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Determinant of metric tensor

  • Thread starter bdforbes
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  • #1
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Homework Statement


Prove that
[tex]\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})[/tex]

where g is the determinant of the metric, and [itex]\Gamma[/itex] are the Christoffel connection coefficients.

The Attempt at a Solution


From the general definition of the coefficients I got:
[tex]\Gamma^\mu_{\mu\lambda}=(1/2)g^{\mu\rho}\partial_\lambda g_{\rho\mu}[/tex]

But I have no idea how to work with the determinant of the metric. I'm not sure if I'm allowed to use this:

det(g)=exp[Tr ln G]

And if I did, would I have to use the GR definition of the trace?

[tex] Tr R = R^\mu_\mu [/tex]

I cleaned it up a little bit with the chain rule:

[tex]\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})[/tex]
[tex]=\frac{1}{2g}\partial_\lambda(g)[/tex]
 

Answers and Replies

  • #2
dextercioby
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The last part you wrote can't be right.

I'd start with

[tex] \delta |g| = \delta \[\exp \(\mbox{Tr} \ln |g|\)\] [/tex]

and then see what the 'delta' of the RHS brings me.
 
  • #3
Hurkyl
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Isn't there an index-notation expression for the determinant?
 
  • #4
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Thanks for the tips. Turns out it's a bit simpler:

1. From Jacobi's formula we get:
[tex] dg=A^{ik}dg_{ik}[/tex]
where [itex]A^{ik}[/itex] are elements of the adjugate of the metric.

2. Standard differential expansion:
[tex]dg=\frac{\partial g}{\partial g_{ik}}dg_{ik}[/tex]
[tex]\Rightarrow A^{ik}=\frac{\partial g}{\partial g_{ik}}[/tex]

3. From Laplace's formula for the determinant we get:
[tex]g^{ik}=(1/g)A^{ik}[/tex]


Putting this all together we have:
[tex]g^{ik}=\frac{1}{g}\frac{\partial g}{\partial g_{ik}}[/tex]

Substitute this into my expression for [itex]\Gamma^\mu_{\mu\lambda}[/itex] above and the result follows.

I'm still working on the proofs to steps 1 and 3 above, but essentially the problem is solved. Feels good to brush up on my linear algebra, haven't used it for years.
 

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