# Find Christoffel symbols from metric

1. Jan 22, 2013

### ck99

1. The problem statement, all variables and given/known data

Find the non zero Christoffel symbols of the following metric

$$ds^2 = -dt^2 + \frac{a(t)^2}{(1+\frac{k}{4}(x^2+y^2+z^2))^2} (dx^2 + dy^2 + dz^2 )$$

and find the non zero Christoffel symbols and Ricci tensor coefficients when k = 0

2. Relevant equations

3. The attempt at a solution

I have tried to use the Lagrangian method here, but it's so long since I was taught this I'm not sure if I'm even half right!

Starting with

$$\frac{d}{dλ} \frac{∂L}{∂\dot{q}} = \frac{∂L}{∂q}$$

and for the case where q = t, so q' = t', I get

$$\frac{∂L}{∂q} = 0 + \frac{2a\dot{a}}{(1+\frac{k}{4}(x^2+y^2+z^2))^2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$$

$$\frac{∂L}{∂\dot{q}} = -2\dot{t}$$

$$\frac{d}{dλ} \frac{∂L}{∂\dot{q}} = -2\ddot{t}$$

Putting it all together, I get

$$-2\ddot{t} = \frac{2a\dot{a}}{(1+\frac{k}{4}(x^2+y^2+z^2))^2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$$

or just

$$\ddot{t} + \frac{a\dot{a}}{(1+\frac{k}{4}(x^2+y^2+z^2))^2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) = 0$$

Compare this to

$$\ddot{X}^a + \Gamma^a_bc \dot{X}^a \dot{X}^b$$

to get

$$\Gamma^0_11 = \Gamma^0_22 = \Gamma^0_33 = \frac{a\dot{a}}{(1+\frac{k}{4}(x^2+y^2+z^2))^2}$$

This all looks OK to me, but I have worked through the rest of the problem using this approach and when I have to calculate Ricci tensor components I can't get the correct answers. I thought I should start by checking my work at the very beginning!

When k = 0 and the metric is much simplified, I get the following non-zero Christoffel symbols by using this approach

$$\Gamma^0_11 = \Gamma^0_22 = \Gamma^0_33 = a\dot{a}$$

$$\Gamma^1_11 = \Gamma^2_22 = \Gamma^3_33 = 2\frac{\dot{a}}{a}$$

and it's at this point that I get the Ricci tensor components all wrong (I know what they should be and I can't get the same answers). Is my Lagrangian method wrong, and have I missed out a load of non-zero Christoffel symbols?

(I didn't want to write out all my other Lagrangian working because this is my first go with LATEX and writing this post has taken me over an hour!)

2. Jan 22, 2013

### CompuChip

I hope that someone will have time to go through your calculation, as unfortunately I don't at the moment. But maybe you will already be helped if I point out that
$$\Gamma_{cab} =\frac12 \left(\frac{\partial g_{ca}}{\partial x^b} + \frac{\partial g_{cb}}{\partial x^a} - \frac{\partial g_{ab}}{\partial x^c} \right) = \frac12\, (g_{ca, b} + g_{cb, a} - g_{ab, c}) = \frac12\, \left(\partial_{b}g_{ca} + \partial_{a}g_{cb} - \partial_{c}g_{ab}\right) \,.$$

That's the way I was taught to calculate them.

3. Jan 22, 2013

### ck99

I have also used that method, and am trying to apply it this problem at the moment as "backup". However, because I have a 4D metric I have 40 possible christoffel symbols even allowing for symmetry. I thought the Lagrangian method would be quicker but I obviously haven't applied it correctly...

4. Jan 22, 2013

### Staff: Mentor

Why don't you use compuchip's standard formula just to spot check some of the coefficients that you calculated with the Lagrangian method? This might also give you a hint as to where you are making your mistake.