Christoffel Symbols: Solving GR Homework | Fred

AI Thread Summary
The discussion centers on a user named Fred seeking help with calculating Christoffel symbols in cylindrical coordinates for a general relativity homework problem. Fred has identified the metric as dR^2 + (R^2)(dtheta^2) + dZ^2 and presented the corresponding metric tensor Gab. He is unsure how to proceed with calculating the Christoffel symbols from the metric and its inverse. Other participants clarify that the calculation involves summing partial derivatives of the metric multiplied by the inverse metric, confirming that Fred has the necessary components to continue. The conversation emphasizes the importance of understanding the relationship between the metric and its inverse in this context.
n1mrod
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Hey Guys,

I'm new here on the forum, and I hope someone can help me out.
I'm solving one of my GR homework exercises and I'm asked to find the christoffel symbols corresponding to cylindrical coordinates.
I'll post my work and please correct if you see mistakes.
I found the metric to be dR^2 + (R^2)(dtheta^2) + dZ^2
therefore
Gab=
1 0 0
0 R^-2 0
0 0 1

Can somebody kind of explain to me how to proceed with these calculations?

Thanks a lot!
Fred.
 
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You'll have to explain how you got that strange result before anyone can help. Where did the e come from?
 
Dick said:
You'll have to explain how you got that strange result before anyone can help. Where did the e come from?

Hey I'm sorry, I was writing the topic and the second part I copied from my other homework assignment, sorry I'm going to fix that.

Until the "matrix" Gab is what I did, but I'm not very sure how to calculate the christoffel from there, can you help me?
 
You calculate the christoffel by evaluating a summation of partial derivatives of the metric times the inverse metric. Again, you are not giving us much to go on.
 
Okay, but I put the metric I found. As you said, to calculate the symbols it's only necessary the Metric and the inverse metric, the inverse metric is calculated straight from the metric right? So I believe we have everything that is needed, but please correct me if I'm wrong.
Thanks.
 
I think so. If you are going wrong, you haven't done so yet.
 
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