# Relativistic Lagrangian

Tags:
1. Feb 8, 2017

### bananabandana

1. The problem statement, all variables and given/known data
Show that

$$\mathcal{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu} = - \frac{1}{2}\partial^{\mu}A^{\nu}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})$$

Where $$F^{\mu \nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$$

2. Relevant equations

3. The attempt at a solution
$$\mathcal{L} = -\frac{1}{4} F^{\mu \nu}F_{\mu \nu} = -\frac{1}{4}(\partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu})(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu})$$
Which expands out to:
$$-\frac{1}{4} \bigg( \partial^{\mu}A^{\nu}\partial_{\mu}A_{\nu} - \partial^{\mu}A^{\nu}\partial_{\nu}A_{\mu} -\partial^{\nu}A^{\mu}\partial_{\mu}A_{\nu}+\partial^{\nu}A^{\mu}\partial_{\nu}A_{\mu} \bigg)$$

So if I just exchange indices on half of the terms, and then take out a factor, I get to the result I want... question is, how am I allowed to do that??

2. Feb 8, 2017

### Orodruin

Staff Emeritus
It does not matter what you call summation indices, in general
$$\sum_\mu V_\mu W^\mu = \sum_\nu V_\nu W^\nu.$$
The only difference in your expression is that the summation convention is being used and therefore we do not write out the sums explicitly but always sum over repeated indices.