these two facts mean that quaternions form "an algebra" . The idea may seem kind of weird -- that you can actually multiply two vectors by each other. You already know how to multiply a vector by a scalar. And you can take the dot and cross products, but neither of those is invertible. But really just multiplying two vectors in a (usually) invertible way might seem weird. And then you realize that you do it all the time with complex numbers, which also form "an algebra". Not to mention matrices. It just so happens that in three dimensions, there are three degrees of freedom in a bivector, and three degrees of freedom in a vector. So when Hamilton discovered quaternions, he was understandably confused about what they represented. His confusion was the whole reason for the vector/quaternion wars of the 1890s . Nowadays, we understand that quaternions and vectors are just two aspects of the same thing: GA. I would argue that this confusion is one of the great tragedies in the history of physics, as Grassmann and Clifford had already developed all the tools necessary to resolve the conflict. We could discuss the name of this thing till the cows come home. But in practice, "Geometric Algebra" is a sub-type of Clifford Algebra , except we assume that in GA the coefficients for our vector space are real numbers, whereas CA can have coefficients from any field -- especially complex numbers. But CA is usually introduced with irrelevant abstractions, and the complex version is almost never necessary for applications in physics (even quantum mechanics!). Spinors in 4-d are sometimes called biquaternions , which are "complexified" quaternions, but that's a very bad path to go down. The complexification is unenlightening, and doesn't really apply to other dimensions. I think it's symptomatic of a tendency to use obscure, accidental features specific to a particular dimension -- as opposed to the intuitive, pedagogical, systematic, and universal approach of GA. The path that spinors and normed division algebras (NDAs) took together splits at dimension four, as the latter head into a dead end (there are no more NDAs after octonions). The spinors in four dimensions do have eight degrees of freedom, like the octonions, but that's just the vector-space property. The other property of algebras, multiplication, can't be the same because octonions are not associative -- but associativity is one of the defining features of GA. So the octonions aren't a particular example of GA. However, it's also worth pointing out that there are other spinor groups even for dimensions ≤3 when you have a non-positive signature. For example, the split-complex numbers are the spinors of a two-dimensional version of Minkowski space. Of course, there's very little need for octonions in physics. John Baez has a typically great introduction to an article about octonions in physics that you can read here , in which he shows that there are applications in supersymmetry / string theory (and pure math, obviously). But that's the most convincing argument I've seen that octonions might ever have any relevant applications in physics -- and I'm certainly not convinced. | {
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169,964 | I have read a fair bit about topological insulators and proximity induced Majorana bound states when placing a superconductor in proximity to a topological insulator. I've also read a bit about cuprates being related to topological superconductivity if that helps. What I cant quite understand is what defines and what is a pure topological superconductor? Or is this simply not the case and is topological superconductivity something that can only be achieved by means of proximity effect arrangements? A general description of what one is would probably be most helpful. | In short, what makes a superconductor topological is the nontrivial band structure of the Bogoliubov quasiparticles. Generally one can classify non-interacting gapped fermion systems based on single-particle band structure (as well as symmetry), and the result is the so-called ten-fold way/periodic table. The topological superconductivity mentioned in the question is related to the class D, namely superconductors without any symmetries other than the particle-hole symmetry. The simplest example in 2D is a spinless $p_x+ip_y$ superconductor: $H=\sum_k c_k^\dagger(\frac{k^2}{2m}-\mu)c_k+ \Delta c_k^\dagger(k_x+ik_y)c_{-k}^\dagger+\text{h.c.}=\sum_k (c_k^\dagger, c_{-k})\left[(k^2/2m-\mu)\tau_z+\Delta k_x\tau_x+\Delta k_y\tau_y\right]\begin{pmatrix}c_k\\ c_{-k}^\dagger\end{pmatrix}$ This Hamiltonian defines a map from the $k$ space (topologically a sphere $S^2$) to a $SU(2)$ matrix $m_k\cdot \sigma$ where $m_k\propto (\Delta k_x, \Delta k_y, \frac{k^2}{2m}-\mu)$ (then normalized), which also lives on a sphere. Therefore such maps are classified by $\pi_2(S^2)=\mathbb{Z}$. If two Hamiltonians belong to the same equivalence class in the homotopy group, it means that one can continuously deform the Hamiltonian from one to another without closing the gap, thus topologically indistinguishable. The integer, called the Chern number $C$, that classifies the class D topological superconductors can be calculated from the Hamiltonian, and in this case it is $C=1$. This idea can be generalized to other symmetry classes and dimensions, basically one needs to understand the map from the momentum space to the appropriate single-particle "Hamiltonian" space (the general case is much more complicated than the $2\times 2$ Hamiltonian). This toy model (and its one-dimensional descendants) is behind all recent proposals of realizing topological superconductors in solid state systems. The basic idea is to combine various mundane elements (semiconductors, s-wave superconductor, ferromagnet, etc): since electrons have spin-$1/2$, one needs to have Zeeman field to break the spin degeneracy and get a non-degenerate Fermi surface (thus effectively "spinless" fermions, really spin-polarized). However, in s-wave superconductors electrons with opposite spins are paired. This is why spin-orbit coupling is necessary since it makes the electron spin "winds" around on the Fermi surface, so that at $k$ and $-k$ electrons can pair up. Putting all these together one can realize a topological superconductor. There are various physical consequences. The general feature is that something peculiar happens on the boundary between superconductors belonging to different topological classes. For example, if the $p_x+ip_y$ superconductor has an edge to the vacuum, there are gapless chiral Majorana fermions localized on the edge. Also if one puts a $hc/2e$ vortex into the superconductor, it traps a zero-energy Majorana bound state. The question also mentioned cuprates. There are some speculations about the possibility of $d+id$ pairing in cuprates, probably motivated by measurement of Kerr rotations which is a signal of time-reversal symmetry breaking. However this is highly debatable and not very well accepted. Notice that $d+id$ superconductor is the $C=2$ case of the class D family. To learn more about the subject I recommend the excellent review by Jason Alicea: http://arxiv.org/abs/1202.1293 . | {
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169,969 | That is may be an easy question, but I am not a professional. The Sun is a star, and when I look at the Sun it is usually yellow. Why are stars in the sky at night white? I suppose it could be due to their distance. What is the explanation? | It has more to do with physiology of the eye rather than the spectrum of light produced by stars. Stars emit light over the full range of visible wavelengths. Hot stars emit more blue/violet light, cool stars emit more red light. The Sun is relatively neutral in that regard, so does not have a strong colouration, but many other stars in the sky have temperatures as low as 3,000K (should appear reddish) to 20,000K or more (should be blueish). However, light received by the eye has to be reasonably bright to trigger the colour sensitive cells (the cones). So whereas the Sun appears to be yellow(ish) (or to be more exact your eyes perceive the mix of wavelengths as yellowish), there are relatively few stars that are bright enough to look distinctly coloured. Betelgeuse is an example that most people see as reddish. Arcturus is distinctly orange (to me). But fainter stars all appear white because they are mainly being seen by the rod cells in your eye, which are not very colour sensitive (and have no sensitivity to light at the extremes of red and blue). This is known as scotopic vision . APPENDIX: I have found a few interesting links that appear to confirm this line of argument. Firstly, one can predict what stellar spectra would look like to the eye under conditions where the cone cells are operational. Here we can see that many stars would appear either pink (K-stars) through orange (M-stars). G-stars like the Sun are just off-white, whilst hot stars are distinctly blue (see also here ). These calculations do not include the effects of the Earth's atmosphere which preferentially absorb and scatter blue light. However, the effect of this on the colours cannot be that serious (it will have no effect at all on rod cells since they are not colour sensitive). Cone cells are sensitive over the range 450-650 nm . The typical atmospheric transmission at zenith for these wavelengths ranges from 85% to 93% (see Fig.1 of Burke et al. 2010 ), but would reduce to perhaps 70% to 85% for an object only 30 degrees above the horizon. Even at this low altitude, the differential effect is quite small. The colours of stars would not be changed by the atmosphere until they got quite close to the horizon. Stars would have different colours, ranging from pale blue through to an orangey-red, if they were bright enough to excite the eye's cone cells. | {
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170,318 | To my understanding, mixed states is composed of various states with their corresponding probabilities, but what is the actual difference between maximally mixed states and maximally entangled states ? | Suppose we have two Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$. A quantum state on $\mathcal{H}_A$ is a normalized, positive trace-class operator $\rho\in\mathcal{S}_1(\mathcal{H}_A)$. If $\mathcal{H}_A$ is finite dimensinal (i.e. $\mathbb{C}^n$), then a quantum state is just a positive semi-definite matrix with unit trace on this Hilbert space. Let's stick to finite dimensions for simplicity. Let's now consider the idea of a pure state: A pure state is a rank-one state, i.e. a rank-one projection, or a matrix that can be written as $|\psi\rangle\langle \psi|\equiv\psi\psi^{\dagger}$ for some $\psi\in\mathcal{H}_A$ (the first being the Dirac notation, the second is the usual mathematical matrix notation - since I don't know which of the two you are more familar with, let me use both). A mixed state is now a convex combination of pure states and, by virtue of the spectral theorem, any state is a convex combination of pure states. Hence, a mixed state can be written as $$ \rho=\sum_i \lambda_i |\psi_i\rangle \langle \psi_i|$$
for some $\lambda_i\geq 0$, $\sum_i \lambda_i=1$. In a sense, the $\lambda_i$ are a probability distribution and the state $\rho$ is a "mixture" of $|\psi\rangle\langle\psi|$ with weights $\lambda_i$. If we assume that the $\psi_i$ form an orthonormal basis, then a maximally mixed state is a state where the $\lambda_i$ are the uniform probability distribution, i.e. $\lambda_i=\frac{1}{n}$ if $n$ is the dimension of the state. In this sense, the state is maximally mixed, because it is a mixture where all states occur with the same probability. In our finite dimensional example, this is the same as saying that $\rho$ is proportional to the identity matrix. Note that a maximally mixed state is defined for all Hilbert spaces! In order to consider maximally entangled states , we need to have a bipartition of the Hilbert space, i.e. we now consider states $\rho\in\mathcal{S}_1(\mathcal{H}_A\otimes \mathcal{H}_B)$. Let's suppose $\mathcal{H}_A=\mathcal{H}_B$ and finite dimensional. In this case, we can consider entangled state. A state is called separable , if it can be written as a mixture $$ \rho =\sum_i \lambda_i \rho^{(1)}_i\otimes \rho^{(2)}_i $$
i.e. it is a mixture of product states $\rho^{(1)}_i$ in the space $\mathcal{H}_A$ and $\rho^{(2)}_i$ in the space $\mathcal{H}_B$. All states that are not separable are called entanglend . If we consider $\mathcal{H}_A=\mathcal{H}_B=\mathbb{C}^2$ and denote the standard basis by $|0\rangle,|1\rangle$, an entangled state is given by $$ \rho= \frac{1}{2}(|01\rangle+|10\rangle)(\langle 01|+\langle 10|)$$
You can try writing it as a separable state and you will see that it's not possible. Note that this state is pure, but entangled states do not need to be pure! It turns out that for bipartite systems (if you consider three or more systems, this is no longer true), you can define an order on pure entangled states: There are states that are more entangled than others and then there are states that have the maximum amount of possible entanglement (like the example I wrote down above). I won't describe how this is done (it's too much here), but it turns out that there is an easy characterization of a maximally entangled state, which connects maximally entangled and maximally mixed states: A pure bipartite state is maximally entangled, if the reduced density matrix on either system is maximally mixed. The reduced density matrix is what is left if you take the partial trace over one of the subsystems. In our example above: $$ \rho_A = tr_B(\rho)= tr_B(\frac{1}{2}(|01\rangle\langle 01|+|10\rangle\langle 01|+|01\rangle\langle 10|+|10\rangle\langle 10|))=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|) $$ and the last part is exactly the identity, i.e. the state is maximally mixed. You can do the same over with $tr_A$ and see that the state $\rho$ is therefore maximally entangled. | {
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170,502 | Most of you if not everybody will agree that the stronger the gravitational pull, the faster an object will fall. For example, on a planet with 50 times the gravity of Earth, any object will hit the ground on that planet much quicker than it would on Earth. So taking all of these into the equation, does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole? | does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole? No. Actually, defining exactly what you mean by the speed an object falls into a black hole is a tricky problem. In relativity you generally find that different observers observe different things. But we can work out what the various observers will see. Let's assume that the black hole is static so the geometry around it is described by the Schwarzschild metric. The task then is to calculate the orbits for objects moving in this spacetime. This is relatively simple by the standards of GR calculations, and you'll find it done in any introductory work on GR , but it's still a bit involved for non-nerds so I'll just quote the results. If you sit a long way from the black hole and watch an object falling into it from far away then the velocity of the object will be related to distance from the black hole by: $$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$ where $r_s$ is the Schwarzschild radius. If we graph the velocity as a function of distance from the black hole we get: The $x$ axis shows distance in Schwarzschild radii while the $y$ axis is the speed as a fraction of the speed of light. The speed peaks at about $0.38c$ then falls as you get nearer to the event horizon and falls to zero at the horizon. This is the source of the notorious claim that nothing can fall into a black hole. An alternative strategy might be to hover at some distance $r$ from the black hole and measure the speed at which the falling object passes you. These observers are known as shell observers . If you do this you find a completely different variation of speed with distance: $$ v = \sqrt{\frac{r_s}{r}}c \tag{2} $$ This time the variation of the speed with distance looks like: and this time the speed goes to $c$ as you approach the horizon. The difference between the two is because time slows down near a black hole, so if you're hovering near the event horizon velocities look faster because your time is running slower. You might be interested to note that the velocity calculated using equation (2) is equal to the Newtonian escape velocity . The event horizon is the distance where the escape velocity rises to the speed of light. The last observer is the falling observer i.e. the one who's falling into the black hole. But here we find something even stranger. The falling observer will never observe themselves crossing an event horizon. If you're falling into a black hole you will find an apparent horizon retreats before you as you fall in and you'll never cross it. You and the horizon will meet only as you hit the singularity. | {
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170,510 | In the spectrum of the blue part in a candle flame , there’s a violet emission at 432 nm due to excited CH* molecules (chemiluminescence). Why 432? Why not 400 or 500? There are emissions at 436, 475 and 520 nm too. Why these numbers? Is it because the energies of the photons emitted correspond to these wavelengths, as E = hc/λ? | does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole? No. Actually, defining exactly what you mean by the speed an object falls into a black hole is a tricky problem. In relativity you generally find that different observers observe different things. But we can work out what the various observers will see. Let's assume that the black hole is static so the geometry around it is described by the Schwarzschild metric. The task then is to calculate the orbits for objects moving in this spacetime. This is relatively simple by the standards of GR calculations, and you'll find it done in any introductory work on GR , but it's still a bit involved for non-nerds so I'll just quote the results. If you sit a long way from the black hole and watch an object falling into it from far away then the velocity of the object will be related to distance from the black hole by: $$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$ where $r_s$ is the Schwarzschild radius. If we graph the velocity as a function of distance from the black hole we get: The $x$ axis shows distance in Schwarzschild radii while the $y$ axis is the speed as a fraction of the speed of light. The speed peaks at about $0.38c$ then falls as you get nearer to the event horizon and falls to zero at the horizon. This is the source of the notorious claim that nothing can fall into a black hole. An alternative strategy might be to hover at some distance $r$ from the black hole and measure the speed at which the falling object passes you. These observers are known as shell observers . If you do this you find a completely different variation of speed with distance: $$ v = \sqrt{\frac{r_s}{r}}c \tag{2} $$ This time the variation of the speed with distance looks like: and this time the speed goes to $c$ as you approach the horizon. The difference between the two is because time slows down near a black hole, so if you're hovering near the event horizon velocities look faster because your time is running slower. You might be interested to note that the velocity calculated using equation (2) is equal to the Newtonian escape velocity . The event horizon is the distance where the escape velocity rises to the speed of light. The last observer is the falling observer i.e. the one who's falling into the black hole. But here we find something even stranger. The falling observer will never observe themselves crossing an event horizon. If you're falling into a black hole you will find an apparent horizon retreats before you as you fall in and you'll never cross it. You and the horizon will meet only as you hit the singularity. | {
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170,962 | I'm wondering if it's possible to send a man to the Moon using equations consistent with Newtonian gravity and without the elaborate tools of Einstein gravity. Are the predictions made by Newtonian gravity sufficiently precise to plan a successful voyage? If not, where would the Newtonian equations fail and how does Einstein gravity correct for these deficits? | The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction. I hope everyone who reads this site will known that gravitational potential energy is given by Newton's law: $$ V(r) = -\frac{GMm}{r} $$ The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy: $$ V_{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$ where $L$ is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we calculate $V_{eff}$ for an object in a Earth-Moon transfer orbit we get a graph like this: The stable circular orbit is at the minimum of the potential i.e. at about 384,400km, which is reassuring as this is the Earth-Moon distance. So far so good. But when we include the effects of General Relativity we find that it modifies the equation for the effective potential. The details are given in the Wikipedia article on Schwarzschild geodesics , but let's skip the details and just give the equation for $V_{eff}$ including relativistic effects: $$ V_{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3} \tag{2} $$ So including relativistic effects just adds a third term in $r^{-3}$. Now we calculate the position of the stable orbit by finding the minimum of $V_{eff}$ i.e. we calculate $dV/dr$, set it to zero and solve the resulting equation for $r$. Doing this for the Newtonian potential (1) gives us: $$ r = \frac{L^2}{GMm^2} \tag{3} $$ Finding the minimum of the relativistic expression (2) is a little more involved as we end up with a quadratic to solve, but some fiddling around ends up with: $$ r = \frac{L^2}{2GMm^2} \left( 1 + \sqrt{1 - \frac{12G^2M^2m^2}{L^2c^2}} \right) $$ and we can approximate the square root using the binomial theorem to get: $$\begin{align}
r &\approx \frac{L^2}{2GMm^2} \left( 1 + 1 - \frac{6G^2M^2m^2}{L^2c^2} \right) \\
&\approx \frac{L^2}{GMm^2} - \frac{3GM}{c^2} \tag{4}
\end{align}$$ And comparing our calculated Newtonian (3) and relativistic (4) distances we find the difference between them is: $$ \Delta r \approx \frac{3GM}{c^2} \approx 1.3 \text{cm} $$ So that's how much difference including general relativity makes to the calculated Earth-Moon transfer orbit - about 1.3cm! | {
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171,790 | After I watched "Particle Fever"--the movie about Large Hadron Collider (LHC) and the successful identification of the Higgs boson--I became a bit concerned with that team's handling of various negative PR incidents. Further, with the amount of money spent and the pressure to produce results, also realized that we may not sometime soon have a way to reproduce any of their results at any other facility with other research teams. With this in mind, how much scientific confidence can we put into things like the mass of the Higgs? Not that a "Higgs-like" particle was found, but the actual calculated mass of the particle known as the "Higgs boson"? This is a question about the scientific method and reproducibility of experimental results. Do we have any similar experiments where we confirm a similar theory without being able to reproduce those results? UPDATE: What is the confidence level of the LHC-calculated mass of the Higgs? How was this confidence level determined? What are the implications of it being wrong? How long will it take before we know if it is wrong? Thank you all for your comments. I do believe this is an important, specific and scientific discussion that can have specific, factual answers. UPDATE 2: I guess I'm not the only one asking these questions, this is very interesting: http://www.sci-news.com/physics/science-techni-higgs-discovery-higgs-boson-02266.html “The current data is not precise enough to determine exactly what the particle is. It could be a number of other known particles,” Dr Frandsen said. A related question might be, based on the Frandsen et al paper, what if it is not the Higgs at all? UPDATE 4/9/15: Came across this re: the reversal of the BICEP2 "discovery" due to having a 2nd team and 2nd set of instruments via the Planck telescope. Without Planck, the BICEP2 team might still be claiming their CMB discovery... perhaps for for a long time to come, potentially leading cosmology research (time and dollars) down the wrong cosmic inflation "rabbit hole." This seems to be an example supporting the importance and relevance to the question I raised here regarding LHC and acceptance of its "discoveries" without a 2nd team using a 2nd set of instruments (ie a 2nd beam and collider): http://physicsworld.com/cws/article/news/2015/feb/03/galactic-dust-sounds-death-knell-for-bicep2-gravitational-wave-claim | Most of the reproduction of results in particle physics comes from two sources: Competing experiments running nearly simultaneously. In this case both ATLAS and CMS got comparable results. Now, they are both using the beam from the LHC, so how do we know the beam is properly understood? Because while they were commissioning those machines they reproduced dozens (literally multiples of twelve) results that don't actually require the full capacity of the LHC. Notably they re-discovered the top quark and re-measured its mass, lifetime and branching ratios. Moreover, CERN has a long and very successful history in characterizing beams. They know how to do that stuff. Commissioning of the next-generation machine. This, of course, doesn't do us any good just now, but it is coming. Indeed, the LHC is the machine answering the same question about the Tevatron (a decade ago you could, after all, have asked something similar about CDF and D0 discovering the top quark). As an aside, don't underestimate the extent to which ATLAS and CMS are different machines, different cultures and differ in almost all the particulars. Nor the degree to which each of these organizations is looking over the other's shoulders: yes they put on a collegial show when announcing their results together but the competition is fierce and they do take each other's reports apart in search of things to criticize. Second aside: a lot of biomedical research has had a problem in recent years with a lack of reproductions at all. And it is a issue when studies are expensive (which is true in particle physics) and money is tight (when isn't it?), so how do we come out ahead in this matter? As far as I can tell it is a feature of building machines that are flexible, and developing a deep pool of talent when it comes to turning the data-set around to examine it from another angle. | {
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171,891 | I can understand the choice of material, silicon 28, but why is it a sphere rather than (say) a cube? Article here I would have thought that a sphere would have been the hardest shape to machine accurately. | If you know the diameter of the sphere, you know everything you need to know about the dimensions. It all comes down to one single value. Any other shape requires multiple dimensions and thus multiple values. Further, measuring a cube or another shape for accuracy is harder than measuring a sphere. Making very accurate spheres is not as difficult as you might think - it's no different than making optical glass or mirrors using grinding techniques, and, in fact, they are measured much the same way with lasers for very high accuracy. This video goes into a little more detail as to why they are doing this, how they achieved it, and how the sphere is made. | {
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172,315 | I think of sand as a lot of very small rocks. Suppose I have a pile of rocks, each about 1cm in size, and the pile is a meter tall. If I pour a bucket of water on the rocks, most of the water will fall through the rocks and form a puddle on the ground. On the other hand, if I have a pile of sand one meter tall and I pour a bucket of water on it, the water will stick to the sand and I'll have a bunch of wet sand, and not much water on the ground. Further, the wet sand acts distinctly differently from dry sand. For example, the angle of repose of wet sand is different from dry sand. An hourglass with wet sand it in might not work, or would run at a different speed from a dry hourglass. But I don't think the angle of repose of a pile of wet 1cm rocks is significantly different from that of dry rocks. So, how small should rocks be before they show wetting properties the way sand does? | For a packing of grains to stay wet up to a height $h$, the gravitational pressure $\rho g h$ needs to be balanced by the capillary pressure $\sigma cos(\theta)/r$. Here, $r$ represents the effective pore radius of the packing, $\theta$ the wetting angle (angle at which the air-water interface meets the sand grains), $\rho$ the water density, $g$ the gravitational acceleration, and $\sigma$ the surface tension associated with the air-water interface. It follows that the hydrostatic head $h$ is given by:
$$h=\frac{\sigma cos(\theta)}{\rho g r}$$
For an order-of-magnitude estimate you can equate the capillary radius $r$ to the grain radius $R_{grain}$, and $cos(\theta)$ to unity (fully water-wet grains). Together with $\sigma \approx 0.1 N/m$, $\rho \approx 10^3 kg/m^3$, and $g \approx 10 m/s^2$, it follows that
$$h \approx \frac{A_{cap}}{R_{grain}}$$
with $A_{cap} \approx 10^{-5}m^2$. As an aside, petroleum engineers use an approach like the above (replacing the air by oil) to determine the transition zone thickness in oil reservoirs. This is the zone over which the oil saturation 'builds up' above the water-bearing zone. As reservoir rock is usually fine-grained, and $\rho$ is replaced by the oil-water density difference, this transition zone can be tens of meters thick. | {
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172,624 | Why is the sky of the moon always dark compared to the sky of the earth, doesn't it have day and night like earth? | The moon does have a night and a day, but this isn't as fully connected to your question as you might think. The moon is tidally locked with the earth, meaning that the same side always faces earth. Since the moon also orbits around the earth (with a period of a lunar month), this means each side changes, over the course of a lunar month, between facing towards the sun and facing away. This is the cause of the phases of the moon, and also describes the day/night cycle for the moon: a full moon is when the side that we can see faces the sun - i.e. "day" for that side - and a new moon is when the side that we can see faces away from the sun - "night".
So the moon does have night and day, but a night/day cycle is one lunar month long. However, this doesn't answer your question of why the sky is always dark when viewed from the moon - even when the sun is above the horizon. The reason that the sky appears bright on Earth is that, even when there are no clouds, the atmosphere scatters the sunlight - meaning that only about 75% of the sunlight that reaches the ground appears to come from the sun, while the rest comes from all over the bright sky. This is called Rayleigh Scattering , and is the reason that the sky appears blue. The moon has no appreciable atmosphere to do this scattering, so the sky appears dark. | {
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172,790 | We know that any object above absolute zero emits electromagnetic radiation. The hotter the object, the shorter the wavelengths. In the electromagnetic radiation spectrum, radio waves have the longest wavelength, then microwaves, then infrared, then visible light, then ultraviolet, then x-rays and gamma rays. Why do we use microwaves in microwave oven when infrared and visible light are much more energetic, and how do microwaves cook food when they are less energetic than visible light and others? | In microwave ovens what matters is how much energy the radiation carries and how that energy is absorbed by the food. Visible light and IR are rapidly absorbed by most foods, so they would only heat the outer layer of the food. You'd get food with the outside carbonised and the inside raw. Microwaves are far less strongly absorbed by foods, so they penetrate deep into the food and can heat the interior. Even so large objects won't be heated throughout, which is one reason why microwave cooking instructions frequently advise a multi stage process of heating, letting the food stand then heating a final time. Microwave ovens often include IR heating as well as the microwave heating. This is done so you get food with a browned exterior and heated throughout. The answers to Why do microwave ovens use radiation with such long wavelength? give a nice discussion of why the exact wavelength used was chosen. The frequencies commonly used in microwave ovens are 2.45 GHz (12 cm) for home ovens and 915 MHz (38 cm) for industrial overs. Much higher frequencies are not used due to the cost of the magnetron, while much lower frequencies would not work because the wavelengths would be too big to allow a half wavelength to fit in the oven. Finally, you say: Why do we use microwaves in microwave oven when infrared and visible light are much hotter and how do microwaves cook food when they are cooler than visible light and others. But this is a slight misunderstanding. The wavelength of light emitted is indeed related to the temperature of the source, but light itself doesn't really have a temperature in the sense that matter does. Light transfers energy, and if this energy is absorbed it will heat the food. However the amount of heating is just related to the intensity of the EM radiation and the abosrption cross section. The wavelength makes a difference only insofar as it affects the absorption cross section. | {
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172,939 | Since the Earth is a good conductor of electricity, is it safe to assume that any charge that flows down to the Earth must be redistributed into the Earth in and along all directions? Does this also mean that if I release a million amperes of current into the Earth, every living entity walking barefooted should immediately die? | Electricity isn't a gas that expands out to shock anything in contact with it. Electricity is a flow from high voltage to low voltage. Touching a charged object is only dangerous if you become a current path--if it uses you to get somewhere. Even if the earth had a net charge, you aren't providing it anywhere to go, so you will not be shocked. It's somewhat like a bird on a power line. | {
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173,207 | In our daily life a lot of photons of visible light, infrared and radio etc move around us. We know that light is an electromagnetic radiation. So why doesn't that electromagnetic radiation affect a magnetic compass? | Most electromagnetic radiation is of very high frequency - the magnetic field changes many times per second. This means that the compass just doesn't have time to "follow" the magnetic field changes. The only thing that does affect a compass is a DC magnetic field - usually this is a large piece of iron etc. that gets magnetized (e.g. by the earth's magnetic field) and thus causes distortion; or it can be a DC current loop of some kind. But even the low frequencies of the mains (50 or 60 Hz depending on where you live) are much too fast to affect the compass (although in the presence of a strong source of electromagnetism, such as a large transformer, you can see vibration in the needle as observed by @vsz). Radio starts in the kHz (for long wave) to MHz (FM) or GHz (WiFi etc). And light, with wavelengths around 500 nm and a speed of 3x10$^8$ m/s, has frequencies in the hundreds of THz range. Too fast. UPDATE - adding a bit of math(s): A compass in the earth's field can be thought of as a damped oscillator: on the one hand there's the torque on the needle that is proportional to the displacement from magnetic North, on the other there's the inertia of the needle; and finally, there are damping terms (a good compass is critically damped - meaning that the damping is such that it will go to the right position in the shortest time). We can write the equation of motion as $$I\ddot\theta + \mu\dot\theta + k\theta = 0$$ In this expression, $\mu$ is the damping term (proportional to the angular velocity) and $k$ is the factor that describes how much torque the needle experiences with displacement. This is a general equation for a Simple Harmonic Oscillator (SHO), and we typically recognize three regimes: lightly damped, heavily damped, and critically damped. How such an oscillator responds when you give it a displacement and then let it go depends on the kind of damping - see this graph: In particular, the critically damped oscillator converges to its equilibrium position as fast as possible - which is why it's preferable for things like a compass. Now when you drive a SHO with an oscillating force, you get a response that depends on the frequency of the drive signal and the natural frequency of the system. If you drive at the natural frequency, you get resonance and the amplitude becomes large; as the difference in frequency gets larger, the amplitude of the response gets smaller. For a lightly damped (or underdamped) system*, the amplitude response is given by $$A = \frac{s_0}{\sqrt{\left[1-\left(\frac{\omega_d}{\omega_0}\right)^2\right]^2 +\left[\frac{\omega_d/\omega_0}{Q}\right]^2}}$$ In the limit of large frequencies, the response scales with $$A \propto \left(\frac{\omega_0}{\omega_d}\right)^2$$ where $\omega_0$ is the natural frequency $\sqrt{\frac{k}{I}}$ and $\omega_d$ is the driving frequency. When the driving frequency is many orders of magnitude larger than the natural frequency, the amplitude response will be negligible. As was pointed out in a comment by MSalters, at extremely high frequencies (above 10 GHz) the wavelength of the EM radiation becomes short compared to the length of the compass needle, so the above is further complicated by the fact that different parts of the needle will experience forces in different directions. All of which points in the same direction: the needle won't move. * I am taking the easy way out here... did not find the expression for the critically damped driven oscillator and don't have the intestinal fortitude to derive it right now and trust myself to get it right. But this is 'directionally correct' even for critically damped oscillator | {
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173,256 | I've read that tellurium-128 has an half-life of $2.2 \times 10^{24}$ years, much bigger than the age of the universe. So I've thought that maybe every single isotope of every single atom are radioactive, and isotopes which we call "stable" are actually unstable but their half-life are immensely big (but not infinite), like $10^{100}$ years. Is this a possible theory or are we truly $100$% sure that stable isotopes are really eternal? | If protons decay, then what you say is true: all atomic nuclei are indeed unstable, and a so-called "stable" nucleus simply has too long a half-life for its decay to be observed. The most tightly bound nucleus is $^{62}$Ni, with a binding energy per nucleon of 8.79 MeV [ source ], which is less than 1% of the mass of a nucleon. On the other hand, the decay of a proton through a process such as $$p \to e^+ + \pi^0$$ results in the loss of most of the mass of the proton. So if the proton can decay then it's pretty clear that an atomic nucleus always has more much more mass than a hypothetical final state in which some or all of the protons have decayed. In other words, while neutrons do not decay inside "stable" atomic nuclei because of the binding energy of the nucleus, protons cannot be so protected because their decay would be much more energetically favourable (than that of a neutron to a proton). The question of whether protons do decay is still unresolved, as far as I know. If protons do not decay, then the $^1$H nucleus, by definition, is stable, so there is at least one stable nucleus. Now, you might be wondering how we can establish that a nucleus is stable (assuming no proton decay). We make the assumption that energy is conserved, and it's impossible for a nucleus to be created if there isn't enough energy in the system to make up its rest mass. Given that assumption, say we have a nucleus. If we know the masses of the ground states of all nuclei with an equal or smaller number of nucleons, then we can rule out the possibility of there being a state that the given nucleus can transform into with less total mass. That in turn guarantees that the given nucleus is stable, since it can't decay into a final state with greater mass without violating conservation of energy. For a simple example, consider a deuteron, $^2$H. Its minimal possible decay products would be: a proton plus a neutron; two protons (plus an electron and an electron antineutrino) two neutrons (plus a positron and an electron neutrino) a diproton (plus an electron and an electron antineutrino) a dineutron (plus a positron and an electron neutrino) But all of those states have higher mass than the deuteron, so the deuteron is stable; it has no decay channel. Of course, you might wonder whether there are possible daughter nuclei whose masses we don't know because we've never observed them. Could, say, the "stable" $^{32}$S decay into $^{16}$P (with 15 protons and 1 neutron) and $^{16}$H (with 1 proton and 15 neutrons)? After all, we don't know the masses of these hypothetical nuclei. But if nuclei so far away from the drip line actually have masses low enough for that to happen, then there would have to be some radically new, unknown nuclear physics that would allow this to happen. Within anything remotely similar to existing models, this simply isn't possible. | {
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173,281 | Using Newton's universal equation and some circular motion equation, the orbiting object's mass cancels out. But can someone please explain why this is without using pure algebra? | But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass ; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given force, is inversely proportional to the inertial mass; the more inertial mass, the less the acceleration. If it is the case that the gravitational mass and inertial mass are equal (so that we speak only of the particle's mass), the gravitational and inertial mass cancel and the gravitational acceleration of a particle is then dependent only on the strength of the gravity at the place the particle is. But, in the Newtonian context, it is observationally the case that gravitational mass and inertial mass are equal. In the particular case of circular motion, the distance from the gravitational source is constant and, thus, the (inwardly, radially directed) gravitational acceleration of the particle is constant (and independent of the particle's mass). | {
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173,477 | I've read a few articles that say that astronauts have already brought guns in space and that shooting bullets in space is possible. But won't the recoil of the gun be too strong? Law of conservation of momentum: $m_{gun}v_{gun}=m_{bullet}v_{bullet}$ $v_{gun}=\dfrac{m_{bullet}}{m_{gun}}v_{bullet}$ I've found some values on Google: $m_{bullet}=0.03$ kg $m_{gun}=1$ kg $v_{bullet}=800$ m.s$^{-1}$ And therefore we get: $v_{gun}=24$ m/s $v_{gun}=87$ km/h $v_{gun}=54$ mph That's the typical speed of a car on a freeway. It's an uncontrollable recoil. The gun would be pushed back so powerfully that it would very probably damage the astronaut's spacesuit and kill him. But Russian Soyuz capsules wouldn't carry firearms if using them would lead to an instantaneous death for their own user. So I must be missing something. Still even if the astronaut could somehow hold extremely tightly to his gun and absorb all the momentum, he would still be ejected at a speed of around $1.5$ km/h $=$ $1$ mph (and maybe he'll even get stuck spinning for a while like in the movie Gravity ), which would make guns not very effective in space since it would take time to find a way to re-stabilize yourself and re-aim in order to fire a second bullet. Note: I know nothing about guns (I don't live in the US), so forgive me if I missed something obvious. | You've calculated the speed of a remote-triggered gun after it fires the bullet, true. However, there's actually nothing about space in your calculation, as @ACuriousMind noted. In theory, a gun fired on Earth could fly off just as fast, at least for a second. What you should use is not $m_\mathrm{gun}$ but $m_\mathrm{gun} + m_\mathrm{person}$. The gun never gets up to that speed because I start acting against it immediately and continuously. In some sense the "problem" cancels itself out--the gun seems to go so fast because it doesn't weigh much! But in reality the fact that it doesn't weigh much also means I can keep it under control. The problem in space you have is that the momentum you do get, even after calculating with both masses, will stay with you, and you start drifting away. Worse, you probably didn't fire along a line intersecting your center of mass, which means that you now have some crazy rotational motion. Whether or not this is dangerous depends on your zero-gravity shooting range's particular setup, but it doesn't sound comfortable. | {
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173,751 | Mine line of thoughts goes like this: A propeller is effectively pushing itself away from molecules of air. The best any propeller can do is to create total vacuum in the front of itself. The maximum suction pressure of vacuum you can get is equal to the opposite of the atmospheric pressure around, right? The speed of the air flowing into the vacuumed space in front of the propeller is equal to the speed of sound. Obviously we are extrapolating the friction of the body of the airplane etc. Is my line of thoughts at least partially plausible? | Don't forget that the aeroplane will be moving forward, so it's not relying on a vacuum filling ahead of the propellor to supply the latter with air. Now I daresay there are good engineering reasons why propellors are not efficient and even impracticable for supersonic flight, but I don't think there is a fundamental physics theoretical reason ruling them out. A propellor, from a theoretical standpoint, is not greatly different from a gas turbine jet or even a rocket insofar that it is simply "throwing stuff backwards", thus thrusting off the air it throws and being pushed forward by dint of Newton's third law. If it can be supplied enough air to throw backwards (and I think my first sentence shows there is probably no shortage of supply) and if it can impart a high enough impulse to the air, then there is no in principle limit to how fast the air is thrown backwards by the propellor. What happens if it is thrown backwards at faster than the speed of sound? Well in this case there would be an overpressure , meaning there would be a buildup of air there, the air thus becomes denser and "stiffer", and the local speed of sound behind the propellor can thus be much higher that that of the surrounding air. As the propellor does this, the air will undergo a sharp adiabatic temperature rise. Carrying this idea to its logical extreme, observe that rocket engines throw gas out behind them at roughly 10 times the speed of sound. It's simply a matter of how much you accelerate the gas - in principle there's no difference whether this acceleration is achieved by chemical energy or a great big bat whacking the air backwards. | {
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174,150 | In a first course on statistical mechanics the partition function is normally introduced as the normalisation for the probability of a particle being in a particular energy level.
$$p_j=\frac{1}{Z}\exp\left(\frac{-E_j}{k_bT}\right),$$
$$Z=\sum_j \exp\left(\frac{-E_j}{k_bT}\right).$$
Through various manipulations (taking derivatives and so on) we can recover the macroscopic thermodynamic variables of the system. It seems a little fortuitous to me that without specificing more than $Z$ we can recover so much information about our system especially when it is introduced as normalisation. Is there a better way to view the partition function other than the normalisation of probabilites? It just seems pretty amazing that it has so much information encoded about the system when really all it does is ensure that $\sum_j p_j=1.$ | The partition function is strongly related to a very useful tool in probability theory called the moment generating function(al) of the probability distribution. For any probability distribution $p$ of some random variable $X$, the generating function $\mathcal{M}(z)$ is defined as being: \begin{equation}
\mathcal{M}(z) \equiv \left\langle e^{zX}\right\rangle
\end{equation} so that we have for instance: \begin{equation}
\left(\frac{\partial \mathcal{M}}{\partial z}\right)_{z = 0} = \langle X \rangle,
\end{equation} \begin{equation}
\left(\frac{\partial^2 \mathcal{M}}{\partial z^2}\right)_{z = 0} = \langle X^2 \rangle,
\end{equation}
and in general
\begin{equation}
\mathcal{M}^{(n)}(0) = \langle X^n \rangle
\end{equation} Now, in statistical mechanics the canonical ensembles (with the exception of the microcanonical ensemble) have an exponential form with respect to their corresponding fluctuating thermodynamic random variables (the energy $E$ for the canonical ensemble, the energy $E$ and the number of particles $N$ for the grand canonical ensemble and the energy $E$ and the volume $V$ for the isobaric ensemble, to name a few) so that the probability distribution itself has a form like this \begin{equation}
p_t(X) = f(X)e^{tX}
\end{equation}where $t$ is a real number corresponding to one of the intensive thermodynamic variables. The moment generating function for probabilities like these will look like \begin{equation}
\mathcal{M}(z) =\left \langle e^{zX}\right\rangle = \int \mathrm d\mu(x) \:f(x) e^{tx} e^{zx}
\end{equation} It is quite easy to realize that if we define a partition function as being \begin{equation}
Z(t) \equiv \int \mathrm d\mu(x) \: f(x)e^{tx},
\end{equation} we find that \begin{equation}
\mathcal{M}(z) = Z(t+z)
\end{equation}so that \begin{equation}
\mathcal{M}^{(n)}(0) = Z^{(n)}(t)
\end{equation} In general in statistical mechanics, we prefer looking at the log of the partition function (which is also incidentally the logarithm of the moment generating function) as it allows to generate the cumulants of the distribution instead of the moments by applying successive derivatives. | {
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174,479 | They say that for a rolling body, the velocity of the contact point is zero. I'm not getting this. How can it be zero when it's in continuous motion? | The wheel is moving all the time, but each point on its circumference accelerates and decelerates all the time and when it is in contact with the ground it stops. You can have a clear and convincing explanation watching this animation showing how each point on the wheel describes a cycloid update But remember that the ever-changing acceleration of each point is just an illusion created in the frame of reference of the road, which is at rest. This is due to the fact that the value $k$ of the translational forward-velocity of the wheel $k$ coincides with the circumference of the wheel $ k = 2\pi r\rightarrow v_w = 2\pi r $ m/s: If you can, just imagine the car moving at the same speed and the wheel spinning at the same angular velocity but not touching the ground. Or imagine the wheel of a landing plane: as soon as it touches the ground the wheel synchronizes itself to $v= 2\pi r$ m/s. -- An illusion of a completely different kind can be experienced by stroboscopic (or wagon-wheel) effect** | {
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174,521 | Is $0 \, \mathrm m = 0 \, \mathrm s = 0 \,\mathrm {kg} = 0$? How do we define $[0 \, \mathrm m]$? I once was given an assignment where I was asked to deduce and write down some physical quantity. It turned out that this quantity was $0$ in one unit or another, so I decided to drop the physical units because I figured that it didn't matter anyhow. But I got to thinking: If $0 \, \mathrm m = 0$, then we could add units of different dimensions (an operation not normally defined) like so: $5 + 0 \,\mathrm m = 5$. Does $[ \cdot ]$ map $0 \, \mathrm m$ to both $0$ and $L$? Is there an established regime of thinking about physical quantities as an abstract algebra? Perhaps as a vector space where formal sums such as $5 \, \mathrm m + 10 \, \mathrm s$ are allowed? Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper? | This is really not that big of a deal. In essence, though, the dimensions of zero are ill-defined, and $[0]$ (i.e. the dimensions of zero) is (a) not meaningfully defined, and (b) never used in practice. Let me start by making one thing clear: Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper? Yes, this is perfectly OK, and it is standard practice. The dimensionality map $[\,·\,]$ has multiple different conventions , but they all work in much the same way. The key fact is that physical quantities form a vector space under multiplication, with exponentiation (over the field $\mathbb Q$) taking the role of scalar multiplication. (This is the essential reason why dimensional analysis often boils down to systems of linear equations, by the way.) The different base dimensions - mass, length, time, electric charge, etc. - are assumed to be algebraically independent and to span the space, and the dimensionality map $[\,·\,]$ reads out the 'coordinates' of a given physical quantity in terms of some pre-chosen canonical basis. This only works, however, if you exclude zero from the game. The quantity $1\,\mathrm m$ has a multiplicative inverse, but $0\,\mathrm m$ does not, so if you included it it would break the vector space axioms. This is in general OK - you're not forced to keep those properties - but it does preclude you from using the tools built on that vector space, most notably the dimensionality map. Thus $[0]$ doesn't map to anything. Since you explicitly asked for it, here's one way you can formalize what I said above. (For another nice analysis, see this nice blog post by Terry Tao.) A positive physical quantity consists of a 8-tuple $(x,m,l,t,\theta,c,q,i) \in\mathbb R^\times\times \mathbb Q^7$, where $\mathbb R^\times=(0,\infty)$ is the real multiplicative group. This is usually displayed in the form
$$ x\,\mathrm{kg}^m\mathrm{m}^l\mathrm s^t\mathrm K^\theta\mathrm A^c \mathrm{mol}^q \mathrm{cd}^i.$$ The multiplication of two physical quantities $p=(x,m,l,t,\theta,c,q,i)$ and $p'=(x',m',l',t',\theta',c',q',i')$ is defined as
$$pp'=(xx',m+m',l+l',t+t',\theta+\theta',c+c', q+q',i+i').$$
The multiplicative identity is $1=(1,0,0,0,0,0,0,0)$, and the multiplicative inverse of $p$ is $1/p=(1/x,-m,-l,-t,-\theta,-c,-q,-i)$. The exponentiation of a physical quantity $p$ to an exponent $r\in\mathbb Q$ is defined as $$p^r=(x^r,rm,rl,rt,r\theta,rc,rq,ri).$$ You can then easily check that these two operations satisfy the vector space axioms. The above construction is, in fact, a specific instantiation of the abstract vector space $\mathcal Q$ of physical quantities, but it suffices to take one specific example to show that this works. As an aside, the choice for $\mathbb Q$ as the scalar field is because (a) it's essential for the vector space structure, and (b) it's still somewhat reasonable to have things like $\mathrm {m}^{-3/2}$ (e.g. the units of a wavefunction). On the other hand, things like $\mathrm {m}^{\pi}$ cannot be made to make sense. The dimensionality map $[\,·\,]$ is, first and foremost, an equivalence relation, that of commesurability. That is, we say that for $p,p'\in\mathcal Q$, $$[p]=[p']\Leftrightarrow p/p'=(x,0,0,0,0,0,0,0)\text{ for some }x\in\mathbb R^\times.$$
This is in fact all you really need to do dimensional analysis, as I argued here , but it's still useful to go on for a bit. The really useful vector space, if you want to do dimensional analysis, is the vector space of physical dimensions: the space of physical quantities once we forget about their numerical value. This is the quotient space of $\mathcal Q$ over the commesurability equivalence relation:
$$\mathcal D=\mathcal Q/[\,·\,]=\{[p]\,:\,p\in\mathcal Q\}.$$
(Here I've abused notation slightly to make $[p]$ the equivalence class of $p$, i.e. the set of all physical quantities commensurable to $p$.) The vector space of physical dimensions, $\mathcal D$, has the same operations as in $\mathcal Q$: $[p][p']=[pp']$, and $[p]^r=[p^r]$. It is easy to check that these definitions do not depend on the specific representatives $p$ and $p'$, so the operations are well-defined. Dimensional analysis takes place in $\mathcal D$. From the definitions above, you can prove that the seven base units give rise to a basis $\{[1\,\mathrm {kg}], [1\,\mathrm m],\ldots,[1\,\mathrm{cd}]\}$ for $\mathcal D$. More physically, though, the seven base units are algebraically independent, which means that they cannot be expressed as multiples of each other, and they are enough to capture the dimensions of all physical quantities. These are the key physical requirements on a set of base units for the abstract space $\mathcal Q$. After this, you're all set, really. And it should be clear that there's no way to make zero fit into this scheme at all. | {
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174,541 | According to the BIPM and Wikipedia, "amount of substance" (as measured in moles) is one of the base quantities in our system of weights and measures. Why? I get why the mole is useful as a unit . In fact, my question isn't really about the mole at all; I just mention it because what little information I could find generally talked about moles, not about "amount of substance". Nor am I asking about why it's chosen as a base quantity and not a derived quantity. I get that any particular choice of bases is more or less arbitrary. I don't understand why it's a dimensional quantity at all . It is, after all, just a count of things; every student is taught to think of it as "like 'a dozen', only more sciencey". Can't we just call it a dimensionless number? No, says SI; molar mass doesn't just have dimensions of $\mathsf{M}$, it has dimensions of $\mathsf{M}\cdot\mathsf{N}^{-1}$; and Avogadro's number isn't just a number, it's got units of "per mole" (or dimensions of $\mathsf{N}^{-1}$). Contrast this with an "actual" dimensionless quantity, plane angle (and its unit the radian). Now, you might say that it's dimensionless because radians are defined as arc length over radius, and so plane angle is just $\mathsf{L}\cdot\mathsf{L}^{-1}$; cancel out and you have no dimensions. That strikes me as arbitrary. We could just as easily argue that arc length is "really" a quantity of $\mathsf{a}\cdot \mathsf{L}$ (where $\mathsf{a}$ is plane angle), because it's the measurement of a quantity that subtends $\mathsf{a}$ at distance $\mathsf{L}$. But this isn't needed; plane angle isn't even a derived quantity, it's a non -quantity. Plane angle is accepted as dimensionless. Why isn't amount of substance? As I said, I've found very little on this question. From the Wikipedia article on the mole , I found a PDF of an interesting IUPAC article on atomic weight . It acknowledges the argument (as does the Wikipedia article), but dismisses it out of hand by saying (essentially) "of course counting things is a way of measuring things, so of course we need a unit of measurement for it". Other than that, Wikipedia (as far as I can tell) touches on eliminating the mole only in the context of eliminating other units (as for example in natural systems of units). The Unified Code for Units of Measure blithely cuts moles from the base units as being "just a count of things", but doesn't go into why SI says it is necessary. Is there any official rationale for the inclusion of "amount of substance" as a dimension? Failing that, can anyone provide, or point me to, some good reasons why it's so special? EDIT: Thank you all for your input. The more I've thought about it, the more I've come to feel that there's no reason why "count of stuff" shouldn't be a dimension (it's clearly different from, say, a dimensionless number included as a scale factor), and that my unease with the idea comes from simple habit: in any case not involving moles, it tends to get left out. Really, I'm now more wondering why angles are considered dimensionless... Reading before coming here: Consultative Committee for Amount of Substance (CCQM) (I thought maybe a committee with this name would be in charge of explaining why amount of substance is needed, but no; it seems they just deal with standardising it) Unified Code for Units of Measure Wikipedia: Amount of substance International System of Quantities (including a PDF from BIPM linked at bottom, which only confirms that "amount of substance" is one of the base quantities) Mole (unit) (including the PDF from IUPAC mentioned above) | So, here's the thing. The chemistry that underlies molar mass ratios dates back at least to 1805. We've known that if you divide by a certain "relative mass" number you can get whole-number ratios for atoms in a pile of stuff, for that long. It took us about 60 more years to get a handle on how large atoms were with the estimations of Loschmidt, who worked out that atoms are much smaller than the wavelengths of visible light -- too small to ever "see". This gave a rough count of how many atoms there were in a confined space, too -- but we weren't able to connect these two different quantities (atomic relative masses, count of atoms) together to figure out the mass of a single atom until some work done by Einstein on diffusion in Brownian motion (1905) and some concrete numbers could finally be rolled in with Millikan's oil-drop experiment (1910). So due to history and convenience, the chemists are basically at the level of saying, "okay, we have N grams of this stuff, our mass spectrometer says that it's M grams per mole, so we've got N/M moles, that includes N/M moles of nitrogen and 15 N/M moles of hydrogen due to the known atomic composition, ..." and so on. You never have to add the uncertainty in Avogadro's number to these calculations; the "size" of a mole isn't important. It's only important when you start to want to know things that are "beyond" historical chemistry approaches, like counting actual numbers of atoms. With all that said, you'll be heart-warmed to know that there is a unit revision being considered by the SI organization, and one of the proposals is to fix the number of atoms in 1 mole. But of course they will still use as a guideline that "1 mole of carbon-12 has exactly 12 grams of mass"; it will just transition from what is now "exactly" to what will be "almost exactly." | {
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175,153 | For most physical measurements, zero is the same regardless of the units used for the measure: $0 \mathrm{mi} = 0 \mathrm{km}$ $0 \mathrm{s} = 0 \mathrm{hr}$ but for absolute temperatures, different systems have different zeros: $0 ^\circ\mathrm{C} \neq 0\,\mathrm{K}$ Are there any other physical, measurable quantities (other than temperature) that have different zero points? I'm looking for measurable quantities that are applicable anywhere -- things like voltage or temperature, not local quantities like "distance from the Empire State Building". | Time, in which case each system's zero point is often called its epoch : http://en.wikipedia.org/wiki/Epoch_%28reference_date%29 | {
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175,834 | I have a pair of shoes, which seem to isolate me from the ground. In effect I'm gathering static charge and every time i grab an aluminum door handle, that current discharges and that hurts. Ouch. I invented a way to workaround that: when I'm going to touch door handle, first I take my Skeletool (a stainless steel multitool) and touch the door handle with it. Sometimes I even see the small spark and hear the discharge and then I can touch the handle myself unharmed. But what interests me is: why discharging through the multitool does not hurt? This is an electrical current flow what hurts and the current flows the same way even if I hold the tool in my hand. Tool has surely less resistance than human body, so it shouldn't change anything. But it does not hurt :) Why? | When discharging without a tool, the whole charge exits your body through a small skin surface area, say $0.1$ mm $^2$ . When you hold a tool that surface is much bigger; perhaps $100$ cm $^2=10,000$ mm $^2$ . That means that the current flowing through neurons in that area is much lower, and perhaps low enough as to not be felt. Pretty much the equivalent of spreading the same electric current through $100,000$ cables instead of $1$ . | {
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175,985 | I was asked by an undergrad student about this question. I think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance. Am I correct? Veritasium/Minutephysics video on Youtube . | No . To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ . Angular momentum and angular position form one too. $$\Delta L \Delta \Theta \geq \frac{\hbar}{2}$$ This doesn't guarantee that angular momentum and angular position will be non-zero. It is an uncertainty - The actual values can be anything, including $0$ . But it does prevent you from arranging them both so the pencil stays upright. Furthermore, if you ask what the probability of finding both values very close to $0$ , you find that it is very small. In the limit, infinitely improbable. If it turns out that $L = \Theta = \sqrt{\hbar}$ , and you plug in reasonable values for the mass and length of the pencil, you will find it falls over in a few seconds. Belated update I was waiting until the weekend to add an update. By the time it got here, Floris had left very little to add. And he did a better job than I would have. Good answers. A number of users felt that an ideal pencil sharpened to an atomic tip was not realistic. The pencil should have a flat on the bottom. My own thought is that the pencil should be mounted on one of those massless, frictionless pulleys that seem to be so common in high school physics classrooms. Never the less, a pencil with a flat can be treated semi classically. Because of the uncertainty principal, the pencil has an initial momentum, and therefore an initial energy. This will cause the pencil to tip. Which in turn will cause the pencil to rotate about an edge of the flat. The center of mass will rise until it is directly over the edge of the flat. If the initial "uncertainty" energy is larger than the energy needed to raise the center of mass, the pencil will tip over. A quantum mechanical treatment would treat the region where the center of mass is over the interior of the flat as a potential well. There is a probability of tunneling out. Both of these scenarios are treated in full detail (with diagrams in case my description is unclear) here . I found this link by following Floris' "interesting post that calculates the same thing." That post had some comments at the bottom. The very last comment contains the link. | {
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175,994 | Vector C has a magnitude 23.4 m and is in the direction of the negative y-axis. Vectors A and B are at angles α = 44.4° and β = 27.7° up from the x-axis respectively. If the vector sum A+B+C = 0, what are the magnitudes of A and B? Using logic I've figured out some stuff:
since A+B+C = 0 then Ax+Bx+0 = 0 (since Cx = 0) and Ay+By-23.4 = 0.
Ax and Bx must be equal with opposite signs.
Ay and By have a sum of positive 23.4.
Since A is in quadrant 1 both Ax and Ay are positive,
and B is in Quadrant 2 so Bx is negative and By is positive. I don't exactly know what to do now... Fairly sure it's something to do with the two angles I haven't used yet but... I have no idea... Edit: Using Lami's Theorem I got the answers of A = 21.9 and B = 17.5, which are the correct answers. | No . To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ . Angular momentum and angular position form one too. $$\Delta L \Delta \Theta \geq \frac{\hbar}{2}$$ This doesn't guarantee that angular momentum and angular position will be non-zero. It is an uncertainty - The actual values can be anything, including $0$ . But it does prevent you from arranging them both so the pencil stays upright. Furthermore, if you ask what the probability of finding both values very close to $0$ , you find that it is very small. In the limit, infinitely improbable. If it turns out that $L = \Theta = \sqrt{\hbar}$ , and you plug in reasonable values for the mass and length of the pencil, you will find it falls over in a few seconds. Belated update I was waiting until the weekend to add an update. By the time it got here, Floris had left very little to add. And he did a better job than I would have. Good answers. A number of users felt that an ideal pencil sharpened to an atomic tip was not realistic. The pencil should have a flat on the bottom. My own thought is that the pencil should be mounted on one of those massless, frictionless pulleys that seem to be so common in high school physics classrooms. Never the less, a pencil with a flat can be treated semi classically. Because of the uncertainty principal, the pencil has an initial momentum, and therefore an initial energy. This will cause the pencil to tip. Which in turn will cause the pencil to rotate about an edge of the flat. The center of mass will rise until it is directly over the edge of the flat. If the initial "uncertainty" energy is larger than the energy needed to raise the center of mass, the pencil will tip over. A quantum mechanical treatment would treat the region where the center of mass is over the interior of the flat as a potential well. There is a probability of tunneling out. Both of these scenarios are treated in full detail (with diagrams in case my description is unclear) here . I found this link by following Floris' "interesting post that calculates the same thing." That post had some comments at the bottom. The very last comment contains the link. | {
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176,033 | I understand that places on the Earth's surface get hotter in summer, and in the middle of the day rather than morning or evening, because the surface of the Earth is presented 'face-on' to the Sun at those times, rather than at a slant. Simple trigonometry (or a simple drawing) shows that the same amount of radiation is spread over a smaller or larger area depending on the angle. (I learnt this in grade school, please correct me if I am wrong.) Analogously, I find that, in the heat of the middle of the day with the sun overhead, I get sunburnt on the top of my head (I am bald), the top of my nose, or my shoulders, far more than any part of my body which is a 'vertical surface', such as my face. However, in the evening or the morning, when the sun is low, I don't feel experience lots of heat and sunburn on body parts which are facing the sun directly. Why isn't the effect of the low sun on individual vertical surfaces just as strong as the effect of noonday sun on horizontal surfaces? | In general, the sun's light (particularly the UV that causes sunburn) has to pass through a lot more atmosphere (or a greater amount of air mass ) in the morning and evening to get to a vertical surface than it does when it is at zenith to a horizontal surface. An example is shown in the generalised image below (all graphs are obviously generalised): The caption from the source ( The Daily Cycle of Sunlight: Part 1 ) is (the relevant statements are in bold): When the sun is close to the horizon, its rays pass through a thicker slice of the atmosphere , which warms the color of sunlight (orange arrow). During the middle portion of the day, the sun’s rays pass through a thinner slice of the at-mosphere (white arrow). We define noon sunlight as being neutral white—neither warm nor cool. As such, the solar radiation is absorbed and scattered more in the atmosphere in the morning or evening. The net effect can be shown on the following diagram: Source: It's Coldest After Dawn In terms of sunburn, the erthemal dose rate (related to UV radiation) follows the same pattern, as seen below: Image source: NOAA , where the define erythemal dose rate as indicates the instantaneous amount of skin damaging UV radiation. | {
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176,442 | I know that when an object exceeds the speed of sound ($340$ m/s) a sonic boom is produced. Light which travels at $300,000,000$ m/s, much more than the speed of sound but doesn't produce a sonic boom, right? Why? | I know that when an object exceeds the speed of sound[340 m/s] a
asonic boom is produced .Light which travels at 300000000m/s [much
more than the speed of sound] doesn't produce a sonic boom right? Why? The answer is already in your own question: just because light is not an object . Sound " is a vibration that propagates as a typically audible mechanical wave of pressure and displacement, through a medium such as air or water " and must propagate itself by compressing particles (atoms/ molecules). The sonic boom is, as you rightly say, sound produced by compression of air molecules by an object , and is also propagating through the air. In this animation is represented a: .. sound source traveling at 1.4 times the speed of sound (Mach
1.4). Since the source is moving faster than the sound waves it creates, it leads the advancing wavefront. The sound source will pass
by a stationary observer before the observer hears the sound it
creates. the shock wave on the edge Light is an electromagnetic wave that propagates also in vacuum modifying electric and magnetic fields. These fields do not interact with air enough to compress them and produce sounds. | {
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176,537 | What causes the surface of the lake to appear darker in some places? Firstly, I know that it's not fishes that cause it. Secondly, the dark surfaces move from time to time and thus are not associated with the depth of the water or the structure of the lake bottom. Lastly, I may say they are due to a kind of ripple distortion. What causes this distortion and what is its mechanism? Please, provide more explanation than "the wind", for example. In this photo, look at the far area; it's darker somehow. | It is an interplay between the wind and the shoreline, and basic laws of reflection. As you can see in your photo, where the water surface is still, you see a reflected image of the skyline - lighter for the sky, darker for the buildings. Where the water surface ripples, you get reflections "from everywhere" - some from the sky, some from buildings, etc. that average is darker than the bright sky. Another factor relating to this: when you look at a rippled surface, the average angle that you see is lower than if you are looking at a smooth surface. Imagine a smooth sine wave ripple: when you look from the side, you see more of the surface "facing you" and less of the surface "facing away". Now Fresnel's equations tell us about the coefficient of reflectivity as a function of angle - and if we ignore polarization, the reflection goes up at the angle of incidence gets more oblique. On the rippled surface the average angle of incidence is less oblique and you see less reflection from the sky. This is why these patches look darker: For larger angles $\theta$ the reflection coefficient is larger - but when you look at the ripples, you can see that you "favor" the water that is facing you more directly - this means you see less reflection, and you are looking more "through the surface at the deep below" than "at a reflection of the sky". Incidentally, the Fresnel equations can be found on this Wikipedia page , where you also find this diagram. Here, $R_p$ and $R_s$ are reflection coefficients for different polarization angles. Although polarization will change in the sky throughout the day (single Rayleigh scattering leaves the light polarized), on average you can ignore its effect for the purpose of understanding this explanation: | {
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176,650 | Moving through the other three dimensions necessitates energy. But why doesn't moving through time necessitate energy? | Moving through space at a uniform pace does not require energy, or force (Newton's 1. law), but accelerating through space does (Newton's 2. law). Similarly, moving through time at a uniform pace does not require a force, but if you're accelerating, your time will change wrt. a non-accelerating observer, so in a way you might say that you accelerate through time. For instance, if you throw yourself in a black hole, free-falling toward the horizon, your time will pass "normally". But when your realize your mistake and use your jetpack to accelerate up of the potential well, back to civilization, you will find that what took ten minutes for you, took 100 years at Earth, so you have increased your speed through time. | {
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177,049 | My understanding has always been that it does from conventional science courses, but really thinking about it, I was wondering if this is really the case. To my limited understanding there is a theory that there are gravitons that act as particles to pull two different masses together. If these gravitons really are the physical particles of gravity, then a so called "vacuum" that had gravity wouldn't be a vacuum at all. A real vacuum should lack these particles, and thus, lack gravity? Anything in the vacuum should then implode due to its own gravitational attraction within itself? If this is the case, could we say in a real vacuum, external gravity does not exist? | Your intuition is good, but you're mixing up some quantum and classical phenomena. In classical (i.e. non-quantum) physics, a vacuum is a region of space with no matter. You can have electromagnetic fields in a vacuum, so long as the charges creating the fields are in a different region. By the same token you can have gravitational fields in a vacuum, generated by masses somewhere else in space. In this classical description of the universe, there are no such things as photons or gravitons, and everything (for the most part) works out. In quantum physics, the story is not so easy. As you say, now our force fields are particles, too (photons and gravitons), so maybe a "quantum vacuum" shouldn't include them either? Unfortunately, it turns out that in quantum mechanics (as rob pointed out) it is impossible to have a perfect vacuum, a state with no particles in it at all. One way to see this is through the energy-time uncertainty principle: $\Delta E \ \Delta t > \hbar/2$. A perfect vacuum, a state with no particles at all, must have exactly zero energy. If the energy is exactly zero, then it is completely certain, and $\Delta E = 0$ which violates the uncertainty principle. So the quantum vacuum is not a state with zero particles, it is a state with probably zero particles. And in different situations you may find useful to alter your definition of "probably," so there are a lot of different things physicists will call a "vacuum" in quantum mechanics. This idea, that quantum mechanically there are always some particles around in any region of space, has some cool consequences that we've verified in the lab! One is the Casimir Effect . This is a force that shows up when you move two objects in a vacuum so close together the pressure from these "virtual" photons causes them to attract. Another is the particle they discovered at the LHC, the Higgs Boson . The Higgs field has a "vacuum expectation value," a perfect quantum vacuum will have a non-zero Higgs field throughout it. Excitations of this field are the Higgs particles found at the LHC! | {
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177,588 | I have just started studying QM and I got into some trouble understanding something: Let's say there is a wave function of a particle in a 1D box ($0\leq x\leq a$): $$\psi(x,t=0) = \frac{i}{\sqrt{5}} \sin\left(\frac{2\pi}{a}x\right) + \frac{2}{\sqrt{5}} \sin\left(\frac{5\pi}{a}x\right)$$ Then if we measure the energy, the probability of getting the energy associated with $ \sin(\frac{2\pi}{a}x) $ is $\left| \frac{i}{\sqrt{5}} \right|^2 = \frac{1}{5}$ and the probability of measuring the energy associated with $\sin\left(\frac{5\pi}{a}x\right)$ is $\left| \frac{2}{\sqrt{5}}\right|^2 = \frac{4}{5}$. So the magnitude of $ \frac{i}{\sqrt{5}} , \frac{2}{\sqrt{5}} $ determines the probability, but what is the meaning of the phase? To me, as someone who measures energy, I'll get the same thing if $$\psi(x,t=0) = \frac{-1}{\sqrt{5}} \sin\left(\frac{2\pi}{a}x\right) + \frac{2}{\sqrt{5}} \sin\left(\frac{5\pi}{a}x\right) $$ So why does the phase matter? If it matters, how do I know to which phase the wave function collapsed after the measurement? | This is an important question. You are correct that the energy expectation values do not depend on this phase. However, consider the spatial probability density $|\psi|^{2}$ . If we have an arbitrary superposition of states $\psi = c_{1} \phi_{1} + c_{2} \phi_{2}$ , then this becomes $|\psi|^{2} = |c_{1}|^{2}|\phi_{1}^{2} + |c_{2}|^{2} |\phi_{2}|^{2} + (c_{1}^{*} c_{2} \phi_{1}^{*} \phi_{2} + c.c.)$ . The first two terms do not depend on the phase, but the last term does. ( $c_{1}^{*}c_{2} = |c_{1}||c_{2}|e^{i (\theta_{2} - \theta_{1})}$ ). Therefore, the spatial probability density can be heavily dependent on this phase. Remember, also, that the coefficients (or the wavefunctions, depending on which "picture" you are using) have a rotating phase angle if $\phi_{1,2}$ are energy eigenstates. This causes the phase difference $\theta_{2} - \theta_{1}$ to actually rotate at the energy difference , so that $|\psi|^{2}$ will exhibit oscillatory motion at the frequency $\omega = (E_{2} - E_{1})/\hbar$ . In summary, the phase information in a wavefunction holds information, including, but not limited to, the probability density. In a measurement of energy this is not important, but in other measurements it certainly can be. | {
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177,767 | Gas molecules go at an insane velocity, and though they are miniscule, yet there is a LOT of them. Of course, because of all these molecules hurtling around, there is air pressure ; yet if you envision a lot of bullets flying around, they don't really "apply pressure": they smash stuff . So why aren't things being destroyed by these mini-torpedoes? I sense the reason they don't wreak havoc is because they are not coordinated, i.e. they are random. Also things may not work out microscopically as they do macroscopically. | When you say "why aren't things being destroyed ", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon bond is $348\,\rm kJ/mol$, which works out to $5.8 \cdot 10^{-19}\,\rm J$ per bond. If an impacting gas molecule is to break this bond, it must (in a simplified collision scenario) have at least that much energy to break the bond. If the average molecule has that much energy, we can calculate what the temperature of the gas must be: $$E_\text{average} = k T$$
$$T = \frac{5.8 \cdot 10^{-19}\,\rm J}{1.38 \cdot 10^{-23}\,\rm m^2 kg\, s^{-2} K^{-1}}$$
$$T = 41,580\rm °C$$ That's pretty hot! Now, even if the average molecule doesn't have that energy, some of the faster-moving ones might. Let's calculate the percentage that have that energy at room temperature using the Boltzmann distribution for particle energy: $$f_E(E) = \sqrt{\frac{4 E}{\pi (kT)^3}} \exp\left(\frac{-E}{kT} \right)$$ The fraction of particles with energy greater than or equal to that amount should be given by this integral: $$p(E \ge E_0) = \int_{E_0}^{\infty} f_E(E) dE$$ In our situation, $E_0 = 5.8 \cdot 10^{-19}\,\rm J$, and this expression yields $p(E \ge E_0) = 1.9 \cdot 10^{-61}$. So, the fraction of molecules at room temperature with sufficient kinetic energy to break a carbon-carbon bond is $1.9 \cdot 10^{-61}$, an astoundingly small number. To put that in perspective, if you filled a sphere the size of Earth's orbit around the sun with gas at STP, you would need around 16 of those spheres to expect to have even one gas particle with that amount of energy. So that's why these "torpedoes" don't destroy things generally - they aren't moving fast enough at room temperature to break chemical bonds! | {
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178,233 | Can we find the exponential radioactive decay formula from first principles? It's always presented as an empirical result, rather than one you can get from first principles. I've looked around on the internet, but can't really find any information about how to calculate it from first principles. I've seen decay rate calculations in Tong's qft notes for toy models, but never an actual physical calculation, so I was wondering if it's possible, and if so if someone could link me to the result. | If you want to be very nitpicky about it, the decay will not be exponential. The exponential approximation breaks down both at small times and at long times: At small times, perturbation theory dictates that the amplitude of the decay channel will increase linearly with time, which means that the probability of decay is at small times only quadratic, and the survival probability is slightly rounded near $t=0$ before of going down as $e^{-t/\tau}$. This should not be surprising, because the survival probability is time-reversal invariant and should therefore be an even function. At very long times, there are bounds on how fast the bound state amplitude can decay which are essentially due to the fact that the hamiltonian is bounded from below, and which I demonstrate in detail below. Both of these regimes are very hard to observe experimentally. At short times, you usually need very good time resolution and the ability to instantaneously prepare your system. At long times, you probably wouldn't need to go out that far out, but it is typically very hard to get a good signal-to-noise ratio because the exponential decay has pretty much killed all your systems, so you need very large populations to really see this. However, both sorts of deviations can indeed be observed experimentally. At long times, the first observation is Violation of the Exponential-Decay Law at Long Times. C Rothe, SI Hintschich and AP Monkman. Phys. Rev. Lett. 96 163601 (2006) ; Durham University eprint . (To emphasize on the difficulty of these observations, they had to observe an unstable system over 20 lifetimes to observe the deviations from the exponential, by which time $\sim10^{-9}$ of the population remains.) For short times, the first observations are Experimental evidence for non-exponential decay in quantum tunnelling. SR Wilkinson et al. Nature 387 no. 6633 p.575 (1997) . UT Austin eprint , which measured tunnelling of sodium atoms inside an optical lattice, and Observation of the Quantum Zeno and Anti-Zeno Effects in an Unstable System. MC Fischer, B Gutiérrez-Medina and MG Raizen. Phys. Rev. Lett. 87 , 040402 (2001) , UT Austin eprint (ps) . To be clear, the survival probability of a metastable state is for all practical intents and purposes exponential. It's only with a careful experiment - with large populations over very long times, or with very fine temporal control - that you can observe these deviations. Consider a system initialized at $t=0$ in the state $|\psi(0)⟩=|\varphi⟩$ and left to evolve under a time-independent hamiltonian $H$. At time $t$, the survival amplitude is, by definition,
$$
A(t)=⟨\varphi|\psi(t)⟩=⟨\varphi|e^{-iHt}|\varphi⟩
$$
and the survival probability is $P(t)=|A(t)|^2$. (Note, however, that this is a reasonable but loaded definition; for more details see this other answer of mine .) Suppose that $H$ has a complete eigenbasis $|E,a⟩$, which can be supplemented by an extra index $a$ denoting the eigenvalues of a set $\alpha$ of operators to form a CSCO , so you can write the identity operator as
$$1=\int\mathrm dE \mathrm da|E,a⟩⟨E,a|.$$
If you plug this into the expression for $A(t)$ you can easily bring it into the form
$$
A(t)=\int \mathrm dE\, B(E)e^{-iEt},\quad\text{where}\quad B(E)=\int \mathrm da |⟨E,a|\varphi⟩|^2.
$$
Here it's easy to see that $B(E)\geq0$ and $\int B(E)\mathrm dE=1$, so $B(E)$ needs to be pretty nicely behaved, and in particular it is in $L^1$ over the energy spectrum. This is where the energy spectrum comes in. In any actual physical theory, the spectrum of the hamiltonian needs to be bounded from below, so there is a minimal energy $E_\text{min}$, set to 0 for convenience, below which the spectrum has no support. This looks quite innocent, and it allows us to refine our expression for $A(t)$ into the harmless-looking
$$
A(t)=\int_{0}^\infty \mathrm dE\, B(E)e^{-iEt}.\tag1
$$
As it turns out, this has now prevented the asymptotic decay $e^{-t/\tau}$ from happening. The reason for this is that in this form $A(t)$ is analytic in the lower half-plane. To see this, consider a complex time $t\in\mathbb C^-$, for which
\begin{align}
|A(t)|
& =\left|\int_0^\infty B(E)e^{-iEt}\mathrm dE\right|
\leq\int_0^\infty \left| B(E)e^{-iEt}\right|\mathrm dE
=\int_{0}^\infty \left| B(E)\right|e^{+E \mathrm{Im}(t)}\mathrm dE
\\ & \leq\int_{0}^\infty \left| B(E)\right|\mathrm dE=1.
\end{align}
as $\mathrm{Im}(t)<0$. This means that the integral $(1)$ exists for all $t$ for which $\mathrm{Im}(t)\leq 0$, and because of its form it means that it is analytic in $t$ in the interior of that region. This is nice, but it is also damning, because analytic functions can be very restricted in terms of how they can behave. In particular, $A(t)$ grows exponentially in the direction of increasing $\mathrm{Im}(t)$ and decays exponentially in the direction of decreasing $\mathrm{Im}(t)$. This means that its behaviour along $\mathrm{Re}(t)$ should in principle be something like oscillatory, but you can get away with something like a decay. What you cannot get away with, however, is exponential decay along both directions of $\mathrm{Re}(t)$ - it is simply no longer compatible with the demands of analyticity. The way to make this precise is to use something called the Paley-Wiener theorem which in this specific setting demands that
$$
\int_{-\infty}^\infty \frac{\left|\ln|A(t)|\right|}{1+t^2}dt<\infty.
$$
That is, of course, a wonky integral if anyone ever saw one, but you can see that if $A(t)\sim e^{-|t|/\tau}$ for large times $|t|$ ($A(t)$ must be time-reversal symmetric), then the integral on the left (only just) diverges. There's more one can say about why this happens, but for me the bottom line is: analyticity demands some restrictions on how fast $A(t)$ can decay along the real axis, and when you do the calculation this turns out to be it. (For those wondering: yes, this bound is saturated. The place to start digging is the Beurling-Malliavin theorem, but I can't promise it won't be painful.) For more details on the proofs and the intuition behind this stuff, see my MathOverflow question The Paley-Wiener theorem and exponential decay and Alexandre Eremenko's answer there , as well as the paper L. Fonda, G. C. Ghirardi and A. Rimini. Decay theory of unstable quantum systems. Rep. Prog. Phys. 41 , pp. 587-631 (1978) . §3.1 and 3.2. from which most of this stuff was taken. | {
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178,371 | So, because I'm a hardcore person, I risked all this afternoon by going out in the wind, the rain and the cold to construct a willow den. Yes, it seems a menial task, but it was actually quite thought-provoking: some pieces of willow were harder to push into the ground than others, and to try and force them in I'd sometimes end up hanging off of them, but other times I'd just push with all my strength, placing my body above the hole and pushing down, my feet still firmly planted on the ground. Now, I was wondering, which of these two methods is better for sticking the stick into the ground? Ignoring the fact that by hanging from it, it is likely to topple, which method help me be more efficient in my future den-building exploits? In other words, can you ever exert more downwards force than your weight? If so, then I'm guessing pushing down would be the better method, and if not, then hanging from the stick, putting all your weight into it, would exert the most possible downwards force. Here's what a willow den looks like: (source: raisingsparks.com ) | The force you can exert is your mass times your acceleration. By the equivalence principle, just standing still is equivalent to accelerating at 9.8 m/s 2 , which is where the force of your weight comes from when you just stand still. But it is easy to accelerate more - like when you jump. The force is only limited by your ability to push yourself off (transfer force to) the willow shoot. Imagine that you lie down next to the shoot, holding it in both hands. If you now pulled yourself up rapidly (the way some circus acrobats can pull themselves up a rope while appearing to hang horizontally) then you apply all your weight to the willow - and if you are strong enough to accelerate yourself while doing this, you could apply a force greater than your weight. However, as you probably realize, there are other far more effective means to drive a stick into the ground. The key is to convert momentum into force - the equation is $$m\Delta v = F \Delta t$$ This equation tells us that the change in momentum ($\Delta(mv) = m\Delta v$) is determined by the integral of force and time ($\int F\cdot dt=F\Delta t$ if F is constant). This is a direct consequence of the equation $F=ma$, which you can integrate with respect to time to get $\int F\cdot dt = \int m\ a\ dt = m \Delta v$. When you use a hammer etc, you give it momentum during a long swing (small F, large t); but it slows down and comes to a stop in a very short interval, meaning that for that short time the force is much greater. A post driver is the tool people use to try to replicate this on the scale of large sticks being driven into the ground (hard to hammer the top of a tall thin stick). It may not be possible to use in your particular situation - but in general, it will allow you to apply a force much greater than your weight (bot for a shorter time). This is also the principle behind pile drivers etc . All these methods require the object to be driven to be strong enough to support the force you use to drive them into the ground... | {
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178,382 | I was wondering what would happen if we were to calculate electric field due to a finite line charge. Most books have this for an infinite line charge. In the given figure if I remove the portion of the line beyond the ends of the cylinder. I believe the answer would remain the same. Also if I imagine the line to be along the $x$ -axis then would it be correct to say that electric field would always be perpendicular to the line and would never make any other angle (otherwise the lines of force would intersect)? Image source: Electric Field of Line Charge - Hyperphysics | The force you can exert is your mass times your acceleration. By the equivalence principle, just standing still is equivalent to accelerating at 9.8 m/s 2 , which is where the force of your weight comes from when you just stand still. But it is easy to accelerate more - like when you jump. The force is only limited by your ability to push yourself off (transfer force to) the willow shoot. Imagine that you lie down next to the shoot, holding it in both hands. If you now pulled yourself up rapidly (the way some circus acrobats can pull themselves up a rope while appearing to hang horizontally) then you apply all your weight to the willow - and if you are strong enough to accelerate yourself while doing this, you could apply a force greater than your weight. However, as you probably realize, there are other far more effective means to drive a stick into the ground. The key is to convert momentum into force - the equation is $$m\Delta v = F \Delta t$$ This equation tells us that the change in momentum ($\Delta(mv) = m\Delta v$) is determined by the integral of force and time ($\int F\cdot dt=F\Delta t$ if F is constant). This is a direct consequence of the equation $F=ma$, which you can integrate with respect to time to get $\int F\cdot dt = \int m\ a\ dt = m \Delta v$. When you use a hammer etc, you give it momentum during a long swing (small F, large t); but it slows down and comes to a stop in a very short interval, meaning that for that short time the force is much greater. A post driver is the tool people use to try to replicate this on the scale of large sticks being driven into the ground (hard to hammer the top of a tall thin stick). It may not be possible to use in your particular situation - but in general, it will allow you to apply a force much greater than your weight (bot for a shorter time). This is also the principle behind pile drivers etc . All these methods require the object to be driven to be strong enough to support the force you use to drive them into the ground... | {
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178,417 | I know that I cannot do this because of conservation of energy, so I am looking for an answer as to why this will not work. So by my understanding of Einstein's whole famous $E=mc^2$ thing it is possible to turn matter into energy, and energy into matter. So what if you were to create a space station way up high and run a wire down to Earth, then pump some energy up it. The space station would then turn that energy into matter and let it drop down to Earth. People down at Earth would harvest the kinetic energy somehow, then turn the matter back into energy, pump it back up, then repeat the process. The result would be the energy they started with plus the kinetic energy. Where is the flaw in this reasoning? (I'm guessing it might take energy for the electrons to fight gravity up to the space station). | [5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment] This is actually the paradox that led Einstein to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up to the Space Station, where they are converted back to a positron and electron. These are dropped back to Earth where the kinetic energy is harvested. Einstein found a way out of the paradox: Gravity must affect energy just like matter. Light must lose energy if it climbs against gravity. Likewise, it must gain energy if it drops from the Space Station to Earth. Minor point - the space station has a large horizontal velocity. Let us pretend that it is stationary at the top of a very tall tower. Let us also ignore the rotation of the Earth. This leads to another of Einstein's thought experiments. Unlike other forces, gravity attracts all particles and therefore must affect all energy. Einstein could not think of an experiment that could distinguish between a freely falling elevator in a uniform gravitational field and an elevator floating in space where there is no gravity. (There are some caveats. The elevator must be small or the effects of a non-uniform field will be noticeable. Looking outside the elevator is cheating. See this for more details.) He concluded the laws of physics in a freely falling elevator are the same as those in an inertial frame of reference without gravity. Furthermore, the laws in a rocket in empty space accelerating at 1 g are the same as the laws in a uniform 1 g gravitational field. On Earth the ground pushes you upward with enough force to accelerate you at 1 g. This is completely equivalent to a rocket. This is called the equivalence principle. In a rocket, you can use an inertial frame of reference. You take into account for your acceleration, and then you see that particles with no forces move in straight lines, particles at rest remain at rest. You can also work in a frame of reference where the rocket is at rest. To do this, you say that a force toward the tail of the rocket acts on everything in the universe. This kind of force is an accounting trick you must use if you want to work in a non-inertial frame of reference. It is called a pseudo force. The equivalence principal is a statement that gravity is a pseudo force, and that a freely falling frame of reference is inertial. Unlike a rocket, the pseudo force of gravity is not always uniform. We can use the Equivalence Principal to calculate the effect of gravity on light. Suppose you are in a rocket at rest a distance L from a space station far from Earth. You start accelerating toward it at 1 g just as a beam of light leaves the station toward you. You would see a blue Doppler shift, a higher frequency. At low speeds, the Doppler shift is related to velocity as below. $\nu = \nu_0 \left[1 + \dfrac{v}{c}\right]$ Neglecting the distance you travel, light takes time $L/c$ to reach you. During this time you would accelerate to velocity $v= gL/c$ . So $\nu = \nu_0 \left[1 + \dfrac{gL}{c^2}\right]$ Since gravity and the acceleration of a rocket are completely equivalent, you see the same Doppler shift on Earth. (See this for more info.) This seems very odd when you think about it. The laws of physics are the same when you don't look out of the elevator or rocket. When you do look out, you see an obvious difference. The acceleration of the rocket gives the universe a velocity, the velocity causes time dilation and length contraction, and these generate the Doppler sift. On Earth, the upward acceleration keeps you from falling. The universe stays still. If the frequency of light increases, then its period decreases. If you measure a shorter period on Earth than the space station finds, it is because your clock is slower. The acceleration of gravity causes the same time dilation as the acceleration of the rocket. $t = \dfrac{t_0} {\left[1 + \dfrac{gL}{c^2}\right]}$ This is valid for points separated by a small L. To compare time intervals on Earth to time intervals not influenced by gravity far from Earth, we would have to use $g = GM/r^2$ , take into account length changes, and integrate. The result of a more exact calculation is $t = \dfrac{t_0}{\sqrt{1 - \dfrac{2GM}{rc^2}}}$ From quantum mechanics, the energy of a photon is related to its wavelength. $E = h \nu = \dfrac{h c}{\lambda}$ . So light that climbs against gravity is red shifted and has a drop in energy. I need to be careful about statements like "Light loses energy as it climbs against gravity." Light doesn't slow down as it climbs. Light travels at the speed of light. You also cannot follow a photon and watch its energy change. There is no frame of reference that follows a photon. You get singularities if you try to look at the universe from the point of view of a photon. All you can do is measure the frequency or energy of a photon as it leaves and as it arrives. You can measure a frequency shift or energy difference between two points. If you aim a laser upwards, many photons in identical states climb against gravity. You can measure the frequency at various altitudes with a diffraction grating or measure the recoil when a particle is struck. If the diffraction grating or particle is at rest with respect to earth, you will find that there is a frequency and energy drop as the beam climbs. If the diffraction grating or particle is fired from a cannon on earth with enough speed to coast up to the space station and stop there, you would find no frequency or energy drop. The values of the entire beam would be the as same measured at the space station. In both cases, the energy is not some "stuff" inside the photon that is lost as the photon rises. If is a result of an interaction between the photon and the diffraction grating or particle. In the stationary with respect to Earth case, the rate of the clock of the diffraction grating or particle changes with altitude, and so do the measured values. In the coasting case, the diffraction grating or particle are at rest in an inertial frame of reference. Their clocks run at the same rate as the space station's. | {
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178,601 | When we send satellite into space using a rocket, logically we consider the shortest path which is a straight line perpendicular to Earth's surface. My question is do rocket takes the shortest path to reach outer space or stray off intentionally at an angle from Earth's surface? | It sounds like you are imagining that what satellites do is go up through the atmosphere, break through into outer space, and hang there. That is not right. If you simply go straight up to outer space (say 300 km above Earth's surface), gravity will pull you right back down, even if you've left the atmosphere, and you'll crash back into the Earth. Gravity is only about 10% weaker at 300km (which is well above the bulk of the atmosphere) than it is on the surface of the Earth Satellites are not only ~300 km above the surface of the Earth, they are also orbiting . For a low-Earth orbit, satellites need to travel some 7,000 meters/sec horizontally (17,000 mph) in order to orbit. Because getting to an orbit is a combination of getting through the atmosphere, getting up to the desired height, and getting the desired orbital velocity, (all while their mass is changing because they carry their fuel with them) rockets do not simply go straight up. Early on in a satellite launch, rockets usually go roughly straight up because the atmosphere is thick near the surface of the Earth and they're trying to get through it as quickly as possible. Later on, rockets tilt over and perform a combination of horizontal and vertical motion as they reach their orbit. The optimal path is a combination of avoiding drag and mixing height and orbital velocity together appropriately. This is also why most satellites orbit in the same direction the Earth spins; by orbiting that way, they get a bit of free sideways velocity from Earth's rotation during launch. | {
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178,612 | My friend asked me,''which is more basic or fundamental? constancy of speed of light which makes space-time behave dynamically OR dynamical behavior of space-time that makes light to travel with constant speed? I think 1st one because space-time is just a human-created model...''? I answered ,''space-time is more fundamental ,because speed of light is constant only locally,where space-time can be taken as flat ,so constancy of speed of light is a property of space-time ,so space-time is more fundamental(?).'' he replied,''speed of light is always constant ,its only different in different coordinate system''. So what should be the correct ans? | It sounds like you are imagining that what satellites do is go up through the atmosphere, break through into outer space, and hang there. That is not right. If you simply go straight up to outer space (say 300 km above Earth's surface), gravity will pull you right back down, even if you've left the atmosphere, and you'll crash back into the Earth. Gravity is only about 10% weaker at 300km (which is well above the bulk of the atmosphere) than it is on the surface of the Earth Satellites are not only ~300 km above the surface of the Earth, they are also orbiting . For a low-Earth orbit, satellites need to travel some 7,000 meters/sec horizontally (17,000 mph) in order to orbit. Because getting to an orbit is a combination of getting through the atmosphere, getting up to the desired height, and getting the desired orbital velocity, (all while their mass is changing because they carry their fuel with them) rockets do not simply go straight up. Early on in a satellite launch, rockets usually go roughly straight up because the atmosphere is thick near the surface of the Earth and they're trying to get through it as quickly as possible. Later on, rockets tilt over and perform a combination of horizontal and vertical motion as they reach their orbit. The optimal path is a combination of avoiding drag and mixing height and orbital velocity together appropriately. This is also why most satellites orbit in the same direction the Earth spins; by orbiting that way, they get a bit of free sideways velocity from Earth's rotation during launch. | {
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178,710 | Does that mean that electrons are infinitely stable? The neutrinos of the three leptons are also listed as having a mean lifespan of infinity. | Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference frames, and it's easiest/most limiting to talk about the electron's rest frame. In this frame, you have no kinetic energy, only rest mass energy equal to about 511 keV. So whatever you decay to has to have less rest mass than that--you might decay to a 300 keV particle, and give it 100 keV of kinetic energy, but you can't decay to a 600 keV particle. (There's no way to offset this with kinetic energy--no negative kinetic energy.) Unfortunately, every other charged lepton and every quark is heavier than that. So what options does that leave us? Well, there are massless particles (photon, gluon, graviton). There are also the neutrinos, which are all so close to massless that it took until very recently for anyone to tell that this was not the case. So you can decay to neutrinos and force carriers, maybe. Except then you run into a problem: none of these have any electric charge, and your decay has to conserve charge. You're stuck. tl;dr: Electrons are the lightest negatively charged particle and therefore cannot decay into lighter particles without violating charge conservation. | {
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178,718 | We're a bunch of friends building an RC boat, this may sound stupid but I had an argument with a friend where he insisted that his design will work. Basically, he's saying that by putting a powerfull fan and a piece of tissue as a sail, when the fan will blow air into it, the boat will move. What do you think? | Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference frames, and it's easiest/most limiting to talk about the electron's rest frame. In this frame, you have no kinetic energy, only rest mass energy equal to about 511 keV. So whatever you decay to has to have less rest mass than that--you might decay to a 300 keV particle, and give it 100 keV of kinetic energy, but you can't decay to a 600 keV particle. (There's no way to offset this with kinetic energy--no negative kinetic energy.) Unfortunately, every other charged lepton and every quark is heavier than that. So what options does that leave us? Well, there are massless particles (photon, gluon, graviton). There are also the neutrinos, which are all so close to massless that it took until very recently for anyone to tell that this was not the case. So you can decay to neutrinos and force carriers, maybe. Except then you run into a problem: none of these have any electric charge, and your decay has to conserve charge. You're stuck. tl;dr: Electrons are the lightest negatively charged particle and therefore cannot decay into lighter particles without violating charge conservation. | {
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178,826 | So today I was trying to derive an expression for the number of radioactive atoms remaining after a time $t$ if I began with $N_0$ atoms in total. At first I tried to assume that they had an average lifetime and work from there, but my friend dropped a hint and I found it much easier to assume that each atom had a specific chance in a small time element $dt$ to decay. After some manipulation, I arrived at an exponential decay formula, which was great. However, it got me to thinking about the concept of 'average lifetime'. (I also managed to find an expression for average lifetime relating it to the chance a single atom will decay in a differential time element). If I had a group of atoms that have an 'average lifetime' of say 5 seconds, after 5 seconds has elapsed, what is the 'average lifetime' of the remaining atoms? I don't think I can arbitrarily choose some reference time to begin ticking away at the atoms' remaining time, does that mean at any point of time that their 'average lifetime' or expected lifetime is always a constant, and never actually diminishes as time goes on? | Congratulations on deriving the exponential law for yourself, one learns a great deal about science working like this. Now to your last question: If I had a group of atoms that have an 'average lifetime' of say 5 seconds, after 5 seconds has elapsed, what is the 'average lifetime' of the remaining atoms? I don't think I can arbitrarily choose some reference time to begin ticking away at the atoms' remaining time, does that mean at any point of time that their 'average lifetime' or expected lifetime is always a constant, and never actually diminishes as time goes on? Yes indeed the average lifetime is constant. And the exponential distribution you have derived is the unique lifetime distribution with this property. Another way of saying this is that the decaying particle is memoryless : it does not encode its "age": there is nothing inside the particle that says "I've live a long time, now its time to die". Yet another take on this - as a discrete rather than continuous probability distribution - is the geometric distribution of the number of throws before a coin turns up heads, and the observation that a coin has no memory that counters the famous gambler's fallacy. To understand this uniqueness, we encode the memorylessness condition into the basic probability law $$p(A\cap B) = p(A) \, p(B|A)$$ Suppose after time $\delta$ you observe that your particle has not decayed (event $A$). If $f(t)$ is the propability distribution of lifetimes, then the probability the particle has lasted at least this long, i.e. the probability that it does not decay in time interval $[0,\,\delta]$ is: $$p(A) = 1-\int_0^\delta f(u)du$$ The a priori probability distribution function that the particle will last until time $t+\delta$ and then decay in the time interval $dt$ (event $B$) is $$p(B\cap A) = f(t+\delta) dt$$. This is events $B$ and $A$ observed together, which is the same as plain old $B$ since the particle cannot last unti time $t + \delta$ without living to $\delta$ first! Therefore, the conditional probability density function is $$p(B|A) = \frac{f(t+\delta)\,dt}{1-\int_0^\delta f(u)du}$$ But this must be the same as the unconditional probability density that the particle lasts a further time $t$ measured from any time, by assumption of memorylessness. Thus we must have: $$\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta),\;\forall \delta>0$$ Letting $\delta\rightarrow 0$, we get the differential equation $f^\prime(t) = - f(0) f(t)$, whose unique solution is $f(t) = \frac{1}{\tau}\exp\left(-\frac{t}{\tau}\right)$. You can readily check that this function fulfills the general functional equation $\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$ for any $\delta > 0$ as well. As Akhmeteli's answer says, true memorylessness is actually incompatible with simple quantum models. For example, one can derive the exponential lifetime for an excited fluorophore from a simple model of a lone excited two state fluorophore equally coupled to all the modes of the electromagnetic field. The catch is that the derivation rests on approximating an integral over positive energy field modes by an integral over all energies, both positive and negative. This of course is unphysical, but an excellent approximation since only modes near to the two state atom's energy gap will be excited: the fluorophore "tries" to excite all modes equally, but destructive interference prevents significant coupling to modes of greatly different energy than the difference between the energies of the states on either side of the transition. I show how this analysis is done in this answer here and here . Linewidths are mostly extremely narrow compared to the frequencies of the photons concerned, so I find it surprising and quite wonderful that Ahkmeteli cites a paper giving experimental evidence of the nonconstant lifetime. | {
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179,082 | Most of us have heard of Einstein's amazing equations which describe the universe around us, yet only some of us understand what the equations are actually saying. What are these equations actually saying, and is there a simple (relatively) way to derive them? Here they are, from Wikipedia : $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}$$ I have a vague notion of what a tensor is (it describes things as an array and higher orders define more complex transformations), but I don't understand what all of these tensors are doing. And why is there a $c^{4}$ in the equation!? | Einstein's equations can be loosely summarized as the main relation between matter and the geometry of spacetime . I will try to give a qualitative description what every term in the equation signifies. I will, however, have to warn potential readers that this will not be a short answer. Furthermore, I will refrain from trying to derive the equations in "elementary" manner, as I certainly don't know of any. Matter On the right hand side of the equation, the most important thing is the appearance of the energy-momentum tensor $T_{\mu\nu}$ . It encodes exactly how the matter---understood in a broad sense, i.e. any energy (or mass or momentum or pressure) carrying medium---is distributed in the universe. For understanding how to interpret the subscript indices of the $T$ , see my explanation of the metric tensor below. It is multiplied by some fundamental constants of nature $\Big($ the factor $\frac{8\pi G}{c^4}\Big)$ but this isn't of any crucial importance: One can view them as book-keeping tools that keep track of the units of the quantities that are related by the equation. In fact, professional physicists typically take the liberty to redefine our units of measurements in order to simplify the look of our expressions by getting rid of pesky constants such as this. One particular option would be to choose "reduced Planck units", in which $8\pi G=1$ and $c=1$ , so that the factor becomes $1$ . Differential geometry On the left hand side of Einstein's equations, we find a few different terms, which together describe the geometry of spacetime. General relativity is a theory which uses the mathematical framework known as (semi-)Riemannian geometry . In this branch of mathematics, one studies spaces which are in a certain sense smooth , and that are equipped with a metric . Let us first try to understand what these two things mean. The smoothness property can be illustrated by the intuitive (and historically important!) example of a smooth (two-dimensional) surface in ordinary three-dimensional space. Imagine, for instance, the surface of an idealized football, i.e. a 2-sphere. Now, if one focuses ones attention to a very small patch of the surface (hold the ball up to your own face), it seems like the ball is pretty much flat. However, it is obviously not globally flat. Without regards for mathematical rigor, we can say that spaces that have this property of appearing locally flat are smooth in some sense. Mathematically, one calls them manifolds. Of course, a globally flat surface such as an infinite sheet of paper is the simplest example of such a space. In Riemannian geometry (and differential geometry more generally) one studies such smooth spaces (manifolds) of arbitrary dimension. One important thing to realize is that they can be studied without imagining them to be embedded in a higher-dimensional space, i.e. without the visualization we were able to use with the football, or any other reference to what may or may not be "outside" the space itself. One says that one can study them, and their geometry, intrinsically . The metric When it comes to intrinsically studying the geometry of manifolds, the main object of study is the metric (tensor). Physicists typically denote it by $g_{\mu\nu}$ . In some sense, it endows us with a notion of distance on the manifold. Consider a two-dimensional manifold with metric, and put a "coordinate grid" on it, i.e. assign to each point a set of two numbers, $(x,y)$ . Then, the metric can be viewed as a $2\times 2$ matrix with $2^2=4$ entries. These entries are labeled by the subscripts $\mu,\nu$ , which can each be picked to equal $x$ or $y$ . The metric can then be understood as simply an array of numbers: $$\begin{pmatrix} g_{xx}&g_{xy}\\ g_{yx}&g_{yy}\end{pmatrix}$$ We should also say that the metric is defined such that $g_{\mu\nu}=g_{\nu\mu}$ , i.e. it is symmetric with respect to its indices. This implies that, in our example, $g_{xy}=g_{yx}$ .
Now, consider two points that are nearby, such that the difference in coordinates between the two is $(\mathrm{d}x,\mathrm{d}y)\;.$ We can denote this in shorthand notation as $\mathrm{d}l^\mu$ where $\mu$ is either $x$ or $y\;,$ and $\mathrm{d}l^x=\mathrm{d}x$ and $\mathrm{d}l^y=\mathrm{d}y\;.$ Then we define the square of the distance between the two points, called $\mathrm{d}s\;,$ as $$\mathrm{d}s^2= g_{xx}\mathrm{d}x^2+g_{yy} \mathrm{d}y^2 + 2 g_{xy}\mathrm{d}x \mathrm{d}y= \sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu$$ To get some idea of how this works in practice, let's look at an infinite two-dimensional flat space (i.e. the above-mentioned sheet of paper), with two "standard" plane coordinates $x,y$ defined on it by a square grid. Then, we all know from Pythagoras' theorem that $$\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2= \sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu$$ This shows that, in this case, the natural metric on flat two-dimensional space is given by $$g_{\mu\nu}=\begin{pmatrix} g_{xx}&g_{xy}\\ g_{xy}&g_{yy}\end{pmatrix}= \begin{pmatrix} 1&0\\0&1\end{pmatrix}$$ Now that we known how to "measure" distances between nearby points, we can use a typical technique from basic physics and integrate small segments to obtain the distance between points that are further removed: $$L=\int \mathrm{d}s= \int \sqrt{\sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu}$$ The generalization to higher dimensions is straightforward. Curvature tensors As I tried to argue in the above, the metric tensor defines the geometry of our manifold (or spacetime, in the physical case). In particular, we should be able to extract all the relevant information about the curvature of the manifold from it. This is done by constructing the Riemann (curvature) tensor $R^{\mu}_{\ \ \ \nu\rho\sigma}$ , which is a very complicated object that may, in analogy with the array visualization of the metric, be regarded as a four-dimensional array, with each index being able to take on $N$ values if there are $N$ coordinates $\{x^1,\dots x^N\}$ on the manifold (i.e. if we're dealing with an $N$ -dimensional space). It is defined purely in terms of the metric in a complicated way that is not all too important for now. This tensor holds pretty much all the information about the curvature of the manifold---and much more than us physicists are typically interested in. However, sometimes it is useful to take a good look at the Riemann tensor if one really wants to know what's going on. For instance, an everywhere vanishing Riemann tensor ( $R^\mu_{\ \ \ \nu\rho\sigma}=0$ ) guarantees that the spacetime is flat. One famous case where such a thing is useful is in the Schwarzschild metric describing a black hole, which seems to be singular at the Schwarzschild radius $r=r_s\neq 0$ . Upon inspection of the Riemann tensor, it becomes apparent that the curvature is actually finite here, so one is dealing with a coordinate singularity rather than a "real" gravitational singularity. By taking certain "parts of" the Riemann tensor, we can discard some of the information it contains in return for having to only deal with a simpler object, the Ricci tensor: $$ R_{\nu\sigma}:=\sum_{\mu\in \{x^1,\dots x^N\}} R^\mu_{\ \ \ \nu\mu\sigma}$$ This is one of the tensors that appears in the Einstein field equations. the second term of the equations features the Ricci scalar $R$ , which is defined by once again contracting (a fancy word for "summing over all possible index values of some indices") the Ricci tensor, this time with the inverse metric $g^{\mu\nu}$ which can be constructed from the usual metric by the equation $$\sum_{\nu\in\{x^1,\dots,x^N\}}g^{\mu\nu}g_{\nu\rho}= 1\ \text{if }\mu=\rho\ \text{and }0\ \text{otherwise}$$ As promised, the Ricci scalar is the contraction of the Ricci tensor and inverse metric: $$R:=\sum_{\mu,\nu\in\{x^1,\dots x^N\}}g^{\mu\nu}R_{\mu\nu} $$ Of course, the Ricci scalar once again contains less information than the Ricci tensor, but it's even easier to handle. Simply multiplying it by $g_{\mu\nu}$ once again results in a two-dimensional array, just like $R_{\mu\nu}$ and $T_{\mu\nu}$ are. The particular combination of curvature tensors that appears in the Einstein field equations is known as the Einstein tensor $$ G_{\mu\nu}:=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu} $$ The cosmological constant There is one term that we have left out so far: The cosmological constant term $\Lambda g_{\mu\nu}$ . As the name suggests, $\Lambda$ is simply a constant which multiplies the metric. This term is sometimes put on the other side of the equation, as $\Lambda$ can be seen as some kind of "energy content" of the universe, which may be more appropriately grouped with the rest of the matter that is codified by $T_{\mu\nu}$ . The cosmological constant is mainly of interest because it provides a possible explanation for the (in)famous dark energy that seems to account for certain important cosmological observations. Whether or not the cosmological constant is really non-zero in our universe is an open issue, as is explaining the value observations suggest for it (the so-called cosmological constant problem a.k.a. "the worst prediction of theoretical physics ever made", one of my personal interests). PS. As pointed out in the comments, if you enjoyed this you may also enjoy reading this question and the answers to it, which address that other important equation of general relativity, which describes the motion of "test particles" in curved spacetimes. | {
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179,388 | Let me start by saying that I understand the definitions of the Lie and covariant derivatives, and their fundamental differences (at least I think I do). However, when learning about Killing vectors I discovered I don't really have an intuitive understanding of the situations in which each one applies, and when to use one over the other. An important property of a Killing vector $\xi$ (which can even be considered the definition) is that $\mathcal{L}_\xi\, g = 0$, where $g$ is the metric tensor and $\mathcal{L}$ is the lie derivative. This says, in a way, that the metric doesn't change in the direction of $\xi$, which is a notion that makes sense. However, if you had asked me how to represent the idea that the metric doesn't change in the direction of $\xi$, I would have gone with $\nabla_\xi g = 0$ (where $\nabla$ is the covariant derivative), since as far as I know the covariant derivative is, in general relativity, the way to generalize ordinary derivatives to curved spaces. But of course that cannot be it, since in general relativity we use the Levi-Civita connection and so $\nabla g = 0$. It would seem that $\mathcal{L}_\xi\, g = 0$ is be the only way to say that the directional derivative of $g$ vanishes. Why is this? If I didn't know that $\nabla g = 0$, would there be any way for me to intuitively guess that "$g$ doesn't change in the direction of $\xi$" should be expressed with the Lie derivative? Also, the Lie derivative is not just a directional derivative since the vector $\xi$ gets differentiated too. Is this of any consequence here? | Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric , whereas $\nabla_Xg = 0$ says something about the connection . Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a pair of tangent vectors along the integral curves of $X$ in such a way that their inner product remains the same. As an example, consider the upper half plane model of the hyperbolic plane. Its metric is $y^{-2}(dx^2 + dy^2)$, so clearly $\partial_x$ is a Killing vector field; its flow, horizontal translation, is an isometry. The fact that $\nabla_{\partial_x}g = 0$ doesn't say anything about $g$, but it does say that Euclidean parallel transport is compatible with this directional derivative of the connection. Now consider $\partial_y$. This of course is not a Killing vector field, since vertical translation is not an isometry. The connection however can be made such (by the theorem of Levi-Civita) that a pair of tangent vectors can be parallel transported in such a way that the inner product is preserved. EDIT I think I have a more illustrative example: consider the sphere embedded in $\Bbb R^3$. Pick an axis and take the velocity vector field $\xi$ associated to rotation around the axis at some constant angular velocity. Also consider a second vector field $\zeta$ that is everywhere (in a neighbourhood of the equator, extend in any smooth way toward the poles) proportional to $\xi$, but that has constant velocity everywhere, something like in this image (downloaded from this page ). Obviously $\xi$ is a Killing field, as it integrates to an isometry. An immediate way to see that $\zeta$ is not, is by noting that curves parallel to the equator remain parallel to the equator under the flow of $\zeta$, hence so do their tangent vectors. What happens to a curve whose tangent vector at the equator points toward a pole, is that the flow of $\zeta$ moves the point at the equator over a smaller angle than a point above the equator, so these two vectors don't remain perpendicular. For parallel transport on the other hand, two perpendicular tangent vectors to a point at the equator will remain perpendicular both under $\xi$ and in $\zeta$, since they only depend on the restriction to the vector fields to the equator, where they are equal. This doesn't say anything about the vector field generating an isometry, i.e. being a Killing vector field. | {
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180,698 | This is a question about the relativistic mass concept which I am having trouble understanding, mainly because of the scenario below. Simple scenario: Suppose 1 gram of matter is accelerated to 99% the speed of light. At this speed, the relativistic mass is 7 times greater than the rest mass. If this matter collides with a stationary quantity of 7 grams of antimatter, will the two masses annihilate completely with each other? Or will the matter just annihilate 1 grams worth of antimatter? If the latter is true then what exactly am I overlooking about the relativistic mass concept that makes the former incorrect? | A sophisticated, yet easy way to see that this the answer must be "No." is to recall that velocity is relative — that there is no absolute notion of velocity. You said the matter was moving and the antimatter still, but that point of view (AKA frame of reference) is not privileged in any way. An observer at rest with respect to the matter has just as much right to conclude that the anti-matter is in motion as you have to conclude that the matter is moving. So you can't rely on a velocity dependent notion of mass to work out the consequences of the scenario. The modern approach to relativity is to define the (only!) mass of a particle or system as the square of its energy-momentum four vector (with appropriate factors of $c$):
$$ m = \frac{\sqrt{E^2 - (pc)^2}}{c^2} \,. $$ The thing that you you've been taught to call "relativistic mass", $\gamma m$, is (to within two factors of $c$) described as the "total energy" of the particle or system. | {
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180,998 | Yesterday we started relativity on our physics class, and my professor taught us a few concepts. We did some examples on how things changed by looking them from different reference systems, and a paradox came to my mind from an example he did on the blackboard: Suppose that we have a bug inside a hole (and suppose that the height of the bug is $0$ ). This hole is $L$ meters deep. There's a man outside the hole that wants to kill the bug with a nail of length $l$ , with $l<L$ . Obviously, it's not possible for him to kill the bug by trying to push the nail in the hole. However, if the nail moves really fast, from it's reference system the hole will have a height $L_n<L$ , and if it moves fast enough then it will happen that $L_n<l$ , so he finally could kill the bug. On the other hand, the bug knows that from his reference system, the nail will have a lenght $l_b<l$ , so it would never reach him. Is it possible to quantify the problem? How is it possible that the nail reach the bug from it's reference system and from the bug's reference system it's even smaller than before? | Imagine a slightly different scenario: two pilots, Alice, and Bob, are in their spaceships. They move towards a tunnel of length $L$ at a velocity $v$, and remain a distance $l'$ apart. Alice is closest to the tunnel and thus enters first, approaching a wall at the end of the length of the tunnel. Just as Bob enters he decelerates, coming quickly to a halt and sending a message to Alice telling her to stop. This message takes a certain amount of time to reach her, and if she is travelling too fast, she will hit the wall before the message arrives. This is analogous to what is happening with the nail and the rivet. The tip of the nail is Alice, and the head of the nail is Bob. This allows the bug to be crushed in its own reference frame, even if $l'< L$, and resolves the paradox. There is an animation of this here . A more rigorous analysis can be found here . Also see Chris White's answer. | {
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182,688 | Imagine, if you will, there exists somewhere out there a supermassive planet. For some reason, this planet is only a shell of its former existence and all that is left is a crust of substantial size to keep it from collapsing in on itself (let's imagine it is stable at 5000' thick). Now, this planet was HUGE! How large, you ask? Large enough that the remaining shell is still able to maintain a gravity well strong enough to keep you anchored to the surface. Pretty neat, huh? Well, since this is a sight to be seen by everyone, you stroll along and, not looking where you are going, you tumble into a chasm and straight down into a hole that permeates the thickness of the shell. As you are falling down and down and down, you think: "At what point do I stop falling down and I start falling up? And would I be able to retain the velocity, since gravity is a conservative force, and fall back through another hole on the surface on the opposite side? Where would the center of gravity even exist? I mean, aaaaahhhhhhh!" What say you, browsers and scientists? A stretch of the imagination, sure, but is it physically possible? Could there exists a configuration similar to this where gravity could be concentrated to a point but still allow free travel up to the focus? | Yes, this is possible. It is perfectly fine for a mass configuration to produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$. The spherical-shell model you describe is the simplest example of this. (On the other hand, it does not describe a physically realizable model from a materials standpoint, as described in Rod Vance's answer.) In addition to this, your spherical shell of a planet will have the additional property that once you are inside the shell, the planet's gravitational attraction will vanish. Objects there will perform uniform linear motion with constant velocity, until of course they reach the edge of the shell. This has been known since the time of Newton and it is one of the standard exercises in electrostatics (which is mathematically identical to newtonian gravity) for a physics undergrad. Schemes like this are fairly common and relatively fun to analyse; you may find the "gravity train" interesting. | {
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182,836 | Let's say we have 2 sticks, both a meter long. We put both of them on the ground vertically. To the top of one of them we attach a weight. Then we tip both of them over and let them fall. Assume that there is no air resistance and that (the lower ends of) the sticks do not slide against the floor. Which one falls first? Would I be correct in guessing that the one with the weight falls slower due to having a higher center of gravity? EDIT: In case you're wondering where this question came from, I was pondering whether having a passenger on the back of a motorcycle would increase or decrease the time it takes to tip over in the case balance is lost (at very low speed). | I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion for the rod we use the rotational analogue to Newton's second law: $$ T = I\frac{d^2\theta}{dt^2} $$ where $T$ is the torque and $I$ is the moment of inertia, and rearranging this gives: $$ \frac{d^2\theta}{dt^2} =\frac{T}{I} \tag{1} $$ If the length of the rod is $\ell$, then the torque is: $$\begin{align}
T &= mg\frac{\ell}{2}\cos\theta + Mg\ell\cos\theta \\
&= (\frac{m}{2} + M)g\ell\cos\theta
\end{align}$$ The moment of inertia of the rod is: $$ I_{rod} = \frac{m\ell^2}{3} $$ and assuming our weight can be approximated as a point mass its moment of inertia is: $$ I_{weight} = M\ell^2 $$ and the total moment of inertia is just the sum of these two: $$ I = \frac{m\ell^2}{3} + M\ell^2 $$ And to get our equation of motion we just substitute our expressions for $T$ and $I$ into equation (1) to get: $$ \frac{d^2\theta}{dt^2} = -\frac{(\frac{m}{2} + M)g\ell\cos\theta}{\frac{m\ell^2}{3} + M\ell^2} $$ There's a minus sign because as I've drawn the diagram the angle $\theta$ decreases with time so the angular acceleration is negative. With some rearranging we get: $$ \frac{d^2\theta}{dt^2} = -\left(\frac{m + 2M}{m + 3M}\right) \frac{3g\cos\theta}{2\ell} \tag{2} $$ Now actually solving this equation of motion would be hard, but we don't need to solve it to answer your question. The left hand side of equation (2) is the angular acceleration, and if we increase the magnitude of the angular acceleration the rod falls faster while if we decrease it the rod falls slower. So your question simplifies to: If we increase the mass M does the magnitude of the angular acceleration increase or decrease? On the right hand side everything outside the brackets is independent of $M$, so we just need to answer whether the term in the brackets increases or decreases if we change $M$. This is easy to answer because we have a term $2M$ on the top of the fraction and a term $3M$ on the bottom, and obviously $3M \gt 2M$. So if we increase $M$ the fraction in the brackets decreases, and therefore the magnitude of the angular acceleration decreases. So the answer is that attaching a mass to the top of the rod does indeed make it fall more slowly. | {
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183,501 | I was watching a documentary on youtube regarding Quantum Physics when it introduced the wavelengths of light emitted. Did a bit more research over the internet and I believe I understand the gist of why each atom gives off different colors. Here's a brief summary of what I've read: An atom of a particular element has several shells of electrons, and each shell is a different energy. When the atoms are heated up, some of the electrons can jump up to shells with higher energy, but they don't stay there very long. When the electrons fall down again to a lower energy shell, they give off that energy as a photon (a "particle" of light). The amount of energy of the photon given off is the difference in energy between the two shells, and determines the wavelength of the light. With all that taken into consideration, why is it that a molecule isn't a combination of its component atoms? More precisely, how is the color of a molecule determined? | I'll do that teacher thing and turn your question around back at you. Why isn't the spectrum of the lithium atom just the spectrum of the hydrogen atom plus the spectrum of the helium atom? And, for that matter, why is the helium spectrum not simply two copies, somehow, of the hydrogen spectrum? Why do atoms have unique spectra in the first place? The answer to my question is that an atom with proton number $Z$ can have a fundamentally different excitation spectrum (and chemistry) from some other atom with proton number $Z+1$ because that extra electron interacts with all the other electrons that were in the atom already. That's also the answer to your question.
In the same way, when you combine two hydrogen atoms and an oxygen atom to make water, there isn't a "hydrogen electron" in the same way as in a bare hydrogen atom, or even a hydrogen molecule. A "hydrogen electron" in a water molecule is interacting with both hydrogen nuclei, and the oxygen nucleus, and the other nine electrons in the system, which makes its excitation spectrum fundamentally different from the spectrum of an electron interacting with a free proton. | {
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183,504 | Will the body rapidly start losing heat or will the water find a seed and solidify around it ? | I'll do that teacher thing and turn your question around back at you. Why isn't the spectrum of the lithium atom just the spectrum of the hydrogen atom plus the spectrum of the helium atom? And, for that matter, why is the helium spectrum not simply two copies, somehow, of the hydrogen spectrum? Why do atoms have unique spectra in the first place? The answer to my question is that an atom with proton number $Z$ can have a fundamentally different excitation spectrum (and chemistry) from some other atom with proton number $Z+1$ because that extra electron interacts with all the other electrons that were in the atom already. That's also the answer to your question.
In the same way, when you combine two hydrogen atoms and an oxygen atom to make water, there isn't a "hydrogen electron" in the same way as in a bare hydrogen atom, or even a hydrogen molecule. A "hydrogen electron" in a water molecule is interacting with both hydrogen nuclei, and the oxygen nucleus, and the other nine electrons in the system, which makes its excitation spectrum fundamentally different from the spectrum of an electron interacting with a free proton. | {
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184,317 | Whenever we switch on a bulb......it takes almost no time to glow up.....But we know that the atoms of a solid are tightly packed and there is a very little space between them.
So how the electrons travel through them irrespective of so much blockages in the conductor??? | The electrons themselves don't move all that fast. The wave energy is the part that moves quickly. Picture it this way. You have 500 meters of pipe, with a small hole at the other end. The pipe is full of water and you increase the pressure at your end. Water will flow out the other end immediately. This is the electrical energy (pressure) and the copper(water). Now add some dye to the water and note how long it takes for the coloured water to come out the hole. The dye represents the new electrons, and they take noticeable time to move through the system. Pressure moves immediately. (ok, in this case pressure moves at the speed of sound in water) | {
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184,659 | The difference in using Distance vs Displacement is demonstrated in this example: Work = Force x Distance If I carry an object to and fro 10 metres, the work done would be Force x 20 metres. and Work = Force x Displacement If I carry an object to and fro 10 metres, the work done would be
Force x 0 metres. In this context, which should be a more accurate representation or formula? I note that Force and displacement are vectors and distance as scalar. | It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. As @lemon pointed out in a comment, this is expressed by writing the work done as the integral: $$W = \int \vec F \cdot d\vec{x}$$ When $F$ is only a function of position and $\vec \nabla\times \vec F = 0$, this integral is independent of the path and depends only on the end points; but if it is a function of direction of motion, you can no longer do the integral without taking the path into account. | {
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184,898 | Everyone who has been interested in modern science has heard explanations (certainly simplifications) of general relativity, mostly that space is curved. The analogy with a rubber sheet is popular. In such an analogy, orbiting planets are said to be naturally following " a straight line in a curved space ". Assuming that is not an oversimplification, would it mean that orbits are loops generated in space by massive objects ? Also, if we consider spacetime as a curved structure, thus lines are not necessarily straight, what would be the meaning of momentum in such a frame? | Calling orbits loops is a dangerous line of thinking. Objects that are not under the influence of other forces follow geodesics , which are the curved space equivalent of straight lines. And, while it's tempting to say that the orbit of a planet is effectively a loop in spacetime, let me try to convince you why such a simplification should be avoided. Yes, for the orbiting planet, it follows a geodesic through spacetime that ends up leading it around the host star in a loop. So for the planet, you could say that the star's mass has warped spacetime such that a straight line is now a loop. However, consider objects with other velocities. For instance, consider a beam of light. Light follows a straight line through space as well. But if you shone a beam of light tangent to the planet's orbit, it would likely not loop around the star. So how can we say that the star's mass has generated a true loop in space if not everything follows this loop? We can't. Spacetime is most assuredly curved, but every observer sees their own curvature. It isn't so simple that we could say a star curves a straight line into a loop when not everything would move around it in a loop. The relationship between gravity, velocity, and curvature is a complicated one and oversimplifying by saying that gravity makes straight paths into loops in space is liable to do more harm than good. As for the meaning of momentum, that does not change. $\sqrt{\frac{E^2}{c^2}-m^2c^2}$ is still the definition of momentum and it still refers to the energy stored in an object's motion through spacetime. | {
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185,110 | I have heard virtual particles pop in and out of existence all the time, most notable being the pairs that pop out beside black holes and while one gets pulled away. But wouldn't this actually violate the conservation of energy? | Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute proofs, mathematically deduced from the axioms. "Existence" for physics means "measurable", for mathematics "possible to be included in the self consistent theory. Modern physics has used mathematical models to describe the measurements and observations in the microcosm of atoms, molecules, elementary particles, adding postulates that connect the mathematical calculations with the physical observables The dominant mathematical model is the field theoretical model that simplifies the mathematics using Feynman diagrams These diagrams represent terms in an expansion of the desired solution, each term has a diminishing contribution to the cross section of the interaction. The diagram below would be the dominant term, as the next one would be more complicated and therefore smaller by orders of magnitude. To each component of the diagram there corresponds one to one a mathematical formula twhich integrated properly will give a prediction for a measurable quantity. In this case the probability of repulsion when one electron scatters on another. This diagram for example, has as measurable quantities the incoming energy and momentum of the electrons ( four vectors ) and of outgoing four vectors . The line in between is not measurable, because it represents a mathematical term that is integrated over the limits of integration, and within the integral energy and momentum are independent variables. The line has the quantum numbers of the photon though not its mass , and so it is called a "virtual photon". It does not obey the energy momentum rule which says that : $$\sqrt{P\cdot P} = \sqrt{E^2 - (pc)^2} = m_0 c^2$$ The photon has mass zero. Through the above relation which connects energy and momentum through the rest mass, the un-physical mass of the virtual line depends on one variable, which will be integrated over the diagram; it is often taken as the momentum transfer. Quantum number conservation is a strong rule and is the only rule that virtual particles have to obey. There are innumerable Feynman diagrams one can write, and the internal lines considered as particles would not conserve energy and momentum rules if they were on mass shell. These diagrams include vacuum fluctuations that you are asking about, where by construction there are no outgoing measurable lines in the Feynman diagrams describing them. They are useful/necessary in summing up higher order calculations in order to get the final numbers that will predict a measurable value for some interaction. Thus virtual particles exist only in the mathematics of the model used to describe the measurements of real particles . To coin a word virtual particles are particlemorphic ( :) ), having a form like particle but not a particle. | {
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"https://physics.stackexchange.com",
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