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## Homework Statement

Show one can obtain Maxwell's equations from $$\partial^{\mu} F_{\mu \nu} = 0\,\,\,; \,\, \partial_{\mu}F_{\nu \rho} + \partial_{\nu} F_{\rho \mu} + \partial_{\rho}F_{\mu \nu} = 0,$$ where ##F_{ij} = \epsilon_{ijk}B_k## and ##F_{i0} = E_i## with ##F_{\mu \nu} = - F_{\nu \mu}##.

From these two equations, show that one also has ##\partial^{\mu}T_{\mu \nu} = 0##, where ##T_{\mu \nu} = F_{\mu \rho}F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}F^{\alpha \beta}##

## Homework Equations

All given in problem statement

## The Attempt at a Solution

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I've done the first part of the question.

My working for the second part is shown below. Insert ##T_{\mu \nu}## and contract with ##\partial^{\mu}##. This gives

$$(\partial^{\mu}F_{\mu \rho}) F_{\nu \sigma} \eta^{\rho \sigma} + F_{\mu \rho}(\partial^{\mu} F_{\nu \sigma}) \eta^{\rho \sigma} - \frac{1}{4}\eta_{\mu \nu}(\partial^{\mu}F_{\alpha \beta})F^{\alpha \beta} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}(\partial^{\mu}F^{\alpha \beta})$$

The first term is zero by Maxwell eqn. In the last term, lower the index on the ##\partial^{\mu}## term and then use the fact that ##\eta_{\mu \nu} \eta^{\mu \gamma} = \delta^{\gamma}_{\nu}## to reduce the last term to ##\frac{1}{4} F_{\alpha \beta} \partial_{\nu} F^{\alpha \beta}##. The same analysis can be applied on third term. Then use the fact that ##F^{\alpha \beta} = F_{\rho \sigma} \eta^{\alpha \rho} \eta^{\beta \sigma}##. The third and fourth terms are then identical and the whole expression reduces to $$F_{\mu \rho} \eta^{\mu \alpha} \partial_{\alpha} F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta} = F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$. I'm unsure of a next step. Possibly rearrange ##\partial_{\alpha}F_{\nu \sigma}## using the second Maxwell eqn above but not sure if this would help.

Many thanks.

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