Problem 2.1(b) in Peskin and Schroeder's Introduction to QFT

Question

In this exercise the author claims that adding $\partial_\sigma K^{\sigma \mu \nu}$ does not affect the divergence of $T^{\mu\nu}$. In other words the author claims that $\partial_\mu \partial_\sigma K^{\sigma \mu \nu}$. To this end, I found:

$\partial_\sigma K^{\sigma \mu \nu} = \partial_\sigma(F^{\mu \sigma}A^\nu) = F^{\mu \sigma}(\partial_\sigma A^\nu) + A^\nu \partial_\sigma F^{\mu\sigma}$. Since $\partial_\sigma F^{\mu\sigma} = 0$ in free space, we have $\partial_\sigma K^{\sigma \mu \nu} = F^{\mu \sigma}(\partial_\sigma A^\nu)$.

Thus we see that $\partial_\mu\partial_\sigma K^{\sigma \mu \nu} = \partial_\mu(F^{\mu \sigma}(\partial_\sigma A^\nu)) = (\partial_\mu F^{\mu\sigma})(\partial_\sigma A^\nu) + F^{\mu\sigma} \partial_\mu\partial_\sigma A^\nu = -(\partial_\mu F^{\sigma\mu})(\partial_\sigma A^\nu) + F^{\mu\sigma} \partial_\mu\partial_\sigma A^\nu = F^{\mu\sigma} \partial_\mu\partial_\sigma A^\nu$.

However, I am not sure why $F^{\mu\sigma} \partial_\mu\partial_\sigma A^\nu = 0$. Can explain with a derivation?

Answer 1

The contraction of a symmetric and anti-symmetric term is zero, and $\partial_\mu\partial_\sigma$ is symmetric (since physical functions are usually assumed smooth).