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Components of Christoffel symbol 
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#1
Oct2809, 12:07 PM

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The metric of Euclidean [itex]\mathbb{R}^3[/itex] in spherical coordinates is [itex]ds^2=dr^2+r^2(d \theta^2 + \sin^2{\theta} d \phi^2)[/itex].
I am asked to calculate the Christoffel components [itex]\Gamma^{\sigma}{}_{\mu \nu}[/itex] in this coordinate system. i'm not too sure how to go about this. it talks about [itex]ds^2[/itex] being the metric but normally the metric is of the form [itex]g_{ab}[/itex] i.e. a 2form but ds^2 isn't a 2form. are these metrics different or do i make [itex]g_{\mu \nu}=ds^2 \omega_{\mu} \omega_{\nu}[/itex] where [itex]\omega_i[/itex] is a 1 form? i think i'm missing some key point here.... 


#2
Oct2809, 12:16 PM

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The metric is often defined according to the equation [itex]ds^2=g_{ab}dx^adx^b[/itex]...In this case, you have [itex]x^a\in\{r,\theta,\phi\}[/itex]....so what are the components of [itex]g_{ab}[/itex]?



#3
Oct2809, 12:44 PM

P: 1,439

[tex]g_{ab}=\left[ \begin {array}{ccc} 1&0&0 \\ 0&r&0
\\ 0&0& {\sin}{\theta}\end {array} \right] [/tex] i didn't know how else to right it. would that work? because say [tex]g_{33}=g_{\phi \phi} = \sin^2{\theta} dx^\phi dx^\phi[/tex] 


#4
Oct2809, 12:58 PM

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Components of Christoffel symbol
Don't you mean:
[tex]g_{ab}=\begin{pmatrix}1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & \sin^2\theta \end{pmatrix}[/tex] 


#5
Oct2809, 03:34 PM

P: 1,439

yeah sorry. okay so that would work out for the formula [itex]ds^2=g_{ab} dx^a dx^b[/itex]
now i guees i'm supposed to use 3.1.30 in Wald: [tex]\Gamma^{\sigma}{}_{\mu \nu}=\frac{1}{2} \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^{\sigma}} \right)[/tex] im confused about how this sum is going to work though. [itex]\sigma,\nu,\mu \in \{ r, \theta, \phi \}[/itex] and so they aren't fixed variables...which is confusing also what values does [itex]\sigma[/itex] take? 


#6
Oct2809, 05:21 PM

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Also, your last term in [itex]\Gamma^{\sigma}{}_{\mu\nu}[/itex] has a typo. So, for example, [tex]\Gamma^{1}{}_{23}=\frac{1}{2} \sum_{\rho} g^{1\rho} \left( \frac{\partial g_{3\rho}}{\partial \theta} + \frac{\partial g_{2\rho}}{\partial \phi}  \frac{\partial g_{23}}{\partial x^{\rho}} \right)=\frac{1}{2} g^{11} \left( \frac{\partial g_{31}}{\partial \theta} + \frac{\partial g_{21}}{\partial \phi}  \frac{\partial g_{23}}{\partial r} \right)=0 [/tex] 


#7
Oct2909, 07:56 AM

P: 1,439

ahh i think i get it. the sum reduces to just the [itex]\rho=1[/itex] term because [itex]g_{12}=g_{13}=0[/itex] which zeroes the whole expression in the cases of [itex]\rho=2[/itex] or [itex]\rho=3[/itex].
so [itex]\Gamma^{\sigma}{}_{\mu \nu}[/itex] will have [itex]3^3=27[/itex] copmonents, correct? i can't write my final answer as a matrix can i? i'd just have to write them out explicitly as: [itex]\Gamma^1_{11}= ...[/itex] [tiex]\Gamma^1_{12}= ...[/itex] etc. that doesn't look very concise though? 


#8
Oct2909, 08:13 AM

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It will look more concise once you realize just how many of those 27 components are zero (Also, you should keep in mind that [itex]g^{ab}[/itex] is the inverse of [itex]g_{ab}[/itex] when doing your calculations)\
As a matter of convention, [itex]\Gamma^1{}_{23}[/itex] is often written as [itex]\Gamma^r_{\theta\phi}[/itex] and so on; which may be what was confusing you earlier. 


#9
Oct2909, 08:18 AM

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Also, my earlier matrix contains a typo, it should be:
[tex]g_{ab}=\begin{pmatrix}1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \end{pmatrix}[/tex] 


#10
Oct2909, 08:27 AM

P: 1,439

why is [itex]g^{ab}[/itex] the inverse of [itex]g_{ab}[/itex] and how will that be useful?
also when u say [itex]\sigma \in \{ 1,2,3 \}[/itex] and [itex]x^1=r, x^2= \theta, x^3 = \phi[/itex] this means that if [itex]\sigma=1[/itex] then [itex]\sigma=r[/itex] and that's why we can write [itex]\Gamma^{1}{}_{23}=\Gamma^r{}_{\theta \phi}[/itex] doesn't that imply that [itex]1=r[/itex] rather than [itex]x^1=r[/itex]? 


#11
Oct2909, 08:49 AM

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It's useful, because you will need to know the components of [itex]g^{ab}[/itex] to compute the Christoffel symbols; and you can get those components just by taking the inverse of [itex]g_{ab}[/itex] 


#12
Oct2909, 09:04 AM

P: 1,439

so because [itex]1=r[/itex] and [itex]x^1=r[/itex], don't you mean to write that [itex]\sigma \in \{ x^1,x^2,x^3 \}[/itex]?
i used maple to quickly get [itex]g^{ab}=\left[ \begin {array}{ccc} 1&0&0\\ \noalign{\medskip}0&{r}^{2}&0 \\ \noalign{\medskip}0&0&{\frac {1}{{r}^{2}{\sin}^{2}\theta}} \end {array} \right] [/itex] 


#13
Oct2909, 09:53 AM

P: 1,439

scratch that above post.
i couldn't think of any quick way to do it so i just did all 27 calculations and found the non zero terms are : [itex]\Gamma^1{}_{22}=r[/itex] [itex]\Gamma^1{}_{33}=r \sin^2{\theta}[/itex] [itex]\Gamma^2{}_{12}=r^3[/itex] [itex]\Gamma^2{}_{21}=r^3[/itex] [itex]\Gamma^2{}_{33}=\frac{1}{2}r^4 \sin{2 \theta}[/itex] [itex]\Gamma^3{}_{13}=r^3 \sin^4{\theta}[/itex] [itex]\Gamma^3{}_{23}=r^4 \sin^3{\theta} \cos{\theta}[/itex] [itex]\Gamma^3{}_{31}=r^3 \sin^4{\theta}[/itex] [itex]\Gamma^3{}_{32}=r^4 \sin^3{\theta} \cos{\theta}[/itex] i'm not sure if there's a pattern i was supposed to spot so i could save myself some time in working out the copmonents or what? anyway, when it asks for the components of the Christoffel symbol, do i just leave it as a list of the non zero ones like i have done above or am i missing how to write the whole thing neatly as a matrix or something? 


#14
Oct2909, 09:56 AM

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#15
Oct2909, 10:01 AM

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[tex]\Gamma^r=\begin{pmatrix}0 & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & r\sin^2\theta\end{pmatrix}[/tex] [tex]\Gamma^\theta=\begin{pmatrix}0 & \frac{1}{r} & 0 \\ \frac{1}{r} & 0 & 0 \\ 0 & 0 & \sin\theta\cos\theta\end{pmatrix}[/tex] [tex]\Gamma^\phi=\begin{pmatrix}0 & 0 & \frac{1}{r} \\ 0 & 0 & \cot\theta \\ \frac{1}{r} & \cot\theta & 0\end{pmatrix}[/tex] Again, this is somewhat sloppy notation, but is still fairly common in the literature. 


#16
Oct2909, 10:04 AM

P: 1,439

are they ok then?



#18
Oct2909, 10:34 AM

P: 1,439

i dont see how im missing those factors. take for example
[itex]\Gamma^2{}_{12}=\frac{1}{2} g^{22} \left( \frac{\partial g_{22}}{\partial r} + \frac{\partial g_{12}}{\partial \theta}  \frac{\partial g_{12}}{\partial \theta} \right) = \frac{1}{2} r^2 \frac{\partial}{\partial r} \left(r^2 \right) = r^3[/itex] i can't see where i'm missing this factor? 


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