Is the Christoffel Symbol Zero for Diagonal Metrics?

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SUMMARY

The discussion centers on the Christoffel symbol \(\Gamma^a_{bc}\) and its behavior when the metric \(g_{ab}\) is diagonal. The formula \(\Gamma^a_{bc} = \frac{1}{2} g^{ad}(\partial_b g_{dc} + \partial_c g_{bd} - \partial_d g_{bc})\) is confirmed as applicable for deriving the components of the Christoffel symbol. Participants clarify that while some components of the Christoffel symbol may be zero, particularly the off-diagonal ones, it is not universally true for all components. The importance of recognizing the structure of the metric and substituting zero for off-diagonal terms is emphasized.

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Homework Statement


I am trying to show that the connection \Gamma^a_{bc} is equal to 0 when the metric g_ab is diagonal. Will the formula
\Gamma^a_{bc} = 1/2 g^{ad}(\partial_bg_{dc} + \partial_cg_{bd} - \partial_dg_{bc}) 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

 
Last edited:
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Why do you think the connection is zero if the metric is diagonal? It's not even true. Look at spherical coordinates.
 
It comes right out of my book!
 

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No it doesn't; your book doesn't ask you to prove all of the components are zero: just some of them.
 
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.
 
If \Gamma is meant to be the Riemann - Christoffel (or the metric) connection, then, yes, that's the formula you should use.
 
Yes. It is the Christoffel symbol of the second kind. So am allowed to do this:

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

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:

\Gamma^a_{bc} = 1/2 (\partial_bg^{ad}g_{dc} + \partial_cg^{ad}g_{bd} - \partial_dg^{ad}g_{bc})
?
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.
 
Last edited:
I see. You just replace d with a.
 
  • #10
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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  • #11
Did you try differentiating it?
 

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