It is well known that in general relativity, the metrics $g_{\mu \nu}$ and $g_{\mu \nu} + \epsilon L_\xi g_{\mu \nu}$ are physically equivalent, where $L_\xi g_{\mu \nu}$ is the Lie derivative of the metric with respect to a vector field $\xi$ and $\epsilon$ is an infinitesimal number. The reason for that is that one can view $g_{\mu \nu}$ and $g_{\mu \nu} + \epsilon L_\xi g_{\mu \nu}$ as differing only by a coordinate change. So, when solving the Einstein equations, we notice that they have some degeneracy and we often eliminate that degeneracy by fixing $\xi ^{u}$ (for example, we can put it equal to zero).

There are occasions though, where we eliminate the degeneracy by fixing some component of the metric. I was wondering if there's a geometric way of understanding this form of gauge fixing (perhaps in terms of diffeomorphism invariance).

  • $\begingroup$ How else are you gonna "fix the metric" if not by using the (gauge) freedom the diffeomorphism invariance grants? $\endgroup$ – ACuriousMind May 21 '15 at 12:36
  • $\begingroup$ @ACuriousMind What I outlined above was a way of seeing how fixing a certain vector $\xi^{\mu}$ eliminates the freedom in considering the diffeomorphism transformation $g_{\mu \nu}$ $\rightarrow$ $g_{\mu \nu} + \epsilon L_\xi g_{ \mu \nu} $. When you fix a metric component, you eliminate the freedom in doing exactly which diffeomorphism transformation? $\endgroup$ – LeastSquare May 21 '15 at 12:45
  • $\begingroup$ That you can find out yourself - look at how the metric transforms under an arbitrary diffeomorphism, and then just impose the condition you want to fix and see if you can solve for the diffeomorphism. $\endgroup$ – ACuriousMind May 21 '15 at 13:05
  • $\begingroup$ @ACuriousMind thanks. Perhaps in practice that might be a little hard to do. I'll see if I can do some examples. $\endgroup$ – LeastSquare May 21 '15 at 13:21

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