2
$\begingroup$

I'm trying to confirm that the $\Gamma^1_{01}$ Christoffel symbol of the FRW metric is $\dot{a}/a$.

I have the FRW metric:

$$ds^2=-dt^2+a(t)^2\left[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta\ d\phi^2)\right]$$

I have an equation for the Christoffel symbols:

$$\Gamma^\sigma_{\mu \nu}=\frac{1}{2}g^{\sigma \rho}\left[\frac{\partial g_{\nu\rho}}{\partial x^\mu} + \frac{\partial g_{\rho \mu}}{\partial x^\nu} - \frac{\partial g_{\mu \nu}}{\partial x^\rho}\right]$$

As far as I can see the Christoffel symbol that I want should be given by:

$$\Gamma^1_{01}=\frac{1}{2}g^{11}\left[\frac{\partial g_{11}}{\partial x^0} + \frac{\partial g_{10}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^1}\right]$$

$$\Gamma^1_{01}=\frac{1}{2}.\frac{a^2}{1-kr^2}.\frac{2a\dot{a}}{1-kr^2}$$

Where have I gone wrong?

Ok - I now understand that $$g^{\sigma \rho}=\frac{1}{g_{\sigma \rho}}$$

So that

$$\Gamma^1_{01}=\frac{1}{2g_{11}}\left[\frac{\partial g_{11}}{\partial x^0} + \frac{\partial g_{10}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^1}\right]$$

$$\Gamma^1_{01}=\frac{1}{2}.\frac{1-kr^2}{a^2}.\frac{2a\dot{a}}{1-kr^2}=\frac{\dot{a}}{a}$$

$\endgroup$

closed as off-topic by ACuriousMind, Kyle Kanos, John Rennie, Martin, JamalS May 24 '15 at 20:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Kyle Kanos, John Rennie, Martin, JamalS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ You've set $\rho$ to 1, but it should be summed over it. $\endgroup$ – MBN May 21 '15 at 13:05
  • 2
    $\begingroup$ Just a minor comment, because I see it a lot. $g^{\sigma\rho}=1/g_{\sigma\rho}$ is easy to write, but depending on what index notation you are using (component or abstract), it may be nonsense. If you mean that "each component of this particular tensor is 1 / ( the component of this other particular tensor )", then I'm fine, but that's only true for the metric if the metric is diagonal. The real relation is $g^{\sigma\rho}g_{\sigma\rho}=1$, true in both abstract and component notation. $\endgroup$ – levitopher May 21 '15 at 13:24
1
$\begingroup$

You set $\rho$ equal to one for no reason. In detail the expression for $\Gamma$ is:

$$\Gamma^1_{01}=\frac{1}{2}\sum_\rho g^{1\rho}\left[\frac{\partial g_{1\rho}}{\partial x^0} + \frac{\partial g_{\rho 0}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^\rho}\right].$$

And since $g_{01}$ equals zero (since your metric is diagonal), all four partial derivatives $\frac{\partial g_{01}}{\partial x^\rho}$ are zero. And as for the four terms with $g_{\rho 0}$ only $\rho=0$ survives (since your metric is diagonal) but $g_{00}=-1$ so its partial derivative is zero. Finally for the four terms with $g_{1 \rho}$ only $\rho=1$ survives (since your metric is diagonal) and that's why what you wrote works.

As for your expression for $g^{\sigma\rho}$ that also only worked for a diagonal metric. In general you need $\sum_\rho g^{\sigma\rho}g_{\rho\beta}$ to equal 1 if $\sigma=\beta$ and zero otherwise. Sometimes written $\sum_\rho g^{\sigma\rho}g_{\rho\beta}=\delta^\sigma_\beta$. That equation is just like the equation for a matrix inverse so the both upper and the both lower versions of $g$ have components that taken all together are matrix inverses of each other.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.