In the laboratory reference frame (LRF), a horizontally moving (with constant speed) flat capacitor would have a different size of plates therefore resulting in different capacity $C'$, namely $$ C' = \frac{1}{\gamma}C, $$ where C is a capacity in its own reference frame. The energy of a capacitor is $$ W' = \frac{q^2}{2C'} = \gamma W. $$ Here I assumed the value of a charges remain constant in different inertial reference frames (otherwise we could distinguish one reference frame from another). So if the capacitor is closed on a resistor in his reference frame then for me in LRF would be seen like there was more heat produced on a resistor since $Q=W'$ no matter what current was. So, wouldn't it be the way I distinguish one inertial reference frame from another?
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2Have you transformed the electronmagnetic field tensor? If the answer is "No" then whatever you think you know about the system is almost certainly wrong. – dmckee --- ex-moderator kitten Mar 24 '18 at 21:28
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1@dmckee what you said is cool, doesn't point out on a logical mistake in my post though. I deliberately wasn't using any field characteristics, but should I understand your comment as "C = q/U" doesn't work in another (apart from its own) inertial reference frame? – Ice-Nine Mar 24 '18 at 21:36
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But you are using field properties. That voltage is related to the field in the capacitor, right? – dmckee --- ex-moderator kitten Mar 24 '18 at 21:39
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You now have moving charges, I.e. current, so you cannot ignore the B field. That’s @dmckee ‘s point about transforming the fields. There’s energy in both parts. – Bob Jacobsen Mar 26 '18 at 09:51
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1@BobJacobsen The point is that the energy in the electric field alone is already naively too much. Adding the magnetic field makes the problem worse, not better. – knzhou Mar 26 '18 at 11:00
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How exactly are you measuring the heat produced on the resistor in the moving frame? I have a feeling there's no contradiction here, because energy, in general, is not Lorentz-invariant. – probably_someone Mar 26 '18 at 13:57
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@probably_someone I think just the same, for example if the heat would disengage as EM-waves in laboratory frame we would see different wavelength, therefore different energy. – Ice-Nine Mar 26 '18 at 15:32
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Ok, but energy isn't Lorentz-invariant anyway, so I don't see why you'd expect it to be the same. If your argument is that you can use it to "distinguish one inertial reference frame from another," you don't need heat to do that. You can distinguish two inertial reference frames just be noting which objects are moving at which velocities. But this is completely in line with the Equivalence Principle, because the Equivalence Principle says that the laws of physics should be the same in all inertial reference frames, not that the measured quantities should be the same. – probably_someone Mar 26 '18 at 15:47
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Put properly, the Equivalence Principle says that the laws of physics are covariant. The equations don't change under coordinate transformations, but the quantities that they act on are allowed to change; the specific way in which they change is governed by the Lorentz transformation. – probably_someone Mar 26 '18 at 15:48
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@probably_someone yeye I wasn't confronting your statement:) I agree – Ice-Nine Mar 26 '18 at 15:50
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1It should also probably be noted that measuring the radiation coming from an object in a moving frame is probably a bad way to measure its temperature. As the object passes by you, the radiation goes from being blueshifted to being redshifted, so your measurement results change with time within any given moving frame. Ideally you'd find an observable that doesn't change with time in such a way. – probably_someone Mar 26 '18 at 15:55
4 Answers
At lowest order, there's no contradiction. Using the relativistic field transformation, the field in the capacitor goes up by a factor of $\gamma$, while the volume of the capacitor goes down by a factor of $1/\gamma$, so the total field energy goes up by a factor of $\gamma^2/\gamma = \gamma$. This is all dissipated in the resistor.
Meanwhile, let the four-momentum given to the resistor in the rest frame be $p^\mu = (U, 0, 0, 0)$. Then the four-momentum given to the resistor in the boosted frame is $p'^\mu = (\gamma U, \gamma v U, 0, 0)$.
That is, in both cases we expect the energy dissipated in the resistor to be $\gamma$ larger than in the rest frame, so there is no contradiction.
However, in the boosted frame there's also a magnetic field which contributes energy at order $v^2$, and I haven't been able to figure out where that energy goes. There's a possibly related subtlety, which is that something must hold the capacitor plates apart. This means there is a pressure somewhere in the system, which can be Lorentz boosted into an energy, and must be counted. This is absolutely essential to account for the case where the boost is perpendicular to the plates, and might fix up the magnetic field energy contribution in this case.
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1A moving capacitor loses kinetic energy when drained, and a moving resistor gains kinetic energy when heated? And when we accelerate a charged capacitor the energy it gains is just kinetic energy? – stuffu Mar 26 '18 at 08:04
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@probably_someone Well I interpreted the maths like this: In the boost the energy got multiplied by gamma, because this much kinetic energy was added: delta E = delta gamma * rest energy. When heat energy moves, it has that amount of kinetic energy. Same story for the energy of the electric field. – stuffu Mar 26 '18 at 14:59
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@stuffu "When heat energy moves, it has that amount of kinetic energy." What does this mean, exactly? Since when does energy have kinetic energy? Do you say that time has kinetic energy too, now, because of time dilation? More to the point, you can't interpret the multiplication of total energy by $\gamma$ as just the addition of kinetic energy, especially when heat and electromagnetic fields are involved. – probably_someone Mar 26 '18 at 15:01
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@probably_someone If I said "hot objects have extra mass, and momentum and and kinetic energy", that would not sound as funny as "moving heat has kinetic energy". But there's actually nothing wrong with the funny formulation – stuffu Mar 26 '18 at 16:36
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@stuffu So you're using a non-standard notion of kinetic energy, without explaining that you're doing so. That's what wrong here. It's fine if you use non-standard terms, but you must indicate that you are doing so. By your definition, "just kinetic energy" encompasses a variety of non-trivial quantities not necessarily associated with direct mechanical motion, so now I don't see what your actual objection is. – probably_someone Mar 26 '18 at 17:47
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@probably_someone, and knzhou there's still something counter intuitive here, for example if I someone touches the resistor and don't have his skin burnt, in rest frame I could possibly see him burning it? – Ice-Nine Mar 29 '18 at 18:54
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@Ice-Nine Is the person touching the resistor in the rest frame of the resistor? – probably_someone Mar 29 '18 at 19:00
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@probably_someone. ok he touches it in his frame. Suppose he holds it until current drops to zero. Suppose he didn't burn his skin in his frame. In the rest frame we also would see him touching the resistor. – Ice-Nine Mar 29 '18 at 19:18
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@Ice-Nine How does he hold the resistor if he's not in the resistor's rest frame? – probably_someone Mar 29 '18 at 19:19
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@probably_someone IM SORRY for calling Laboratory Reference Frame as a rest frame, my mistake. He touches it in his rest frame, of course. – Ice-Nine Mar 29 '18 at 19:20
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@Ice-Nine Let's be careful here. Your system consists of an observer, a man, and a resistor. The man is touching the resistor. Who is moving relative to the resistor, and who is not? – probably_someone Mar 29 '18 at 19:22
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@probably_someone the man touches the resistor. They are moving with the same speed. An observer beholds it from laboratory reference frame. – Ice-Nine Mar 29 '18 at 20:00
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Related: Is the inertia of light equal to the inertia of mass under $E=mc^2$?
The source of inertia of the capacitor is its stress-energy, just like the source of gravity of the capacitor is its stress-energy.
When we accelerate a box that contains light, the box must have reduced inertia, because the light has increased inertia - because the box as a whole should have normal inertia.
In the case of a capacitor we have charges under pressure and containers of charges under stress. The charges have increased inertia, the containers have decreased inertia.
When some charges move from the positive plate to the negative plate, stress in both plates decreases. Which means that the inertia defect decreases in both plates, in other words inertia of plates increases.
As the plates gain inertia while keeping the same velocity, the plates gain momentum and kinetic energy. The momentum and the kinetic energy come from the electric field.
Magnetic fields are just relativistic effects on electric fields, so magnetic fields have not been in any way left out or forgotten in the above analysis.
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So if capasitor is closed on a resistor in his reference frame then for me in LRF would be seen like there was more heat produced on a resistor since $Q=W'$ no matter what current was. So, wouldn't it be the way I distinguish one intertial reference frame from another?
In the frame where the capacitor with resistor are at rest, all EM energy in the capacitor eventually transforms into increase in internal energy and manifests as increased temperature and increased mass of the system.
In the frame where the capacitor with resistor move, EM energy available is higher, but here not all EM energy needs to transform into increase of internal energy; some of it may transform into increased kinetic energy.
So different frames imply different division of available energy. This does not mean, as I see it, any indication that some reference frames are preferred in the sense of ether theories.
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If we somehow contract the plates of a charged capacitor, the charges get closer to each other and the repulsive Coulomb forces between charges become stronger. The voltage of the capacitor increases. The capacitance of the capacitor decreases.
If we somehow contract the plates of a charged capacitor, and also the fields of the charges, the repulsive Coulomb forces between charges stay the same. The voltage of the capacitor stays the same. The capacitance of the capacitor stays the same. This is what happens when we accelerate or boost a capacitor.
Let's say we have a spherical fleet of spaceships, every ship is positive charged. Adding a positive ship to the fleet takes lot of energy. Now let's say the fleet accelerates to high speed keeping it's dimensions unchanged, like Bell's spaceships. Now it's easy to add a ship to the end of the formation, because the electric fields are contracted, and it's easy to add a ship to the side of the formation because magnetic attraction is helping. The sphere seems to have a larger capacitance when it moves.
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