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Suppose I look at a parallel plate capacitor in its rest frame and calculate the electrostatic energy, $E$.

Next, I look at the same capacitor in a primed frame boosted in the direction perpendicular to the plane of the plates. In this frame, the $E$-field is the same strength, there is no magnetic field, and the volume over which the $E$-field extends is less by a factor $1/\gamma$. This suggests $E' = \frac{1}{\gamma} E$, but relativity states that energy transforms as $E' = \gamma E$.

Where is the missing energy?

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Surely in the primed frame there are large magnetic fields generated at the leading and trailing edges of the capacitor from the $\frac{\partial \vec{E}}{\partial t}$ term and from the plates themselves from the $\vec{J}$ term, no? –  dmckee Feb 2 '13 at 18:59
    
@dmckee I did think of that, but am not sure how to make it work. By "no magnetic field", I just meant that if you transform the constant E-field, it doesn't create a B-field. –  Mark Eichenlaub Feb 2 '13 at 19:04
    
    
I already did that search and found those notes you linked. Why are you linking to them when they don't answer the question? –  Mark Eichenlaub Feb 3 '13 at 16:37
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Rindler and Denur wrote a paper on this paradox in 1987 in the AJP: "A simple relativistic paradox about electrostatic energy". –  Larry Harson Mar 4 '13 at 17:07
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3 Answers

up vote 4 down vote accepted

First of all, thanks for this question because it made me think about relativity which was always fun!

It's true that $E'=\frac{1}{\gamma} E$. You say that relativity states that the energy should increase by a factor of $\gamma$. This is certainly true for a massive particle whose energy is $\gamma mc^2$, but why would you expect this to hold for the energy in the fields in this situation? I think the answer simply is that there is no contradiction; the energy in the fields transforms by a factor of $\frac{1}{\gamma}$ and that's that!

Actually, not quite! (as Mark argued in the comments)

After the discussion in the comments below, I realized that perhaps "that's that" was both premature and doesn't get at the heart of Mark's question. So I dug deeper (namely I scoured Jackson's EM) and I found an answer that is significantly more complete.

The definition of the energy and momentum densities in the fields given by the $\Theta^{00}$ and $\Theta^{0i}$ components of the (symmetric-traceless version of the) stress tensor (see Jackson 12.114) $$ \Theta^{00} = \frac{1}{8\pi}(\mathbf E^2+\mathbf B^2), \qquad \Theta^{0i} = \frac{1}{4\pi}(\mathbf E\times\mathbf B)^i $$ leads to the following candidate for the electromagnetic four-momentum: $$ P_\mathrm{cand}^\mu=\left(\int d^3 x\,\Theta^{00}, \int d^3x\, \Theta^{0i}\right) $$ Unfortunately, this quantity does not transform as a four-vector should in the presence of sources. The basic reason this is that $$ \partial_\alpha\Theta^{\alpha\beta} = -F^{\beta\lambda}J_\lambda/c \neq 0 $$ and the spatial integrals of $\Theta^{0\alpha}$ yield a four-vector only if the four-divergence of the tensor vanishes identically. To remedy this one needs to add a term $P^{\mu\nu}$ to the stress tensor that takes into account the so-called Poincare stresses of the sources; $$ S^{\mu\nu} = \Theta^{\mu\nu} + P^{\mu\nu} $$ This new tensor does have vanishing four-divergence provided the Poincare stresses are chosen appropriately for the system at hand, and therefore the spatial integrals of the $S^{0\mu}$ are the components of a four-vector. Jackson indicates that the Poincare stresses should be thought of as the contributions to the energy of the system that come from the non-electromagnetic forces necessary to ensure the stability of electric charges.

From this vantage point, the answer to the question is that the extra energy that seems to go missing is the energy present in the sources.

Perhaps this is begging the question in the sense that I have nowhere attempted to write down the Poincare stresses present in the parallel plate capacitor system, but for the time being, I'm more satisfied, and hopefully, Mark, you are too.

BTW see Ch. 16 in Jackson for many more details including the explicit calculation of Poincare stresses for a charged shell of uniform density.

Cheers!

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Energy is a component of a four-vector. It must transform as a four-vector. Imagine putting the capacitor in a black box, which we examine in the rest frame of the box. Coming out of the box are leads that we can use to charge the capacitor. If we charge the capacitor, we increase the mass of the black box according to E = mc^2. But now this is just a box with that extra mass added to it. Its energy must transform the same way as that of a particle. –  Mark Eichenlaub Feb 4 '13 at 5:00
    
I'm not convinced that the energy we are talking about here (that obtained by integrating the energy density $T^{00}$ of the fields) is the time component of a four-vector. In fact, see the bottom of page 607 in Jackson's EM where he writes "the integrals in 12.106 do not appear to have the transformation properties of a 4-vector. For source-free fields they do in fact transform properly, but in general do not." The integrals he is referring to are $\int d^3 x\,T^{00}$ and $\int d^3 x\, T^{0i}$. –  joshphysics Feb 4 '13 at 16:29
    
Which simply indicates that the energy is somewhere else; the black box argument holds. –  Mark Eichenlaub Feb 4 '13 at 19:30
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Great, much clearer, thanks! –  Mark Eichenlaub Feb 5 '13 at 0:08
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Np Mark. Thanks again for this (in my opinion) great question. I must admit I've been thinking about it constantly, and it has really added to my understanding of EM! Also I'm glad you stuck to your guns in the comments. –  joshphysics Feb 5 '13 at 0:14
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The field energy in the capacitor rest frame: $$ E_0 = \frac{1}{8 \pi} \int \left[ \boldsymbol{E}^2+\boldsymbol{B}^2 \right] d^3x$$ does not have the character of the time-component of a 4-vector, as required for an energy. Instead it's the 00 component of the EM stress tensor $\Theta$ times a non-invariant volume element. One cannot apply this formula to boosted frames.

One way to generalize to any frame is to re-write this quantity as the contraction of the stress tensor with a time-like 4-vector: $$ E = \int \Theta^{0 \beta} \eta_\beta \, d^3 \sigma_i$$ where $\eta = (1,0,0,0)$ and the invariant volume element $d^3 \sigma = d^3 x$ in the capacitor rest frame.

In this particular case, an equivalent but more appealing approach is to note that $\boldsymbol{B}=0$ in the capacitor rest frame, so one can write: $$ E = \gamma \int \frac{\boldsymbol{E}^2 - \boldsymbol{B}^2}{8 \pi} d^3 \sigma $$ where $\gamma$ is referenced to the capacitor rest frame (and hence is unity there). The corresponding field momentum is 0 in this frame, again because $\boldsymbol{B}=0$.

The minus sign is unsettling, but the beautiful feature here is that $\left[\boldsymbol{E}^2 - \boldsymbol{B}^2 \right]$ is Lorentz-invariant, so its value at any space-time point in any frame is the same as at that point in the capacitor rest frame, where it is always non-negative because $\boldsymbol{B}=0$.

Combining, one finds: $$E = \gamma m_e c^2$$ where $$m_e = \frac{1}{8 \pi c^2} \int \left[\boldsymbol{E}^2-\boldsymbol{B}^2 \right] d^3\sigma $$ is a frame-invariant electromagnetic mass, just as one would expect for boosts from a rest frame.

Ref Jackson Chapter 17.

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+1: Yeah I saw this in Jackson as well, but I feel that the viewpoint of including Poincare stresses (slightly earlier in Jackson) is a bit more physical and speaks more to the spirit of the question. –  joshphysics Feb 6 '13 at 3:41
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I also came up with this problem a month ago and wrote a post in my blog. But I resolved it in a much different way than any other answers posted here. I'm still not quite sure about my argument here but it seems plausible and interesting to me.

The total field energy in the capacitor's rest frame is

$U=\int \frac{\epsilon_0}{2}E^2dV=\frac{\epsilon_0E^2Ad}{2} $

Now an important point to note is that the capacitor plates are attracting one another, and they cannot simply stay there without crashing into each other. So let's say there's a rigid massless rod between the plates to hold the plates in place.

enter image description here

In the capacitor's rest frame, we can calculate the magnitude forces acting on left and right plates.

$F=\int (\frac{E}{2})\sigma dA=\frac{\epsilon_0 E^2 A}{2}$

In this frame obviously these forces do not provide any work. However in the primed frame, the rod plays a role as an energy transmitter. I mean, first imagine that there is no rod between the plates. Since the plates are attracting each other, the right plate will slow down and the left plate will speed up. Now if there is a rigid rod between them, their velocities will not change at all. In other words, the rod is taking energy from the left plate at a rate $F.v$ and transfer it into the right plate to account for the attraction. But, remember that the energy can’t teleport from one plate to the other plate instantaneously. Thus perhaps some of it has not reach the right plate yet, and still located between the plates. Or we can also say that the rod's mass is increased.

*

In the capacitor's rest frame, we can safely say that the event $1$ “force start acting on the left plate” and event $2$ “force start acting on the right plate” must happen simultaneously due to symmetry.

However, in the primed frame there's a loss of simultaneity. Event $1$ happens $\Delta t=\gamma \frac{vd}{c^2}$ seconds before event $2$. During this time, the rod steals an amount of energy $\Delta U$ from the left plate without paying any energy to the right plate.

$\Delta U=F.v \Delta t=\frac{\gamma \epsilon_0 E^2 Ad}{2} \frac{v^2}{c^2}$

if we take into account this "hidden energy" to the total energy in the primed frame

$U'=U/\gamma+\Delta U=\frac{\epsilon_0E^2Ad}{2\gamma}+\frac{\gamma \epsilon_0 E^2 Ad}{2} \frac{v^2}{c^2}$

$U'=\frac{\gamma\epsilon_0E^2Ad}{2}=\gamma U$


* EDIT:

The arguments starting from * until the horizontal rule can be replaced with an alternate way of viewing as suggested by Larry Harson:

Now suppose that the whole rod suddenly disappear simultaneously in the capacitor's rest frame. Thus the event $1$ “force stop acting on the left plate” and event $2$ “force stop acting on the right plate” must happen simultaneously.

However, in the primed frame there's a loss of simultaneity. Event $1$ happens $\Delta t=\gamma \frac{vd}{c^2}$ seconds before event $2$. During this time, the rod has done extra an amount of work $\Delta U$ to the right plate without returning any energy to the left plate.

$\Delta U=F.v \Delta t=\frac{\gamma \epsilon_0 E^2 Ad}{2} \frac{v^2}{c^2}$

That means the same amount of energy was contained in the rod before disappearance. if we take into account this "hidden energy" to the total energy in the primed frame

$U'=U/\gamma+\Delta U=\frac{\epsilon_0E^2Ad}{2\gamma}+\frac{\gamma \epsilon_0 E^2 Ad}{2} \frac{v^2}{c^2}$

$U'=\frac{\gamma\epsilon_0E^2Ad}{2}=\gamma U$

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What events are we talking about? The capacitor can just sit there and never accelerate at all. We just look at it in a boosted frame. The concept of "not reaching the right plate yet" doesn't seem to make sense if the capacitor is just sitting still, and always has sat still. –  Mark Eichenlaub Feb 26 '13 at 4:54
    
I was't saying that the capacitor is accelerated. I mean the plates are attracting each other through electrostatic forces, so if there is nothing that counters these forces(i.e. a rigid rod) the plates will crash into each other. Now if there is a rod between the plates everything looks the same as in the problem description, I know that we are just viewing it in a moving frame. But the rod plays a role here. In the frame where the capacitor appears to be moving, the rod does work on the capacitor. And the total work turned out to negative, as a result the rod's mass is increased. –  Emitabsorb Feb 26 '13 at 5:20
    
But what events are you talking about? That's what I asked. What event are you referring to with 'event 1 “force start acting on the left plate”'. There is no stopping or starting involved; the capacitor just sits there. –  Mark Eichenlaub Feb 26 '13 at 6:19
    
I think it is not quite a problem, we can always recreate the starting condition. Suppose that initially there is already a rod between the plates, call this rod 1. Then at $t=0$, this rod suddenly disappears and rod 2 suddenly appears to replace it. From the point of view of the rest frame of capacitor, doing this won't change the energy of the system. And thus it must be so in any other frame. In the moving frame, rod 2 does not receive work from the left plate and the right plate simultaneously. Thus some energy is transferred into rod 2. And this energy must be taken into account. –  Emitabsorb Feb 26 '13 at 7:26
    
As best I can tell, you are trying to argue that rod 2 is going to make physical changes after it pops into existence. But if it pops into existence in exactly the same state as rod 1 was, then clearly nothing will change. –  Mark Eichenlaub Feb 26 '13 at 8:30
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