Derivative of metric and log identity

robousy
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Has anyone seen this identity:

g^{ab}\nabla g_{ab}=\nabla ln|g|

I've seen it used, but want to figure out where it comes from.

Does anyone know a name or have any ideas??
 
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You will have to show the exact formula. The way it's written, it doesn't make sense - covariant derivative of the metric is zero in GR because the connection is chosen 'metric compatible'.
 
ok, thanks, I'll check the paper when I get back to my office tomorrow and post.

Rich
 
robousy said:
Has anyone seen this identity:

g^{ab}\nabla g_{ab}=\nabla ln|g|

I've seen it used, but want to figure out where it comes from.

Does anyone know a name or have any ideas??

I think you mean

g^{\alpha \beta} \partial_\mu g_{\alpha \beta} = \partial_\mu \ln \left|g\right|.

See pages 12-13 of Poisson.
 
Thanks Greorge. I Don't have that book but I'll see if I can find someone with it. And thanks for pointing out the correction.
 
Lots of have books probably have this, but I won't be able to tell you any others until Monday.

Maybe Carroll.
 
robousy said:
Has anyone seen this identity:

g^{ab}\nabla g_{ab}=\nabla ln|g|

I've seen it used, but want to figure out where it comes from.

Does anyone know a name or have any ideas??

Differentiate the matrix identity

ln[(det.M)] = Tr[(ln M)]

and put

M = g_{\mu \nu}


regards

sam
 
I'm not seeing it Sam.

\partial_\mu ln|g^{ab}|=\partial_\mu ln[(det.g^{ab})]=\partial_\mu Tr[ln g^{ab} ] =<br /> \partial_\mu (ln g^{00}+lng^{11}+...)<br /> <br />
 
robousy said:
I'm not seeing it Sam.

\partial_\mu ln|g^{ab}|=\partial_\mu ln[(det.g^{ab})]=\partial_\mu Tr[ln g^{ab} ] =<br /> \partial_\mu (ln g^{00}+lng^{11}+...)<br /> <br />


\partial (ln |G|) = Tr ( \partial ln G) = (G^{-1} \partial G)^{\mu}_{\mu}

thus

\partial (ln|g|) = g^{\mu\nu}\partial g_{\mu\nu}
 
  • #10
robousy said:
Thanks Greorge. I Don't have that book but I'll see if I can find someone with it. And thanks for pointing out the correction.

http://books.google.com/books?id=v-Uw_uzbw7EC
search the book for: ln (natural log)
 
  • #11
samalkhaiat said:
robousy said:
\partial (ln |G|) = Tr ( \partial ln G) = (G^{-1} \partial G)^{\mu}_{\mu}

thus

\partial (ln|g|) = g^{\mu\nu}\partial g_{\mu\nu}

Great, got it! Thanks a bunch.
 
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