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:

[itex]\delta(det[g_\mu\nu])[/itex].

Homework Equations

The answer is

[itex]\delta(det[g_\mu\nu])[/itex] =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
[itex]\delta(det[g_\mu\nu])[/itex]
=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

 

[paranormal]

The variation of the determinant of the metric tensor is an important concept in tensor calculus and differential geometry. It measures the change in the determinant of the metric tensor under a small variation in the components of the tensor.

Your attempt at the solution is on the right track, but there are a few things that need to be corrected. First, the correct expression for the variation of the determinant is:

\delta(det[g\mu\nu]) = det[g\mu\nu] g\mu\nu \delta g^{\mu\nu}

Note that the indices on the variation are raised, indicating that it is a contravariant quantity. Also, the inverse of the metric tensor should be used, not the trace.

To see why this is the correct expression, let's break it down step by step. First, we can use the fact that the determinant of a matrix is equal to the product of its eigenvalues to write:

\delta(det[g\mu\nu]) = det[g\mu\nu] \delta(\lambda_1 \lambda_2 ... \lambda_n)

Where \lambda_i are the eigenvalues of the matrix [g\mu\nu]. Then, using the chain rule, we can write:

\delta(\lambda_1 \lambda_2 ... \lambda_n) = \delta\lambda_1 \lambda_2 ... \lambda_n + \lambda_1 \delta\lambda_2 ... \lambda_n + ... + \lambda_1 \lambda_2 ... \delta\lambda_n

Next, we can use the fact that the variation of an eigenvalue is equal to the trace of the variation of the matrix divided by the eigenvalue. This gives us:

\delta(det[g\mu\nu]) = det[g\mu\nu] \left( \frac{\text{tr}(\delta[g\mu\nu])}{\lambda_1} + \frac{\text{tr}(\delta[g\mu\nu])}{\lambda_2} + ... + \frac{\text{tr}(\delta[g\mu\nu])}{\lambda_n} \right)

Finally, substituting in the eigenvalues of the metric tensor, which are just the components of the tensor, we get:

\delta(det[g\mu\nu]) = det[g\mu\nu] \left( \frac{\text{tr}(\delta[g\mu\nu