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:
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
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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
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).