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.

• Thanks for trying to clarify. I know that's the conservation law, but I was looking for a deeper explanation than "we've defined it to give us that answer". In Florian Scheck, the expression for the energy-stress tensor is derived for a general Lagrangian density without explicit dependence on x. I don't see anything in his derivation that would lead me to think that the current density should not be included
– nic
Mar 17, 2019 at 17:35
• 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, 2019 at 18:18
• And the bit in the summation being equal to the derivative of the Lagrangian, so if we include the whole lagrangian, it'd all go to 0.
– nic
Mar 17, 2019 at 18:20
• 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 $\nabla_\mu(T^{\mu\nu}_{\rm Maxwell}+T^{\mu\nu}_{\rm matter})=0$. Mar 17, 2019 at 20:24
• Yes, I think you're right as to why the derivation doesn't still hold. I think I may have found a hint of an answer in Jackson to my question. It was towards the end of the chapter. What do you think? "The discussion above focused on the electromagnetic field, with charged particles only mentioned as the soures of the 4-current density. A more equitble tretment of the combined system of particlesand fields involves a Lagrangian having three terms, a free-field Lagrangian, a free particle Lagrangian and an interaction Lagrangian that involves both field and particle degrees of freedom.
– nic
Mar 22, 2019 at 0:42

The form of the field stress tensor does not change, but its value does. The terms that you discuss are part of the matter stress tensor. The total stress tensor is the sum of the field and the matter part.

• I don't quite follow what you're saying. The current is part of the Lagrangian for the fields. If you don't include it when varying the action then you wouldn't get the right equations of motion for the fields. What's different here?
– nic
Mar 17, 2019 at 17:14
• The current is a matter, not an em field phenomenon. Mar 17, 2019 at 17:24