Connectedness of boundary points in Schwarzschild Penrose chart

[PeterDonis]

TL;DR Summary

The Penrose chart for Schwarzschild spacetime appears to show all of the boundary points "at infinity" as being connected. This thread is to investigate/discuss whether and to what degree that is actually the case.

In the Penrose chart for Schwarzschild spacetime, the boundary "at infinity" appears to be connected all the way around. I want to explore what that means physically and whether particular boundary points that appear to be connected on the chart actually are.

I am using the following notes as a reference for the Penrose chart on Schwarzschild spacetime since it contains expressions for the Penrose coordinates in terms of Kruskal coordinates that are well suited for this discussion:

https://cosmo.nyu.edu/yacine/teaching/GR_2018/lectures/lecture24.pdf

The key expressions I will use for the Penrose coordinates $\tilde{T}$ and $\tilde{X}$ in terms of the Kruskal coordinates $T$ and $X$ (I am using $X$ instead of the $R$ that appears in the notes to avoid any confusion with the Schwarzschild areal radius $r$) are as follows (the first is equation 15 in the notes, the second is derived from the second part of equation 13 in the same way as equation 15 is derived from the first part of equation 13):

$$
\tan \tilde{T} = \frac{2 T}{1 + X^2 - T^2}
$$

$$
\tan \tilde{X} = \frac{2 X}{1 + T^2 - X^2}
$$

My first question is regarding the second formula above. Outside the horizon, we have $1 + T^2 - X^2 < 0$, so this formula would imply that ouside the horizon, positive $X$ corresponds to negative $\tilde{X}$ and vice versa. That seems odd. It would seem more natural for the denominator of the second formula to be the same as that of the first, since $1 + X^2 - T^2 > 0$ everywhere (the singularities are where this value approaches zero from above), so the signs would always match. But I can't get that result from what is given in the paper (and making that change would also create issues for at least the singularity limit given below).

Ignoring the sign issue just mentioned in everything that follows from here on, let's now look at the various limit points "at infinity" at the boundary of the Penrose diagram. I will briefly describe my understanding of how the Penrose coordinates of these points are obtained. I will only discuss the "future" cases for the timelike and null infinities, since the "past" cases are just the obvious time reverses of them.

$i^+$, or "future timelike infinity", is obtained by taking the limit $T \to \infty$ while holding $X^2 - T^2$ constant (which corresponds to holding the Schwarzschild areal radius $r$ constant). This obviously gives $\tan \tilde{T} \to \infty$, which in turn gives $\tilde{T} \to \pi / 2$, and the same for $\tilde{X}$.

$i^0$, or "spacelike infinity", is obtained by taking the limit $X^2 - T^2 \to \infty$ while keeping $T = 0$; at least, that's how it's implicitly presented in the paper, by analogy with the explicit procedure described for MInkowski spacetime, if we take the Minkowski $r$ to correspond with the Schwarzschild $r$. It seems to me that, since all spacelike lines are supposed to end on $i^0$, we should do something more like keeping $T / X$ constant (which corresponds to looking at spacelike lines with a constant "slope", instead of just limiting to the horizontal line with zero slope). However, in either case, we obtain $\tan \tilde{T} \to 0$ and $\tan \tilde{X} \to 0$. This implies $\tilde{T} \to 0$ (as the paper has it this would be true by assumption), but $\tilde{X} \to \pi$; we cannot have $\tilde{X} \to 0$ in this limit because that is inconsistent with $X^2 - T^2 \to \infty$.

$\mathscr{I}^+$, or "future null infinity", is obtained by taking the limit $X + T \to \infty$ while holding $X - T = K$, where $K$ is some positive constant (it must be positive because we are outside the horizon). (The usual interpretation of this constant would be the null coordinate that labels outgoing null lines.) Each value of $K$ labels a distinct point on $\mathscr{I}^+$. This limit implies $\tilde{T} \to \pi / 2 - \tilde{K}$ and $\tilde{X} \to \pi / 2 + \tilde{K}$, where $\tilde{K} = \arctan K = \arctan (X - T) = - \arctan (T - X)$ (the last equation shows how the signs of the $\tilde{K}$ terms are obtained from equation 13 in the paper) now labels the outgoing null lines and covers the range $( 0, \pi / 2)$.

The singularity, $r = 0$, is obtained by taking the limit $X^2 - T^2 \to -1$, or $1 + X^2 - T^2 \to 0^+$ as it is stated in the paper. This limit implies $\tilde{T} \to \pi / 2$ but allows $\tilde{X}$ to assume any value in the range $( - \pi / 2, \pi / 2 )$.

The horizon is the line $T = X$ in the Kruskal diagram, which becomes the line $\tilde{T} = \tilde{X}$ in the Penrose diagram. It covers the range $( - \pi / 2, \pi / 2)$ for both coordinates.

Now for the connectedness questions.

First: does the horizon connect with $i^+$? The diagram makes it seem like it does, and if we consider the horizon to be the limiting case $K = 0$ of the family of outgoing null lines that we analyzed for $\mathscr{I}^+$ above, it ends up at $i^+$. But physically, the horizon never gets to "infinity"--that's the whole point of the horizon, that it is the boundary of the region that can't send light signals to infinity.

Second: does the singularity connect with $i^+$? Again, the diagram makes it seem like it does, and if we consider the limiting case $X \to \infty$ on the singularity, i.e., with $1 + X^2 - T^2 \to 0^+$ (which requires us to also take $T \to \infty$), we end up at $i^+$ (i.e., $\tilde{X} \to \pi / 2$). But again, physically, the singularity is inside the horizon and should not be able to connect to infinity.

 

[Orodruin]

What do you mean by ”connect”?
Remember that, just as the singularity is not part of the Schwarzschild spacetime, neither are the ”infinity points”.

 

[martinbn]

But physically, the horizon never gets to "infinity"--that's the whole point of the horizon, that it is the boundary of the region that can't send light signals to infinity.

I think you need to emphasize that the horizon never gets to null infinity, and the diagram shows that.

 

[PeterDonis]

What do you mean by ”connect”?

I mean the topological concept of connectedness, evaluated in the extended spacetime depicted in the Penrose chart. In this sense, for example, an outgoing null line outside the horizon is connected to future null infinity in the same sense as a line drawn across a circular region on a plane is connected to the circular boundary of the region.

just as the singularity is not part of the Schwarzschild spacetime, neither are the ”infinity points”.

Yes, but those points are part of the extended spacetime depicted in the Penrose chart. The whole point of Penrose charts is to make the "infinity points" into actual "places" at which one can evaluate tensors and tensor equations without having to take limits, in order to be able to have precise definitions of concepts like asymptotic flatness, ADM and Bondi energies, etc.

(Granted, asking about connectedness of the singularity is stretching the above since tensors and tensor equations are not well-defined on the singularity; but all tensors and tensor equations are perfectly valid on the horizon so the question about the connectedness of the horizon should not involve any issues.)

 

[PeterDonis]

I think you need to emphasize that the horizon never gets to null infinity, and the diagram shows that.

I agree, but I am specifically asking about whether the horizon connects to future timelike infinity. The diagram makes it appear that it does, but for the reasons I gave, that seems odd to me. It's possible that this is just a mathematical artifact of the way the Penrose chart is constructed and doesn't correspond to anything physical.

 

[DrGreg]

But physically, the horizon never gets to "infinity"--that's the whole point of the horizon, that it is the boundary of the region that can't send light signals to infinity.

I agree, but I am specifically asking about whether the horizon connects to future timelike infinity. The diagram makes it appear that it does, but for the reasons I gave, that seems odd to me.

I'm no expert in Penrose charts (so I can't guarantee what I'm about to say is correct), but as I understand it, future timelike infinity $i^+$ represents a finite distance after an infinite amount of time i.e. anything that persists forever but remains within a finite distance. So that would include an event horizon (a classical, non-evaporating one). For something to never reach $i^+$ or $\mathscr{I}^+$, it would have to cease to exist after a finite time?

 

[PeterDonis]

as I understand it, future timelike infinity $i^+$ represents a finite distance after an infinite amount of time i.e. anything that persists forever but remains within a finite distance.

As I understand it, the limit that is taken to reach $i^+$ does not just involve "a finite distance" (constant Schwarzschild $r$), it involves a finite distance outside the horizon. The very name of this infinity, future timelike infinity, indicates that: curves of constant $r$ are only timelike outside the horizon. The horizon itself is null, not timelike.

 

[PeterDonis]

Outside the horizon, we have $1 + T^2 - X^2 < 0$, so this formula would imply that ouside the horizon, positive corresponds to negative and vice versa.

Actually I misstated this. At the horizon, $T^2 - X^2 = 0$, so $1 + T^2 - X^2 = 1$, not $0$. The quantity $1 + T^2 - X^2$ remains positive until the Schwarzschild $r$ reaches twice the Schwarzschild radius, at which point $T^2 - X^2 = -1$. As that point is approached from the left (i.e., from smaller values of $r$, i.e., in the limit $1 + T^2 - X^2 \to 0^+$), $\tan \tilde{X} \to \infty$, so $\tilde{X} \to \pi / 2$.

For $\tilde{X} > \pi / 2$, we have $1 + T^2 - X^2 < 0$, but in that range we also have $\tan \tilde{X} < 0$ even though $\tilde{X} > 0$, because we are now on the next branch of the $\tan$ function. So the sign switch is not actually an issue; the signs of $\tilde{X}$ and $X$ do remain the same. Only the sign of $\tan \tilde{X}$ flips. So that question from the OP is resolved.