Null geodesics in FRW metric: why angular coordinates are constant? Consider a ray passing through $r=0$ in the FRW metric
$ds^2 = -dt^2 +a(t)^2(\frac{dr^2}{1-kr^2} + r^2(d\theta^2 + \sin{\theta}^2d\phi^2))$
The geodesic curve is parametrized by the affine parameter $\lambda$,  such that $x^\mu = (T(\lambda),R(\lambda),\Theta(\lambda),\Phi(\lambda)$.
How is it possible to show, in a mathematically convincing way, that $\Theta(\lambda)$ and $\Phi(\lambda)$ are constant?  
 A: I would make an argument using Killing Vector Fields. Since the metric is not dependent on $\phi$, the vector $\left(\frac{\partial}{\partial \phi} \right)^a$ is a KVF. That is, the quantity
$L = u_a \left(\frac{\partial}{\partial \phi} \right)^a = g_{ab} u^b \left(\frac{\partial}{\partial \phi} \right)^a = g_{\phi \phi} u^{\phi} = r^2 \sin^2{\theta} \Phi'(\lambda)$
is conserved along a geodesic (where $u^a$ is the tangent vector to the geodesic). Since your geodesic passes through $r=0$ (presumably at the parameter $\lambda = 0$), we have $r = 0$ at some point along the geodesic. Since $L$ is conserved, we may evaluate it at this point ($\lambda = 0$), and we find $L = 0$ for all points along the geodesic. So we must have $\Phi'(\lambda) = 0$ for all points along the geodesic (or $\sin^2{\theta} = 0$, but this is a trivial case). This is equivalent to the statement that the coordinate $\Phi(\lambda)$ is constant.
To show that $\Theta(\lambda)$ is constant, we need to be a little more careful (since we don't have an equivalent Killing Vector Field). We use the fact that $x^{\mu}$ is a geodesic. So we have
$u^a \nabla_a u^b = 0.$
$u^a (\partial_a u^b + \Gamma^{b}_{ac} u^c) = 0.$
If you compute out the Christoffel tensor components and solve this equation, you can easily show (though with a fair amount of computation), that $\Theta(\lambda)$ does not change with respect to $\lambda.$
