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Christoffel Symbols homework

  • Thread starter n1mrod
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
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I think the element in G is R^2, not R^-2. Once you find the metric, you can calculate the Christoffel symbol by using the direct formula (which involves derivatives of Gab). Or you can use the change of coordinate formula for the Christoffel symbol.
 
  • #3
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I think the element in G is R^2, not R^-2. Once you find the metric, you can calculate the Christoffel symbol by using the direct formula (which involves derivatives of Gab). Or you can use the change of coordinate formula for the Christoffel symbol.
Hello!
It absolutely is, I just saw I forgot to write, I wrote Gab as being the inverse metric already, so there I believe it is R^-2, shouldn't it?
 
  • #4
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Yes.
 
  • #5
cristo
Staff Emeritus
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I tried to calculate the Christoffel 2 3,3 and I found it to be -2R / e^(-2R)
Where's the e come from here?

Incidentally, by G_ab, in your post, you mean g_ab; the metric tensor. You should use the correct notation from the beginning of your studies, as you will find another important tensor denoted G_ab later on!
 
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  • #6
CompuChip
Science Advisor
Homework Helper
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Also note the difference between g_ab and g^ab - one is the metric tensor (which?) and the other one is its inverse. Difference is a power of (-1) (if it's diagonal) - not entirely unimportant. If you use the right one but write down the wrong one, people will get confused. If you use the wrong one and write that down correctly... better :smile:
 
  • #7
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Where's the e come from here?

Incidentally, by G_ab, in your post, you mean g_ab; the (inverse) metric tensor. You should use the correct notation from the beginning of your studies, as you will find another important tensor denoted G_ab later on!
Hey, the e was wrong, I mispelled when I typed in, sorry.

Also, I used G_ab as the inverse. I'm sorry I'm just not used how to type these things here on the forum, but I used G_ab as the inverse, that's why its R^-2 .
 
  • #8
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Also note the difference between g_ab and g^ab - one is the metric tensor (which?) and the other one is its inverse. Difference is a power of (-1) (if it's diagonal) - not entirely unimportant. If you use the right one but write down the wrong one, people will get confused. If you use the wrong one and write that down correctly... better :smile:
thanks! I know that I did a mistake there, the one I wrote is the inverse already, I'll fix my notations for next time. But still, I don't know how to proceed to calculate the symbols once I have the inverse metric.
 
  • #9
cristo
Staff Emeritus
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Hey, the e was wrong, I mispelled when I typed in, sorry.
Ahh, ok.
Also, I used G_ab as the inverse. I'm sorry I'm just not used how to type these things here on the forum, but I used G_ab as the inverse, that's why its R^-2 .
Yea, the point of my post wasn't to be picky and say "you should use an _" but, more importantly, that the metric tensor is denoted by a small g. When most people look at G_ab they immediately think of the Einstein tensor. Anyway, it may have been a little bit of a pedantic point!
 
  • #10
12
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Ahh, ok.

Yea, the point of my post wasn't to be picky and say "you should use an _" but, more importantly, that the metric tensor is denoted by a small g. When most people look at G_ab they immediately think of the Einstein tensor. Anyway, it may have been a little bit of a pedantic point!
ahhh okay, so my mistake was even bigger heheh. Hey, please I asked people to correct me, you weren't being picky, you were correcting me :D
I'm just starting with General Relativity and I really want to learn, but as you know it's not very easy to get used to the nomenclature since it's very different from "regular" Physics.
So I believe what I wrote is g_ab right? The inverse of the metric?
 
  • #11
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According to my books,

[tex]g_{\mu\nu}[/tex] is the metric.
[tex]g^{\mu\nu}[/tex] is the inverse metric.

Christoffels are
[tex]\Gamma^{m}_{ab} = \frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})[/tex]

with summation over k, and the ,n means differentiated wrt [tex]x^n[/tex]

If you click on the text above, you'll get a window with the Tex source.
 
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  • #12
cristo
Staff Emeritus
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I have no idea what I was doing this morning! Of course, the metric tensor is denoted g_ab, and g^ab is the inverse metric tensor. Thus, what the OP has written in the opening post should g^ab=(..and then the matrix he writes out...)

Apologies for that, and thanks, Mentz for spotting it!
 
  • #13
12
0
According to my books,

[tex]g_{\mu\nu}[/tex] is the metric.
[tex]g^{\mu\nu}[/tex] is the inverse metric.

Christoffels are
[tex]\Gamma^{m}_{ab} = \frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})[/tex]

with summation over k, and the ,n means differentiated wrt [tex]x^n[/tex]

If you click on the text above, you'll get a window with the Tex source.
I have no idea what I was doing this morning! Of course, the metric tensor is denoted g_ab, and g^ab is the inverse metric tensor. Thus, what the OP has written in the opening post should g^ab=(..and then the matrix he writes out...)

Apologies for that, and thanks, Mentz for spotting it!
Okay now I understand it, I went to my professor this morning on my class, and also with the help you guys gave I was able to figure out.. now I can move forward =P
thanks a lot for the help!
 

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