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  1. I saw people write $\frac{\partial( F^{ab} F_{ab})}{\partial g^{ef}}$ as $\frac {\partial (g^{ca}g^{db}F_{cd}F_{ab})}{\partial g^{ef}}$ in a way that exposes the dependence on the metric. but exactly what it means?

  2. So we have $$\frac {\partial (g^{ca}g^{db}F_{cd}F_{ab})}{\partial g^{ef}}=\frac{\partial g^{ca}}{\partial g^{ef}}g^{db}F_{cd}F_{ab}+\frac{\partial g^{db}}{\partial g^{ef}}g^{ca}F_{cd}F_{ab}+\frac{\partial F_{cd}}{\partial g^{ef}}g^{ca}g^{db}F_{ab}+\frac{\partial F_{ab}}{\partial g^{ef}}g^{ca}g^{db}F_{cd}.$$ Is it correct?

  3. $$\frac{\partial g^{ca}}{\partial g^{ef}}=\frac{1}{2}(\delta^c_{e} \delta^a_{f}+\delta^c_{f} \delta^a_{e})$$ Is it correct?

  4. What's happening for $\frac{\partial F_{cd}}{\partial g^{ef}}$?

  5. Also I have no idea about varying a vector field as $N_\mu(x^\nu)$ with respect to metric or$\nabla_\mu N_\nu$ with respect to the metric.

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    $\begingroup$ 1. I don't understand what you want to know. 2. & 3. Check-my-work questions are off-topic 4. & 5. seems to be off-topic as homework-like. $\endgroup$ – ACuriousMind Apr 21 '15 at 10:57
  • $\begingroup$ what is the variation of $N_\mu$ and $\nabla_\nu N_\mu$ with respect to inverse metric? $N_\mu$ is unit time like vector field. i'm confused!!! $\endgroup$ – the_doors Apr 21 '15 at 11:20
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    $\begingroup$ 3. Is answered here: physics.stackexchange.com/q/149066 $\endgroup$ – innisfree Apr 21 '15 at 15:06
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When we vary $F^{ab}F_{ab}$ with respect to the metric, we must also specify what we are holding fixed. Assuming that the context is that of electromagnetism, we consider the four-potential $A_b$ as an independent variable, and therefore under variations of other variables (such as the metric), it is held fixed, as is $F_{ab} = \partial_a A_b - \partial_b A_a$.

But $F^{ab} = g^{ac}g^{bd}F_{cd}$, and it is therefore dependent on the metric. Thus, when varying $$F^{ab}F_{ab} = g^{ac}g^{bd}F_{cd}F_{ab}$$ with respect to the metric while holding $F_{ab}$ fixed, it is convenient to explicitly express the dependence on the metric.

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  • $\begingroup$ so how compute this : $\frac{\partial F_{ab}}{\partial g^{ef}}.$ $\endgroup$ – the_doors Apr 21 '15 at 11:09
  • $\begingroup$ It's zero: there is no dependence of $F_{ab}$ on the metric, for these variations. $\endgroup$ – AV23 Apr 21 '15 at 11:15
  • $\begingroup$ my problem is exactly here , why there is no dependence on metric ?. for example , if $F_{12}=r^4$ and $g^{33}=r^2$ then we will have 2 r. $\endgroup$ – the_doors Apr 21 '15 at 11:38
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    $\begingroup$ We're dealing with partial derivatives here, not full derivatives. $\endgroup$ – AV23 Apr 21 '15 at 12:15
  • $\begingroup$ so is this correct : $\frac{\partial (N^\mu N_\nu)}{\partial g^{\rho\zeta}} = \frac{\partial g^{\mu\sigma}}{\partial g^{\rho\zeta}} (N_\sigma N_\nu) $ $\endgroup$ – the_doors Apr 21 '15 at 13:02

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