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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?

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  • $\begingroup$ It's a consequence of spherical symmetry. The tangent vector to the geodesic when it passes through the origin chooses some direction on the sphere, and any deflection from this direction would break the symmetry. $\endgroup$ – Holographer Apr 21 '15 at 21:14
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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 $\lambda.$

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  • $\begingroup$ thanks for your answer! I would just like to ask: in the $\Theta$ argument, why do you say that $u^a \nabla_a u^b$ should be equal to 0? $\endgroup$ – johnhenry Apr 21 '15 at 21:57
  • $\begingroup$ This comes from the definition of an affinely-parameterized geodesic. See Wald p. 41 $\endgroup$ – user35736 Apr 21 '15 at 23:38

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