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The FLRW metric which I am using is $$ds^2 = dt^2 - \frac{a(t)^2}{c^2} \left( dx^2 + dy^2 + dz^2 \right)$$ where $a(t)$ is the so-called 'scale factor'. I did not want to calculate the Christoffel symbols directly by using the metric and its derivatives. Instead, I took the Lagrangian, $$\mathcal{L} = g_{\mu \nu} \frac{\mathrm{d}x^{\mu}(s)}{\mathrm{d}s}\frac{\mathrm{d}x^{\nu}(s)}{\mathrm{d}s}$$

and then applied the Euler-Lagrange equations for each coordinate, i.e. $t,x,y,z$. Note that the scale factor $a(t)$ is a function of time, $t$, which is a function of $s$, that is, $t(s)$. I think I reach my problem when I compute the term: $$\frac{\partial}{\partial s}\frac{\partial \mathcal{L}}{\partial \left( \frac{\mathrm{d} x(s)}{\mathrm{d}s} \right)}$$

It leads to the equation:

$$\frac{\mathrm{d}^2x(s)}{\mathrm{d}s^2} + \frac{2 \dot{x} a'(t)}{a(t)}=0$$

where $\dot{x}$ is short-hand for the derivative with respect to $s$, and the $a'$ is short-hand for the derivative of $a(t)$ with respect to $s$. I do not see the derivative of another one of the coordinates in the second term, so when attempting to discern the Christoffel symbols by comparing the equation to the geodesic equation, I do not know what the $\gamma$ index should be:

$$\frac{\mathrm{d}^2 x^{\alpha}(s) }{\mathrm{d}s^2} + \Gamma^{\alpha}_{\beta \gamma} \frac{\mathrm{d}x^{\beta}(s)}{\mathrm{d}s}\frac{\mathrm{d}x^{\gamma}(s)}{\mathrm{d}s}$$

$\alpha=1$ because of the second derivative of $x$ in the equation, $\beta=1$ because of the first derivative of $x$ in the second term, but I do not know what $\gamma$ is. There is no other derivative.

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You get a $\dot{t}$ by the chain rule: $da/ds = da/dt \times dt/ds$. Remember, $a$ is a function of the spacetime coordinate $t$. –  Michael Brown May 30 '13 at 8:57
    
Thank you very much! Solved my problem. –  Anonymous May 30 '13 at 10:13
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