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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
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@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

4 Answers 4

up vote 5 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
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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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This is correct, but hard to follow, especially without a copy of Jackson open in front of you. The reference to Poincare stresses also seems to me to unnecessarily obscure; that term has connotations of referring to specific confusions in electrodynamics ca. 1900, i.e., pre-Einstein. In the present context there is nothing mystical about these stresses; they are simply the mechanical stresses that allow the capacitor to hold its shape. I've written an answer that I hope demystifies all of this a little bit. –  Ben Crowell Nov 2 at 18:15

My result is that there is no missing energy; the correct answer for the field energy $\mathcal{E}'$ in the boosted frame relative to that in the rest frame $\mathcal{E}$ is indeed: $$ \mathcal{E}'= \gamma \mathcal{E} \,\,\, , \, \gamma=1/\sqrt{1-(v/c)^2}$$

For simplicity, I assume massless capacitor plates, held fixed in some fashion (say by a rigid frame) so they can't accelerate. (The usual electrostatic setup.)

I want to form the 4-momentum for this ideal capacitor as the contraction of the stress tensor with the capacitor 4-velocity $n_\nu$, integrated over a 3-volume of simultaneity defined by $n_\nu$:

$$ p^\mu = \int \Theta^{\mu \nu} n_\nu \, d^3 \sigma $$

where $n_\nu = (1,0,0,0)$ in the rest frame and the invariant volume element $d^3 \sigma$ is defined such that $d^3 \sigma = d^3x$ in the capacitor rest frame.

Then the energy seen by an observer with 4-velocity $u^\mu$ is just $$ \mathcal{E}=p^\mu u_\mu $$

Our observer will always be in the capacitor rest frame, so $ u^\mu=(1,0,0,0) $ and $ \mathcal{E}=p^0 $.


A seemingly fatal objection at this point, from Ben Crowell, is that one cannot form a conserved 4-momentum solely from the electromagnetic stress-energy tensor $\Theta$ because it is not divergence-free.

[For a divergence-free tensor field, one can show that the integral of $\Theta$ over the 3-dimensional surface of a large 4-dimensional volume (chosen to contain the system in question) vanishes, by converting the surface integral into a 4-volume integral of the divergence of $\Theta$ via Gauss' theorem.]

This objection is certainly valid in general, but for the case of an electrostatic system (like this one), where the conductors' positions' are all fixed relative to one another, and stay that way, it can be finessed, as follows.

The electromagnetic stress tensor divergence is: $$ \partial_\alpha \Theta^{\alpha \beta} = - F^{\beta \lambda} J_\lambda = - \gamma \left( \boldsymbol{J \cdot E} , \rho \boldsymbol{E} + \boldsymbol{J \times B} \right) $$ where $\rho$ is the charge density, $\boldsymbol{J}$ the current, $\boldsymbol{E}$ the electric field, and $\boldsymbol{B}$ the magnetic field.

Crucially, for an electrostatic system like this one, there is a privileged frame in which all the elements are at rest, and so in which $\boldsymbol{J}$ and $\boldsymbol{B}$ are both 0.

In fact, if $\boldsymbol{v}$ is the velocity of an arbitrary frame relative to the privileged one, we can write: $$ \boldsymbol{J}= \rho \boldsymbol{v} \, \, , \, \boldsymbol{B} = \boldsymbol{v \times E}/c^2 $$

The 4-volume integral over the divergence is then: $$ \int{dt \, d^3x \, \partial_\alpha \Theta^{\alpha \beta} } = - \gamma \int{dt\, d^3x \, \left( \boldsymbol{v \cdot} \rho \boldsymbol{E} , \rho \boldsymbol{E} + \boldsymbol{v \times} \left( \boldsymbol{v \times} \rho \boldsymbol{E} \right)/c^2 \right)} $$

Now $\rho \boldsymbol{E}$ is just the electric force on a charge element $\rho d^3x$, so for an isolated electrostatic system, (in which the sum of the electric forces over a finite region of the rest frame = 0), we can choose our 4-volume to fully enclose that region over the period of interest, and then the spatial integral of $\rho \boldsymbol{E}$ vanishes, and so the four-integral vanishes as well. (The same applies to a boosted frame, where the field components parallel to the boost are unchanged, while the perpendicular elements are all scaled by $\gamma.$)

We can then use Gauss' theorem to establish the conservation of the 4-momentum $p^\mu$ for the 3-volumes bounding the 4-volume.


With conservation established, we proceed to calculate the energy.

For simplicity, assume the only field in the rest frame is the constant electric field $E$ between the capacitor plates (of area $A$, separation $d$). (A more elaborate calculation gives the same result if fringing fields are included, or indeed for any isolated electrostatic system.) In the rest frame the energy $\mathcal{E}$ is: $$\mathcal{E}= \frac{E^2}{8 \pi} Ad$$

Because the boost is parallel to the field, the only field in the boosted frame is the same constant electric field. Therefore in the boosted frame, where the capacitor 4-velocity is $ \eta=\gamma(1,v/c,0,0) $ the field energy $\mathcal{E}'$ is:

$$ \mathcal{E}' = \int {\Theta^{0\nu} n_\nu d^3 \sigma } = \int { \frac{E^2}{8 \pi} \gamma \left( \gamma d^3x \right) } = \gamma^2 \frac{E^2}{8 \pi} \int { d^3x } = \gamma^2 \frac{E^2}{8 \pi} \left( A \frac{d}{\gamma} \right) = \gamma \frac{E^2}{8 \pi} Ad = \gamma \mathcal{E} $$

as claimed.

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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
    
I don't think this is right. What you claim to prove here is actually false. If one considers, as you do, only the electromagnetic field's contribution to the stress-energy tensor, then the stress-energy tensor has a nonvanishing divergence. For such a stress-energy tensor, it is not true that the energy-momentum obtained by integrating over a surface of simultaneity transforms like a four-vector. The proof of that transformation property depends on two things: (1) the divergence-free property of $T$ (which you're violating), and (2) the fact that it's an isolated system (which you never use). –  Ben Crowell Nov 2 at 18:53
    
@BenCrowell, thank you for this comment (and sorry for my earlier snark). I think there is a way around the divergence-issue for electrostatic systems, which I have added to my answer (in a major re-write). I hope you can take a look. Thanks again. –  Art Brown Nov 5 at 17:46

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

This is a nice example of one of the foundational issues in relativity: how do we know that energy-momentum transforms like a four-vector, or, essentially, how do we know that $E=mc^2$? A historical overview is given in [Ohanian 2008] and [Ohanian 2009]. As Ohanian points out, there are logical problems if one tries to do what Einstein did in 1905 and prove this without making use of the stress-energy tensor $T^{\mu\nu}$. In relativity, we always assume the following properties of the stress-energy tensor:

  1. $T$ is a sum over contributions from every "matter field" that is present. This includes all fields except for the gravitational field.

  2. Each term in this sum is directly observable (so, e.g., it can't depend on a choice of gauge).

  3. $T$ is a symmetric tensor, $T^{\mu\nu}=T^{\nu\mu}$.

  4. Local conservation of energy momentum is expressed by the vanishing of the four-divergence $\partial T^{\mu\nu}/\partial x^\mu=0$ (with the Einstein summation convention for $\mu$).

From these assumptions, one can prove that for an isolated system, the total energy-momentum vector $p^\mu=\int T^{\mu\nu} dS_\nu$ is conserved and transforms like a four-vector. Here the integral is over a hypersurface of simultaneity according to a particular observer, and $dS_\nu$ is the three-volume covector. For proofs, see section 9.3.4 of my SR book.

For the electromagnetic field, which is massless, we also expect that:

  1. The stress-energy tensor is traceless, $T^\mu_\mu=0$.

With these fundamentals in place, it becomes straightforward to analyze the problem of the capacitor.

enter image description here

In figure 1, there is pressure trying to make the capacitor plates explode laterally, and also tension trying to make them collapse together against each other. From the assumptions above, one can show that in this example, in the rest frame of the capacitor, the electromagnetic field's stress-energy tensor has components $T_{(em)}^{tt}=(1/8\pi k)E^2$ (energy density) and $T_{(em)}^{yy}=-(1/8\pi k)E^2$ (tension in the $y$ direction, parallel to the field). It's easy to see that this has a nonvanishing four-divergence, since $\partial T_{(em)}^{yy}/\partial y\ne0$ at the plates, and there are no other terms in the stress-energy tensor that could compensate for this.

There is nothing surprising here; only the total stress-energy tensor $T$ has to be divergenceless, not $T_{(em)}$. It would violate the laws of physics if the capacitor were to remain in equilibrium like this without some force to counter the electromagnetic tension. Let's say that this force is provided by a spring, as in figure 2. The spring has its own contribution $T_{(s)}$ to the stress-energy. For convenience, let's imagine making the spring filled in (rather than a hollow cylinder) and fattening it up so that it fills the entire interior volume of the capacitor. Then to achieve static equilibrium in the rest frame, we need the pressure in the spring to cancel out the pressure in the electric field. We therefore have $T^{yy}=0$ for the total stress-energy tensor.

If we now apply the tensor transformation law to the stress-energy tensor, we find that the stress-energy tensor in the boosted frame contains a mass-energy density $T^{t't'}$ that depends only on $T^{tt}$ and $T^{yy}$. (There also has to be an xx component to keep the plates from exploding laterally, but that doesn't enter here.) But we have $T^{yy}=0$, so the problem is exactly the same as transforming a lump of nonrelativistic matter, and we know that that calculation comes out OK.

[Rindler 1988] shows that this still works out if we drop the simplifying assumption that the spring fills the entire interior volume of the capacitor.

References

Ohanian, "Einstein's $E = mc^2$ mistakes," 2008, http://arxiv.org/abs/0805.1400

Ohanian, "Einstein's Mistakes: The Human Failings of Genius," 2009

Rindler and Denur, "A simple relativistic paradox about electrostatic energy," Am J Phys 56 (1988) 9.

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+1: This helped me understand my own answer better. Thanks especially for the references. –  joshphysics Nov 3 at 16:04

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