2
$\begingroup$

I'm going through my GR notes and came across the following.

gauge transformation

Besides the missing integral sign in the first line I don't get the step. I thought (product rule) $$\delta_\xi (g_{\mu\nu} \dot{x^\mu}\dot{x^\nu})=\delta_\xi(g_{\mu\nu})\dot{x^\mu}\dot{x^\nu}+ 2g_{\mu\nu}\dot{x^\mu}\delta\dot{x^\nu}=\partial_\rho g_{\mu\nu}\delta\dot{x^\rho}\dot{x^\mu}\dot{x^\nu}+2g_{\mu\nu}\dot{x^\mu}\delta\dot{x^\nu}.$$

Why is there a third term ($\delta_\xi(g_{\mu\nu})\dot{x^\mu}\dot{x^\nu}$) in the first line?

Reference:GR Script, Hohm, p.24

$\endgroup$
0

1 Answer 1

1
$\begingroup$

The third term in eq. (183) in Hohm's notes is the (infinitesimal gauge) transformation $\delta_{\xi}g_{\mu\nu}$ of the metric field $g_{\mu\nu}$ itself. In the very next eq. (184) it is concluded that it is the Lie derivative. [All the other terms in eq. (183) are changes due to transformation $\delta_{\xi}x^{\mu}$ of the point particle position $x^{\mu}$.]

$\endgroup$

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.