Calculation of Christoffel-Symbols for a given specific metric I am stuck with the following problem.
We are given the following metric $$ds^2 = c^2dt^2 - a^2(t)\bar{g}_{ij}dx^idx^j$$ and I want to calculate its Christoffe symbols but I keep getting the wrong answers. For example $$ \Gamma^{\mu}_{00} $$ is supposed to be 0 (greek indices run from 0 to 3, and latin indices from 1 to 3), whereas I keep getting mixed metric terms. Is there some tricks someone could share easing the pain when calculating these?
Thank you
 A: $$\Gamma^\mu_{00}= g^{\mu\tau} [0 0,\tau]$$
with
$$2[0 0,\tau]= g_{\tau 0, 0} + g_{0\tau, 0}- g_{0 0, \tau}  $$
Now $g_{00}= c^2$ is constant and $g_{\tau 0}= g_{0 \tau}$ is either constant (when $\tau = 0$) or $=0$ otherwise -- so every derivative is $=0$, too. So $\Gamma^\mu_{00}=0 $ (no mixed term). If you arrive at something else write down how you got there. 
A: The Christoffel symbols can be obtained from the equations of motion using the rule : 
$$ \frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = 0$$
To obtain the equations of motion first write a Lagrangian based on your metric, 
$$ L = -c^2 t'^2 + a^2 \bar{g}_{ij} x'^i x'^j, $$
where the $'$ denotes differentiation with respect to a parameter $\lambda$ and $\dot{}$ will represent differentiation with respect to $t$.. 
The Euler-Lagrange equation for the time component of the geodesic equation is then, 
$$\frac{\partial L}{\partial t} = \frac{d}{d\lambda} \frac{\partial L}{\partial t'} $$
$$ 2a \dot{a} \bar{g}_{ij} x'^i x'^j = \frac{d}{d\lambda} (-2c^2t')$$
$$ 2a \dot{a} \bar{g}_{ij} x'^i x'^j = -2c^2t''$$
$$ c^2 t'' + a \dot{a} \bar{g}_{ij} x'^i x'^j = 0$$
$$  t'' + \color{blue}{\frac{1}{c^2}\left( \frac{\dot{a}}{a}\right)  a^2 \bar{g}_{ij} }x'^i x'^j = 0$$
From this we can see that, 
$$ \Gamma^t_{ij} = \frac{1}{c^2}\left( \frac{\dot{a}}{a}\right)  a^2 \bar{g}_{ij} $$
