On the gravitational collapse of a massive shell

[Mattergauge]

The "singularities" at the horizon of Schwarzschild spacetime are coordinate singularities only, not "artificial". They are not singularities of the spacetime geometry.No "replacement" is necessary; the Schwarzschild spacetime already only has a singularity of the spacetime geometry at $r = 0$. See above. (Note that this singularity is not "at the center", however; it's a moment of time, not a place in space.)

Can we say how the black hole manifold looks like objectively? I mean, without putting coordinates on it?

 

[ergospherical]

What does "ingoing" stand for? Freely falling?

Nope. There is a tortoise coordinate $r_{\star}$ defined for $r > 2M$ as\begin{align*}
r_{\star} = r + 2M \mathrm{log}\left( \dfrac{r-2M}{2M} \right)
\end{align*}from which follows $dr_{\star} = (1- 2M/r)^{-1} dr$. Radial-travelling light rays follow null trajectories with $ds^2 = 0 \implies dr_{\star}/dt = \pm 1$. Introduced are the null coordinates\begin{align*}
v &= t+r_{\star} \\
u &= t - r_{\star}
\end{align*}for which constant $v$ corresponds to the ingoing geodesics and constant $u$ corresponds to the outgoing geodesics. The coordinates $(v,r,\theta,\phi)$ are the ingoing EF coordinates, and by analytic continuation across $r=2M$ they cover the black hole interior region as well as the exterior. The coordinates $(u,r,\theta,\phi)$ are the outgoing EF coordinates, and by analytic continuation across $r=2M$ they cover the white hole interior region as well as the exterior.

 

[Mattergauge]

Nope. There is a tortoise coordinate $r_{\star}$ defined for $r > 2M$ as\begin{align*}
r_{\star} = r + 2M \mathrm{log}\left( \dfrac{r-2M}{2M} \right)
\end{align*}from which follows $dr_{\star} = (1- 2M/r)^{-1} dr$. Radial-travelling light rays follow null trajectories with $ds^2 = 0 \implies dr_{\star}/dt = \pm 1$. Introduced are the null coordinates\begin{align*}
v &= t+r_{\star} \\
u &= t - r_{\star}
\end{align*}for which constant $v$ corresponds to the ingoing geodesics and constant $u$ corresponds to the outgoing geodesics. The coordinates $(v,r,\theta,\phi)$ are the ingoing EF coordinates, and by analytic continuation across $r=2M$ they cover the black hole interior region as well as the exterior. The coordinates $(u,r,\theta,\phi)$ are the outgoing EF coordinates, and by analytic continuation across $r=2M$ they cover the white hole interior region as well as the exterior.

Ah yes. So there will not be a singularity anymore for R? Is a white hole necessarily included?

 

[PeterDonis]

Can we say how the black hole manifold looks like objectively? I mean, without putting coordinates on it?

If you can describe "what it looks like" without coordinates, sure. For example, you could describe the tidal gravity in coordinate-free terms.

 

[PeterDonis]

Yes. Viz, ingoing Eddington-Finkelstein coordinates.

It would help if you said what this is in response to.

 

[PeterDonis]

So there will not be a singularity anymore for R?

If by $R$ you mean the horizon, Eddington-Finkelstein coordinates do not have a coordinate singularity at the horizon.

Is a white hole necessarily included?

The two different kinds of Eddington-Finkelstein coordinates (ingoing and outgoing) are two different coordinate charts, covering two different patches of the maximally extended Schwarzschild spacetime. Ingoing E-F coordinates cover the exterior region and the black hole region; outgoing E-F coordinates cover the exterior region and the white hole region.

For a chart that covers all of maximally extended Schwarzschild spacetime, I suggest looking at Kruskal-Szekeres coordinates.

 

[ergospherical]

Ah yes. So there will not be a singularity anymore for R?

Recall that the Schwarzschild metric in terms of $(t,r,\theta,\phi)$ has two singularities: $r=0$ and $r=2M$. The latter is in fact merely a coordinate singularity, as can be established by consideration of the Kretschmann scalar $K \sim M^2/r^6$ for the Schwarzschild solution (it does not diverge at $r=2M$, in contradistinction to its behaviour at $r=0$). Re-writing the metric in terms of ingoing Eddington-Finkelstein coordinates removes the coordinate singularity at $r=2M$.

Is a white hole necessarily included?

If you use ingoing EF coordinates, the cross term in the metric is $+2dvdr$. If you use outgoing EF coordinates, the cross term in the metric is $-2dudr$. In outgoing EF coordinates, ingoing null geodesics which start inside the horizon move outward toward the horizon, whilst ingoing null geodesics which start outside the horizon cannot ever cross the horizon and reach the singularity at $r=0$. In other words, the metric describes the interior of a white hole.

 

[Dale]

Can we say how the black hole manifold looks like objectively? I mean, without putting coordinates on it?

Yes. You can simply say that it is the unique spherically symmetric vacuum manifold.

 

[Mattergauge]

Yes. You can simply say that it is the unique spherically symmetric vacuum manifold.

But how does it look? How can a black hole be vacuum? Flat vacuum is spherically symmetric too.

 

[Mattergauge]

Ah! There is no birdview possible...

 

[Dale]

But how does it look? How can a black hole be vacuum? Flat vacuum is spherically symmetric too.

Oops, you are right, I forgot to mention that it was curved with a length scale $R_s$. It isn’t unique otherwise. Minkowski spacetime is the one with $R_s=0$

The point is that we don’t have English words for this so we have to describe such shapes using math. If we want to describe it using math without coordinates then we do so by setting up the equations that it obeys. Of course, in GR they all obey the EFE, so we just describe the stress energy tensor, any symmetries, and the boundary conditions. In this case it is sufficient to mention the spherical symmetry, vacuum, and curvature length scale. That uniquely describes the shape.

Ah! There is no birdview possible...

It is possible, but not necessary.