1
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ If you are using the metric connection the covariant derivative of the metric vanishes. $\endgroup$ Jul 5 '20 at 21:30
  • $\begingroup$ Actually the whole point of GR is having non-costant g**s so that their derivatives don't necessarily vanish. $\endgroup$ Jul 7 '20 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. $\endgroup$ Jul 7 '20 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. $\endgroup$ Jul 7 '20 at 17:01
  • $\begingroup$ Good thing is Einstein, it wasn't hard to find a free translation. $\endgroup$ Jul 7 '20 at 21:56
0
$\begingroup$

Note that:

$g_{\mu\nu}g^{\mu\nu}=4$

meaning that:

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

$\endgroup$
1
  • $\begingroup$ I wish it was that simple. I double checked that it's the metric tensor multiplied by its derivative. $\endgroup$ Jul 9 '20 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.