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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.

Qmechanic
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the_doors
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1 Answers1

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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.

AV23
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