Studying Einstein's original Die Grundlage der allgemeinen Relativitätstheorie published in 1916's Annalen Der Physik, I came across Equations 66) and 66a) regarding the electromagnetic contribution to the stress-energy-momentum-tensor:

$$ \begin{align} x_{\sigma} = \frac{\partial T_{\sigma}^{\;\nu }}{\partial x_{\nu}} - \frac{1}{2}g^{\tau\mu} \frac{\partial g_{\mu \nu}}{\partial x_{\sigma}}\,T^{\;\nu}_{\tau} \tag{66} \end{align} $$

$$ \begin{align} T^{\; \nu}_{\sigma} = -F_{\sigma \alpha}F^{\nu \alpha} + \frac 14 \delta_{\sigma}^{\; \nu}\; F_{\alpha \beta}F^{\alpha \beta} \tag{66a} \end{align} $$

According to the discussion in the previous pages, by substituting 66a) into 66) you should get three terms, but even before doing the actual computation, it's plain to see that you also get a fourth term which doesn't seem to vanish:

$$ -\frac 18\;F_{\alpha\beta}F^{\alpha\beta}\;g^{\mu\nu}\;g_{\mu\nu,\sigma} $$

Am I missing something?

  • $\begingroup$ If you are using the metric connection the covariant derivative of the metric vanishes. $\endgroup$ Commented Jul 5, 2020 at 21:30
  • $\begingroup$ Actually the whole point of GR is having non-costant g**s so that their derivatives don't necessarily vanish. $\endgroup$ Commented Jul 7, 2020 at 7:45
  • $\begingroup$ I guess you ignored the word covariant, of course the derivative of g is non-zero, but the covariant is, if the connection is the metric connection. Also as you know German and your reference is in german I think you should ask this in the german site. $\endgroup$ Commented Jul 7, 2020 at 15:11
  • $\begingroup$ @DiógenesFigueroa bear with me but I don't see the covariant derivative in the extra term. F is a tensor representing the electromagnetic field components.Unfortunately I don't understand German well enough, I've linked the German source because hopefully it should be the "official" one with the correct equations. $\endgroup$ Commented Jul 7, 2020 at 17:01
  • $\begingroup$ Good thing is Einstein, it wasn't hard to find a free translation. $\endgroup$ Commented Jul 7, 2020 at 21:56

1 Answer 1


Note that:


meaning that:

$\partial_\alpha(g_{\mu\nu}g^{\mu\nu})=2(\partial_\alpha g_{\mu\nu})g^{\mu\nu}=0$

  • $\begingroup$ I see what you mean, I'm not sure that the derivative of the contravariant metric tensor is the same as the derivative of the covariant one though. There might be some extra terms involved. $\endgroup$ Commented Mar 25, 2023 at 22:36

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