If one tries to get Einstein's equations $R_{\mu\nu} - \frac12g_{\mu\nu}R = T_{\mu\nu}$ from variational principle it turns out that one can define then covariant form of Energy-Momentum Tensor $T_{\mu\nu}$: $$\delta S_m = \frac12\int d^4x \sqrt{-g} T_{\mu\nu}\delta g^{\mu\nu}.$$ I tried to get expression for $T_{\mu\nu}$ of electromagnetic field. I started from invariant action $$\int d^4x\sqrt{-g}L_{em};\quad L_{em} = -\frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu},$$ where $F_{\mu\nu}$ tensor of electromagnetic field: $$F_{\mu\nu} = \frac{\partial A_{\nu}}{\partial x^\mu}-\frac{\partial A_{\mu}}{\partial x^\nu}.$$ If I choose $F_{\mu\nu}F^{\mu\nu} = F_{\mu\nu}g^{\mu\sigma}g^{\nu\tau}F_{\sigma\tau}$ then variation gives: $$-\frac{1}{16\pi}\int d^4x\sqrt{-g}\Bigl\{2g^{\alpha\beta}F_{\alpha\mu}F_{\beta\nu}-\frac12g_{\mu\nu}F^{\epsilon\zeta}F_{\epsilon\zeta}\Bigl\}\delta g^{\mu\nu}.$$ So I conclude $$T^{(em)}_{\mu\nu} = -\frac{1}{4\pi}\Bigl(g^{\alpha\beta}F_{\alpha\mu}F_{\beta\nu}-\frac14g_{\mu\nu}F^{\epsilon\zeta}F_{\epsilon\zeta}\Bigr).$$ Let us keep in mind that $$T_{\mu\nu}\delta g^{\mu\nu} = -T^{\mu\nu}\delta g_{\mu\nu}.$$ Now the main thing. What if I chose for $F_{\mu\nu}F^{\mu\nu}$ the following expression: $$F_{\mu\nu}F^{\mu\nu} = F^{\mu\nu}g_{\mu\sigma}g_{\nu\tau}F^{\sigma\tau}.$$ By acting same as above we get $$-\frac{1}{16\pi}\int d^4x\sqrt{-g}\Bigl\{2g_{\alpha\beta}F^{\mu\alpha}F^{\nu\beta} + \frac12 g^{\mu\nu}F_{\delta\epsilon}F^{\delta\epsilon}\Bigr\}\delta g_{\mu\nu},$$ which forces me to conclude that $$T^{\mu\nu} = \frac{1}{4\pi}\Bigl(g_{\alpha\beta}F^{\mu\alpha}F^{\nu\beta} + \frac14 g^{\mu\nu}F_{\delta\epsilon}F^{\delta\epsilon}\Bigr).$$ But if I lower indexes I get $$g_{\mu\sigma}g_{\nu\tau}T^{\sigma\tau} = \tilde{T}_{\mu\nu} = \frac{1}{4\pi}\Bigl(g_{\mu\sigma}g_{\nu\tau}g_{\alpha\beta}F^{\sigma\alpha}F^{\tau\beta} + \frac14g_{\mu\nu}F_{\delta\epsilon}F^{\delta\epsilon}\Bigr),$$ which is not $T_{\mu\nu}$ due to first term. What am I wrong? Why covariant components of vector-potential $A$ is more useful here than contravariant?


I believe your error is that $$ \frac{\delta F^{\alpha \beta}}{\delta g_{\mu \nu}} \neq 0. $$ The lowered-index version of the the field-strength tensor is independent of $g_{\mu \nu}$, since it is defined as \begin{align*} F_{\mu \nu} &= 2 \nabla_{[\mu} A_{\nu]} \\ &= \partial_\mu A_\nu + \Gamma^{\rho} {}_{\mu \nu} A_\rho - \partial_\nu A_\mu - \Gamma^{\rho} {}_{\nu \mu} A_\rho \\ &= 2 \partial_{[\mu} A_{\nu]}, \end{align*} since the Christoffel symbols have the property that $\Gamma^{\rho} {}_{\mu \nu} = \Gamma^{\rho} {}_{\nu \mu}$. In other words, the lowered-index version of the field strength is independent of the metric (or, properly speaking, independent of the connection.)

We could try something similar with the raised-index version of the field strength tensor. Viewing $A^{\mu}$ as our fundamental field now, this would be \begin{align} F^{\mu \nu} &= 2 \nabla^{[\mu} A^{\nu]} \\ &= g^{\mu \rho} \nabla_{\rho} A^{\nu} - g^{\nu \rho} \nabla_{\rho} A^{\mu} \\ &= g^{\mu \rho} (\partial_{\rho} A^{\nu} - \Gamma^\nu {}_{\rho \sigma} A^\sigma) - g^{\nu \rho} (\partial_{\rho} A^{\mu} + \Gamma^\mu {}_{\rho \sigma} A^\sigma) \\ &= 2 \partial^{[\mu} A^{\nu]} - 2 g^{\rho [\mu} \Gamma^{\nu]} {}_{\rho \sigma} A^\sigma. \end{align} But there is no reason that the second term in this last equation would vanish; Christoffel symbols do not have any particular symmetry between their first two indices (AFAIK). Thus, if you were to vary the metric (and its associated connection), a term would arise from the variation of the covariant derivative operator implicit in your definition of $F^{\mu \nu}$.

If you were to take this term into account, it would give rise to terms that would be derivatives of $g^{\mu \nu}$ and $A^{\rho}$. I suspect that these terms would combine to form another copy or two of the field strength tensor, and in the end the sign of the aberrant term in your question would work out fine. But the proof is left as an exercise for the questioner. :-)

  • $\begingroup$ I agree. Really, I did not take into account dependence of $F^{\mu\nu}$ on metric tensor. THX. $\endgroup$ – LRDPRDX Nov 12 '16 at 4:40

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