Answer: Learn Asymptotic Flatness for General Relativity

[gnieddu]

Homework Statement

This is not really part of some homework, but I'd rather post it here than annoy other people on the more general threads.

I'm trying to self study the topic of asymptotic flatness related to General Relativity. The textbook I'm using tries to explain the concept by introducing a flat spacetime M with a metric, in spherical coordinates, and then modifying it.

Homework Equations

It all starts with the metric:

$$g = dt^2-dr^2-r^2(d\theta^2+sin^2{\theta}d\phi^2)$$

and expresses it in terms of a retarded system (u, r, theta, phi), where u=t-r:

$$g = du^2-2du dr-r^2(d\theta^2+sin^2{\theta}d\phi^2)$$

In order to have points at infinity, it then introduces a new coordinate system $$(u,\Omega,\theta,\phi)$$, where $$\Omega=r^{-1}$$, and gets:

$$g = \Omega^{-2} (\Omega^2du^2-2du d\Omega-d\theta^2-sin^2{\theta}d\phi^2)$$

In this way, M gets extended to a new manifold $$\hat{M}$$, which has a well-defined boundary at infinity for $$\Omega=0$$.

The metric is not yet fine at the boundary, so it gets rescaled by defining a new metric:

$$\hat{g} = \Omega^2g = \Omega^2du^2-2du d\Omega-d\theta^2-sin^2{\theta}d\phi^2$$

The Attempt at a Solution

So far, so good. Now the text says that (I report it literally here):

"from the form of the metric $$\hat{g}$$, we see that:

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega = \Omega^2$$

and

$$\hat{\nabla}^a\Omega\hat{\nabla}_au = -1$$"

and this really beats me! The best I was able to obtain, by some manipulation is:

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega=\hat{\nabla}_a\Omega\hat{\nabla}_b\Omega\hat{g}^{ab}=(\hat{\nabla}_a\Omega\hat{\nabla}_b\Omega)\Omega^{-2}g^{ab}=(\nabla_a\Omega\nabla_b\Omega)\Omega^{-2}g^{ab}=(\nabla_a\Omega\nabla^a\Omega)\Omega^{-2}$$

from which one would get that:

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega=\nabla_a\Omega\nabla^a\Omega=0$$

which is not true in general, but I can't see where I'm making a mistake...

Any help is really appreciated.

 

[nickjer]

Are you sure the book didn't make a mistake and it actually meant to use $$u$$ and not $$\Omega$$. Especially since the 2nd equation you listed with the $$u$$ and $$\Omega$$ makes sense.

Edit: I am not an expert in GR, so take what I say with a grain of salt :)

 

[gnieddu]

Are you sure the book didn't make a mistake and it actually meant to use $$u$$ and not $$\Omega$$. Especially since the 2nd equation you listed with the $$u$$ and $$\Omega$$ makes sense.

That's unlikely: in the same section, the text notes that, at the border of $$\hat{M}$$, where $$\Omega=0$$, $$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega = 0$$. This makes sense only if you start from $$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega = \Omega^2$$.

Thanks for the suggestion, anyway...

 

[George Jones]

I hope I'm not too late.

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega=\hat{\nabla}_a\Omega\hat{\nabla}_b\Omega\hat{g}^{ab}=(\hat{\nabla}_a\Omega\hat{\nabla}_b\Omega)\Omega^{-2}g^{ab}=(\nabla_a\Omega\nabla_b\Omega)\Omega^{-2}g^{ab}=(\nabla_a\Omega\nabla^a\Omega)\Omega^{-2}$$

This looks okay.

from which one would get that:

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega=\nabla_a\Omega\nabla^a\Omega=0$$

This isn't correct. It is tempting, but incorrect, to think that

$$\hat{\nabla}_a\Omega = \nabla_a\Omega$$

implies

$$\hat{\nabla}_a\Omega\hat{\nabla}^a\Omega=\nabla_a\Omega\nabla^a\Omega.$$

To get what you want, use your result above together with $\Omega = 1/r$ and the fact that connections act as partial derivatives on scalar functions.