# Variation of the action in a non-flat spacetime

In Padmanabhan's Quantum Field Theory, The Why, What and How, in chapter $$3$$, section $$3.1.6$$, in the paragraph between equations ($$3.75$$) and ($$3.76$$), he states regarding the Belinfante tensor that:

On the other hand, we can easily determine$$^{28}$$ the energy momentum tensor by introducing a $$g^{ab}$$ in to the electromagnetic action and varying with respect to it.

Consider the usual free Maxwell field in an arbitrary spacetime, $$\mathcal{S} = -\frac{1}{4}\int d^{4}x \sqrt{-g}\ F^{\mu\nu}F_{\mu\nu}$$ Varying this with respect to the metric gives me, $$\delta\mathcal{S} = -\frac{1}{4}\int d^{4}x \Big(\delta(\sqrt{-g})F^{\mu\nu}F_{\mu\nu} + 2\sqrt{-g}\ F^{\mu\nu}F^{\alpha\beta}g_{\alpha\mu}\delta g_{\beta\nu}\Big)$$

The determinant gives me $$\delta\sqrt{-g} = \sqrt{-g}\underline{g}^{-1}\delta \underline{g}/2 = \sqrt{-g}g^{\mu\nu}\delta g_{\mu\nu}/2,$$ where the underline represents a matrix. This gives me, $$\delta\mathcal{S} = -\frac{1}{4}\int d^{4}x \sqrt{-g}\ \delta g_{\beta\nu} \Big(\frac{1}{2}g^{\beta\nu}F^{\mu\alpha}F_{\mu\alpha}+ 2F^{\mu\nu}F^{\alpha\beta}g_{\alpha\mu}\Big) = -\frac{1}{2}\int d^{4}x \sqrt{-g}\ \delta g_{\beta\nu} \Big(F^{\beta\mu}F^{\nu}_{\phantom{-}\mu}+\frac{1}{4}g^{\beta\nu} F^{\mu\alpha}F_{\mu\alpha}\Big)$$ This would mean that the Belinfante tensor comes out to be, $$T^{\beta\nu}=F^{\beta\mu}F^{\nu}_{\phantom{-}\mu}+\frac{1}{4}g^{\beta\nu} F^{\mu\alpha}F_{\mu\alpha}$$ this is wrong since there should be a relative sign difference in the two terms when written this way. Here, I took the variation with respect to $$g_{\mu\nu}$$ instead of $$g^{\mu\nu}$$ (which does have the relative sign as $$\sqrt{-g}g^{\mu\nu}\delta g_{\mu\nu}/2 = - \sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}/2$$).

This suggests that the metric variation must always be with respect to $$g^{\mu\nu}$$ - is this correct? And if so, what is the reason? In flat spacetime, this probably does not matter, but this seems to be an issue in curved spacetimes.

On a similar note - does this hold for other fields as well (like $$A_{\mu}$$)? I was looking at a paper (I asked a question on it here), and one of the equation seems to be different depending whether the variation is with respect to $$A_{\mu}$$ or $$A^{\mu}$$ - I am unsure if it was my mistake or something else.

Any help would be appreciated!

• Did you make sure to use $\delta g^{\mu\nu} = \boldsymbol{-} g^{\mu\alpha} g^{\nu\beta} \delta g_{\alpha\beta}$? Commented Jan 23 at 13:51
• Also, in your variation, it looks like you assumed that $F^{\mu\nu}$ does not depend on the metric. However, it is $F_{\mu\nu}$ that is independent of $g$ and $F^{\mu\nu} = g^{\mu\alpha} F^{\nu\beta} F_{\alpha\beta}$. Commented Jan 23 at 13:53
• @stringynonsense Yeah, I took care of that negative (it is the reason for the relative sign when doing the variation with respect to $g^{\mu\nu}$). I feel like you are trying to say that the first variation is not allowed because $F_{\mu\nu}$ is the "proper" quantity (independent of the metric) and the first method disregards that - is my understanding correct? In that case, though, I wonder why $F_{\mu\nu}$ is the one we consider - if I am not mistaken, we usually identify $A^{\mu}$ as the potential leading to the $E,\ B$ fields (and not $A_{\mu}$). Commented Jan 23 at 15:00
• It has nothing to do with $A^\mu$ vs $A_\mu$. $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ whereas $F^{\mu\nu} = \nabla^\mu A^\nu - \nabla^\nu A^\mu$. The former has only partial derivatives (no dependence on metric). The latter covariant derivatives which depends on the metric through the Christoffel symbols. Commented Jan 23 at 15:28
• @stringynonsense So, if I understand this right, we vary with respect to $g^{\mu\nu}$ because of the indepence of $F_{\mu\nu}$ on the metric? The same thing does not happen when varying with $A^{\mu}$ or $A_{\mu}$, I guess, and both are equivalent? Commented Jan 23 at 15:43