- #1
spaghetti3451
- 1,344
- 33
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
$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = 8\pi T_{\mu\nu}$$
$$\nabla_{\mu}F^{\mu\nu} = 0.$$
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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?
$$S = \frac{1}{16\pi}\int d^{4}x\sqrt{-g}\ (R-F^{\mu\nu}F_{\mu\nu})$$
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
$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = 8\pi T_{\mu\nu}$$
$$\nabla_{\mu}F^{\mu\nu} = 0.$$
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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?