Where's the energy in a boosted capacitor?

Question

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?

Answer 1

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!

Answer 2

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:

5. 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.

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.

Answer 3

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.

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

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

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

Answer 4

Let's accelerate a charged plate capacitor by pulling one of it's plates while carefully adjusting the pulling force so that:

A) The distance between plates stays constant in the capacitor frame

B) The distance between plates stays constant in the capacitor's original frame

What are the energies of the E-fields? (not counting kinetic energy)

Case A: $\frac{1}{\gamma} E$

Case B: $E$

Now let's say that in both cases we used energy E to pull the plate, and let's go to capacitor frame and see how the pulling operation looks like from there:

Case A: Energy k*E is used to accelerate the plates

Case B: Energy k*E is used to pull the plates apart and to accelerate the plates

Now it should be clear that capacitor has higher velocity in case A. And there we have an answer to the question "were is the missing energy in case A". Kinetic energy of capacitor plates is what became of the missing energy.

Answer 5

It's actually very simple: Energy in motion has kinetic energy.

If energy is E in the frame where the energy is not moving, then in the frame where the energy is moving the energy is $\gamma E$

People here have calculated that the energy of a very fast moving E-field without the kinetic energy is very small. OK, but the energy of a very fast moving E-field with the kinetic energy is still $\gamma E$

Let me point out that a fast moving object emits light that is time dilated, so that the energy is $\frac{1}{\gamma} E$. That energy is the energy of the light in the same way as $\frac{1}{\gamma} E$ is the energy of a moving E-field. The light and the E-field have total energy $\gamma E$