# Understanding maximally extended Schwarzschild solution

1. Jan 1, 2009

### Hurkyl

Staff Emeritus
I'm going to use the notation in this Wiki article, and refer to the diagram therein. I assume the information here is essentially correct, aside from the fact the axes on the diagram should be T and R.

Firstly, I wonder what sort of uniqueness properties this space-time is supposed to have. Suppose I consider the general extension problem
Let M be a space-time. Do there exist other space-times N such that M is a submanifold of N?​
If M is region I -- i.e. the exterior Schwarzschild solution, then the maximally extended Schwarzschild solution (call it E) is obviously a solution to the extension problem. And E is maximal in the sense that the extension problem for E has no solutions.

However, there is another extension of region I, constructed as follows:
(2) Remove the origin
(3) Identify region I with region III. (Specifically, by identifying (T, -R, theta, phi) in region I with (T, R, pi - theta, phi + pi) in region III)
Call this extension F.

(Question: do I need to remove the origin? I can't convince myself either way)

F appears to be another maximal extension in the sense that the extension problem for F has no solutions. However, F is clearly not a "best" extension, because it is constructed as a quotient space of (a subset of) E.

This prompts a question: are all extensions of region I actually quotient spaces of subsets of E? If not, then what if we invoke other constraints (such as being vacuum solutions)?

Okay, the hard mathematical question is out of the way. I have some more lowbrow questions.

My first observation from the diagram is that in these coordinates, region I bears a very strong resemblance to a diagram one might draw for a family of uniformly accelerating special relativistic observers.
In this space-time:
(1) Curves of constant Schwarzschild time are lines through the origin
(2) Curves of constant Schwarzschild radii are hyperbolae asymptotic to the event horizon
(3) Along a line of constant t, the spatial separation between two hyperbolae of constant r is fixed
Uniformly accelerating SR observers:
(1) Lines of simultaneity are lines through the origin
(2) Observers' worldlines are hyperbolae asymptotic to the light cone from the origin
(3) Along a line of constant t, the spatial separation between two hyperbolae of constant r is fixed

Of course, the spatial distance between hyperbola differ between the two settings. I had previously intuited the perspective of an observer hovering at a fixed distance above an event horizon as looking very much like that of a uniformly accelerating observer in SR, and the above seems to cement that. So my question is is my intuition here actually correct? I want to make sure I'm not misleading myself!

Physically, this space-time looks like it's describing a white-hole that has been collapsing since the beginning of time... and when it finally collapses, it immediately forms a black hole that remains for the rest of eternity.

It seems obvious that a 'real' black holes aren't required to have matching white holes. (right?) I would expect a 'real' black hole to look more or less like the solution F I constructed above, but instead of having region IV in the past, you instead have region I (= region III) continuing backwards. (right?)

But this does prompt a hypothetical: can there be a space-time with a black hole with the following properties?
(1) There isn't a region IV
(2) Regions I and III can communicate before the formation of the black hole
(3) Regions I and III are forever separated once the black hole forms

Hrm, I thought I had more questions, but I can't think of them. So I'll settle for just asking these ones.

Last edited: Jan 4, 2009
2. Jan 1, 2009

### JesseM

I can't help with the more technical questions, but one interesting thing I learned about the Schwarzschild solution is that from the outside it is more like a "gray" hole in the sense that an outside observer at constant radius can see objects coming out of it at the same time he sees things falling in (although of course he never actually sees anything cross the horizon). This observer might see particles passing him on the way out at the rate of one per second while at the same time he sees particles falling in from a larger radius passing him at a rate of one per second, for example. But from the ingoing particle's point of view, as it nears the horizon it passes outgoing particles more and more rapidly, so that an ingoing particle manages to pass all the infinite number of outgoing particles that will ever pass the outside observer before it reaches the event horizon. Likewise, an outgoing particle will pass an infinite number of ingoing particles between crossing the horizon and reaching the outside observer at a finite distance. You can see how this works if you draw marks at equal proper time intervals along the hyperbola representing the worldline of the outside observer in the kruskal diagram, and then draw diagonals slanting up and down from each mark, representing ingoing and outgoing photons that pass the outside observer at a constant rate. So, this gives some insight into how the horizon that the ingoing particles cross is actually a different horizon than the one the outgoing particles cross, despite the fact that the outside observer sees only a single spherical object that the particles are coming from and going into.

Also, it would be interesting to draw lines of simultaneity in Schwarzschild coordinates in a Kruskal diagram--since the two coordinate systems define simultaneity differently they obviously wouldn't be horizontal in the Kruskal diagram, and I'm pretty sure all lines of simultaneity which cross the worldline of the outside observer would converge on the center of the diagram, meaning that according to the Schwarzschild coordinates definition of simultaneity, outgoing particles came out of the white hole horizon infinitely far in the past, and ingoing particles will cross the black hole horizon infinitely far in the future. Maybe this stuff is obvious to anyone who's studied the diagrams a bit, but as I said it gave me some insight into how their can be two horizons and two interior regions in a single Schwarzschild spacetime.
If you have access to MTW's Gravitation there's a Kruskal diagram of a more realistic black hole that forms at a finite time on p. 848 (I can scan and upload it if you don't). There's no white hole interior region in the diagram, and one other thing I notice is that the diagram cuts off down the middle at the 0 of the space coordinate u, not showing anything to the left of the vertical u=0 axis, I don't know if that was done just to simplify the diagram or if it's somehow incorrect to continue the diagram to the left (maybe that would imply negative radius or something?) P. 413 of Wald's General Relativity also shows a Penrose diagram (which I think is just like a slightly distorted Kruskal diagram) of a black hole that both forms and evaporates at finite times, again featuring only a single interior region, and again cutting off down the middle.
Would this just be an artifact of the way we draw the diagrams with only one spatial dimension? It would be helpful to see what a Kruskal diagram would look like if you increased the number of spatial dimensions to two--the naive guess would be that region II and IV would just look like two cones with the tips touching, but that can't be right since it would mean there'd be a simple path around the event horizon that takes you from region I to region III, I assume if the seeming separation between I and III were nothing more than an artifact of using only one spatial dimension than physicists writing about these diagrams would comment on it and not describe them as separate "universes".

Last edited: Jan 1, 2009
3. Jan 1, 2009

### atyy

Last edited: Jan 1, 2009
4. Jan 2, 2009

### Hurkyl

Staff Emeritus
It took a bit to wrap my head around it, but this is how I am picturing a spatial slice of the maximally extended Schwarzschild solution.

The key thing to realize is that the exterior solution, region I, describes a space with the topology of a spherical shell. Normally it's drawn with outer radius infinity, but for this exercise, it is helpful to use a finite outer radius, so it's easier to visualize the shell shape.

Spatial slices of region II (T > 0) and region IV (T < 0) not through the singularity also describe a spherical shell. If the slice is through the singularity, then it describes two disjoint spherical shells, one contained in the other.

The event horizons, describe the spheres that form the boundary between the regions.

So now for T != 0, we can picture the entire spatial slice: it is a nested collection of spherical shells representing (orienting increasing R to the outside)
(1a) A spatial slice of region III in the shape of a spherical shell
(1b) The event horizon, a sphere forming the outer boundary of 1a and inner boundary of 1c
(1c) A spatial slice of region II (or IV) in the shape of a spherical shell, possibly with a spherical shell hole in the middle
(1d) The event horizon, a sphere forming the outer boundary of 1c and inner boundary of 1e
(1e) A spatial slice of region I in the shape of a spherical shell

At T = 0, we have the same picture, except 1c is nonexistant, and the spheres 1b and 1d coincide.

If we have a picture where region IV doesn't exist, then the spherical shepps 1a and 1e would merge into one continuous region in pre-black hole spatial slices. But once the black hole has formed, it forms an impenetrable barrier between the region I and region III spherical shells.

I do not.

Unless I'm seriously misunderstanding something, E is a vacuum solution if and only if F is, due to the way F is constructed as a quotient of E.

5. Jan 2, 2009

### Hurkyl

Staff Emeritus
A correction: what I called F isn't even a manifold! The glueing procedure I use to construct F must also be applied to regions II and IV (the black and white holes). And as such, I need to cut out the origins of every spatial slice, not just the T=0 slice.

6. Jan 2, 2009

### JesseM

OK, here are the diagrams from MTW's Gravitation and Wald's General Relativity (once you open them you can click them again to enlarge). The first pair of diagrams from Gravitation show a black hole that forms from a collapsing star (the star shown in gray) in both Schwarzschild coordinates and Kruskal-Szekeres coordinates (the Kruskal-Szekeres diagram includes a bunch of lines corresponding to surfaces of constant t in Schwarzschild coordinates, I guess the event horizon at t=infinity and r=2M is the main one of interest since it divides the spacetime into two regions). The diagram from General Relativity shows a "conformal diagram" (which I think is the same as a Penrose diagram) of a black hole that forms from a collapsing star then evaporates at some later time (though I'm told this diagram is fairly speculative since black hole evaporation isn't possible in pure GR).

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7. Jan 2, 2009

### JesseM

Is the gluing procedure you're thinking of just a mirror reflection across a vertical line through the center of the diagram? If so, a different sort of gluing procedure, which identifies region II with IV as well as I with III, is discussed here:

He says that one objection to this involves a type of "conical singularity" at the t=0 point (which is also the point where the two white hole horizons meet and transform into the two black hole horizons, see this animation from this page on the same site)...I wonder if this means you'd have to remove this point from the spacetime in order to avoid a violation of the equivalence principle there.

8. Jan 3, 2009

### JesseM

I noticed that on p. 276 of the relativity book by Rindler that atyy linked to, Rindler wrote that "The R in the Kruskal diagram (Fig. 12.5) tells us the radius of the 2-sphere R = const of full Kruskal space through the event in question. In fact, each point in the diagram can be regarded as representing such a 2-sphere." When you say the exterior radius "describes a space with the topology of a spherical shell", are you also referring to the idea that every point on the diagram represents a 2-sphere, or are you saying something different about the topology of the entire 4D manifold (or the topology of a 3D spacelike slice)?

Either way, the idea that each point in the Kruskal diagram represents a 2-sphere has helped me better understand those embedding diagrams in which the Schwarzschild spacetime is represented as a "wormhole" connecting the two "universes" (region I and region III). Here's an example of such a diagram:

http://www.wishop.com/science/Astronomy/Black%20Holes/Black%20holes%20Inside%20'em_files/wormhole.gif [Broken]

In this diagram we see a series of 5 horizontal spacelike slices through the Kruskal diagram, labeled A-E; each spacelike slice of 4D spacetime is supposed to represent a curved 3-space, but since we can't visualize curved 3D surfaces, one of the spatial dimensions is suppressed to give an http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag.html [Broken] showing the curvature of a 2D surface in this space. In each embedding diagram, we see both radial lines which would correspond to the radial Schwarzschild coordinate, and a series of concentric circles which would correspond to an angular coordinate, call it theta. If we pick a particular radial line with a fixed value of theta in each diagram--say, theta=0--we can see that each point on that radial line would have a corresponding circle on the embedding diagram, a circle defined by picking a fixed radius and allowing theta to vary from 0 to 2pi; and since an embedding diagram suppresses one dimension, we know this means that each point on a radial line is "really" associated with a 2-sphere defined by picking a fixed radius and allowing two angular coordinates theta and phi (if we're using spherical coordinates) to vary. So, if we map the horizontal lines representing spacelike slices on the Kruskal diagram to a radial line on the embedding diagram for that spacelike slice, this makes it easier to see the meaning of the statement that each point on the Kruskal diagram is associated with a 2-sphere.

(an animated diagram with the event horizons drawn in can be seen here, from this page which I linked to earlier...this page uses a different series of spacelike slices of the Kruskal diagram, they aren't all horizontal like in the diagram posted above)

With this in mind, it's interesting to think about what the embedding diagrams would look like if we took spacelike slices through the Kruskal diagram for a more realistic black hole that forms at some finite time-coordinate from a collapsing sphere of matter. I couldn't find any diagrams like this online or in books I have, but I have a speculation on what it would look like (approximately) which I drew up and included as an attachment. Do you think it looks about right? If it is, it would explain why in the case of a realistic black hole, it doesn't make sense to continue the Kruskal diagram past 0 into negative values of the space coordinate u...before the formation of the singularity, u=0 would just be the bottom of a gravity well in an infinite plane (in the embedding diagram), not the middle of the "throat" of a wormhole as in the middle slices of the Kruskal diagram for an eternal Schwarzschild black hole (where negative u just means going further through the wormhole, past the middle of the throat). To verify that my speculation is right one would actually have to calculate what the correct embedding diagram would be for a spacetime containing a realistic black hole that forms out of collapsing matter, of course.

edit: one minor error I just spotted in my diagram was that I wrote that the singularity was at u=0 in the top embedding diagram, but the u coordinate in a Kruskal diagram is not the same as the radial coordinate in Schwarzschild coordinates, the singularity in the top spacelike slice I had drawn was actually at a larger u-coordinate...this could be easily fixed by just picking a slightly lower spacelike slice which intersects with the curve representing the singularity at u=0.

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9. Jan 4, 2009

### yuiop

With reference to the Wiki article, would you agree that regions I and II cover all time and all space in this universe? In other words they cover the space region from R = 0 to R = +infinity and the time extends from T = -infinity to T = +infinity. Therefore regions III and IV do not represent anything in our universe now or in the infinite past or the infinite future.

To quote from mathpages:

"However, the physical applicability of this analytically complete solution is highly dubious, because no known physical process would lead to such a result. The “black holes” to be discussed in Section 7, hypothesized to result from the gravitational collapse of stars, do not entail this complete solution, so the global topology of the complete Schwarzschild solution, as exhibited by the Kruskal coordinates, is presumably of only theoretical interest."

See http://www.mathpages.com/rr/s6-04/6-04.htm

The construction of regions III and IV is entirely imaginary and arbitrary. Various geometries can be made depending on whether you want to make the construction by an arbitrary rotation or by mirror imaging the real coordinates. What you end up with entirely fanciful. You could just as usefully fill regions III and IV with the words "Beyond here be dragons (or pink unicorns)"

Last edited: Jan 4, 2009
10. Jan 4, 2009

### Hurkyl

Staff Emeritus
Note: to align with what seems to be more standard notation, I'm going to use (u, v) to refer to what the wiki article calles (R, T). And I'm going to ignore the wiki article's (U, V).

I'm happy with the Kruskal-Szekeres diagram 32.1(b) from Gravitation; it's how I imagined things would look. Just to make sure things are clear:
(A) outside of the star, the metric is exactly that of the exterior Schwarzschild solution, right?
(B) The metric is still smooth at u=0 inside the star?

In the Kruskal-Szekeres coordinates, it looks like the surface of the star doesn't meet the singularity at u=0, but instead at some positive value for u. Is this correct? If so, is the limiting volume of a spatial slice still zero, or is it actually a positive value? (Or do something strange?)

I assume this diagram is fairly stable; small changes to the initial conditions will result in roughly the same overall shape. Is that correct?

In terms of glueing, yes I was thinking that each point in the diagram represents a 2-sphere. I've since realized a more 'visual' way of producing a full spatial slice:

The (spatial slices of the) manifold I call F is formed by taking one horizontal line from the Kruskal-Szekers diagram, placing it as a line in R^3 so that it's origin is at the origin of 3-space, fix the origin in place, and then rotate it around in all possible directions, so that each point traces out a 2-sphere. This identifies region I with III, some points of II with other points of II, and introduces a singularity at the origin. (which I promptly cut out.

Spatial slices of E can be formed by the same process, but by first deforming the diagram by a transformation like $u' = e^u$ so that each horizontal line lives entirely within the interval $(0, +infty)$. In this diagram, no extra singularities are introduced, and "infinity" for region III lies at the origin of 3-space. ('infinity' for region I lies at the 'infinity' of 3-space in both visualizations)

So, F is indeed different from the mirror wormhole you mention. (Does it really count as a wormhole if you can't exit after entering? Incidentally, his rebuttal to the second objection isn't strictly correct; a man could meet himself going in reverse if he originated in the white-hole portion. I had initially rejected this glueing, because orientation is reversed at the interface... but I suppose that's merely an oddity, not a technical limitation....

I've been trying to imagine if the 'realistic' black hole could be lifted to a parallel space-time diagram. I think you can do it if you remove u=0 from all spatial slices. But the only contact between the two space-times would be through the origin, which no longer exists because we removed it! Of course, at this point it would seem silly to do so. However, I can imagine a slight variation on things that could be described as a traversable wormhole between two parallel universes that collapses into a black hole.

11. Jan 4, 2009

### yuiop

You do realise that that origin of the Kruskal-Szekeres diagram is at the event horizon where r=2m in Schwarschild coordinates?

Is it a good idea to introduce a singularity at r=2m, when the whole point of Kruskal-Szekeres coordinates is to show that there is no singularity at r=2m?

Oh wait a minute, you cut it out. Is that the mathematical equivalent of sweeping something under the carpet?

12. Jan 4, 2009

### Hurkyl

Staff Emeritus
Only in the v=0 slice. In the other slices, it's the t=0 part of the interior Schwarzschild solution. (And note that your proposal to just look at regions I and II also omits that point.)

It means that those particular points of Kruskal-Szekeres coordinates simply doesn't refer to a point in space-time F -- just like the $v^2 - u^2 \geq 1$ region. I don't know if it is strictly necessary to exclude them; I haven't worked out if the gluing procedure would really introduce a singularity or not. (And even if it is singular, it is interesting to consider if F has an extension that fills in the hole at u=0)

13. Jan 4, 2009

### yuiop

Interior Schwarzschild solution usually refers to the spacetime inside a region that is not a vacuum. I assume here you are using “interior solution” to mean the spacetime of a vacuum below r=2m?

I get the impression in this thread that the Kruskal-Szekeres metric is being interpreted as describing the temporal evolution of a black hole. If that is the case, I would like to point out that the Kruskal-Szekeres metric is a simple transformation of the Schwarzchild metric which describes the motion of a massless point particle in the static spacetime of a fully formed black hole and neither system of coordinates describes a temporal evolution of the spacetime.

The point I was making about regions III and IV is that there is no corresponding spacetime for those regions in Schwarzschild coordinates. Do you agree with that point?

14. Jan 4, 2009

### Hurkyl

Staff Emeritus
Yes, that is what I meant, sorry about that.

15. Jan 4, 2009

### JesseM

Can't any slicing of a spacetime into a series of spacelike slices parametrized by a time coordinate (a foliation) be called a type of "temporal evolution", even if the spacetime is stationary in the sense that it is possible to find a foliation where the metric at each point in space doesn't change over time? No foliation can be more correct than any other in an absolute sense, since that would imply absolute simultaneity. Also, this page says it is only the exterior region of a Schwarzschild spacetime that can be called static, is that incorrect?
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.

edit: although I'm not sure if Schwarzschild coordinates actually cut out regions III and IV, or if they're "degenerate" in the sense that they can assign a point in III the same coordinates as a point in I, or a point in II the same coordinates as a point in IV.

Last edited: Jan 4, 2009
16. Jan 5, 2009

### yuiop

I can't see where it actually says that in the link you provided. Could you quote it?

First some definitions. These are the normal definitions used in most texts:

(a) Interior Schwarzchild solution:
The non-vacuum spacetime. For example the spacetime inside a non-collapsed star would be described by the interior Schwarzschild solution.

(b) Exterior Schwarzschild solution:
The vacuum spacetime. For example the spacetime above the surface of a non-collapsed star would be described by the exterior Schwarzschild solution. In the case of a black hole that has collapsed to a central singularity of infinite density, the exterior solution would describe the spacetime from infinity all the down through the event horizon to the central singularity. The exterior solution is a single equation that describes the spacetime above and below the event horizon as long as there is a vacuum.

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.

By those definitions, it would be reasonable to assume that a non-vacuum region might not be static, as in the case of a collapsing star where the spacetime geometry would be changing over time. Obviously it is possible to have a non-vacuum solution that is static as in the case of a non-rotating, non-collapsing star. The non-vacuum solution below the surface of the Earth would be static if we ignore the rotation of the Earth.

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. I am of course ignoring Hawking radiation which the traditional Schwarschild soltuion does not take into account. This shortcoming can be addressed by considering a black hole that is thermal equilibrium with the Cosmic Microwave Background Radiation, so that is gaining mass from absorbed radiation as fast as it is losing mass due to Hawking radition.

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. The brain of the observer also stops at the event horizon if we assume the brain is governed by the same laws of physics that makes clocks stop. The observer is locked in a moment of time staring at a stationary clock for all eternity. Proper time has indeed stopped at a finite time for that observer.

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? First, I would ask if you are absolutely sure that the event horizon is not a singularity? For example, try integrating the path of a photon from above the event horizon to below the event horizon and see what you get. 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

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 III and IV?

Last edited: Jan 5, 2009
17. Jan 5, 2009

### Hurkyl

Staff Emeritus
What parametrization gives a geometry unchanging over time? I thought about it a little bit, but none were obvious. Sure, the Schwarzschild metric is independent of t, but in region II that only implies that it's spatially uniform.

It sounds like you're interpreting him as talking about an observer that decides to hover exactly on the event horizon. Of course his proper time is going to 'stop', because he's now travelling along a null vector. JesseM was talking about geodesics; if you exclude regions III and IV, then you have a whole slew of geodesics that end for no (local) reason; temporal ones that vanish after being traced back a finite duration, and spatial ones that vanish after finite distance. Having geodesics vanish at the singularity is a problem too, but at least there is a (local) justification for not continuing space-time through the singularity -- the metric would not be differentiable (let alone satisfy the EFE).

(Note there's a singularity at infinity too; but that's not a problem because it takes infinite time/distance for a geodesic to reach)

Integrate what along the path?

18. Jan 5, 2009

### JesseM

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.
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.
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.
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 [Broken] of Stephen Hawking's Brief History of Time:
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:
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.
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.
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.
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.
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 this page 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.

Last edited by a moderator: May 3, 2017
19. Jan 5, 2009

### yuiop

I can only answer that by asking what is different about a thermally stable black hole at time t1 and at a later time t2 if no mass is being added or lost? The no hair theorem states a black hole is completely characterized by its mass, electrical charge and angular momentum. By definition a a Schwarzschild black hole has zero electric charge and zero angular momentum so they are constants so the Schwarzschild black hole is completely characterised by its mass. if the mass of Schwarzschild black hole is unchanging over time then nothing about the Schwarzschild black hole is changing over time. Of course you can say that the coordinate time parameter is advancing anyway even if nothing physical is changing, but that argument applies equally to both regions I and II and that is not in the spirit of the definition of the Schwarzschild solution as being a static solution.

Sorry you have lost me here. (Probably more my fault than yours. :P) Is there any chance of posting a sketch of a vanishing geodesic in Kruskal coordinates?

The velocity of a photon in Schwarzschild coordinates is given by :

$${dr \over dt} = c \left(1-{2m \over rc^2} \right)$$

Invert the above velocity equation and integrate with respect to r to obtain the motion of a photon in terms of r and t coordinates:

$$\int_{r_1}^{r_2}\left(\frac{dt}{dr}\right){dr}=\int_{r_1}^{r_2}c \left( 1-\frac{2m}{rc^2}\right)^{-1} {dr}$$

==>

$$t=\left[(rc^2+2m \,ln(2m-rc^2))/c^3\right]_{r_1}^{r_2}$$

==>

$$t=\frac{(r_{_2}-r_{_1})}{c} + \frac{2m}{c^3} \,ln\left(\frac{2m-r_{_2}c^2}{2m-r_{_1}c^2}\right)$$

from the above equation it can be seen that if r1 and r2 are both greater than 2m or if r1 and r2 are both less than 2m then the argument of the logarithm is positive and the solution is real. If r1 and r2 span the event horizon (eg set r1=3m and r2=m), then the solution has imaginary components. That would suggest (to me anyway) that the equation for the motion of a photon in Schwarschild coordinates is discontinuous across the event horizon.

In your first post you suggested identifying (T,-R) in region I with (T,R) in region III. That suggests to me that you are using a construction that looks like this:

rather than the mirror image version that looks like this:

Just wanted to check we are all singing from the same hymn sheet

20. Jan 5, 2009

### George Jones

Staff Emeritus
I'm afraid I can't devote as much time as I would like (maybe in a few days) to this thread, but I would like to make a few points.
What you suggest be called the "interior region" is usually called the black hole region (of vacuum Schwarzschild solution). The black hole region of a spacetime is the part of the spacetime that is not in the causal past of future null infinity. In other words, event p is in a black hole region of spacetime if there is no future-directed timelike or lightlike curve that extends from p off to inifinity. Of course, this definition can be stated more precisely.
This is wrong. Regions II and IV are neither static nor stationary. In spite of the symbol used, the Killing vector $\partial / \partial t$ is a spacelike, not timelike, killing vector in regions II and IV. Unfortunately, much confusion has been caused by some very poor notation; $\left( t, r , \theta, \phi \right)$ are coordinate labels for disjoint charts, and it would make more sense, and would be better pedagogically, if different coordinate labels were used in the different charts. See also

What I wrote above is relevant. You're trying to follow the worldline of an observer by using disjoint charts that don't contain all the events of the observer's worldline. Note that "$r=2M$" is not part of either chart.