Black hole matter accumulation

[?????]

A lot of talk here about how to experimentally verify what happens at the EH. This is the same old problem. Let me restate the problem by quoting an old Chinese proverb: If the grasshopper jumps halfway to the wall every time, how many jumps does it take him to get to the wall?

If the EH did exist, then this is a priori proof that you can never experimentally prove that it exists. The notion that some theoretical device could be lowered past the EH is irrelevant. If it could go past the EH, the rest of the universe would have lived out it lifetime and would no longer exist when the passage was finally made. Forget whether or not the coordinates match up. The Schwarzschild Radius EH is logically impossible on its face. Nothing can exist past the end of the life of the universe.

 

[Q-reeus]

...I agree. Again, the metric and the coordinates are not the same thing at all. The metric must be continuous, and every individual coordinate chart must also be continuous, but two different coordinate systems need not be continuous with each other anywhere.

I have never disagreed with that as trivially true for e.g. 'patching' between spherical and Cartesian co-ords, but it still ducks the point imo. The chart may not be the territory, but it had better darn well properly describe it in the regime intended. But I'll hammer that out in another thread.

 

[Q-reeus]

It doesn't. The metric inside the shell is "flat" in the sense that the metric coefficients are independent of the coordinates, so that it can be written in the Minkowski form. But the coordinates themselves are scaled differently than they are at infinity, so that what you are calling the "potential" matches up across the thin spherical shell, and the length contraction and time dilation factors do too.

Careful - this is in a sense basically agreeing with my argument. So you think there is finite length contraction in the interior Minkowski region. I believe that is the actual case, but it does not accord with a physically consistent transition from SM to MM, as per #138. In this case though, you are, like it or not, opting for an inexplicable contraction of the tangent (i.e. azimuthal) spatial components in traversing the shell from outer to inner radius. One cannot have it both ways. If the tangent spatial components remain independent of potential and thus invariant, interior flatness demands the radial component jump back to the potential-free value. Or vice versa as you imply, the contracted radial component changes but little, and the tangent components undergo a contraction jump. Makes no physical sense to me either way. An exterior metric ('chart' if you insist) that is isotropic in *all* components does - the transition is smooth, slight, and then and only then imo anomoly free.

Remember also that an "infinitely thin" spherical shell is unphysical if it has nonzero mass; there has to be *some* region of finite thickness where the stress-energy tensor is nonzero, and in that region the metric coefficients will change too.

Of course, and I have never suggested otherwise. A thin shell is simply perhaps the best way of manifesting the anomoly I maintain exists.

 

[Q-reeus]

Because, at times, you have been treating one set of coordinates as specially meaningful.

Well this gets down to whether SM (standard OR isotropic) is to be treated as a unique description of the spacetime surrounding a stationary spherically symmetric mass, as determined by coordinate values. You do not agree that Birkhoff's theorem http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity) is saying just that? You do acknowledge that I have posed a problem pertinent to just that regime, and so for sure SM is 'specially meaningful' in that context? The issue should be thrashed out and settled in that regime and the transition issue to Minkowski regime.

Some features you complain about don't exist in other coordinates. Therefore they are coordinate artifacts, not physical predictions.

Could you detail which such features?

The point of the wiggly line analogy was to emphasize the difference between physical predictions versus coordinate values. However you want to construct the clock paths, I could come up with coordinates that say the clock that shows more time at the end, for example, matched the other clock for 99% of the the time on this other clock; then, shot forward only as the lower clock was brought up. I will agree those would be strained coordinates, but the point is, it is only the comparison at the end that is a true physical prediction.

I agree it is a strained argument. By using my two-runs method, one can easily and unambiguously determine the relative clock rates - as precisely given by the SM temporal component. Which is a mathematical statement of how spacetime regions temporally relate, and no mere arbitrarily chosen 'chart'. How on Earth would GPS function as well as it does if this sort of thing was up for grabs?

In several places you talk about an observer making physical statements about what is true at a distance. In SR, this can be made to work because there exist global frames for inertial observers. In GR all such statements have no unique validity.

Ditto my remarks above re GPS.

As to your so call main problem, I just don't see the problem. The total solution for a shell is simply the Schwarzschild geometry outside, a non vacuum solution through the shell, and Minkowski inside. There would be no discontinuities.

That depends on what you mean by 'discontinuities'. Obviously there is a sharp transition in the potential gradient ("g"), but for sure the potential itself hardly alters, as is true imo of all the spatial and the temporal metric components.

If the shell were transparent, and and light were emitted inside the shell of some frequency observed locally, it would red shifted when received by an outside observer slightly more than a similar signal from the surface of the shell (the difference being due to shell). Despite their being no formal, general, potential in GR, this experiment would behave very similar to the Newtonian case. Almost all of these properties follow directly from Birkhoff's Theorem, which is rigorously proven for GR.

Well fine, you have committed to that the interior temporal metric component is depressed. Kindly commit to whether iyo the interior spatial metric components have depressed values, and would that be for all components equally, and equal to the temporal component?

 

[Q-reeus]

This is a truly fundamental misunderstanding. The isotropic coordinates are just different coordinates for the same geometry, as are the Kruskal, Eddington, Gullestrand, etc. They all represent the same metric as a geometric object, they all make the same physical predictions. It is no different that polar versus rectilinear versus logarithmic coordinates on a plan. Does drawing different lines on a plane change its geometry? No. However, the Euclidean metric expression can only be used with rectilinear coordinates. Using the standard tensor transformation law, you would derive different metric expressions for the other coordinates that would yield all the same geometric facts.

This seems to me to be trivializing the value and/or uniqueness of a coordinate system AS a true and faithful representation of some metric, within it's regime of applicability. How can standard SM and ISM for example be making the same predictions? Both claim to be a mathematically correct mapping of gravitationally depressed metric components to that of a distant coordinate observer. So when said observer checks for redshift, or determines whether a 'test sphere' remains perfectly spherical under the telescope (duly allowing for light bending etc), there is not a problem? If SSM predicts the test sphere will be viewed as oblate owing to radial length contraction, while ISM predicts the sphere will remain perfectly spherical (that is after all what 'isotropic' implies), I see a problem. And how is that sensibly resolved?

 

[Q-reeus]

PAllen is correct here, different coordinates are like different fishnet stockings on a woman's leg, the leg does not change!

Depends how tight the stockings are! Seriously though, I disagree but you will have seen that elsewhere by now.

 

[Q-reeus]

...Forget whether or not the coordinates match up. The Schwarzschild Radius EH is logically impossible on its face. Nothing can exist past the end of the life of the universe.

Actually, penta-questionmark, it very much matters whether the coordinates 'match up'. If as I maintain the 'mismatch' is a pathological feature of SM and thus the EFE's, EH's and BH's turn out to be literally non-entities so then just quit worrying about a non-issue, period.

 

[Dale]

The chart may not be the territory, but it had better darn well properly describe it in the regime intended.

And all charts do.

 

[Dale]

In this case though, you are, like it or not, opting for an inexplicable contraction of the tangent (i.e. azimuthal) spatial components in traversing the shell from outer to inner radius.

It is perfectly explicable. You are changing from anisotropic coordinates to isotropic coordinates.

An exterior metric ('chart' if you insist)

Here is further evidence of this confusion in your mind. There is a huge difference between the metric and a chart. They are not synonymous or interchangeable.

 

[PeterDonis]

Well, I am hand waving but I think I see some potential problems, mainly to do with light travel time issues.

The light travel time would affect how long it took the faraway observer to receive the time stamped radio signals, but that's why I specified they are time stamped; they contain information that tells the faraway observer what the infalling observer's clock reading was when the signal was emitted. That is what verifies that the GR calculation for proper time for an infalling observer is correct. The time of reception of the signal is not the primary piece of experimental data; it just affects how long it takes for the experimental data to be collected.

 

[Passionflower]

The light travel time would affect how long it took the faraway observer to receive the time stamped radio signals, but that's why I specified they are time stamped; they contain information that tells the faraway observer what the infalling observer's clock reading was when the signal was emitted. That is what verifies that the GR calculation for proper time for an infalling observer is correct. The time of reception of the signal is not the primary piece of experimental data; it just affects how long it takes for the experimental data to be collected.

Yes I realize that Peter, I was more hinting at how long it takes to get updates.

But why don't we make an attempt to calculate it?

And please the next on who comes around claiming it is simple should come with formulas. :)

 

[PeterDonis]

If it could go past the EH, the rest of the universe would have lived out it lifetime and would no longer exist when the passage was finally made. Forget whether or not the coordinates match up. The Schwarzschild Radius EH is logically impossible on its face. Nothing can exist past the end of the life of the universe.

This is an interesting comment. It is true that there is a sense in which the entire region of spacetime inside the horizon is "in the future" for every event outside the horizon. However, that does not mean it is logically impossible for the region inside the horizon to exist. First, a clarification: if the universe has an end point in time, i.e., if it is closed, then what I just said, and what you said in the above quote, is no longer true; the region of spacetime inside the EH of any black hole inside the closed universe gets caught up in the "big crunch" that ends the universe, just like everything else, so it is no longer true that, from the viewpoint of someone inside the EH, the outside universe "would have lived out its lifetime and would no longer exist".

Second, if we assume the universe exists forever into the future, so that a black hole can be "eternal" and the region inside remains hidden behind the EH forever, then the region inside the horizon is indeed "in the future" for any event outside the horizon--but that doesn't mean the region inside the horizon is logically impossible. If it were logically impossible, there could be no consistent mathematical models containing a region inside the horizon, and there are; any coordinate chart that is nonsingular at the horizon constitutes such a consistent mathematical model, and several have been named in this thread.

You could still claim that the region inside the horizon was *physically unreasonable*, but I've already laid out what that claim requires: it requires you to believe that physics suddenly starts working completely differently at the horizon, for no apparent reason.

 

[PeterDonis]

Careful - this is in a sense basically agreeing with my argument. So you think there is finite length contraction in the interior Minkowski region. I believe that is the actual case, but it does not accord with a physically consistent transition from SM to MM

Why not? Remember that there is a shell of finite thickness in between, where the stress-energy tensor is nonzero. What does the "potential" look like from the outer to the inner surface of that shell?

If the tangent spatial components remain independent of potential and thus invariant

Do they inside the substance of the shell (i.e., between its outer and inner surfaces)?

 

[PeterDonis]

But why don't we make an attempt to calculate it?

I agree it would be an interesting calculation; if I have time in the next day or two I'll try to post one.

 

[PAllen]

This is an interesting comment. It is true that there is a sense in which the entire region of spacetime inside the horizon is "in the future" for every event outside the horizon. However, that does not mean it is logically impossible for the region inside the horizon to exist. First, a clarification: if the universe has an end point in time, i.e., if it is closed, then what I just said, and what you said in the above quote, is no longer true; the region of spacetime inside the EH of any black hole inside the closed universe gets caught up in the "big crunch" that ends the universe, just like everything else, so it is no longer true that, from the viewpoint of someone inside the EH, the outside universe "would have lived out its lifetime and would no longer exist".

Second, if we assume the universe exists forever into the future, so that a black hole can be "eternal" and the region inside remains hidden behind the EH forever, then the region inside the horizon is indeed "in the future" for any event outside the horizon--but that doesn't mean the region inside the horizon is logically impossible. If it were logically impossible, there could be no consistent mathematical models containing a region inside the horizon, and there are; any coordinate chart that is nonsingular at the horizon constitutes such a consistent mathematical model, and several have been named in this thread.

You could still claim that the region inside the horizon was *physically unreasonable*, but I've already laid out what that claim requires: it requires you to believe that physics suddenly starts working completely differently at the horizon, for no apparent reason.

One more thing: the interior region is in the infinite future from the point of view any external observer, using a typical simultaneity convention. From the point of view of an interior observer, their history from the horizon to the singularity corresponds to perfectly finite (and often short) period of external history. Further, nothing prevents an external observer from ignoring arguably optical effects and using Kruskal simultaneity to establish a relationship between interior events and exterior events. This would be analogous to an accelerating rocket choosing to ignore that light from distant objects can't catch them, and use Minkowski coordinates rather than Rindler coordinates. They simply say, I saw Earth until last light time T_final, I believe it's still there, evolving normally, rather than vanished or frozen.

 

[PAllen]

This seems to me to be trivializing the value and/or uniqueness of a coordinate system AS a true and faithful representation of some metric, within it's regime of applicability. How can standard SM and ISM for example be making the same predictions? Both claim to be a mathematically correct mapping of gravitationally depressed metric components to that of a distant coordinate observer. So when said observer checks for redshift, or determines whether a 'test sphere' remains perfectly spherical under the telescope (duly allowing for light bending etc), there is not a problem? If SSM predicts the test sphere will be viewed as oblate owing to radial length contraction, while ISM predicts the sphere will remain perfectly spherical (that is after all what 'isotropic' implies), I see a problem. And how is that sensibly resolved?

If this is the point you want to argue, I am sorry, there is no point in further discussion. There is no such thing as a unique correct coordinate system. There are some that make a class of calculations easier, or make certain aspects of reality more apparent (but others less apparent). In particular, Schwarzschild coordinates are extremely misleading for describing the local experience of a near horizon observer; yet they have many advantages for other purposes. You need to take this to the math forum and explain that differential geometry is invalid.

 

[Dale]

Well this gets down to whether SM (standard OR isotropic) is to be treated as a unique description of the spacetime surrounding a stationary spherically symmetric mass, as determined by coordinate values. You do not agree that Birkhoff's theorem http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity) is saying just that?

Here again, you are confusing the metric with the coordinates. Birchoff's theorem shows that the Schwarzschild spacetime is the unique manifold for a spherically symmetric mass. It says nothing about Schwarzschild coordinates, which are not some uniquely valid coordinates.

Could you detail which such features?

Specifically the anisotropy. It is an artifact of the standard Schwarzschild coordinates only. When you change coordinates to an isotropic system, like Minkowski, then of course it disappears.

If you were sailing you might measure horizontal distance in nautical miles and vertical distances in fathoms. In such a coordinate system the metric would be anisotropic. When you got off the boat and started walking around on land you might measure all distances in meters. In such a coordinate system the metric would be isotropic and it would be scaled differently from the other metric.

Would you claim that there is some big tear in spacetime or some huge logical failure in physics because of those changes? If not, then perhaps you can understand why the rest of us are underwhelemed by the "gravity" of the problem.

 

[PAllen]

Well fine, you have committed to that the interior temporal metric component is depressed. Kindly commit to whether iyo the interior spatial metric components have depressed values, and would that be for all components equally, and equal to the temporal component?

Ignoring the shell, you can see how to fit Schwarzschild geometry to flat spacetime by looking at the isotropic coordinates. Using the conventions you can look up in wikipedia, if you simply 'freeze' r1 at the shell value, treating it constant inside, you have an interior metric of the form:

c^2 d tau ^2 = b^2 c^2 dt^2 - a^2 (dx^2 + dy^2 + dz^2)

where a and b are constants computed from shell r1 and other constants. This metric form obviously smoothly joins exterior metric, is diagonal, and flat (all metric derivatives are zero). Further, if we simply scale t, x, y, and z based on constants b and a, we recover the Minkowski metric inside. This re-scaling can be done globally, and now you have the natural point of view of an interior observer: I'm all Minkowski inside, the asymptotic infinity of the outside is also flat but scaled. In the original coordinates, it is the asymptotic infinity observer who is normal Minkowski, and the interior that is scaled Minkowski. The outside observer can also, of course, transform the outside to the more common Schwarzschild coordinates, to make some calculations easier, and some of their perceptions more apparent. No physical predictions are changed by any of these coordinate transforms. For that, I ask you to read any intro to differential geometry, as covered in typical intro to GR books (no need to study the full formality of math text treatment).

 

[Passionflower]

I agree it would be an interesting calculation; if I have time in the next day or two I'll try to post one.

That would be great as I think there are a few challenges.

For starters we cannot use an observer at infinity as it would take an infinite time for any signal to reach this observer. If we pick a very large r value we still would have to make sure this observer accelerates albeit very lightly.

I could propose a scenario:

Black hole: rs=1 (M = 1/2) to simplify our formulas.
Observer station S: Stationary at r=100,000 (if that is far enough? We could take something farther away or closer if we worry about the time delay)
Probe B: Free falling from r=100,000 and instantly decelerating to stationary at r=1.01

Of course feel free to change any of those numbers if you think they are unsuitable.

Would that be a good starting point for you?
If so, what do you propose next? The probe to send messages at fixed intervals? Or alternatively we could have a small rocket being sent back to the observer station as well.

I am flexible but not with "we assume the observer station is at infinity and does not need to accelerate" because obviously then it would take forever to get any signal from the probe to the observer station.

Makes sense?
Others want to pitch in as well?

 

[PeterDonis]

I could propose a scenario:

Black hole: rs=1 (M = 1/2) to simplify our formulas.
Observer station S: Stationary at r=100,000 (if that is far enough? We could take something farther away or closer if we worry about the time delay)
Probe B: Free falling from r=100,000 and instantly decelerating to stationary at r=1.01

In general this looks reasonable, except that I would want the probe to be in free-fall always, no acceleration to a stop above the horizon. The key thing to determine is, if the probe emits a light signal at fixed intervals of its own proper time, what is the last r value > 1 (i.e., above the horizon) where a signal is emitted?

I don't know about the exact numerical values; I would first work the problem leaving them undetermined, calling the observer's radius r_O and the probe's radius r_P (the latter would of course be a function of time). I agree that r_O has to be finite.

Setting rs=1 is fine as that basically just scales the r and t coordinates as r / 2M, t / 2M, which is often done in the GR formulas anyway (a lot of the analysis in MTW is done this way, IIRC).

 

[Passionflower]

In general this looks reasonable, except that I would want the probe to be in free-fall always, no acceleration to a stop above the horizon.

No problem, but it will make the calculations slightly more difficult.

The key thing to determine is, if the probe emits a light signal at fixed intervals of its own proper time, what is the last r value > 1 (i.e., above the horizon) where a signal is emitted?

Well that is not that hard. I think think that only depends on the r-value we pick for the stationary space station. The probe traveling from the space station to the EH will send a given number of signals, unless the last signal is exactly at r=rs all signals will eventually be received by the space station. (However with the caveat that the signals will be dimmer as well and if the signal is below the Planck value the space station would not pick it up but I assume you want a classical answer to the question).

Also I hope you agree the space station must accelerate because otherwise it starts to move towards r=rs as well. And I suspect it will make a difference when we consider limit conditions.

Setting rs=1 is fine as that basically just scales the r and t coordinates as r / 2M, t / 2M, which is often done in the GR formulas anyway (a lot of the analysis in MTW is done this way, IIRC).

Yes, that makes reading those formulas a lot easier.

 

[Passionflower]

One more thing: the interior region is in the infinite future from the point of view any external observer, using a typical simultaneity convention. .

Could you explain this a bit more.

Lets have two observers at r=100,000 (rs=1) one is free falling and another one is stationary, what is exactly in the infinite future according to you for each observer?

 

[PeterDonis]

Also I hope you agree the space station must accelerate because otherwise it starts to move towards r=rs as well.

Yes. The space station is supposed to "hover" at r = r_O.

 

[PeterDonis]

Lets have two observers at r=100,000 (rs=1) one is free falling and another one is stationary, what is exactly in the infinite future according to you for each observer?

I'll pitch in here since I used the "infinite future" bit too. The crucial thing is not a particular observer's state of motion, per se, but what simultaneity convention is used--what set of spacelike lines (or hypersurfaces, if we include the angular coordinates) count as "lines of simultaneity".

If the simultaneity convention is that of exterior Schwarzschild coordinates, then the horizon and the interior region are in the "infinite future" for both the observers you mention. That's because all the lines of simultaneity with finite values of the Schwarzschild t coordinate are in the exterior region; on a Kruskal chart, they are lines emanating from the "center point" where the two null "horizon" lines cross, and spreading out into the exterior region at various angles up to 45 degrees. The limit of those lines is the future horizon itself, which has a "t coordinate" of plus infinity, loosely speaking.

If the simultaneity convention is that of Kruskal coordinates, however, then the lines of simultaneity are simply horizontal lines on the Kruskal chart, and those cover the future interior region as well as the exterior. We could also use lines of constant Painleve "time" or ingoing Eddington-Finkelstein "time" as lines of simultaneity; those look more complicated on a Kruskal chart but they also cover the future interior region as well as the exterior. So in all these cases events in the future interior region have finite "time" coordinates. So with this convention, both of the observers you mention would *not* see the horizon and the future interior as being in the "infinite future".

If we wanted to draw a distinction between the "time" perceived by the two observers, we could say that the Schwarzschild lines of simultaneity are the "natural" ones for the hovering observer, while the Painleve lines of simultaneity are "natural" to the infalling observer. In *that* case, then the future interior region would be in the "infinite future" for the hovering observer but not for the infalling one. But each observer could, in principle, choose a simultaneity convention other than the "natural" one for his state of motion. The physics is the same.

 

[Q-reeus]

Originally Posted by Q-reeus:
"In this case though, you are, like it or not, opting for an inexplicable contraction of the tangent (i.e. azimuthal) spatial components in traversing the shell from outer to inner radius."
It is perfectly explicable. You are changing from anisotropic coordinates to isotropic coordinates.

So you have elected to answer on PeterDonis's behalf here. Alright, but firstly note I was above making a basically rhetorical comment to the effect such azimuthal variations would be 'voodoo'. But regardless, to answer your comment, how on Earth do you arrive at your conclusion of anisotropic -> isotropic? As far as I'm concerned, one sticks with spherical coordinate system, but one finds that the metric components either do or do not undergo physically real change in traversing the shell (as predicted by SM, that is). No chopping and changing of coordinate system.

Originally Posted by Q-reeus: "An exterior metric ('chart' if you insist)"

Here is further evidence of this confusion in your mind. There is a huge difference between the metric and a chart. They are not synonymous or interchangeable.

That criticism has been repeated now so often, decided to do a little searching and found this: http://casa.colorado.edu/~ajsh/schwp.html

"Schwarzschild metric

Schwarzschild's geometry is described by the metric (in units where the speed of light is one, c = 1)
ds² = - (1-r_s/r)dt²+(1-r_s/r)^-1dr²+r²do² .
The quantity ds denotes the invariant spacetime interval, an absolute measure of the distance between two events in space and time, t is a `universal' time coordinate, r is the circumferential radius, defined so that the circumference of a sphere at radius r is 2pi*r, and do is an interval of spherical solid angle."

Well is this right or wrong then, because seems clear enough SM here is described entirely in a slightly compact form of SC's, just as I thought was so.
And that exprerssion is clearly showing anisotropy of spatial components - of the metric, just as I thought it should.
Much earlier on I pointed out that the physically significant redshift formula lifts straight out of SC's. But you will maintain it is meaningless because coordinates are just chalk lines drawn on the ground and in no way tell us what the 'real metric' is all about? This response btw is to cover your #192 also.

 

[Q-reeus]

Why not? Remember that there is a shell of finite thickness in between, where the stress-energy tensor is nonzero. What does the "potential" look like from the outer to the inner surface of that shell?

It drops by a typically small fractional value (depending naturally on the shell thickness) in a smooth way. I see no issue there.

Do they inside the substance of the shell (i.e., between its outer and inner surfaces)?

Why would there be - there is explicitly zero dependence on potential everywhere exterior to the shell according to SC's. And there is some strange physical reason that should change within the shell wall?

 

[Q-reeus]

Ignoring the shell, you can see how to fit Schwarzschild geometry to flat spacetime by looking at the isotropic coordinates...

Just so, and all along I have argued that isotropic and only isotropic external coords (as accurately reflecting the underlying metric) will give an anomaly free transition to the flat interior. IMO ISC's are a half-way house in that the essentially equal dependence on potential for both temporal and spatial components increasingly diverges in strong gravity, for what I consider no good plausable reason. But thanks for indicating the procedure of matching up - I just believe anomaly free necessarily means a totally isotropic exterior metric, which neither SM or ISM are. As far as there being no physical difference between the predictions of various 'standard' coords, well my next post will try and put that to a simple test procedure.

 

[Q-reeus]

All right, faced with a solid wall of consensus opinion telling me coords are mere conveniences for calculation and do not reflect the intrinsic properties of any underlying metric, here's a proposal. Taking the shell arrangement, I would like answers to the following:

Assume a modest gravitational potential such that r_s/r << 1, and a thin shell such that any 'delta' in potential from exterior to interior is negligible to first order. What then is the mathematically correct transformation expressions that a distant coordinate observer must apply to account for observed distortion of a test sphere (perfectly spherical in asymptotically flat coordinate frame) located a:) just outside the shell. b:) just inside the shell. Obviously there are just two components to consider - scale factor for azimuthal, and radial directions.
Similarly for temporal change of a test clock. Assume 'g' forces, light bending have all been accounted for - only changes reflecting the metric components are being considered.

 

[?????]

Actually, penta-questionmark, it very much matters whether the coordinates 'match up'. If as I maintain the 'mismatch' is a pathological feature of SM and thus the EFE's, EH's and BH's turn out to be literally non-entities so then just quit worrying about a non-issue, period.

I understand perfectly what you are saying. You have no way to respond to my issue, so would I please go away. Yes, I will. Continue with your mathematical analysis and find an answer that seems pleasing. But, just because you can prove by legitimate mathematical derivation that 1=0, it still is not so. In the end, no matter how perfect the derivation, one does not equal zero.

 

[PeterDonis]

It drops by a typically small fractional value (depending naturally on the shell thickness) in a smooth way.

This much is fine; but the relationship of the radial and tangential metric components to the potential is what I was getting at. See below.

Why would there be - there is explicitly zero dependence on potential everywhere exterior to the shell according to SC's. And there is some strange physical reason that should change within the shell wall?

Yes, because the shell is not vacuum. However, after thinking this over, I may have approached this wrong by focusing on the tangential metric components. Let me try a different tack.

First, I want to be clear about how the Schwarzschild "r" coordinate is defined. It is defined such that the area of a sphere at "radius" r is $4 \pi r^{2}$. Now consider two spherical "shells" in Schwarzschild spacetime, one at radius r and the other at radius r + dr. These "shells" are not actual physical objects; they are just a way of helping to visualize the physics involved. The area of the inner "shell" is $4 \pi r^{2}$, and that of the outer is $4 \pi \left( r + dr \right)^{2}$. Suppose we put a ruler between the two "shells" and measure the distance between them. What will it be? If space were "flat", it would simply be dr; but because of the way the Schwarzschild metric works, the actual distance will be

$$\frac{dr}{1 - \frac{2M}{r}}$$

(assuming that dr << r, so the metric at r is, to a good enough approximation, the metric at r + dr as well).

So the "anisotropy" you are talking about is "real", in the sense that there is something non-Euclidean about the space between the shells. (However, see my footnote about this at the end of this post.) We can carry this all the way down to the outer surface of the actual shell, the non-vacuum region. Call that outer radius $r_{o}$. The area of that outer surface is $4 \pi r_{o}^{2}$. If we imagine a "shell" (an imaginary one this time) at radius $r_{o} + dr$, its area would be $4 \pi \left( r_{o} + dr \right)^{2}$, but the distance between the two "shells" would be

$$\frac{dr}{1 - \frac{2M}{r_{o}}}$$

However: now imagine a spherical "shell" slightly *below* the outer surface of the non-vacuum region, at $r_{o} - dr$. Its area will be $4 \pi \left( r_{o} - dr \right)^{2}$ What will the distance between this "shell" and the outer surface of the non-vacuum region be? It will still not be the "Euclidean" distance dr, but something larger; but it will be slightly "less larger" than it would have been if the two shells had been separated by vacuum. If we then continue down through the non-vacuum region, down to its inner radius $r_{i}$, the distance between "shells" at radius r and radius r + dr, inside the non-vacuum region, will continue to get "less larger" than the "Euclidean" value.

Finally, we reach the inner surface of the non-vacuum region, at $r_{i}$. The area of that inner surface is $4 \pi r_{i}^{2}$, and the area of a "shell" at a slightly larger radius, $r_{i} + dr$, would be $4 \pi \left( r_{i} + dr \right)^{2}$. The distance between these shells, as measured with a ruler, will be just *slightly* larger than dr.

And now, if we consider a "shell" in the vacuum region just inside the inner surface of the non-vacuum region, at radius $r_{i} - dr$, its area will be $4 \pi \left( r_{i} - dr \right)^{2}$, *and* the distance, measured with a ruler, between it and the inner surface of the vacuum region will be exactly dr--no "correction" factor any more. This tells us that the vacuum region inside the inner surface is now "flat"--space there is Euclidean. However, the "potential" there is going to be the same as it is on the inner surface of the shell (because the potential has to be constant throughout the inner vacuum region), and we know that potential is somewhat *less* than that at the outer surface of the vacuum region (as you've already agreed). So the potential in the inner vacuum region is indeed "redshifted" compared to that at infinity. That potential difference no longer shows up in the spatial parts of the metric, but if we compared the rate of time flow in the inner vacuum region to that at infinity, we would find it to be slower, by exactly the same factor as on the inner surface of the non-vacuum region. Another way of saying this is to say that, to put the metric in the interior vacuum region into the standard Minkowski form, we would have to re-scale the time coordinate, compared to that "at infinity", by the "time dilation factor" on the inner surface of the non-vacuum region.

You'll note that I didn't change anything about the tangential metric components at all during any of this; each sphere at "radius" r had the same area as a function of r. However, the metric coefficient g_rr did change, meaning that the relationship between tangential distances and radial distances, expressed as a function of the coordinate r, changed as well, in just the right way to make the metric "flat" in the interior vacuum region.

Footnote: Everything I've said above depends not only on a particular definition of the "r" coordinate, but on a particular definition of simultaneity; basically, what I said above applies in a "surface of constant time" picked out of the global spacetime, and it depends on a particular way of picking out that "surface of constant time", the way that Schwarzschild coordinates pick it out. If we chose a different way of picking surfaces of constant time, we would find different spatial geometries in those surfaces, and the above analysis would proceed differently. For example, if we chose surfaces of constant time the way Painleve coordinates do, the surfaces of constant time in the exterior vacuum region, at least, would be spatially flat--the distance between two "shells" at r and r + dr, where "r" is still defined as the square root of (area of the sphere at "r" divided by 4 pi), would be the "Euclidean" value, dr. That's because the surfaces of constant Painleve time are not the same as the surfaces of constant Schwarzschild time.

 

[Passionflower]

So the "anisotropy" you are talking about is "real", in the sense that there is something non-Euclidean about the space between the shells.

Yes, although some may insist that the space remains Euclidean but the rulers are shrunk there.

By the way the same happens with the volume between the two spheres (or shells as you call them). There is more volume as one would expect if the space would be Euclidean.

For example, if we chose surfaces of constant time the way Painleve coordinates do, the surfaces of constant time in the exterior vacuum region, at least, would be spatially flat--the distance between two "shells" at r and r + dr, where "r" is still defined as the square root of (area of the sphere at "r" divided by 4 pi), would be the "Euclidean" value, dr. That's because the surfaces of constant Painleve time are not the same as the surfaces of constant Schwarzschild time.

That is a way of seeing it, however 'Painleve' time is simply the time for a free falling (at escape velocity) observer. If you apply the LT before integrating the distance in Sch. coordinates you indeed get that the distance r2-r2 is exactly r2-r1. Conversely we can do the same by transforming the free faller in PG coordinates into a stationary observer and obtain the distance r2-r1 for stationary observers identical to Sch. coordinates.

 

[PeterDonis]

By the way the same happens with the volume between the two spheres (or shells as you call them). There is more volume as one would expect if the space would be Euclidean.

Yes, agreed.

That is a way of seeing it, however 'Painleve' time is simply the time for a free falling (at escape velocity) observer. If you apply the LT before integrating the distance in Sch. coordinates you indeed get that the distance r2-r2 is exactly r2-r1. Conversely we can do the same by transforming the free faller in PG coordinates into a stationary observer and obtain the distance r2-r1 for stationary observers identical to Sch. coordinates.

Yes, all this is another way of saying that the surfaces of constant Painleve time are different than the surfaces of constant Schwarzschild time. The Lorentz transformation you speak of, at any given event, transforms between Painleve time and Schwarzschild time at that event (and, of course, it also transforms the distances appropriately); but from the viewpoint of the local inertial frame at that event, the transformation simply "tilts" the time and (radial) space axes between the "Painleve" axes and the "Schwarzschild" axes at that event; and the radial axis combined with the angular coordinates gives a local spacelike "surface of constant time", with whichever "tilt" you choose. Fit together the local "constant time" surfaces with either "tilt" at all events, and you get global surfaces of constant Painleve or Schwarzschild time.

 

[Dale]

how on Earth do you arrive at your conclusion of anisotropic -> isotropic?

Same way you did. Schwarzschild coordinates are anisotropic, Minkowski coordinates are isotropic.

As far as I'm concerned, one sticks with spherical coordinate system, but one finds that the metric components either do or do not undergo physically real change in traversing the shell (as predicted by SM, that is). No chopping and changing of coordinate system.

If you are no longer changing coordinate systems then what are you still fretting about?

Well is this right or wrong then, because seems clear enough SM here is described entirely in a slightly compact form of SC's, just as I thought was so.
And that exprerssion is clearly showing anisotropy of spatial components - of the metric, just as I thought it should.

Yes, that is all correct.

Much earlier on I pointed out that the physically significant redshift formula lifts straight out of SC's. But you will maintain it is meaningless because coordinates are just chalk lines drawn on the ground and in no way tell us what the 'real metric' is all about?

I certainly would never say it was meaningless. There is nothing wrong with Schwarzschild coordinates, nor is there anything wrong with any other coordinates. Redshift is a feature of all coordinate systems.

 

[zonde]

I would like pick up this line in discussion as it is much closer to the point where I see the problem.

I'll pitch in here since I used the "infinite future" bit too. The crucial thing is not a particular observer's state of motion, per se, but what simultaneity convention is used--what set of spacelike lines (or hypersurfaces, if we include the angular coordinates) count as "lines of simultaneity".

Yes, simultaneity is the key.

If the simultaneity convention is that of exterior Schwarzschild coordinates, then the horizon and the interior region are in the "infinite future" for both the observers you mention.

That is the point where I see problem with black hole interior.
I say that your statement is wrong. It's EH that is in the "infinite future" for external observer. Black hole interior is "beyond" infinite future. And because we define infinity as a limit where nothing is beyond that limit we get contradiction in terms.
Therefore there is no black hole interior in Schwarzschild coordinates. And it is not because of chosen coordinate system but because of chosen simultaneity.

So my statement implies that simultaneity is not just convenience but rather physical fact.
And yes that is so as we know from SR. Simultaneity is defined in such a way as to get isotropic speed of light. And there is only one "correct" way how to define simultaneity for any state of motion.
But that's not all. It should be possible to convert consistently between reference frames that correspond to different states of motion.

And I think there is complete chaos in GR regarding different coordinate charts and different (contradictory) simultaneity conventions they implement.

 

[PAllen]

Zonde,

please think carefully about some of the points I raised in my post #128 (with references). In particular, consider what is observed if a large globular cluster happened to undergo collapse to a supermassive black hole (there is no evidence of this happening, but there is no reason it could not). The key thing here is that for supermassive BH, you have the event horizon forming with very modest matter density. The stars are still many millions of miles apart when the event horizon forms. Key characteristics for an outside observer:

1) They see normal brightening as star density increases, at first.
2) The event horizon expands from the center out. First they see innermost stars slowing and reddening, the becoming invisible as too little light is emitted. This phenomenon expands out from the center until the whole cluster has effectively vanished from inside out. All that is left is a completely dark horizon, after finite time (rate of photon emission too low to observe; darker than any normal body in space, if all this occurred isolated, away from dust).

If you saw this, what would you conclude? It seems very hard to (for me) to see any possible interpretation than that the globular cluster still exists inside the event horizon, and is undergoing some history we cannot see (e.g. catastrophic collapse of some nature; probably not as neat as idealized solutions).

The other thing I point out in #128 is that the question of what is normally inside the horizon may be fully testable if there are exception to the cosmic censorship hypothesis. I give a number of references, including proposals for how this could be looked for.

The upshot, to me, is that accepting GR really does seem to require accepting the horizon interior as real, and this may actually be fully testable physics.

If you are in doubt about my description of what would be seen in the globular cluster collapse, part of this is discussed in MTW. The rest is discussed in some nice links George Jones can provide (but I can't lay hands on now). They discuss the formation and growth of the event horizon for collapsing dust, but the same would apply for the collapsing cluster. [The advantage of the cluster is that you can see inside it, at least ideally.]