Help computing Christoffel coefficient

[S.P.P]

Not so much a homework problem, more needing help understanding where something comes from. I've attached a jpg file with what I need help on.

I've done a general relativity course at uni but can't seem to work out what should be a very simple problem.

 

[Mentz114]

I get this,

\Gamma^i_{0j} = \frac{1}{2}g^{i\nu}\partial_0(g_{\nu j}) with \nu running over 1,2,3.

Differentiating
\Gamma^i_{0j} = \frac{1}{2}g^{i\nu}(2aa'\eta_{\nu j} + 2aa'h_{\nu j} + a^2\partial_0 h_{\nu j})

which gives
\Gamma^i_{0j} = (\eta^{i\nu}-h^{i\nu})(\frac{a'}{a}\eta_{\nu j} + \frac{a'}{a}h_{\nu j} + \frac{1}{2}\partial_0 h_{\nu j})


and so
\Gamma^i_{0j} = \frac{a'}{a}g^{i\nu}g_{\nu j} + \frac{1}{2}g^{i\nu}\partial_0 h_{\nu j})

summing we get,
\Gamma^i_{0j} = \frac{a'}{a}\delta^i_j + \frac{1}{2}h'^i_j


Which is not quite the right answer !

 

[S.P.P]

That's a bit closer than I can get. Does \delta^i_j = \delta_{ij}?

 

[Mentz114]

\delta^i_j = g^{ik}\delta_{kj}

which requires a summation. I hope you don't mind me asking, but do you know how to raise and lower indexes ? I notice in your equ 5 you've lost the dummy index.

 

[S.P.P]

I'm starting to think that the answer I've been given has a typo, and should read \delta^i_j instead of \delta_{ij}.

I've lost the dummy indices as I've summed over v. When v=0 everything equalled zero, so the next index was j (where j = 1,2,3). Is this not the way to proceed?

Also, I understand that g_{iv}g^{vj}=\delta^i_j, but (\eta_{iv}+h_{iv})*(\eta^{vj}-h^{vj})=0?

 

[Mentz114]

Aha - yes, I made an error, so my answer should have \delta_{ij}.
Don't know why I wrote a mixed index there. That still leaves the sign and the prime on h.

 

[S.P.P]

there should be a prime on h, that was a typo on my part.

 

[Mentz114]

Well, if you can track down the why the sign disagrees, it's mission accomplished.

I'm busy right now but I'll check it a little later.

 

[Mentz114]

There's another typo in your doc. The expression you quote as the answer is wrong. The first delta must have one upper and one lower index like the other terms.

So I stand my earlier result because

g^{ik}g_{kj} = g_j^i = \delta_j^i

I haven't tracked down the sign yet.

 

[S.P.P]

aha! got it.

\Gamma^i_{0j} = \frac{a'}{a}\delta^{ij}+\frac{1}{2}g^{i\nu}\partial_0(h_{\nu j}).

If I expand the last term out,

\frac{1}{2}(\eta^{i\nu}-h^{i\nu})\partial_0(h_{\nu j}).

Ignore higher order terms (ie h^2), using the metric signiture (+,-,-,-) coupled with the fact that only the spatial part is non zero, then,

\eta^{i\nu} = \eta^{ij}=-\delta^{ij}

Thanks for the help

 

[Mentz114]

I'm glad you're content. I don't follow your last step, but I suppose it's time to move on ...