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The Wikipedia article on the derivation of the Schwarzschild solution (https://en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution) lists 4 assumptions. The second of which is:

A static spacetime is one in which all metric components are independent of the time coordinate $t$ (so that $\tfrac\partial{\partial t}g_{\mu \nu}=0$) and the geometry of the spacetime is unchanged under a time-reversal $t \rightarrow -t$.

How do we know that this assumption is physical? How do we know that spacetime doesn't invert under a time-reversal instead? Is there any experimental data to support it?

aepryus
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    The shwarzschild solution doesn't describe real astrophysical black holes in the first place. – Triatticus Jan 06 '23 at 00:34
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    Building on what @Triatticus said, that specific boundary condition is actually not physical, because it leads to a white hole, which never appear if you require a black hole to form from gravitational collapse of matter. In fact, for a long time black holes were thought not to exist, basically because the Schwarzschild spacetime was thought to be a singular situation requiring a lot of unrealistic symmetry. What really convinced people to take black holes seriously were the Hawking-Penrose singularity theorems that showed singularities were a generic consequence of gravitational collapse. – Andrew Jan 06 '23 at 00:56
  • @Andrew So, the Schwarzschild solution is wrong? What would the effect of inverting this boundary condition be? Would it fix it? – aepryus Jan 06 '23 at 01:22
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    @aepryus No, the Schwarzschild spacetime is not wrong; it is a perfectly valid solution of Einstein's equations. It's just that real black holes actually had to form from the collapse of matter. They aren't an eternal part of the Universe (which is what that boundary condition assumes). So, in effect, "part" of the Schwarszchild spacetime is a good approximation for real black holes after they collapse, but you can't take the full spacetime back to the infinite past absolutely seriously from a physics point of view. – Andrew Jan 06 '23 at 01:23
  • I think Andrew has the makings of a good answer here, I'm not well versed enough in GR but that's kind of what I was hinting at that it's really a good example of exactly solving the Einstein field equations given some assumptions. We also know the spacetime isnt static since black holes can accrete matter and can also decay which shows the spacetime can change with time. It's just really cool because it doesn't require any matter to make this solution, recall that it is also a vacuum solution to the EFE. – Triatticus Jan 06 '23 at 01:38
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    It is completely normal in mathematical physics to make nonsensical assumptions (point particles, plane waves, rigid bodies, ...) in order to get equations we can actually solve. In a real physical situation, then, you must deal with the fact that your math is always formally wrong when you look at the problem in sufficient detail. Still, idealized math can often be useful for modeling the phenomena. – John Doty Jan 06 '23 at 15:54
  • "All models are wrong, but some are useful". – Tom Jan 07 '23 at 15:13

2 Answers2

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This is a very long-winded answer, so let me state the punchline up front: the boundary condition you pointed out really is physically suspect, and probably not realized in Nature -- however using it anyway leads to a very useful exact solution of Einstein's equations!

The full Schwarzschild solution (sometimes called the maximally extended Schwarzschild solution) is the metric which solves Einstein's equations subject to the boundary conditions you cite. The solution actually describes 4 interconnected regions of spacetime, as can be seen in this diagram (which I took from https://jila.colorado.edu/~ajsh/insidebh/penrose.html):

enter image description here

First, a note on the rules for how to read this diagram.

  • Time flows from bottom to top.
  • "Left/right" motion is essentially "radial" motion toward or away from the center of the spacetime.
  • The diagram is 2 dimensional but spacetime is 4 dimensional; in fact, each point on this diagram represents a 2-sphere at the same time and radius.
  • Light rays travel at 45 degree lines. Any massive particle must travel on a path whose tangent is always less than 45 degrees.

With that in mind, we can see there are 4 distinct regions. Let's start with the right diamond, labeled "Universe." This corresponds to an observer "outside" the black hole. A particle starts at the bottom of the diamond, in the infinite past, and eventually can reach the top of the diamond in the infinite future. That is, unless the particle crosses the horizon, and enters the top region, labeled Black Hole. Once an observer crosses the horizon, there is no way to get back out; one must hit the singularity (wave line) at $r=0$. These two regions correspond to a black hole that you have likely heard of.

However, essentially because of the time reversal symmetry you cite, there are additional regions in this spacetime. To the left, notice that there is a parallel universe, which can also throw things into the black hole. However, we can never cross from our Universe into the parallel universe, because doing so would require faster than light travel. However, the fact that there is a connection between these regions means this spacetime has an Einstein-Rosen bridge, or wormhole. In this case, it is a non-traversable wormhole, since no light or matter can make it from one side to the other. (Although, two observers from the two different universes could both jump into the black hole and meet there before hitting the singularity).

Finally, perhaps the most bizarre region is the bottom of the diagram, representing the white hole. Any matter that starts inside the white hole must eventually leave, and end up in one of the two Universes. If a black hole "eats" everything that falls into it, a white hole "vomits" everything that started inside.

The white hole is associated with a major theoretical issue, which is one reason why the maximally extended Schwarzschild solution is widely considered unphysical. Einstein's equations don't tell us what to do with a singularity; they equations break down there. It's possible that a quantum theory of gravity would tell us what happens near a singularity, but so far that is conjecture. For a black hole, this issue doesn't matter, because the singularity is hidden behind the horizon. From our perspective in the Universe, once something falls into the black hole, we know it hits the singularity and "something" happens, but we never need to deal with what the "something" is, because nothing can escape the horizon. (Except for Hawking radiation, but that's a whole other story). However, for a white hole, we have exactly the opposite situation. Is matter supposed to be coming out of the singularity? Is the singularity supposed to be doing nothing? We have no idea, but in order to solve the equations in the Universe part of the diagram, we would need to know what was happening in the white hole, including at the singularity. This makes the solution ill-defined.

However, a realistic, astrophysical black hole must form from the collapse of matter. (At least, most physicists don't believe there are "eternal" black holes that have always existed). The solution for a collapsing black hole is quite different (from here: https://jila.colorado.edu/~ajsh/talks/heraeus_ajsh_19/penrosecollapse_trad.html):

enter image description here

Here, the gray region represents a cloud of dust that collapsed due to its own gravity and eventually formed a black hole. As you can see, the "parallel Universe" and "white hole" parts of the diagram (as well as the anti-horizon) do not exist in this picture. Relatedly, the time reversal symmetry you pointed out is no longer present -- there was matter in the past that collapsed into a black hole; this fact breaks the time reversal symmetry.

Finally, it should be said that even this final diagram assumes a perfectly spherically symmetric collapse (it is called "Oppenheimer-Snyder collapse"). Because of all the symmetries involved in constructing these various solutions, physicists in the 1950s-60s or so widely thought that black holes were an artifact of assuming too much symmetry, and not a generic prediction of general relativity. (I take this sort of attitude to be the thrust behind your question -- you are in good company with some very good physicists in those days!).

What really started to convince people to take black holes seriously were the Penrose-Hawking singularity theorems, which showed that singularities do generically form in solutions of Einstein's equations, without needing to make any special symmetry assumptions. Penrose won the Nobel Prize for this work in 2020 (sharing it with observers who actually provided observational evidence for the existence of black holes).

The modern point of view is that the Schwarzschild solution (at least the "Universe" and "Black Hole" parts) are a good approximation to the equilibrium state of a non-spinning black hole after it has formed. This is also confirmed by numerical simulations.

Andrew
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    This answer is much appreciated. It will take me a while to consume it. "I take this sort of attitude to be the thrust behind your question" - black holes aren't bothering me; I'm thinking about the dilation of falling (and blasting) clocks https://physics.stackexchange.com/questions/743771/a-clock-falling-at-escape-velocity-another-blasting-off-at-escape-velocity and https://physics.stackexchange.com/questions/130323/a-clock-in-freefall. – aepryus Jan 06 '23 at 03:03
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    @aepryus Oh interesting. I'll take a look at those later :). If you want to calculate things like "what is the proper time for an observer to fall into a non-spinning black hole," the Schwarzschild metric is fine to use and none of these subtleties in my answer will really come up. – Andrew Jan 06 '23 at 03:13
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    Penrose's and Hawking's work was abstract and mathematical, not physical. It didn't really address the issue in practice. However, we found astronomical objects that fit the "black hole" description, and nothing else. For example, Phil Morrison's "spinar" model faltered as observations of active galactic nuclei bushed the bounds on their properties to the point where black holes and spinars were essentially the same thing. – John Doty Jan 06 '23 at 17:04
  • @JohnDoty All I can say is that if you don't believe that Penrose's and Hawking's work had deep physical insight, then I disagree. – Andrew Jan 06 '23 at 17:10
  • @Andrew I find it mathematically insightful, but what experiment can test it? Physics is not mathematics: it is experimental science. It doesn't happen on a whiteboard, it happens in the universe. – John Doty Jan 06 '23 at 17:14
  • Could one see the big bang as a kind of white hole? – Paŭlo Ebermann Jan 07 '23 at 17:20
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    @PaŭloEbermann In fact, yes, that is a valid way of looking at things. The big bang singularity is a singularity in our past light cone, similar to the white hole singularity. And the fact that objects cross the white hole "anti-horizon" and cannot return is similar to how we will never see galaxies that cross beyond our cosmological horizon again. – Andrew Jan 07 '23 at 17:31
  • @JohnDoty Of course physics is an experimental science. But I think knowing that general relativity predicts that singularities and black holes form as a generic consequence of gravitational collapse is a physically insightful statement that informs our interpretation of observational data. I agree mathematics is just a tool, and people can mistake math for physics, but it is also possible to derive physical insight using math. I'm trying to understand your position, but it's hard for me to understand how it doesn't reduce to "theory does not contribute to physics," which would be absurd. – Andrew Jan 07 '23 at 17:48
  • @Andrew Theory most certainly contributes to physics. LIGO, for example, needed GR and the theory of black holes to get its results. However, theory these days is obsessed with mathematical phenomena, like singularities, that range from impractical to impossible to observe. Meanwhile, we don't even understand electricity very well: instead we've redefined the words, so the original phenomenon is "triboelectricity", and the part we understand better is "electricity". But then, we're not even teaching students that: https://www.youtube.com/watch?v=aIhk9eKOLzQ. – John Doty Jan 07 '23 at 17:57
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The spacetime be static is not a necessary assumption for obtaining the Schwarzschild solution. Birhoff's theorem tells us that any spherically symmetric vacuum solution to the Einstein equation is also static and asymptotically flat, and therefore must be (part of) the Schwarzschild solution.

Of course as Andrew's answer explains physical solutions cannot be vacuum everywhere, and probably are not exactly spherical either. The Penrose-Hawking singularity theorems, however, tell us formation of black holes will happen generically. The No hair theorems ensure that after collapse happens the result will asymptote to a member of the Kerr family of solutions.

Consequently, working with the Kerr solution (or its special non-spinning) case as actually a reasonable physical assumption when dealing with a black hole that has had sufficient time to settle down.

TimRias
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    You can't fix doubt about a strange mathematical physics result using more math: you have to let the phenomena decide. And that's just what we did. – John Doty Jan 06 '23 at 17:06