Help computing Christoffel coefficient

1. Apr 13, 2007

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.

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Last edited: Apr 13, 2007
2. Apr 13, 2007

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 !

Last edited: Apr 13, 2007
3. Apr 15, 2007

S.P.P

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

4. Apr 15, 2007

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.

5. Apr 15, 2007

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 $$(\eta_{iv}+h_{iv})*(\eta^{vj}-h^{vj})=0$$?

Last edited: Apr 15, 2007
6. Apr 15, 2007

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.

7. Apr 15, 2007

S.P.P

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

8. Apr 15, 2007

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.

9. Apr 15, 2007

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.

10. Apr 16, 2007

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

Last edited: Apr 16, 2007
11. Apr 16, 2007

Mentz114

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