Variation of the metric tensor determinant

[InsertName]

Homework Statement

This is not homework but more like self-study - thought I'd post it here anyway.

I'm taking the variation of the determinant of the metric tensor:

$\delta(det[g_\mu_\nu])$.

Homework Equations

The answer is

$\delta(det[g_\mu\nu])$ =det[g_\mu\nu] g^$\mu\nu$ $\delta$g_$\mu\nu$

Here, g_$\mu\nu$ is the metric tensor, [g_$\mu\nu$] is the matrix of the components of the metric tensor, and $\delta$ is a variation.

The Attempt at a Solution

I have managed to get close to the answer, I hope, with
$\delta(det[g_\mu\nu])$
=det[g_\mu\nu] tr([g_\mu\nu]^-1$\delta$([g_\mu\nu]))

The problem, in my view, is the trace. I cannot see how to remove it.

Also, if someone could kindly describe how to tidy the LaTeX up, I will do that.

Thank you!

 

[clamtrox]

You can't use the tags in latex. Instead, write g_{\mu \nu}, g^{\mu \nu} etc.

I have managed to get close to the answer, I hope, with
$\delta(\det[g_{\mu\nu}]) = \det[g_{\mu\nu}] \mathrm{Tr}([g_{\mu\nu}^{-1}\delta([g_ {\mu\nu}]))$

This is a little bit tricky, as you don't really keep track of how your matrices are multiplied together. If you do it with more care, you should find something like
$$\delta \det(g) = \det(g) \mathrm{Tr}( g \cdot \delta(g) ),$$
with $(g \cdot \delta(g))_{\mu \nu} = g_{\mu \lambda} \delta ({g^\lambda}_\nu)$
Now it's easy to see what the trace is:
$$\mathrm{Tr}( g \cdot \delta(g) ) = (g \cdot {\delta(g))_\mu}^{\mu} = g_{\mu \lambda} \delta (g^{\lambda \mu})$$

 

[andrien]

use
g=(1/4!)ε^αβγδε^μvρσg_αμg_βvg_γρg_δσ
gg^αμ=ε^αβγδε^μvρσg_βvg_γρg_δσ
now when you apply δ on g you can use the second of the above to get whole result.(i am not going to do it in full,because of the requirement of homework section)

 

[andrien]

Hmm,may be tough.
so use Tr (ln M)=ln(det M)
variation of it yields,
Tr(M^-1 δM)=(1/det M)δ(det M)
JUST PUT M=g_μv