# Variation of the metric tensor determinant

1. Dec 12, 2012

### InsertName

1. The problem statement, all variables and given/known data

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:

$\delta(det[g\mu\nu])$.

2. Relevant equations

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

3. 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!

2. Dec 13, 2012

### clamtrox

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

This is a little bit tricky, as you dont 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:
$$\mathrm{Tr}( g \cdot \delta(g) ) = (g \cdot {\delta(g))_\mu}^{\mu} = g_{\mu \lambda} \delta (g^{\lambda \mu})$$

3. Dec 13, 2012

### andrien

use
g=(1/4!)εαβγδεμvρσgαμgβvgγρgδσ
ggαμαβγδεμvρσgβvgγρgδσ
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)

4. Dec 14, 2012

### 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=gμv