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PeterDonis

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- 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

##\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.

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.