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Relativistic Lagrangian

  1. Feb 8, 2017 #1
    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. jcsd
  3. Feb 8, 2017 #2

    Orodruin

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    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.
     
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