# Electromagnetic energy stress tensor with non zero current

I have read through Jackson and also Classical field theory (Florian Scheck) on this topic and neither of them address my question. In both of them we are told that the canonical stress energy tensor is given by

$$T^{\alpha \beta}=\frac{\partial \mathscr{L}}{\partial\left(\partial_{\alpha} A^{\lambda}\right)} \partial^{\beta} A^{\lambda}-g^{\alpha \beta} \mathscr{L}$$

We go on and derive it for a source free environment so the Lagrangian density contains no current term. From this we also derive the symmetric stress energy tensor. But, when we move to the situation where there is a source current, the symmetric stress energy tensor is exactly the same. What confuses me about this is that it seems that they do not include the current from $$\mathscr{L}=-\frac{1}{16 \pi} F_{\alpha \beta} F^{\alpha \beta}-\frac{1}{c} J_{\alpha} A^{\alpha}$$.

What I'm expecting is

$$\Theta^{\alpha \beta}=\frac{1}{4 \pi}\left(g^{\alpha \mu} F_{\mu \lambda} F^{\lambda \beta}+\frac{1}{4} g^{\alpha \beta} F_{\mu \lambda} F^{\mu \lambda}\right)+\frac{g^{\alpha\beta}}{c}J_{\mu}A^{\mu}$$

but what they're saying is

$$\Theta^{\alpha \beta}=\frac{1}{4 \pi}\left(g^{\alpha \mu} F_{\mu \lambda} F^{\lambda \beta}+\frac{1}{4} g^{\alpha \beta} F_{\mu \lambda} F^{\mu \lambda}\right)$$

What I've done so far is try to show that extra term goes to zero once differentiated so it doesn't effect the conservation laws but haven't been unsuccessful.

What my2cts is saying is that the Maxwell stress tensor is defined so that Maxwell's equations give $$\nabla_\mu {T^{\mu}}_\nu = J^\mu F_{\mu\nu}$$ (some signs depend on metric conventions), the RHS being the Lorentz force on the charged matter. So the $$J^\mu$$ is not included in the stress tensor.
• Thinking about his derivation again, I would even say that if we did include the extra current term in the Lagrangian, then the modified stress-energy tensor would again be zero when we take its divergence (can't actually show it though). This is because $\partial_{\alpha} T^{\alpha \beta}=\sum_{k}\left[\frac{\partial \mathscr{L}}{\partial \phi_{k}} \partial^{\beta} \phi_{k}+\frac{\partial \mathscr{L}}{\partial\left(\partial_{\alpha} \phi_{k}\right)} \partial^{\beta}\left(\partial_{\alpha} \phi_{k}\right)\right]-\partial^{\beta} \mathscr{L}$ – nic Mar 17 '19 at 18:18
• When the current $J^\mu$ is external it does depend explicitly on $x$ through. If $J^\mu$ comes from another field, such as a charged scalar, then the matter part gives rise to a $T^{\mu^\nu}_{\rm matter}$ and the entire conservation law is $\nabla_\mu(T^{\mu\nu}_{\rm Maxwell}+T^{\mu\nu}_{\rm matter})=0$. – mike stone Mar 17 '19 at 20:24