Is the Christoffel Symbol Zero for Diagonal Metrics?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Connection
Click For Summary

Homework Help Overview

The discussion revolves around the properties of the Christoffel symbols in the context of diagonal metrics. The original poster attempts to demonstrate that the connection \(\Gamma^a_{bc}\) is zero for diagonal metrics, referencing a specific formula for the Christoffel symbols.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of the formula for the Christoffel symbols and question the assumption that all components are zero for diagonal metrics. There is a focus on distinguishing between off-diagonal and diagonal components.

Discussion Status

The conversation is ongoing, with participants providing guidance on the use of the formula and discussing the nature of the components of the Christoffel symbols. Some participants suggest specific manipulations of the formula while others question the validity of the assumptions made by the original poster.

Contextual Notes

There is a mention of the original poster's textbook, which may not require proving that all components are zero, but rather focuses on specific components. The discussion also touches on the implications of using spherical coordinates and the nature of the metric being diagonal.

ehrenfest
Messages
2,001
Reaction score
1

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

 
Last edited:
Physics news on Phys.org
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!
 

Attachments

  • hobson exercise 36.jpg
    hobson exercise 36.jpg
    3.9 KB · Views: 467
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 [itex]\Gamma[/itex] 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:

[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.
 
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?
 

Attachments

  • hobson page 66.jpg
    hobson page 66.jpg
    27.6 KB · Views: 471
  • #11
Did you try differentiating it?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
10K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 62 ·
3
Replies
62
Views
7K