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
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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