kev said:
I can't see where it actually says that in the link you provided. Could you quote it?
Under "Examples of a static spacetime", number 1 is "The (exterior) Schwarzschild solution." I assume whoever wrote this was referring to the exterior of the event horizon in a Schwarzschild black hole, not the region beyond the surface of a stable spherical mass like a star. Of course wikipedia authors frequently make mistakes, so if you think this is such a case you could be right.
kev said:
In the context of this thread I think it would be better to call definition (a) the "non-vacuum" solution and (b) the "vacuum solution" and reserve "interior solution" to mean the region below the event horizon and exterior solution to mean the region above the event horizon. That is not the formal use of those words but it seems to be the least confusing and most natural way of using those phrases in the context of this thread.
Yes, this is what I meant when I referred to the interior region, the region of a black hole spacetime inside the event horizon. I don't think this is a nonstandard use of terminology though, if you do a google search using the terms "interior region black hole" or the terms "exterior region black hole" you find plenty of papers by physicists using the terms "interior region" and "exterior region" this way, like
this one.
kev said:
So now we come to the interesting case of the interior region below the the event horizon at r=2m. This is a little tricky. It is impossible for a particle to remain stationary at r>0 in this region but does that mean the Schwarzschild spacetime is not static in this region? I don't think it does. If we imagine a fully formed black hole that is not accumulating mass then the spacetime of the black hole is pretty much the same ten minutes in the past as it is ten minutes in the future. The geometry is not changing over time and is therefore defined as static.
You may be right that the entire spacetime is stationary and/or static (static spacetimes being a subset of stationary ones), but I'm not sure--does a stationary spacetime require that you be able to foliate the spacetime into a series of
spacelike hypersurfaces such that the curvature remains constant at any given space coordinate on different hypersurfaces, or does it just require that you be able to slice the spacetime into a stack of
any type of hypersurfaces (not necessarily spacelike) such that this is true? If you divide the Schwarzschild black hole spacetime into a stack of surfaces of constant t in Schwarzschild coordinates, then I imagine it's true that for any given R, theta, phi coordinates the curvature remains constant from one surface to another, but these surfaces are only "spacelike" in the exterior region, since in the interior region t is no longer a timelike coordinate.
Also, you didn't address my point that even if a spacetime is stationary, so it's
possible to find a foliation where curvature is unchanging from one slice to another, this foliation is not physically preferred, you can always choose a different foliation where the curvature is changing from one surface to another, like foliating the Schwarzschild black hole spacetime using surfaces of constant time-coordinate in Kruskal-Szekeres coordinates.
JesseM said:
If we look at just the combination of region I and II, this spacetime would have the weird property that there are potential geodesics that just "end" at some finite proper time without hitting a curvature singularity (consider the worldline of a particle that has been moving away from the event horizon since t=-infinity in Schwarzschild coordinates)...I think this would mean a failure of the spacetime to be "maximally extended", see pages 84-85 here.
kev said:
I am not quite sure what you are getting at here. For the sake of discussion let us consider replacing your particle with an observer that is holding a clock. At the apogee the observer stops rising and falls back to the event horizon in a finite proper time. If the observer were to stop at at the event horizon, then it would seem strange that her proper time suddenly ends at a finite time. However if you consider that the clock of the observer also stops at the event horizon it does not seem so strange. Proper time is measured by a local co-moving clock. If the clock in the hand of the observer has stopped advancing then proper time has indeed stopped for you. The gravitational time dilation factor is proper time t' =t*sqrt(1-2gm/(rc^2)) and t' goes to zero when r=2gm/c^2. There it is in plain black and white maths. Proper time stops at the event horizon.
No, no modern physicist would agree with that statement. The gravitational time dilation factor isn't absolute, it just relates the time of an observer hovering at a certain radius from the horizon with the time of an observer at arbitrarily large distances from the black hole. The distant observer will never
see the falling observer cross the horizon due to time dilation, but there is no reason to privilege the distant observer's perspective over the infalling observer's perspective, the infalling observer will cross the horizon at some finite proper time according to his own clock and then continue on towards the singularity. Any book by a physicist which discusses black holes will tell you this. For example, consider http://64.233.169.104/search?q=cache:_yDFPsUuWbgJ:irealitylib.hit.bg/Stephen%2520Hawking/A%2520Brief%2520History%2520of%2520Time/e.html+%22would+not,+in+fact,+feel+anything+special+as+he+reached+the+critical+radius%22&hl=en&ct=clnk&cd=1&gl=us of Stephen Hawking's
Brief History of Time:
Gravity gets weaker the farther you are from the star, so the gravitational force on our intrepid astronaut’s feet would always be greater than the force on his head. This difference in the forces would stretch our astronaut out like spaghetti or tear him apart before the star had contracted to the critical radius at which the event horizon formed! However, we believe that there are much larger objects in the universe, like the central regions of galaxies, that can also undergo gravitational collapse to produce black holes; an astronaut on one of these would not be torn apart before the black hole formed. He would not, in fact, feel anything special as he reached the critical radius, and could pass the point of no return without noticing it. However, within just a few hours, as the region continued to collapse, the difference in the gravitational forces on his head and his feet would become so strong that again it would tear him apart.
More textbook sources will usually mention the distinction between coordinate singularities (where infinities are only artifacts of a particular badly-behaved choice of coordinate system on a given spacetime) and physical singularities (where infinities are actually physical, like infinite curvature and infinite tidal forces). For example, from pp. 820-823 of MTW's
Gravitation:
The Schwarzschild spacetime geometry ... appears to behave badly near r = 2M; there g_{tt} becomes zero, and g_{rr} becomes infinite. However, one cannot be sure without careful study whether this pathology in the line element is due to a pathology in the spacetime geometry itself, or merely to a pathology of the (t, r, \theta , \phi ) coordinate system near r=2M. (As an example of a coordinate-induced pathology, consider the neighborhood of \theta = 0 on one of the invariant spheres, t=const and r=const. There g_{\phi \phi} becomes zero because the coordinate system behaves badly; however, the intrinsic, coordinate-independent geometry of the sphere is well-behaved there.
...
The worrisome region of the Schwarzschild geometry, r=2M, is called the "gravitational radius," or the "Schwarzschild radius," or the "Schwarzschild surface," or the "Schwarzschild horizon," or the "Schwarzschild sphere." It is also called the "Schwarzschild singularity" in some of the older literature; but that is a misnomer since, as will be shown, the spacetime geometry is not singular there.
To determine whether the spacetime geometry is singular at the gravitational radius, send an explorer in from far away to chart it. For simplicity, let him fall freely and radially into the gravitational radius ... Of all the features of the traveler's trajectory, on stands out most disturbingly: to reach the gravitational radius, r=2M, requires a finite lapse of proper time, but an infinite lapse of coordinate time ... Of course, proper time is the relevant quantity for the explorer's heart-beat and health. No coordinate system has the power to prevent him from reaching r=2M. Only the coordinate-independent geometry of spacetime could possibly do that ... Let the explorer reach r=2M, then. What spacetime geometry does he measure there? Is it singular or nonsingular? Restated in terms of measurements, do infinite tidal gravitational forces tear the traveler apart as he approaches r=2M, or does he feel only finite tidal forces which in principle his body can withstand?
...
The payoff of this calculation: according to equations (31.6), none of the components of Riemann in the explorer's orthonormal frame become infinite at the gravitational radius. The tidal forces the traveler feels as he approaches r=2M are finite; they do not tear him apart ... The gravitational radius is a perfectly well-behaved, nonsingular region of spacetime, and nothing there can prevent the explorer from falling inward.
By contrast, deep inside the gravitational radius, at r=0, the traveler must encounter infinite tidal forces, independently of the route he uses to reach there. One says that "r=0 is a physical singularity of spacetime."
...
Since the spacetime geometry is well behaved at the gravitational radius, the singular behavior there of the Schwarzschild metric components, g_{tt} = -(1 - 2M/r) and g_{rr} = (1 - 2M/r)^{-1}, must be due to a pathology there of the Schwarzschild coordinates t, r, \theta , \phi. Somehow one must find a way to get rid of that pathology--i.e., one must construct a new coordinate system from which the pathology is absent.
In subsequent pages the authors discuss various other coordinate systems for describing the same Schwarzschild spacetime in which the coordinate time to reach the horizon is finite, including Kruskal-Szekeres coordinates.
kev said:
You state that is OK for proper time to end abruptly at a finite time as long as the observer/particle meets a curvature singularity. I guess that argument is OK because by definiton "anything goes" in a singularity, right?
It's not an argument original to me, physicists have always defined "maximally extended" spacetimes as ones where all timelike curves either go on for infinite proper time or "end" at some finite proper time because they run into a singularity, but where they never end at finite proper time for any other reason. I'm not sure they'd say it was "OK" for curves to end at singularities, they'd probably expect something more subtle to happen in a theory of quantum gravity, it's just that in pure GR there is no meaningful way to continue such curves beyond the point they hit a singularity.
kev said:
First, I would ask if you are absolutely sure that the event horizon is not a singularity?
I suppose it depends how you define "a singularity", but I have read countless physicists saying there is no
physical singularity there (no locally measurable physical quantities go to infinity), if anything goes to infinity at the horizon it's an artifact of a choice of coordinate systems which is badly-behaved at the horizon.
kev said:
The path is discontinuous across r=2m. Kevin Brown of Mathpages states: "This is not surprising, because
the t coordinates are discontinuous at r = 2m, so we cannot unambiguously “carry over” the labeling of the t coordinates in the region r > 2m to the region r < 2m." Ref:
http://www.mathpages.com/rr/s6-04/6-04.htm
Note he doesn't refer to a physical discontinuity, only a coordinate discontinuity. As an analogy, suppose we start with an inertial coordinate system (x,y,z,t) in Minkowski spacetime, then consider a non-inertial coordinate system defined by the following transformation:
x' = x
y' = y
z' = z
t' = (1 second^2)/(t - 2 seconds)
So, as you approach t=2 seconds in the original inertial coordinate system, the t' coordinate of the new non-inertial system approaches infinity. Does this infinity have any physical significance? Of course not, it's just an artifact of a weird choice of coordinate systems on flat spacetime.
kev said:
I am not sure how you think including regions III and IV fix any of the problems you perceive. Even if you consider the trajectory of a falling particle to continue across the event horizon and fall all the way to the central singularity (which is the conventional interpretation) then that eventuality is covered by region II. Why the need to introduce regions II and IV?
That's why I used the example a particle which has been rising
away from the event horizon for infinite time in Schwarzschild coordinates (i.e. in the limit as the t-coordinate goes to negative infinity, the radius of the particle approaches r=2M in Schwarzschild coordinates), not a particle falling in. Even though the particle has been rising away from the horizon for infinite coordinate time, if you trace its proper time backwards from some point on its worldline outside the horizon (like the moment it passes r=3M), you find that only a finite proper time has elapsed since it crossed the horizon. Since there's no physical singularity at the horizon, if you want the spacetime to be "maximally extended" you need to include a spacetime region where its worldline passed through at earlier proper times before it crossed the horizon, back to the white hole singularity that it emerged from (region IV on the Kruskal diagram). Note that even if there has been a constant rain of infalling particles from t=-infinity in the exterior region I, a new one passing r=3M every second as measured by an observer sitting there, then for every one of those infinite infalling particles that passed r=3M
before the outgoing particle reached r=3M, the outgoing particle will have crossed paths with that infalling particle
after it crossed the horizon. So unless you do some interesting topological identification that allows the outgoing particle to cross paths with the infalling particle at two distinct points on its worldline (and where the "first" time they crossed paths from the outgoing particle's perspective was actually the "second" time from the ingoing particle's perspective and vice versa--see http://casa.colorado.edu/~ajsh/schwm.html#kruskal I linked to earlier), the outgoing particle won't have crossed paths with any of the infinite infalling particles during its outward trip from the singularity to the event horizon, which gives an idea of why the interior region the outgoing particle passed through must be different than the interior region all the infalling particles passed through.