3
$\begingroup$

In short: what is the manifold in discussion for Schwarzschild metric $$ ds^2 = -(1-\frac {2M}r)dt^2 + \frac1{1-\frac{2M}r} dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)$$ and Bertotti-Robinson metric (cf. Thorne and Blandford, "Applications of Classical Physics" http://www.pma.caltech.edu/Courses/ph136/yr2011/1025.1.K.pdf, Exercise 24.2) $$ ds^2 = Q^2(-dt^2 + \sin^2 t dz^2 + d\theta^2 + \sin ^2 \theta d\phi^2)$$

(where $Q=$const, $0\leq t \leq \pi, -\infty < z < +\infty, 0\leq \theta \leq \pi, , 0\leq \phi \leq 2\pi$)?

More description:

It seems that often, in general relativity, spacetimes are discussed purely by describing their metric tensors but not the manifold in question. This is very troublesome for me, since as far as I know, a manifold must be defined before describing metric tensor on it. For example, if one is to talk about metric tensor of a sphere $ds^2 = r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$ before defining what a sphere is, then it would be ambiguous to see what exactly $\theta$ and $\phi$ means (unless one is judging from the context, which somewhat works in this case because $\theta,\phi$ usually means angle)

So when I saw Schwarzschild metric and Bertotti-Robinson metric suddenly showing up without giving any description of which manifold we are talking about, I became very confused (and still am, having been unable to find any fruitful answer to such question of mine).

My guess for Schwarzschild is that, judging from the context, $r$ is "radius", $\theta,\phi$ are "angles" so it is "obviously true" that Schwarzschild metric is defined on the manifold $\mathbb R^4$ endowed with spherical polar coordinates for the spatial part. For Bertotti-Robinson, I'm guessing that $z,\theta,\phi$ together describe a somewhat alternative spherical polar coordinate where $z$ is height and so on, and thus that Bertotti-Robinson is also defined on $\mathbb R^4$.

$\endgroup$
  • $\begingroup$ Locally, any manifold is $\mathbb{R}^4$, and GR almost always quietly supposes these local patches are large enough that we don't have to switch between them when doing physics. $\endgroup$ – ACuriousMind Dec 7 '14 at 14:01
  • $\begingroup$ So does that mean that, in theoretical settings, manifold in discussion is generally $\mathbb R^4$? $\endgroup$ – progressiveforest Dec 7 '14 at 14:05
  • $\begingroup$ Also, I'd like to know the following important additional detail: I found a claim that Bertotti-Robinson spacetime is spherically symmetric. How can one see this? The only way that I can think of is via showing that any element in $SO(3)$ induces an isometry on Bertotti-Robinson. However, I failed to find any general, explicit formula for rotation in terms of spherical coordinates. Therefore I am clueless in trying to demonstrate it. $\endgroup$ – progressiveforest Dec 7 '14 at 14:08
  • $\begingroup$ There is some details on the matter on these notes arxiv.org/abs/1403.2371, I suggest you take a look. $\endgroup$ – user1620696 Mar 16 '18 at 17:44
1
$\begingroup$

There are many manifolds that admit the Schwarzschild metric, but the standard manifold to discuss it is the maximally extended Schwarzschild solution, which is $\mathbb R^2 \times S^2$. The easiest way to see this is to pick the Kruskal coordinates, which cover the whole manifold

$$ds^{2} = \frac{32M^3}{r}e^{-r/2M}(-dT^2 + dX^2) + r^2 d\Omega^2$$

with coordinates $(T,X,\theta,\varphi)$, with the angular coordinates describing a 2-sphere while $(T,X)$ indeed describe a region homeomorphic to $\mathbb R^2$. Since we have

$$X \in \mathbb R,\ T^2 - X^2 \in (-\infty, 1)$$

and

$$\frac {r}{2GM}=1+W_{0}\left({\frac {X^{2}-T^{2}}{e}}\right)$$

with $W_0$ the Lambert function, this means that the argument of the Lambert function is in the range $(-e^{-1}, \infty)$, so that $r \in (0, \infty)$, the metric never runs into any singularity (the values $T^2 - X^2 = 1$ corresponding to the actual singularities of the metric).

Once you have the maximally extended Schwarzschild solution, you can define a lot of other manifolds from it, either subsets or quotients. The exterior Schwarzschild solution still has the same topology : it's just a restriction to a quadrant of $(T,X)$, and in Schwarzschild coordinates you can easily see that it is just $\mathbb R^4$ with a sphere removed from every Cauchy surface, which is also equal to $\mathbb R^2 \times S^2$. The same goes for the interior Schwarzschild solution, or the Schwarzschild-Eddington-Finkelstein coordinate patch.

There are weirder topologies for the Schwarzschild solution, such as the elliptic Schwarzschild solution, which corresponds to the quotient by identification of the black hole and white hole solution (which is not causal or time-orientable), which I think has topology $\mathbb R \times S \times S^2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.