The electromagnetic Lagrangian $L=(8\pi)^{-1}(E^2-B^2)$ vanishes for an EM plane wave. Why doesn't the energy-momentum tensor also vanish?

Question

The electromagnetic field Lagrangian is $$L=(8\pi)^{-1}(E^2-B^2)\,\, ,$$ which vanishes everywhere for a plane wave since the electric and magnetic fields have the same magnitude.

$\quad\quad$ By the formula for the energy-momentum tensor

$$T^{\mu\nu} = -2\frac{\partial L}{\partial g_{\mu\nu}}-g^{\mu\nu}L \quad\quad\quad\quad(*)$$

why isn't the energy-momentum tensor also zero?

One can verify $(*)$ directly using $L=-(16\pi)^{-1}F_{\mu\nu}F^{\mu\nu}$, and indeed one obtains $T^{\mu\nu} = (4π)^{-1} ( -{F^μ}_β F^{νβ} + (1/4) g^{μν} F_{αβ} F^{αβ} )$ which is the correct answer.

Is it true that the energy-momentum tensor is zero in a plane wave? That cannot be correct, since EM plane waves definitely carry energy and momentum.

EDIT: If you're going to downvote the question, at least have the decency to make a useful comment.

Answer 1

The Lagrangian does not vanish everywhere. In the case of a free electromagnetic field, it is a function of the field derivatives $\partial^\mu A^\nu$. From these derivatives, you built the field strength tensor which contains the components of $\vec{E}$ and $\vec{B}$.

The Lagrangian doesn't vanish for arbitrary values of $\vec{E}$ and $\vec{B}$, only for the specific values you get when you solve the equations of motion (i.e. the values that minimize the action, which we call the on-shell configuration.)

But when you vary the action, you have to view the arguments of the Lagrangian (i.e. $\partial^\mu A^\mu$ and thus also $\vec{E}$ and $\vec{B}$) as independent variables and allow them to take on any value.

So while $L$ does vanish on-shell, the derivatives of $L$ don't necessarily.

Answer 2

The other answers are correct that $L=0$ everywhere when the metric equals the Minkowski metric does not imply that the derivative of $L$ with respect to the metric (evaluated on the Minkowski metric) vanishes everywhere. Of course the derivative of $L$ with respect to spatial coordinates is zero, but that's not the relevant derivative in the stress energy tensor.

This answer I want to give a slightly different perspective. As you said, the Lagrangian density is proportional to $E^2-B^2$. Of course both $E^2$ and $B^2$ are non-negative, and in fact since a plane wave is present, both terms will be positive at some points in spacetime. However, the minus sign means that there is a chance for the two terms to cancel; and indeed, in appropriate units, a plane wave in vacuum does have $|E|=|B|$. On the other hand, the energy density ($T^{00}$) is proportional to $E^2+B^2$. Since that isn't zero for a plane wave, and neither term can be negative, and they are being added, the net result must be positive.