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ck99
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Homework Statement
Find the non zero Christoffel symbols of the following metric
[tex]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 ) [/tex]
and find the non zero Christoffel symbols and Ricci tensor coefficients when k = 0
Homework Equations
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
[tex]\frac{d}{dλ} \frac{∂L}{∂\dot{q}} = \frac{∂L}{∂q}[/tex]
and for the case where q = t, so q' = t', I get
[tex]\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) [/tex]
[tex]\frac{∂L}{∂\dot{q}} = -2\dot{t}[/tex]
[tex]\frac{d}{dλ} \frac{∂L}{∂\dot{q}} = -2\ddot{t}[/tex]
Putting it all together, I get
[tex]-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) [/tex]
or just
[tex]\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[/tex]
Compare this to
[tex]\ddot{X}^a + \Gamma^a_bc \dot{X}^a \dot{X}^b[/tex]
to get
[tex]\Gamma^0_11 = \Gamma^0_22 = \Gamma^0_33 = \frac{a\dot{a}}{(1+\frac{k}{4}(x^2+y^2+z^2))^2}[/tex]
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
[tex]\Gamma^0_11 = \Gamma^0_22 = \Gamma^0_33 = a\dot{a}[/tex]
[tex]\Gamma^1_11 = \Gamma^2_22 = \Gamma^3_33 = 2\frac{\dot{a}}{a}[/tex]
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!)