Is the Christoffel Symbol Zero for Diagonal Metrics?

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Homework Statement


I am trying to show that the connection [tex]\Gamma^a_{bc}[/tex] is equal to 0 when the metric g_ab is diagonal. Will the formula
[tex]\Gamma^a_{bc} = 1/2 g^{ad}(\partial_bg_{dc} + \partial_cg_{bd} - \partial_dg_{bc})[/tex] be of use? How can I manipulate that equation and use the fact that the metric is diagonal?

Homework Equations


The Attempt at a Solution

 
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It comes right out of my book!
 

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OK. So it only wants me to prove that the off-diagonal components are zero (that statement in parentheses was pretty important). Anyway, I still have the same two questions as in the first post.
 
Yes. It is the Christoffel symbol of the second kind. So am allowed to do this:

[tex]\Gamma^a_{bc} = 1/2 (\partial_bg^{ad}g_{dc} + \partial_cg^{ad}g_{bd} - \partial_dg^{ad}g_{bc})[/tex]
?

Then the first term becomes the Kronecker delta, I think.

If not, how should I use the fact that the metric is diagonal?
 
ehrenfest said:
Yes. It is the Christoffel symbol of the second kind. So am allowed to do this:

[tex]\Gamma^a_{bc} = 1/2 (\partial_bg^{ad}g_{dc} + \partial_cg^{ad}g_{bd} - \partial_dg^{ad}g_{bc})[/tex]
?
Why do you think, for all d, that gad is a constant with respect to the b, c, and d-th variables?



If not, how should I use the fact that the metric is diagonal?
Any way you can imagine. You could substitute zero for the off-diagonal terms. You could decompose the metric into a linear combination of simpler tensors. Et cetera.
 
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I see. You just replace d with a.
 
Now, can someone explain to me where in the world this natural log comes from at the bottom of attached page of the Hobson book?
 

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