# Conservation of electromagnetic energy-momentum tensor

• A
I'm trying to show that $\partial_\mu T^{\mu \nu}=0$ for

$$T^{\mu \nu}=F^{\mu \lambda}F^\nu_{\; \lambda} - \frac{1}{4} \eta^{\mu \nu} F^{\lambda \sigma}F_{\lambda \sigma},$$

with the help of the electromagnetic equations of motion (no currents):

$$\partial_\mu F^{\mu \nu}=0,$$
$$\partial_\mu F_{\nu \lambda}+\partial_\nu F_{\lambda \mu}+\partial_\lambda F_{\mu \nu}=0.$$

Applying the partial derivative directly gives us

$$\partial_\mu T^{\mu \nu} = F^\nu_{\; \lambda} \partial_\mu F^{\mu \lambda}+F^{\mu \lambda} \partial_\mu F^\nu_{\; \lambda}-\frac{1}{4} \eta^{\mu \nu}F_{\lambda \sigma} \partial_\mu F^{\lambda \sigma}-\frac{1}{4} \eta^{\mu \nu}F^{\lambda \sigma} \partial_\mu F_{\lambda \sigma}.$$

The first term on the right-hand side drops out due to the first equation of motion. The last two terms can also be combined, but I'm not sure if it would be right to do this immediately as right now I have three terms with three partial derivatives. So my instinct was to try and put them in a form where I could use the second equation of motion. Unfortunately I couldn't see how to do that, especially with the $\frac{1}{4}$ factor on the last two terms. I then did what every physics student does and played with the expression in hopes of finding something simpler. I got a simpler expression, however I could not see how to proceed from my simpler expression to showing what I want to show. I will post my general thought process below, and would appreciate any help on proceeding beyond my final step.

The first thing I did was lower all indices on the tensors being differentiated, as that is in accordance with the form of my second equation of motion. So we got

$$\partial_\mu T^{\mu \nu} =\eta^{\nu \sigma} F^{\mu \lambda} \partial_\mu F_{\sigma \lambda}-\frac{1}{4} \eta^{\mu \nu} \eta^{\alpha \lambda} \eta^{\beta \sigma} F_{\lambda \sigma} \partial_\mu F_{\alpha \beta}-\frac{1}{4} \eta^{\mu \nu}F^{\lambda \sigma} \partial_\mu F_{\lambda \sigma}.$$

I next noticed that if I lowered all indices on the field tensors then each of my terms would have three metric factors, which looked more promising if I wanted to try and factor things out, so I got

$$\partial_\mu T^{\mu \nu} =\eta^{\nu \sigma} \eta^{\alpha \mu} \eta^{\delta \lambda} F_{\alpha \delta} \partial_\mu F_{\sigma \lambda}-\frac{1}{4} \eta^{\mu \nu} \eta^{\alpha \lambda} \eta^{\beta \sigma} F_{\lambda \sigma} \partial_\mu F_{\alpha \beta}-\frac{1}{4} \eta^{\mu \nu} \eta^{\alpha \lambda} \eta^{\beta \sigma} F_{\alpha \beta} \partial_\mu F_{\lambda \sigma}.$$

I initially thought that if I replace the dummy indices properly I might get something where I can factor out the metric factors, however since $\nu$ is not a dummy index I can't move it around freely, and it doesn't look like it's going to be very easy to factor out the metric factors. I therefore changed my plan to simplify the expression to see if I could get something that I could make sense of. Rather than focusing on making the metric part factorable, I tried to change the $F \partial F$ part into a form that might be factorable. In the first term I made the following replacements: $\sigma \leftrightarrow \alpha,$ $\lambda \leftrightarrow \beta.$ The second term was left alone. On the last terms I interchanged $F_{\alpha \beta} \leftrightarrow F_\lambda \sigma$ which I think is okay since both indices are contracted over. All of this gave me

$$\partial_\mu T^{\mu \nu} =\eta^{\nu \alpha} \eta^{\sigma \mu} \eta^{\delta \beta} F_{\sigma \delta} \partial_\mu F_{\alpha \beta}-\frac{1}{4} \eta^{\mu \nu} \eta^{\alpha \lambda} \eta^{\beta \sigma} F_{\lambda \sigma} \partial_\mu F_{\alpha \beta}-\frac{1}{4} \eta^{\mu \nu} \eta^{\alpha \lambda} \eta^{\beta \sigma} F_{\lambda \sigma} \partial_\mu F_{\alpha \beta}.$$

Combining the last two terms and factoring out what I can gave me

$$\partial_\mu T^{\mu \nu} = \eta^{\sigma \beta} F_{\lambda \sigma} \partial_\mu F_{\alpha \beta} \left( \eta^{\nu \alpha} \eta^{\lambda \mu} - \frac{1}{2} \eta^{\mu \nu} \eta^{\alpha \lambda} \right).$$

I can't see how to simplify this expression to zero, so I assume I made a wrong turn during my calculation somewhere. However, I can't see where. How do I use the equations of motion to show that the energy-momentum tensor is conserved here?

Orodruin
Staff Emeritus