Understanding maximally extended Schwarzschild solution

In summary, the conversation discusses the uniqueness properties of a space-time diagram and explores the concept of general extension problem for a given space-time. The diagram under consideration resembles a family of uniformly accelerating special relativistic observers and raises questions about the physical interpretation of this space-time. It also delves into the concept of a "gray" hole in the Schwarzschild solution and the distinction between the horizons for ingoing and outgoing particles. Finally, the conversation touches on the idea of a hypothetical space-time with a black hole that lacks a region IV and the possibility of communication between regions I and III before the formation of the black hole.
  • #1
Hurkyl
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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:
(1) Start with E
(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! :smile:



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.
 
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  • #2
Hurkyl said:
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.
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.
Hurkyl said:
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?)
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.
Hurkyl said:
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
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".
 
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  • #4
JesseM said:
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".
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.

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).
I do not.

atyy said:
BTW, regarding the vacuum solution constraint you mentioned, is F a vacuum solution?
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
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
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
Hurkyl said:
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.
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:

http://casa.colorado.edu/~ajsh/schwm.html#kruskal

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 http://casa.colorado.edu/~ajsh/schww_gif.html from http://casa.colorado.edu/~ajsh/schww.html 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
Hurkyl said:
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.
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

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 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 http://casa.colorado.edu/~ajsh/schww_gif.html, from http://casa.colorado.edu/~ajsh/schww.html 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
Hurkyl said:
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.


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)"
 
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  • #10
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 [itex]u' = e^u[/itex] so that each horizontal line lives entirely within the interval [itex](0, +infty)[/itex]. 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
Hurkyl said:
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.)

You do realize that that origin of the Kruskal-Szekeres diagram is at the event horizon where r=2m in Schwarzschild 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
kev said:
You do realize that that origin of the Kruskal-Szekeres diagram is at the event horizon where r=2m in Schwarzschild coordinates?
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.)


Oh wait a minute, you cut it out. Is that the mathematical equivalent of sweeping something under the carpet?
It means that those particular points of Kruskal-Szekeres coordinates simply doesn't refer to a point in space-time F -- just like the [itex]v^2 - u^2 \geq 1[/itex] 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
Hurkyl said:
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.)


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 Schwarzschild 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
kev said:
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?
Yes, that is what I meant, sorry about that.
 
  • #15
kev said:
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 Schwarzschild 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.
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?
kev said:
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?
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.
 
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  • #16
JesseM said:
... Also, this page says it is only the exterior region of a Schwarzschild spacetime that can be called static, is that incorrect?

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 Schwarzschild 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 Schwarzschild 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.

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.

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?
 
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  • #17
kev said:
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.
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.



I am not quite sure what you are getting at here.
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 traveling 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)


try integrating the path of a photon from above the event horizon to below the event horizon and see what you get.
Integrate what along the path?
 
  • #18
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 [tex]g_{tt}[/tex] becomes zero, and [tex]g_{rr}[/tex] 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 [tex](t, r, \theta , \phi )[/tex] coordinate system near r=2M. (As an example of a coordinate-induced pathology, consider the neighborhood of [tex]\theta = 0[/tex] on one of the invariant spheres, t=const and r=const. There [tex]g_{\phi \phi}[/tex] 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, [tex]g_{tt} = -(1 - 2M/r)[/tex] and [tex]g_{rr} = (1 - 2M/r)^{-1}[/tex], must be due to a pathology there of the Schwarzschild coordinates [tex]t, r, \theta , \phi[/tex]. 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.
 
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  • #19
Hurkyl said:
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.

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.

Hurkyl said:
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.

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?


Hurkyl said:
Integrate what along the path?

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

[tex] {dr \over dt} = c \left(1-{2m \over rc^2} \right) [/tex]

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:

[tex] \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} [/tex]

==>

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

==>

[tex]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) [/tex]


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 Schwarzschild 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:

stworm.gif


rather than the mirror image version that looks like this:

stmir.gif


Just wanted to check we are all singing from the same hymn sheet :wink:
 
  • #20
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.
kev said:
and reserve "interior solution" to mean the region below the event horizon and exterior solution to mean the region above the event horizon.

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.
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.

This is wrong. Regions II and IV are neither static nor stationary. In spite of the symbol used, the Killing vector [itex]\partial / \partial t[/itex] is a spacelike, not timelike, killing vector in regions II and IV. Unfortunately, much confusion has been caused by some very poor notation; [itex]\left( t, r , \theta, \phi \right)[/itex] 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

https://www.physicsforums.com/showthread.php?p=1146536l#post1146536.
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?

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 "[itex]r=2M[/itex]" is not part of either chart.

The proper time of an observer does not stop. The worldline of no observer follows the event horizon. Crossing is possible, but not following.
 
  • #21
George Jones said:
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 [itex]\partial / \partial t[/itex] is a spacelike, not timelike, killing vector in regions II and IV. Unfortunately, much confusion has been caused by some very poor notation; [itex]\left( t, r , \theta, \phi \right)[/itex] 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

https://www.physicsforums.com/showthread.php?p=1146536l#post1146536.


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 "[itex]r=2M[/itex]" is not part of either chart.

The proper time of an observer does not stop. The worldline of no observer follows the event horizon. Crossing is possible, but not following.

George Jones said:
What you suggest be called the "interior region" is usually called the black hole region (of vacuum Schwarzschild solution).

I'm afraid I contributed to the confusion by referring to the interior region as the interior solution. Here is another set of definitions that I hope will clear things up:

(a)Interior region
The vacuum spacetime below the event horizon or the black hole region as suggested by George.

(b)External region
The vacuum spacetime above the surface of a solid body or above the event horizon of a black hole.

(c)Exterior solution
The Schwarzschild spacetime of a vacuum, external to a massive body. In the case of a fully formed stable black hole the exterior solution is a single equation that completely describes the exterior region AND the interior region as defined in (a) and (b).

(d)Interior solution
The Schwarzschild solution for the non-vacuum spacetime below the surface of a massive body. This is a separate equation (metric) from the exterior solution. Exactly at the surface of the massive body the interior solution mathches the exterior solution.

George Jones said:
This is wrong. Regions II and IV are neither static nor stationary.

I am curious how you define static and stationary (and how they differ). As far as I know the Schwarzschild solution has always been described as a "static solution" and this includes above and below the event horizon. My homeboy defition of static, is something that does not change over time. As I said in an earlier post, how does a thermally stable black hole evolve over time? What does it evolve to? A star is in example of something that is not static. Over time a star burns it nuclear fuel and then it might evolve to become a red giant or a white dwarf and then if it is massive enough it collapses (evolves) to become a black hole. Note that at no stage of the evolution of a star, is it a white hole. No one has ever observed a white hole, or seen any evidence of a white hole. Note, that as far as we know a thermally stable black hole is the end of the evolution and it is not evolving to become anything else. In what sense is a non-rotating black hole in thermal equilibrium not static?


George Jones said:
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 "[itex]r=2M[/itex]" is not part of either chart.

What two charts are you referring to? Do you mean the Kruskal-Szekeres coordinates for the region r>2m and the different set of Kruskal-Szekeres coordinates for the region r<2m? If that is the case, does that not hint at something special about the Schwarzschild radius (r=2m) if it is not part of either chart?

George Jones said:
The proper time of an observer does not stop...

The equation tau = t*sqrt(1-2gm/(rc^2)) suggests proper time tau does stop relative to a clock at infinity when the clock is located at r=2gm/c^2. JesseM pointed out (quite correctly) that most textbooks do not interpret the proper time of a particle to actually slow down as you get closer to the event horizon and that it is just a optical illusion due to delayed time signals and red shift as seen by observers higher up. I would counter that argument by this thought experiment:

Imagine a very massive gravitational body like a neutron star that has a tower on its surface. Half way up the tower are the usual twins and each has there own clock. One twin is sent quasi statically to the top of the tower and the other is sent quasi statically to the base of the tower. They stay separated for a number of years before returning quasi statically to the mid point of the tower and compare notes. The twin that was at the base of the tower is significantly younger than the twin that was at the top of the tower. The qravitational time dilation is not just a an optical illusion or an artifact of delayed or red shifted signals. A clock really does slow down near a black hole relative to a clock higher up, in a very physically real sense.
 
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  • #22
kev said:
The equation tau = t*sqrt(1-2gm/(rc^2)) suggests proper time tau does stop relative to a clock at infinity when the clock is located at r=2gm/c^2. JesseM pointed out (quite correctly) that most textbooks do not interpret the proper time of a particle to actually slow down as you get closer to the event horizon and that it is just a optical illusion due to delayed time signals and red shift as seen by observers higher up.
I never said that, nor do "most textbooks" say it's an "optical illusion"! The time dilation is certainly real in the sense that if two observers stay at different constant radii for a long period of time, and then meet in the middle with the travel time being negligible compared to the time at which they were staying at their constant radius, then the ratio of how much each one aged will be very close to the ratio of their time dilation factors given by the equation above. My point was just that the equation is telling you the time dilation experienced by an observer at a certain radius relative to the time of an observer at an arbitrarily large distance from the black hole--just as in SR, time dilation is always relative, never absolute. The ratio between clock ticks of an observer at r=3M and an observer at r=4M will be different from the ratio between the ticks of the observer at r=3M and an observer at r=infinity, and it's the latter that the formula is giving you, it isn't telling you how much the clock at r=3M is slowed down in "absolute" terms. Also, I believe the formula is only intended to work for clocks hovering at a constant radius, if you have a clock which dips down close to the horizon then immediately climbs back up to your radius, I don't know if you can calculate the elapsed time on that clock in terms of any simplistic integral of the time dilation factors at each radius it passed through during the trip (though I could well be wrong about that).
 
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  • #23
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 metric; the components of [itex]d\tau^2[/itex] are different at the time r=t1 and the time r=t2.


Incidentally, here's another space-time to sink your teeth into. (it's 1+1-dimensional for simplicity) The metric is given by
[tex]d\tau^2 = r dt^2 - \frac{1}{r} dr^2[/tex]
over all [itex]r, t \in \mathbb{R}[/itex]. (Except, of course, for [itex]r = 0[/itex], where the metric is singular)

A sample infalling lightlike geodesic is given in the exterior (r > 0) region by

[tex]r = \frac{1 - \tanh(t/2)}{1 + \tanh(t/2)}[/tex]

Note that it doesn't reach the horizon until [itex]t = +\infty[/itex]. After passing through the horizon, it continues along

[tex]r = \frac{\tanh(t/2) - 1}{\tanh(t/2) + 1}[/tex]

Note also that light inside in the r < 0 region cannot escape through the horizon. r is timelike inside the horizon, but spacelike outside.

What does your intuition say about this space-time? Does proper time stop at the horizon? Is a maximally extended version of it silly to consider? Ponder these questions before reading my spoiler:

The maximally extended version of this space-time is actually plain, ordinary Minkowski space! I simply made an interesting coordinate change to get a coordinate chart for 1+1 Minkowski space-time that shared many of the features of Schwarzschild coordinates.
 
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  • #24
JesseM said:
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...

Here is a diagram showing the lines of simultaneity in Schwarzschild coordinates in a Kruskal diagram. It is an adaptation of a diagram you posted earlier.

http://www.the1net.com/images/realisticBHkruskal.gif

The original diagram describes a collapsing black hole, but the worldline with the light cones attached to it is very similar to the path of a free falling particle into a black hole. Each line of simultaneity in Schwarzschild coordinates is shown in the same colour when transformed to the Kruskal coordinates for easy identification. Corresponding events in the two coordinate systems are identified by the same letter.

An interesting feature can be seen in the dark green line of simultaneity at Schwarzschild coordinate time t = 44.3. The particle is in two widely separated locations simultaneously in Schwarzschild coordinate time at events (d) and (i). This bizarre situation comes about because in the conventional interpretation, the falling particle arrives at the event horizon in infinite Sxhwarzschild coordinate time and then travels backwards in time to arrive at the central singularity in finite proper (and coordinate) time. In the conventional interpretation this does not violate causalty or imply faster than light travel, because conventionally ingoing photons that originate outside the event horizon also travel back in time once they have crossed the event horizon.

The orignal diagram from MTW is desribed as the spacetime of a realistic black hole and it interesting to note that MTW have not included regions III and IV in their diagram. Every point in Schwarzschild coordinates can be mapped one for one onto regions I and II of the Kruskal-Szekeres coordinates. If the Schwarzschild solution is taken to cover all space and all time in a single universe, then there is no place in the same universe for regions III and IV and they would have to exist in some sort of parallel universe.

The maximally extended solution would have be defined across two universes and would look something like the following embedding diagram:

http://www.the1net.com/images/BHWH.gif


Each black hole would have to have a corresponding white hole in its mirror image parallel universe and vice versa. Particles falling into a black hole would be ejected from a white hole into the parallel universe. To avoid a nett loss or gain of mass in a given universe, there would have to be an equal number of white holes as there are black holes in each universe. Athough there is plenty of observational evidence for black holes in our universe, does it not strike anyone as odd that there is not one shred of observational evidence for a white hole in our universe, as implied by the maximally extended kruskal solution?

If mass entering a black hole is ejected into a parallel universe, then it would be difficult to recover that mass in the form of hawking radiation. Quantum information theory says all the information (mass and energy) must be recovered when the black hole evaporates. White holes ejecting mass into a parallel universe, suggests we would have to reject Hawking radiation as a viable notion.

With the Schwarzschild lines of simultaneity clearly identified, it becomes clear that the diagram below does not depict spacelike slices of the Kruskal spacetime, if we define a spacelike slice as a surface of equal coordinate time. Of course I am referring the Schwarzschild coordinate time, but even in Kruskal-Szekeres coordinates, lines of equal time are not horizontal. In regions II and IV the line of equal Kruskal coordinate time are vertical.

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

It was shown above, that all lines of simultaneity in Schwarzschild coordinates extend to the central singularity. Therefore all spacelike slices would include the central singularity and so it would appear that whatever the above diagram is illustrating, are not spacelike slices.

JesseM said:
... My point was just that the equation is telling you the time dilation experienced by an observer at a certain radius relative to the time of an observer at an arbitrarily large distance from the black hole--just as in SR, time dilation is always relative, never absolute. The ratio between clock ticks of an observer at r=3M and an observer at r=4M will be different from the ratio between the ticks of the observer at r=3M and an observer at r=infinity, and it's the latter that the formula is giving you, it isn't telling you how much the clock at r=3M is slowed down in "absolute" terms.

Gravitational time dilation can be expressed like this:

[tex] \tau = t {\sqrt{1-(2GM / R_P)} \over \sqrt{1-(2GM / R_O)}[/tex]

where Rp is the radius at which the clock measuring proper time is located and Ro is the radius at which the observer is located. When Ro is infinity then the equation reduces to the more familiar:

[tex] \tau = t \sqrt{1-(2GM / R_P)} [/tex]

When the clock is located at the event horizon the proper time is given by:

[tex] \tau = t \, {0\over \sqrt{1-(2GM / R_O)}[/tex]

With the exception of observers located at the central singularity or at the event horizon where the relative proper time is undefined, all observers at any Ro agree that the clock located at the event horizon has stopped.

Of course the next argument against my position is that a clock can not be stationary at the event horizon and I will address that issue in my next post.
 
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  • #25
kev said:
The orignal diagram from MTW is desribed as the spacetime of a realistic black hole and it interesting to note that MTW have not included regions III and IV in their diagram.
That's because the diagrams were showing a spacetime containing a realistic black hole that formed at a finite time from a collapsing matter sphere, which is a different spacetime geometry than a Schwarzschild black hole that has existed for infinite time in Schwarzschild coordinates. The realistic black hole spacetime has no other regions besides I and II.

If you want to see diagrams in Schwarzschild and Kruskal coordinates of a Schwarzschild black hole spacetime, I've scanned some from MTW and uploaded them: #1, #2, and #3 (from p. 825, p. 834, and p. 835 of MTW respectively). The first diagram from p. 825 is in Schwarzschild coordinates, and it shows the worldlines of two infalling test particles (F A starting from rest at r=5.2 M and the other starting from rest at infinity) and an infalling photon B; also drawn in are some spacelike and timelike paths that don't correspond to the worldline of any particles--note in particular the horizontal line E' E'', which has constant t-coordinate, but which the note mentions is actually timelike (because it's inside the horizon).

The second diagram from p. 834 shows all four regions of the Schwarzschild geometry in Kruskal-Szekeres coordinates, with lines of constant t and constant r in Schwarzschild coordinates drawn in. They show on the left that Schwarzschild coordinates can only cover half of this spacetime, either region I and II or region III and IV.

The third diagram from p. 835 shows various curves in Kruskal-Szekeres coordinates, and the portions of these curves that lie in region I and II are mapped into the Schwarzschild coordinate system on the left. Note that the paths in the Schwarzschild diagram here are the same ones that were in the first diagram from p. 825, so you can see what they look like in Kruskal-Szekeres coordinates. One simple aspect of Kruskal-Szekeres coordinates is that spacelike curves are always closer to horizontal than 45 degrees, and timelike curves are always closer to vertical than 45 degrees, and light paths are always exactly 45 degrees; you can see, for example, that the line of constant t inside the horizon, E' E'', does look closer to vertical in the Kruskal-Szekeres diagram.
kev said:
Every point in Schwarzschild coordinates can be mapped one for one onto regions I and II of the Kruskal-Szekeres coordinates.
Not one for one, exactly...the set of all points on the event horizon at r=2M and finite t in Schwarzschild coordinates actually corresponds geometrically to only a single event, and in Kruskal coordinates this event is more faithfully represented as a single point, the point at the center where the white hole horizon meets the black hole horizon.
kev said:
If the Schwarzschild solution is taken to cover all space and all time in a single universe
But it can't, not if you want the spacetime to be "maximally extended" in the sense that timelike worldlines are only allowed to "end" when they hit curvature singularities (where the tidal forces become infinite), so ending at the horizon is not allowed (as in my example of a particle which has been rising away from the event horizon since t=-infinity).
kev said:
The maximally extended solution would have be defined across two universes and would look something like the following embedding diagram:

http://www.the1net.com/images/BHWH.gif
That diagram is all wrong, there is no spacelike slice through the spacetime that involves two wormholes in the same slice.
kev said:
Each black hole would have to have a corresponding white hole in its mirror image parallel universe and vice versa.
If you take a series of spacelike slices moving up the Kruskal diagram, what you get is a wormhole with two white hole horizons that are moving towards one another, then as soon as they meet and pass they become two black hole horizons, so in any given slice you either have a white hole connected to a white hole or a black hole connected to a black hole, never a black hole connected to a white hole. If you haven't already, take a look at http://casa.colorado.edu/~ajsh/schww_gif.html from http://casa.colorado.edu/~ajsh/schww.html which I mentioned earlier, it shows the embedding diagrams for a series of spacelike slices, with both the white hole horizons (red) and black hole horizons (pink) drawn onto the embedding diagrams.
kev said:
Particles falling into a black hole would be ejected from a white hole into the parallel universe.
No, there is no way to actually cross from region I to region III, although if you fall into region II from region I you may see light (or meet aliens) that fell into region II from region III before you're all annihilated by the black hole singularity.
kev said:
Athough there is plenty of observational evidence for black holes in our universe, does it not strike anyone as odd that there is not one shred of observational evidence for a white hole in our universe, as implied by the maximally extended kruskal solution?
The maximally extended Schwarzschild black hole spacetime (what you call the 'Kruskal solution' although I think this is incorrect since Kruskal-Szekeres just found a different coordinate system to describe the maximally extended version of the Schwarzschild geometry, not an altogether new GR solution) is completely unrealistic in the real world because it's based on a black hole/white hole that has been there eternally, it never formed from a collapsing star (and this would be equally true for a non-maximally-extended version of the Schwarzschild spacetime). Note that from the outside it does not really appear as either a black hole or white hole, instead it appears as a kind of "gray hole" because an outside observer can simultaneously be watching stuff that came from region IV passing him on its way out, and at the same time be watching stuff falling in towards it that will enter into region II.
kev said:
With the Schwarzschild lines of simultaneity clearly identified, it becomes clear that the diagram below does not depict spacelike slices of the Kruskal spacetime, if we define a spacelike slice as a surface of equal coordinate time.
That is not what "spacelike slice" means in relativity, "spacelike" is a geometric notion that has nothing to do with your choice of coordinate system, a spacelike slice is one where every curve contained in the slice is spacelike at every point (i.e. the integral of ds^2 is positive between any pair of points on the curve). As mentioned in the diagram from p. 825, a surface of constant t-coordinate in Schwarzschild coordinates is actually timelike inside the horizon, not spacelike. One advantage of Kruskal-Szekeres coordinates is that every surface whose slope always stays closer to horizontal than 45 degrees represents a genuine spacelike surface.
kev said:
Gravitational time dilation can be expressed like this:

[tex] \tau = t {\sqrt{1-(2GM / R_P)} \over \sqrt{1-(2GM / R_O)}[/tex]
Yes, but as you say later, this equation only makes sense for a pair of clocks that are hovering at a fixed radius, which is impossible at the horizon r=2m.
 
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  • #26
This thread is very interesting. I have a question about ontological status and coordinate transformations in GR.

If a phenomemon P is predicted by one set of coordinates, but not by another, is that enough to say that P cannot be physical, but is observer dependent ?
 
  • #27
JesseM said:
If you want to see diagrams in Schwarzschild and Kruskal coordinates of a Schwarzschild black hole spacetime, I've scanned some from MTW and uploaded them: #1, #2, and #3 (from p. 825, p. 834, and p. 835 of MTW respectively). The first diagram from p. 825 is in Schwarzschild coordinates, and it shows the worldlines of two infalling test particles (F A starting from rest at r=5.2 M and the other starting from rest at infinity) and an infalling photon B; also drawn in are some spacelike and timelike paths that don't correspond to the worldline of any particles--note in particular the horizontal line E' E'', which has constant t-coordinate, but which the note mentions is actually timelike (because it's inside the horizon).
JesseM said:
The second diagram from p. 834 shows all four regions of the Schwarzschild geometry in Kruskal-Szekeres coordinates, with lines of constant t and constant r in Schwarzschild coordinates drawn in. They show on the left that Schwarzschild coordinates can only cover half of this spacetime, either region I and II or region III and IV.

Which agrees with my earlier claim that all points in Schwarzschild coordinates can be transformed one for one to points in regions I and II of Kruskal-Szekeres coordinates. There is no need for regions III and IV.

JesseM said:
The third diagram from p. 835 shows various curves in Kruskal-Szekeres coordinates, and the portions of these curves that lie in region I and II are mapped into the Schwarzschild coordinate system on the left. Note that the paths in the Schwarzschild diagram here are the same ones that were in the first diagram from p. 825, so you can see what they look like in Kruskal-Szekeres coordinates.

Thanks for uploading the scanned diagrams. I have extended and added to the third diagram from page 825 to illustrate some important points in the diagram below:

http://the1net.com/images/SK.gif

The diagram now includes the full trajectory of a free falling particle in Schwarzschild coordinates. The trajectory is identified by events F3, F4, F, F' and F''. These points are mapped into regions I and II of the Kruskal-Szekeres coordinates. Note that the red segment (F5 to F4) which was part of the original MTW diagram does not exist in Schwarzschild coordinates. It has been added to satisfy the desire to terminate all worldlines at a singularity. However, if you look carefully you will see the trajectory already terminates at a singularity at point F3 and there is no need to add the red segment, as that part of the trajectory is already described by the dark green segment (F3 to F4). The path F3, F4, F is continuous although it does not look like it in the Kruskal diagram. Point F4 appears to defined in two places in the Kruskal diagram, but they are in fact the same "place" as the diagonal line from top left to lower right is all defined by the coordinates r=2m and t=-infinity, (except at the origin of the diagram).

One disappointing aspect of the MTW drawing is that they have got the curvature of the F,F' path wrong. It should be curving the other way like the A,A' curve. The F,F' curve should look like the dotted curve in this diagram from mathpages: http://www.mathpages.com/rr/s6-04/6-04.htm

http://www.mathpages.com/rr/s6-04/6-04_files/image054.gif

Unfortunately mathpages also neglects to plot the outgoing trajectory of the particle (path F3 to F4) in the top left of region I of the Kruskal diagram.

Why do authoritative sources neglect to show the correct path for the F3,f4 segment. I suspect it because it rather embarrassing for the conventional interpretation of black holes, because it clearly shows that particles are escaping the black hole even though light can not. Rather than try to explain this anomaly the texts gloss over this aspect by not mentioning or plotting that part of the curve.


JesseM said:
Not one for one, exactly...the set of all points on the event horizon at r=2M and finite t in Schwarzschild coordinates actually corresponds geometrically to only a single event, and in Kruskal coordinates this event is more faithfully represented as a single point, the point at the centre where the white hole horizon meets the black hole horizon.

All free falling particles and even photons only cross the event horizon at future infinity or past infinity so that is not an issue.


JesseM said:
kev said:
If the Schwarzschild solution is taken to cover all space and all time in a single universe.
But it can't, not if you want the spacetime to be "maximally extended" in the sense that timelike worldlines are only allowed to "end" when they hit curvature singularities (where the tidal forces become infinite), so ending at the horizon is not allowed (as in my example of a particle which has been rising away from the event horizon since t=-infinity).

All the issues you describe here are addressed in my diagram and explanation above.

JesseM said:
That diagram is all wrong, there is no spacelike slice through the spacetime that involves two wormholes in the same slice.

Sorry, I should of made it clear it was illustrating two black holes. Although Schwarzschild coordinates describe a single black hole, in an otherwise empty universe, I hope you will concede that in the real world there is more than one black hole. The diagram I posted was just an informal sketch of an old idea by Hawking about black holes ejecting matter into baby universe, but even Hawking has retracted that idea now.



JesseM said:
No, there is no way to actually cross from region I to region III, although if you fall into region II from region I you may see light (or meet aliens) that fell into region II from region III before you're all annihilated by the black hole singularity.

That is correct. Future light cones from regions III and I overlap in region II. (Not sure about the aliens though ;)


JesseM said:
The maximally extended Kruskal solution is completely unrealistic in the real world because it's based on a black hole/white hole that has been there eternally, it never formed from a collapsing star. Note that from the outside it does not really appear as either a black hole or white hole, instead it appears as a kind of "gray hole" because an outside observer can simultaneously be watching stuff that came from region IV passing him on its way out, and at the same time be watching stuff falling in towards it that will enter into region II.

You make a very astute observation. Not many people are aware of this fact and it not widely publicized. The conventional interpretation predicts that particles can leave the black hole at the same time as particles fall in. You are quite correct to state that the conventional interpretation of a black hole is that it is a black hole and a white hole at the same time (or a gray hole as you call it). In the conventional interpretation, particles can enter or leave the gray hole but strangely photons can only enter. Odd, that.

I accept the definition of the spacetime slice as you have defined it.



kev said:
With the exception of observers located at the central singularity or at the event horizon where the relative proper time is undefined, all observers at any Ro agree that the clock located at the event horizon has stopped.

Of course the next argument against my position is that a clock can not be stationary at the event horizon and I will address that issue in my next post.
JesseM said:
Yes, but as you say later, this equation only makes sense for a pair of clocks that are hovering at a fixed radius, which is impossible at the horizon r=2m.

Take a look at this diagram from this mathpages article: http://www.mathpages.com/rr/s6-07/6-07.htm

image013.gif


It shows that whatever height the particle falls from, the acceleration is always zero at the event horizon r=2m. This is the Schwarzschild coordinate acceleration. The acceleration in proper time at the event horizon might be infinite but that is not surprising if the observer is using a stopped clock. All velocity and acceleration appears infinite or undefined if the proper time of the observer is zero.

Something to ponder: It is easy to show that a photon can stationary at the event horizon. Agree or disagree?
 
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  • #28
Kev, what the heck? If you don't want to discuss the maximally extended Schwarzschild solution, then don't respond to the parts of the thread that are discussing it -- stop pretending that it's the same thing as your mirror geometry.
 
  • #29
Hurkyl said:
Kev, what the heck? If you don't want to discuss the maximally extended Schwarzschild solution, then don't respond to the parts of the thread that are discussing it -- stop pretending that it's the same thing as your mirror geometry.

I find you outburst a bit odd. In my last post I posted two images that clearly show all 4 quadrants of the maximally extended solution. The rest of the post is responses to direct questions put to me by JesseM and they all relate to the maximal extended solution. I am discussing the maximally extended solution in great depth and analysing it critically. I guess that is what making you uncomfortable, because my points are taking you out of your comfort zone.

You on the other hand, have not had the decency to respond to most of the questions I put to you in post #19 or comment on any of the answers I provided to you in that same post in response to questions you put directly to me.

I think it is also fair to point out that I have not once mentioned or even hinted at any personal theories or solutions I might have, beyond asking why we can't live without regions III and IV, which is very relevant to understanding the maximally extended solution, which is the topic of this thread.
 
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  • #30
kev said:
The rest of the post is responses to direct questions put to me by JesseM and they all relate to the maximal extended solution.
No they don't. You're identifying points all over the place; you assert the green and red segments cover the same trajectory, and that all of the points on the v=-u diagonal are actually the same "place".

In the maximal extended solution, if two points have different Kruskal-Szekeres coordinates, then they are different points.
 
  • #31
kev said:
The no hair theorem states a black hole is completely characterized by its mass, electrical charge and angular momentum.
The no hair theorem, as I understand it talks about what quantities an external observer can measure about the black hole -- it says nothing about what an internal observer can measure. And we already see counterexamples: there's a metric inside, internal observers can look at other particles that have fallen in, etc...


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?
http://the1net.com/images/SK.gif
Red+Blue+Pink vanishes twice, as it runs off of the coordinate chart in different directions. (F5 and F'' don't correspond to points of space-time; aren't part of the coordinate chart)
<-- C'' - C' - C --> vanishes twice, as it runs off of the coordinate chart in different directions.
Green vanishes twice, once as it runs off of the coordinate chart, and once at F4. (F3 isn't part of the coordinate chart)

And for fun, in your Schwarzschild picture:
The green geodesic vanishes twice
The blue geodesic vanishes twice
The pink geodesic vanishes twice
(And these are all different geodesics; they are not connected)
The horizontal axis contains two different geodesics: the interval (0, 2) and the interval [itex](2, +\infty)[/itex]. Both geodesics vanish as they run off of the coordinate chart at both ends.

(Yes, let me repeat that for emphasis: the point (r, t) = (2, 0) is not part of the Schwarzschild coordinates. It is part of (r, t)-space, of course, but it is not a part of the coordinate chart)


That would suggest (to me anyway) that the equation for the motion of a photon in Schwarzschild coordinates is discontinuous across the event horizon.
Yes, we already knew that: as the photon passes along the blue & pink geodesic, it passes through the point F' which is not covered by Schwarzschild coordinates.
 
  • #32
kev said:
Which agrees with my earlier claim that all points in Schwarzschild coordinates can be transformed one for one to points in regions I and II of Kruskal-Szekeres coordinates. There is no need for regions III and IV.
Again, there's a "need" for them if you want to ensure that every possible worldline seen in region I and II does not end at finite proper time unless it hits a singularity.
kev said:
Thanks for uploading the scanned diagrams. I have extended and added to the third diagram from page 825 to illustrate some important points in the diagram below:

http://the1net.com/images/SK.gif

The diagram now includes the full trajectory of a free falling particle in Schwarzschild coordinates. The trajectory is identified by events F3, F4, F, F' and F''. These points are mapped into regions I and II of the Kruskal-Szekeres coordinates. Note that the red segment (F5 to F4) which was part of the original MTW diagram does not exist in Schwarzschild coordinates. It has been added to satisfy the desire to terminate all worldlines at a singularity. However, if you look carefully you will see the trajectory already terminates at a singularity at point F3 and there is no need to add the red segment, as that part of the trajectory is already described by the dark green segment (F3 to F4).
Unless you explicitly perform some kind of topological identification like the one discussed http://casa.colorado.edu/~ajsh/schwm.html#kruskal, the dark green worldline F3 to F4 does not connect to the dark blue worldline, it's not the same particle at all! The dark green worldline represents a separate particle that fell into region II from region III (or if you remove region III, it's a particle that just appeared suddenly at an 'edge' of spacetime that is not a singularity). It may look like they connect in the Schwarzschild coordinate diagram, since both approach r=2m in the limit as t goes to -infinity, but once again you have to think in terms of the actual spacetime geometry rather than just the coordinate representation. Physicists surely have some purely geometric definition of what it means for a timelike curve to be continuous, one that doesn't depend on your choice of coordinate system. And since the Kruskal-Szekeres coordinate system is well-behaved everywhere whereas the Schwarzschild coordinate system is badly-behaved on the horizon, if we want to know about the continuity of curves which cross the horizon, I'm sure the Kruskal-Szekeres system is a better guide to what this geometric definition (whatever it is) would say about a given choice of worldline. Now, I'm not an expert in GR or differential geometry or topology so I don't know what this geometrical definition would be, though I'm confident some such definition would exist (google the phrase "continuous timelike curves" in quotes and you get a bunch of relativity papers which use the term, and the wikipedia pages on curve and continuous function may give some insight), and unless you know the definition yourself you have no basis for concluding that the green curve and the blue curve are part of single continuous curve. Does anyone else on this forum know how continuity for timelike curves is defined in GR?
kev said:
The path F3, F4, F is continuous although it does not look like it in the Kruskal diagram. Point F4 appears to defined in two places in the Kruskal diagram, but they are in fact the same "place" as the diagonal line from top left to lower right is all defined by the coordinates r=2m and t=-infinity, (except at the origin of the diagram).
This is not "naturally" true unless you explicitly identify the point you've labeled F4 on the white hole horizon on the bottom right with the point you've labeled F4 on the black hole horizon on the upper left. If you don't, they represent different geometric points on the manifold, points that the Schwarzschild coordinates just can't distinguish because Schwarzschild coordinates are badly-behaved on the horizon (neither point would have a finite R and t coordinate). And if you leave out regions III and IV of the manifold, and don't do any topological identification of the edges, then both points lie on the edge of the manifold where curves just stop despite not hitting a singularity.

Perhaps this would be easier to see if you think about embedding diagrams for a series of spacelike slices through a spacetime that includes only regions I and II with no explici topological identification of the edges. As shown in this diagram from MTW, you don't have to pick horizontal lines in Kruskal-Szekeres coordinates, any spacelike slice corresponds to an embedding diagram:

p528Gravitation.jpg


Now suppose we paint the event horizons onto the embedding diagrams based on where the spacelike slices cross the horizons, as in http://casa.colorado.edu/~ajsh/schww_gif.html posted earlier. If you simply remove regions III and IV without doing any topological identifications, then any spacelike slice which had passed through a horizon into III or IV will now simply end at the point where that horizon was painted onto the embedding diagram--the embedding diagram might look something like a tube sliced down the middle. If you perform the topological identification given http://casa.colorado.edu/~ajsh/schwm.html#kruskal which is what you seem to be doing implicitly, then effectively what you are doing is stitching together open-ended embedding diagrams from earlier spacelike slices with open-ended embedding diagrams from later slices so that each stitched-together version is a continuous surface with no abrupt edges (for example, in the MTW diagram above imagine taking a version of slice A that ends at the right 'white hole' horizon, and stitching it together with a mirror-reflected version of slice C that ends at the left 'black hole' horizon).
kev said:
One disappointing aspect of the MTW drawing is that they have got the curvature of the F,F' path wrong. It should be curving the other way like the A,A' curve. The F,F' curve should look like the dotted curve in this diagram from mathpages: http://www.mathpages.com/rr/s6-04/6-04.htm

http://www.mathpages.com/rr/s6-04/6-04_files/image054.gif
[/URL]
Well, in the caption on the diagram from p. 835 they do say that the shape of the curves in the Schwarzschild diagrams is "schematic only", maybe the same is supposed to be true of the curves in the Kruskal-Szekeres diagram there. I wonder, though, if it's possible that if we consider a family of curves representing particles ejected to different maximum heights (from 2.5m in the mathpages diagram to the 5.2m in the MTW diagram), maybe the shape of the curves would smoothly vary from flexing inward to flexing outsward? There might be some intermediate height, say 3m, where the worldline of a particle that rose to that height and then fell again would just look like a vertical line in the Kruskal-Szekeres diagram. On the other hand, it may just be an error as you say.
kev said:
Why do authoritative sources neglect to show the correct path for the F3,f4 segment.
Because it is not the correct one unless you perform the topological identification mentioned above. If you don't, then the green line represents an infalling particle that came from region III, its worldline is not continuous with the blue worldline of the particle falling from region I. And in this case the proper time of the green worldline should be increasing as the coordinate time t increases in the Schwarzschild diagram, at least if you want your definition of proper time to have the nice property that whenever two timelike worldlines cross at a point, they should always agree which of the two light cones emanating from that points is the future light cone of increasing proper time. In most "normal" spacetimes it's possible to define each timelike worldline's proper time in a consistent way so this will be true, but I believe that if you do the weird topological identification discussed, the only way this can work is if the proper time of either outgoing or ingoing particles is forced to suddenly reverse directions when they cross the horizon...this is a weird features of this topology which suggests that in such a universe you'd get into serious trouble if you tried to imagine both ingoing and outgoing objects as non-equilibrium systems with their own thermodynamic arrow of time and "memory".
JesseM said:
Not one for one, exactly...the set of all points on the event horizon at r=2M and finite t in Schwarzschild coordinates actually corresponds geometrically to only a single event, and in Kruskal coordinates this event is more faithfully represented as a single point, the point at the centre where the white hole horizon meets the black hole horizon.
kev said:
All free falling particles and even photons only cross the event horizon at future infinity or past infinity so that is not an issue.
That's not true, on the Kruskal-Szekeres diagram it's clear you can draw timelike worldlines that cross through the exact center of the diagram, and even light worldlines that are stuck on the horizon so they cross through the center too. If you plotted these wordlines in Schwarzschild coordinates (which would just be a matter of doing a coordinate transformation), the light worldline would just look like vertical line on the horizon, and I imagine the particle worldline would be broken up in two, one part a vertical line that stays on the horizon forever and another part a curve inside the horizon hitting the singularity at some finite time.
kev said:
Sorry, I should of made it clear it was illustrating two black holes. Although Schwarzschild coordinates describe a single black hole, in an otherwise empty universe, I hope you will concede that in the real world there is more than one black hole. The diagram I posted was just an informal sketch of an old idea by Hawking about black holes ejecting matter into baby universe, but even Hawking has retracted that idea now.
OK, but the main part I was objecting to was your claim that for each individual hole, the black hole in one universe connects to a white hole in another. That's not right, any spacelike slice that contains a black hole horizon on one side contains either a black hole horizon on the other, or no horizon (see the diagram from p. 528 I posted above where slice D crosses the black hole horizon on the right side but stays between the horizon and the singularity forever on the left).
JesseM said:
The maximally extended Kruskal solution is completely unrealistic in the real world because it's based on a black hole/white hole that has been there eternally, it never formed from a collapsing star. Note that from the outside it does not really appear as either a black hole or white hole, instead it appears as a kind of "gray hole" because an outside observer can simultaneously be watching stuff that came from region IV passing him on its way out, and at the same time be watching stuff falling in towards it that will enter into region II.
kev said:
You make a very astute observation. Not many people are aware of this fact and it not widely publicized. The conventional interpretation predicts that particles can leave the black hole at the same time as particles fall in.
That's a little imprecise, in the exterior region I you can see particles passing you on their way out simultaneously with particles passing you on their way in, but the region of spacetime the outgoing particles came from (region IV) is different from the region of spacetime the ingoing particles are going to (region II), and it's convention to refer to region II as the black hole interior region and region IV as the white hole interior region.
kev said:
In the conventional interpretation, particles can enter or leave the gray hole but strangely photons can only enter. Odd, that.
What do you mean by that? Both particles and photons can be emitted from region IV, I see no difference in this respect.
kev said:
Take a look at this diagram from this mathpages article: http://www.mathpages.com/rr/s6-07/6-07.htm

image013.gif


It shows that whatever height the particle falls from, the acceleration is always zero at the event horizon r=2m. This is the Schwarzschild coordinate acceleration.
Sure, but that's just a coordinate acceleration, it's no more physical than the coordinate acceleration as you cross the horizon in Kruskal-Szekeres coordinates, which need not be zero. So what physical point are you trying to make here? Do you agree that no matter what coordinate system we use, the proper time to reach the event horizon along a timelike curve (setting its clock to zero at some finite radius) will be finite? Note that the square of the proper time along a curve is just the integral of ds^2 along that curve times the constant -(1/c^2). Also, do you agree that physicists surely have some coordinate-independent geometric definition of what it means for a timelike curve to be continuous, and that presumably it's possible to verify that even in Schwarzschild coordinates the curve representing the infalling particle outside the horizon meets up continuously with the curve representing the same infalling particle inside the horizon?
kev said:
The acceleration in proper time at the event horizon might be infinite but that is not surprising if the observer is using a stopped clock.
What do you mean by "acceleration in proper time"? If you're talking about proper acceleration, i.e. acceleration as measured in the object's instantaneous local inertial rest frame at a particular moment (which is the acceleration they'd feel as G-forces), then if we're talking about a particle following a geodesic the proper acceleration is always zero at every point on their worldline.
kev said:
Something to ponder: It is easy to show that a photon can stationary at the event horizon. Agree or disagree?
"Stationary" in Schwarzschild coordinates but not in Kruskal-Szekeres coordinates, and more physically, still measured to move at c in the locally inertial frame of a freefalling observer crossing the horizon (because in this locally inertial frame, the horizon itself is moving outsward at a speed of c).
 
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  • #33
To everyone who posted - thanks.

I have one question / thing to add. If we have in some hypothetical universe, a 'real' maximally extended Schwarzschild solution, what happens when we toss in some mass in one side? I mean what happens to the _masses_. Say the soln is for mass M and you toss in M/3 into one side. It would then seem that since information can't get to the 'other side', the other side will stay with a mass of M, while this side would be 1.33M.

What does a Kruskal diagram look like when you have M on one side, and say 1.33M on the other? Put another way - if you draw 2 Kruskal diagrams with different masses, is there a way to cut them (eg along some coordinate axis, etc) so that one 'dual mass' drawing would be the result?

It might seem that an entirely new solution of the GR equations would be needed, but since there is no information passed - it is not obvious how each side could be different from the static solution. But if there is a curvature discontinuity of some sort, then the door is open to communicate through the hole, by playing with the mass of the hole.
 
  • #34
JesseM said:
..
Well, in the caption on the http://www.jessemazer.com/images/p835Gravitation.jpg" they do say that the shape of the curves in the Schwarzschild diagrams is "schematic only", maybe the same is supposed to be true of the curves in the Kruskal-Szekeres diagram there. I wonder, though, if it's possible that if we consider a family of curves representing particles ejected to different maximum heights (from 2.5m in the mathpages diagram to the 5.2m in the MTW diagram), maybe the shape of the curves would smoothly vary from flexing inward to flexing outsward? There might be some intermediate height, say 3m, where the worldline of a particle that rose to that height and then fell again would just look like a vertical line in the Kruskal-Szekeres diagram. On the other hand, it may just be an error as you say...

Here is a Kruskal-Szekeres diagram on which I have plotted a family of free falling particle trajectories with apogees at r = 2.001M, 2.1M, 2.5M, 3.0M, 3.5M and 4.0M respectively:

http://the1net.com/images/KT.gif

The light blue curves are the complete Schwarzschild trajectories transformed to K-S coordinates. The curves are not just some sort of artists impression, but are plotted from equations using graphical software. The yellow curves are copies of the particles and their trajectories in some sort of parallel possibly imaginary) universe.

The trend is clear. The curves go from being a straight vertical line at the origin of the KS coordinates to curving more strongly outwards as the apogee gets further away from the event horizon. In other words they curve in the opposite direction to the F.F' curve shown in the MTW drawing. Now I don't think MTW are in the habit of making such gross mistakes, so we we will have to assume that the F,F' curve is the path of an object that is being artificially accelerated inwards such as a rocket. To be fair to MTW they do not claim it is the path of a free falling particle, but perhaps they should of made it clear it is not. On the other hand, the A,A',A'' path in the MTW diagram would appear to be the correct depiction of a free falling particle in Schwarzschild and KS coordinates, with the observation that the apogee of that particle is not placed at the usual t=0, but there is nothing that says it has to be.
 
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  • #35
Hurkyl said:
No they don't. You're identifying points all over the place; you assert the green and red segments cover the same trajectory, and that all of the points on the v=-u diagonal are actually the same "place".

In the maximal extended solution, if two points have different Kruskal-Szekeres coordinates, then they are different points.

I claimed the green segment in http://the1net.com/images/SK.gif" is the transformation of the green segment in the Schwarzschild diagram. The red segment is an arbitrary copy of the green segment in a supposed parallel universe. The second copy of the segment comes about by giving reality to the fact that a square root has two roots, one positive and one negative. For example the length of the diagonal of a right angled trangle has a positive and negative solution and so does the proper time of an object with relative motion but we know from experience that the only the positive root has physical significance in those cases. Maybe in some parallel universe the proper time of a moving object is running backwards relative to the proper time of the copy of the object in our universe.


As for my other claim that the two copies of F4 on the u=-v diagonal of the SK diagram are the same "place" I present the folowing argument. In the diagram below I have made a cut along the u=v axis and rotated region II down to where region IV normally is and glued the two u=-v edges to make some arbitrary rotated version of the normal KS coordinates.


http://the1net.com/images/F4.gif

In the rotated diagram, F4 is unambiguously one place and now it is F' that appears to be in two diffent places but that is obviously just an artifact of where I chosen to make a cut. In the same way, F4 appears to be two different places in normal KS coordinates because it happens to be on the line where the cut is made, in those coordinates.

The yellow lines are future light cones. It can be seen that whatever coordinates you choose the particle moving along the segment F3,F4 is unambiguously moving backwards in time as it moving from its future light cone to its past light cone. As the particle (or photon) travels backwards in time it has to arrive at past infinity before it can cross from region II into region III, but because the big bang is not in the infinite past that option is not open and the particle can not move from region II to region III and a similar argument means a particle or photon can not move from region IV to region I or vice versa. Regions III and IV are completely cut off from our universe (regions I and II) if our universe has only existed for a finite time since the big bang. I think that answers one the questions you posed in your opening post. It also solves the problem of the requirement of all worldlines to terminate at a singularity without introducing regions III and IV because the big bang counts as a singularity.

http://the1net.com/images/PBB.gif

The black triangular regions in the above diagram is time before the big bang going back to past infinty. It effectively cuts off regions I and II from regions III and IV and so regions III and IV have no physical significance in our universe with a finite past.

However, it might be argued that a particle falls into a black hole (or is emitted) in finite proper time and if we take proper time as somehow more physically real than coordinate time, then fact that the universe probably does not have an infinite past is not a problem, but that requires that the trajectory in proper time does not have a corresponding trajectory in coordinate time. For me, if the path does not exist in all possible valid coordinate systems, then it is not a physically real path. This sort of relates to an interesting question posted in this thread by Mentz, that no one has attempted to answer yet.
 
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