Conformal compactification of Minkowski spacetimes

Question

Let's say we compactify the Minkowski space-time through the following coordinate transformation:

$$u=t-r, \\ \Omega=\frac{1}{r},\\ \theta=\theta,\\ \phi=\phi.$$

The conformally rescaled unphysical metric becomes, $$\hat{ds}^2= -\Omega^2 ds^2 = -\Omega^2 du^2 + 2 du d\Omega + d\Omega_2^2,$$ $d\Omega_2^2$ being the metric of a unit 2-sphere.

Now I understand, quite naturally, the conformal factor $\Omega$, goes to zero as $r \rightarrow \infty$. Can someone please give me a proof of how $\nabla_a \Omega = \partial_a \Omega$ is non-zero as $r \rightarrow \infty$? I have also heard claims that if the conformal factor goes like $\Omega = \frac{1}{r^2}$ then $\nabla_a \Omega$ would be zero too but $\nabla_a\nabla_a \Omega$ nonzero instead?

Answer 1

$r$ is a terrible coordinate near infinity, which can lead to these sorts of problems. Let us trade $r$ for $\Omega$ as a new coordinate (which clearly is fine everywhere away from $r=0$). Then it is clear that $$\partial_\Omega \Omega = 1,$$ with $\partial_{\mu} \Omega = 0$ for $\mu = u, \theta, \phi$. Therefore, the derivative does not vanish.

The problem with working in terms of $r$ is that by assumption this coordinate breaks down at infinity, so you need to change coordinates. Notice also that the derivative you're looking at only makes sense in the conformal extension, not in the original spacetime, which is another reason for $r$ not working.