Is the Schwarzschild Metric Valid Inside a Black Hole?

[Chris Hillman]

Back in August 2004, kurious asked:

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?

1. There is considerable observational evidence that astrophysical black holes (think of their definining characteristic as being the presence of an event horizon; an abstract surface which operates a bit like a trapdoor---things and signals can fall in but they can't get out--- but which is nonetheless not a physical surface and is not, according to gtr, anything you could notice from local physics as you passed through it on your way into the hole), do exist in nature, and are well described even in the vicinity of the horizon by gtr,

2. In addition to specific evidence for the existence of black holes (and one should note that other theories besides gtr predicts them), there are various "tests of gravitation theories" such as the three of the "four classical solar system tests" (light bending, precession of pericenters, light delay) which tightly constrain, on the basis of observations, viable gravitation theories to closely mimim the predictions of gtr; the overall thrust of this work is that gtr appears to be distinguished as the simplest viable relativistic classical field theory of gravitation, at least within a large class of competing theories of this kind (the fourth test, gravitational redshift, validates a prediction of a large class of theories including gtr),

3. At present, there is a general expectation, based upon various theoretical arguments, that the interior of astrophysical black holes more closely resembles the Schwarzschild interior than the Kerr vacuum or Reissner-Nordstrom electrovacuum interiors,

4. This does not contradict assertions you might have heard about such as the no hair theorems or Price's theorem.

The Schwarzschild metric is an exterior solution and is derived as such. So when the empty spacetime field equation holds theoretically the Schwarzschild metric is valid.

Say instead: "the Schwarzschild solution is one of many, MANY vacuum solutions of the EFE" (it also arises as a vacuum solution in some other theories).

Recall that Birkhoff's theorem states that spherically symmetric vacuum solutions of the EFE must be static (from which it is easy to see that such a solution must in fact be locally isometric to "the" Schwarzschild solution, or more precisely, one in this one parameter family of solutions.) This does NOT mean that every vacuum solution must be locally isometric to a Schwarzschild solution, just every spherically symmetric vacuum solution. There are plenty of axisymmetric vacuum solutions which are not a Schwarzschild vacuum, for example. (These are Weyl's vacuum solutions, a rather significant generalization.)

In terms of e.g. Eddington-Finkelstein coordinates you can still investigate what happens inside the black hole using the Schwarzschild metric and it predicts (not unreasonable) ingoing null geodesics. But it's more a matter of faith than science because no experiment could be performed to investigate the validity of the Schwarzschild metric inside a black hole.

Actually, according to gtr, experiments could be performed inside a black hole by any intrepid physicist willing to dive into one (at least, if the hole were sufficiently massive for him to survive for some time inside the hole). The problem for science is that he couldn't report his results back to his more timorous colleagues waiting outside. The problem for himself is that once he falls in, his remaining lifetime will be finite and rather short.

Also, if a black hole is made from photons, would it be massless and move at the speed of light?

Since the notions of "geons" and "massless stationary axisymmetric asymptotically flat vacuum solutions" have come up recently in other threads, I should stress that neither of these neccessarily correspond to anything like "a black hole made from photons".

The 2004 thread was long and at times quite bizzarre; the incoherent posts by "Orion 1" would be suitable candidates for withdrawal, I think.

pmb_phy was puzzled by a computation he made:

I then said that when you calculate the Riemann tensor in Schwarzschild coordinates then at least one component is infinite at the horizon and that what that means physically is a bit difficult to interpret.

...
Are you saying that the component above in Scwazchild coordinates is incorrect? Are you saying that this component in these coordinates is finite when r = r_s.

It looks like pmb probably tried to compute the Riemann components with respect to the frame field for a static observer, which of course becomes ill-defined at the horizon. If you compute the Riemann tensor with respect to a frame field which is well defined at the horizon, all components are finite. The easiest way to see this is to observe that the Kretschmann invariant $$R_{abcd} \, R^{abcd}$$ is finite there, in fact blows up only as $$r \rightarrow 0$$. Sophisticates know that this is not a general criterion, but in this case we know that the Schwarzschild vacuum is Petrov type D, so we don't need to worry about gravitational radiation possibly leading to something drastic but unrevealed by curvature scalars.

Chris Hillman

 

[MeJennifer]

The problem for himself is that once he falls in, his remaining lifetime will be finite and rather short.

Actually there is something I do not understand here perhaps you can help.

Please tell me where I am going wrong: The arc distance between the infalling physicist and the center of the black hole is rapidly decreasing even faster than c but the radar distance is obviously not since curvature increases. So how can we say his proper lifetime is finite?

 

[pervect]

Actually there is something I do not understand here perhaps you can help.

Please tell me where I am going wrong: The arc distance between the infalling physicist and the center of the black hole is rapidly decreasing even faster than c but the radar distance is obviously not since curvature increases. So how can we say his proper lifetime is finite?

By "arc distance" do you perhaps mean r, the radial Schwarzschild coordinate? (I'm trying to clarify the question).

 

[Chris Hillman]

Lifetime of infalling astronaut?

Hi, Jennifer,

Please tell me where I am going wrong: The arc distance between the infalling physicist and the center of the black hole is rapidly decreasing even faster than c but the radar distance is obviously not since curvature increases. So how can we say his proper lifetime is finite?

I am not sure I understand what you mean, but two comments might help clarify the situation:

1. The "distance" I had in mind above (I didn't compute it in the post, but I have done so on many other occasions) is the "length" measured along a timelike curve (in fact, a geodesic, the world line of the freely and radially infalling astronaut). This "distance" corresponds physically to the elapsed time measured by an ideal clock carried by our astronaut. When I discussed the "pedometer distance" between two static observers in the exterior region in another post, I computed the distance measured long a spacelike curve (in fact, a geodesic); this distance does not directly correspond to a "timed slow highwalk along a taut wire between the two static observers", but with more effort we could justify this interpretation.

2. Black holes have no "center" in the sense that an ordinary star has a center.

Regardless, I encourage you to obtain a copy of the excellent popular book by Robert Geroch, General Relativity from A to B, which has the purpose of conveying correct geometric intuition for gravitational collapse and I think succeeds admirably. It is extremely readable and has no mathematical (or physical) prerequisites whatever. It's a lot of fun and I think you would enjoy it!

Chris Hillman