Determinant of metric tensor

In summary, the equation \Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g}) can be proven by using Jacobi's formula and Laplace's formula for the determinant of the metric. By substituting g^{ik}=(1/g)A^{ik} into the general definition of the Christoffel connection coefficients, the desired result can be obtained. Using these steps, the problem can be solved without the need for the GR definition of the trace.
  • #1
bdforbes
152
0

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]
 
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  • #2
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
Isn't there an index-notation expression for the determinant?
 
  • #4
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.
 

1. What is the determinant of a metric tensor?

The determinant of a metric tensor is a mathematical quantity that describes the volume scaling factor of a coordinate system. It is denoted by det(g) or g, and it is used in many areas of mathematics and physics.

2. How is the determinant of a metric tensor calculated?

The determinant of a metric tensor is calculated by taking the determinant of the matrix formed by the components of the tensor. This involves multiplying the diagonal elements of the matrix and subtracting the product of the off-diagonal elements.

3. What does the determinant of a metric tensor tell us about the geometry of a space?

The determinant of a metric tensor tells us about the curvature and shape of a space. A positive determinant indicates a Euclidean space, while a negative determinant indicates a non-Euclidean space with curved geometry.

4. How does the determinant of a metric tensor relate to the concept of distance?

The determinant of a metric tensor is directly related to the concept of distance in a space. It is used to calculate the length of a curve in a space and is essential in many physical theories, such as general relativity.

5. Can the determinant of a metric tensor be negative?

Yes, the determinant of a metric tensor can be negative. This indicates a non-Euclidean space with curved geometry. It is important to consider the sign of the determinant when using it in calculations or interpreting its meaning in a physical context.

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