Derivative of metric and log identity

[robousy]

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??

 

[smallphi]

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'.

 

[robousy]

ok, thanks, I'll check the paper when I get back to my office tomorrow and post.

Rich

 

[George Jones]

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.

 

[robousy]

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.

 

[George Jones]

Lots of have books probably have this, but I won't be able to tell you any others until Monday.

Maybe Carroll.

 

[samalkhaiat]

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

 

[robousy]

I'm not seeing it Sam.

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

 

[samalkhaiat]

I'm not seeing it Sam.

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

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

thus

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

 

[robphy]

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)

 

[robousy]

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

thus

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

Great, got it! Thanks a bunch.

 

[robousy]

http://books.google.com/books?id=v-Uw_uzbw7EC
search the book for: ln (natural log)

Useful link, thanks robphy!