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A Minimal coupling in general relativity

  1. Apr 16, 2017 #1
    Consider the Einstein-Maxwell action (setting units ##G_{N}=1##),

    $$S = \frac{1}{16\pi}\int d^{4}x\sqrt{-g}\ (R-F^{\mu\nu}F_{\mu\nu})$$

    where

    $$F_{\mu\nu} = \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$

    This describes gravity coupled to electromagnetism. The equations of motion derived from this action are

    $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = 8\pi T_{\mu\nu}$$
    $$\nabla_{\mu}F^{\mu\nu} = 0.$$

    --------------------------------------------------------------------------------------------------------------------------------------------

    Why does the electromagnetic field tensor ##F_{\mu\nu}## reduce to ##\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}## even in curved spacetime?

    Would this not mean that the equation ##\nabla_{\mu}F^{\mu\nu} = 0## would also reduce to ##\partial_{\mu}F^{\mu\nu} = 0## even in curved spacetime?
     
  2. jcsd
  3. Apr 16, 2017 #2

    dextercioby

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    The first why is simply some calculation. Do it and convince yourself.
     
  4. Apr 16, 2017 #3
    Can you help me get started?
     
  5. Apr 16, 2017 #4

    dextercioby

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    What is ##\nabla_{\mu}A_{\nu}## equal to ?
     
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