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

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

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

## Best Answer

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.