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 2-form but ds^2 isn't a 2-form. 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....
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]?
[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]
Don't you mean: [tex]g_{ab}=\begin{pmatrix}1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & \sin^2\theta \end{pmatrix}[/tex]
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?
No, all of [itex]\mu,\nu,\sigma,\rho\in\{1,2,3\}[/itex], with [itex]x^{1}=r[/itex], [itex]x^{2}=\theta[/itex] and [itex]x^{3}=\phi[/itex] 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]
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?
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.
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]
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]?
Because of its definition; [itex]g^{ab}g_{ac}=\delta^{b}{}_{c}[/itex]....which tells you that multiplying the matrix [itex]g^{ab}[/itex] with the matrix [itex]g_{ac}[/itex] produces the identity matrix....i.e. [itex]g^{ab}[/itex] is the inverse of [itex]g_{ab}[/itex]. 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] That's why that notation is often confusing; nevertheless, it is still the convention....
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]
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?
No, [itex]\sigma \in \{ 1,2,3 \}[/itex] with [itex]x^1=r[/itex], [itex]x^2=\theta[/itex] and [itex]x^3=\phi[/itex]....so the Christoffel symbols should be labeled [itex]\Gamma^1{}_{23}[/itex] etc... But, by convention they are often labeled [itex]\Gamma^r_{\theta\phi}[/itex] etc...it's sloppy notation to do this, but nevertheless, convention is convention.
You seem to be missing a factor of [itex]1/r^4[/itex] And for these you are missing a factor of [itex]1/(r^4\sin^4\theta)[/itex] You could save a little time by remembering that the Christoffel symbols are symmetric in the bottom pair of indices...so you only have to calculate 18 of them. You can't really write them as a single matrix, but they are often written as a set of 3 matrices in the form: [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.
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?
You seem to have used [itex]g^{ab}=g_{ab}[/itex] instead of using the inverse matrix you calculated... [tex]\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} \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \right) = \frac{1}{r}[/tex]
lol. i'm an idiot sometimes...thanks. the next bit asks me to write out the copmonents of the geodesic equaiton in this coordinate system and verify the solutions correspond ot straight lines in Cartesian coordinates. so i guess the eqn tehy're referring to is 3.3.5. [tex]\frac{d^2 x^{\mu}}{dt^2} + \sum_{\sigma, \nu} \Gamma^{\mu}{}_{\sigma \nu} \frac{d x^{\sigma}}{dt} \frac{dx^{\nu}}{dt}=0[/tex] am i required here to solve 17 different differential equations? even if the Christoffelsymbol is zero there will still be that first term equal to zero so none of the equaitons are going to be trivial. or do i just solve for one term of the Christoffel symbol and show the solutino is a straight line?