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Christoffel Symbols

  • #1
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I have that [itex]g=L^2 \left( e^{-2U} \left( e^{2A} \left( -dt^2 + d \theta^2 \right) + R^2 dy^2 \right) + e^{2U} dx^2 \right)[/itex] is the metric on my spacetime.

taking [itex]\{ t, \theta, x , y \}[/itex] as a coordinate system for the manifol M, i can write this in matrix form as

[itex]g_{ab}=L^2 \left( \begin {array}{cccc} -{e}^{2 \left( A-U \right)}&0&0&0
\\ \noalign{\medskip}0&{e}^{2 \left(A-U \right)}&0&0\\ \noalign{\medskip}0&0&{e}
^{2U}&0\\ \noalign{\medskip}0&0&0&{R}^{2}{e}^{-2U}\end {array}
\right)[/itex]

now i need to show the vacuum einstein equations for g are
[itex]\partial_t^2 R - \partial_{\theta}^2 R =0[/itex]
[itex]\partial_t (R \partial_t U ) - \partial_{\theta} ( R \partial_{\theta} U ) =0[/itex]
[itex]\partial_t^2 A - \partial_{\theta}^2 A = ( \partial_{\theta} U )^2 - ( \partial_t U)^2[/itex]

and

[itex]\partial_{\theta} \partial_+ R = ( \partial_+ A)(\partial_+ R) - R ( \partial_+ U)^2[/itex]
[itex]\partial_{\theta} \partial_- R = ( \partial_- A)( \partial_- R) - R ( \partial_- U )^2[/itex]

where [itex]\partial_{\pm} = \partial_{\theta} \pm \partial_{t}[/itex]

so i want to start by computing the christoffel symbols andyway.

this is done using [itex]\Gamma^{\sigma}_{\mu \nu} = \frac{1}{2} \displaystyle \sum_{\rho} g^{\sigma \rho} \left( \frac{ \partial g_{\nu \rho}}{\partial x^{\mu}} + \frac{ \partial g_{\mu \rho}}{\partial x^\nu} - \frac{ \partial g_{\mu \nu}}{ \partial x^\rho} \right)[/itex]

however in previous examples i've worked with, [itex]\sigma, \mu, \nu, \rho \in \{ 1,2,3 \}[/itex] but now i have a problem because of this fourth index due to the presence of time in my metric and i don't know how to deal with it. any advice?
 
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Answers and Replies

  • #2
1,482
3
just run rho from 0 to 3

[itex]
g^{\sigma 0} ..... + g^{\sigma 1} ..... + g^{\sigma 2} ..... + g^{\sigma 3}

[/itex]
 
  • #3
1,444
0
so do i keep [itex]\sigma, \mu, \nu[/itex] running from 1 to 3 only?
 
  • #4
1,444
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so does this look ok so far...

[itex]\Gamma^1{}_{11}=\frac{1}{2} g^{11} ( \partial_{\theta} g^{11} ) = \frac{1}{2} L^2 e^{2(A-U)} \partial_{\theta} (L^2 e^{2(A-U)})[/itex]
[itex]=\frac{1}{2}L^4 e^{2(A-U)} e^{2(A-U)} ( \partial_{\theta} ( 2(A-U)))[/itex]
[itex]=L^4 e^{4(A-U)} ( \partial_{\theta} A- \partial_{\theta} U)[/itex]

[itex]\Gamma^1{}_{12}=\frac{1}{2} g^{11} ( \partial_x (L^2 e^{2(A-U)}))=0[/itex]
as [itex]A,U[/itex] are functions of [itex]\theta,t[/itex] only.

to be honest i don't see how there's ever going to be a situation where we use [itex]\rho=0[/itex] as if [itex]\sigma \in \{ 1,2,3 \}[/itex], then the [itex]g^{\sigma \rho}[/itex] term in front of the brackets in the formula will always be zero when [itex]\rho=0[/itex] will it not?
 
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  • #5
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0
bump.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,791
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so do i keep [itex]\sigma, \mu, \nu[/itex] running from 1 to 3 only?
No, all indices here run from 0 to 3.

(I can't speak for general usage but in Eddington's "The Mathematical Theory of Relativity", he specifically uses Greek letters for indices from 0 to 3, Latin letters for indices from 1 to 3.)
 
  • #7
1,444
0
ok. thanks.
so now that i have the non-zero Christoffel symbols, the next step in getting to the Einstein equations there would be to copmute the Ricci tensor using the eqn

[itex]R_{\mu \rho}= \displaystyle \sum_{\nu} \frac{\partial}{\partial x^{\nu}} \Gamma^{\nu}{}_{\mu \rho} - \frac{\partial}{\partial x^{\mu}} \left( \displaystyle \sum_\nu \Gamma^{\nu}{}_{\nu \rho} \right) + \displaystyle \sum_{\sigma, \nu} \left( \Gamma^{\alpha}{}_{\mu \rho} \Gamma^{\nu}{}_{\alpha \nu} - \Gamma^{\alpha}{}_{\nu \rho} \Gamma^{\nu}{}_{\alpha \mu} \right)[/itex]


then i would write the Ricci tensor as a matrix

then i calculate [itex]R=R_{a}{}^{a}[/itex]

and then i put these into [itex]R_{ab}-\frac{1}{2}Rg_{ab}=8 \pi T_{ab}[/itex]
yeah?

and if these are the vacuum Einstein eqns i can set [itex]T_{ab}=0[/itex], yeah?

thanks.
 
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  • #8
1,482
3
then i would write the Ricci tensor as a matrix

then i calculate [itex]R=R_{a}{}^{a}[/itex]

and then i put these into [itex]R_{ab}-\frac{1}{2}Rg_{ab}=8 \pi T_{ab}[/itex]
yeah?

and if these are the vacuum Einstein eqns i can set [itex]T_{ab}=0[/itex], yeah?
Yeah, that's the plan. But why do you keep the big sigma?

summation convection should get rid of it.
 
  • #9
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ye, i think it was just left in in the book i was working from (Wald's General Relativity).

anyway i found that copmuting, even [itex]R_{00}[/itex] was about 4 pages of wrok and my final answer was about 10 lines long...has something gone wrong or is this typical?
 
  • #10
1,482
3
anyway i found that copmuting, even [itex]R_{00}[/itex] was about 4 pages of work
yup that's normal if you write small

and my final answer was about 10 lines long...has something gone wrong or is this typical?
I'm not sure what the answer is to this metric, but the components of the Ricci tensor usually simplify to simple second order differential equations.

Hobson has a nice step by step calculation on this.
 

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