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In GR, the vector potential is defined as $A^\mu$ which is a contravariant vector. Then lowering the indices requires the metric $A_\mu =g_{\mu\nu} A^\nu$, using this vector one defines the field strength as

$F_{\mu\nu}:=\partial_\mu A_\nu -\partial_\nu A_\mu$

or when written the metric explicitly

$F_{\mu\nu}=g_{\nu\lambda}\partial_\mu A^\lambda-g_{\mu\lambda}\partial_\nu A^\lambda +A^\lambda (\partial_\mu g_{\lambda\nu}-\partial_\nu g_{\lambda\mu} )$

Obviously $\frac{\delta F_{\mu\nu}}{\delta g_{\alpha\beta}}\neq 0$

However, when considered the Maxwell action

$S_{\text{Maxwell}}= \int d^4 x\sqrt{-g}\left[-\frac14 F^{\mu\nu}F_{\mu\nu}\right]$

The derived energy-momentum tensor is

$T_{\mu\nu}=-\frac2{\sqrt{-g}}\frac{\delta S_{\text{Maxwell}}}{\delta g^{\mu\nu}}=F^{\mu\lambda}F^\nu_{~~\lambda}-\frac14 g_{\mu\nu} F^{\alpha\beta}F_{\alpha\beta}$

which is the expected one since it corresponds to the one given in flat space. Here $\frac{\delta F_{\mu\nu}}{\delta g_{\alpha\beta}}= 0$ was actually used to derive the above.

So my question is,

Is the field strength dependent or independent with respect to the metric?

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    $\begingroup$ The premise in your first paragraph is false. The em potential is a 1-form field (leaving the difficult geometrical subtleties aside), not a vector field. The field tensor (called Faraday tensor in electrodynamics) is actually a 2-form field. Their definition and relation is independent of a metric on the spacetime manifold. $\endgroup$ – DanielC Oct 9 at 5:19
  • $\begingroup$ Yes that clears the calculation errors Thank you I suspect that one cannot defined a field strength wrt an actual vector? $\endgroup$ – KayS Oct 9 at 6:05

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