## black hole

 Quote by DaleSpam 0 order effects of the metric are due to the choice of coordinates. For example consider the standard Minkowski metric. $$ds^2=-dt^2+dx^2+dy^2+dz^2$$ And a transformation to anisotropic coordinates, $t=T$, $x=2X$, $y=Y$, $z=Z$: $$ds^2=-dT^2+4dX^2+dY^2+dZ^2$$ They are mere artifacts of the coordinates and say nothing about the physics.
I will now concede after looking again over the way ISC is formulated at http://en.wikipedia.org/wiki/Schwarz...zschild_metric , with the r1 there having a changed meaning to the standard r, that ISC's are indeed just an alternate expression and not different physics to what standard SC's predict. Happy? Given how things have developed here, will let the other thread die with dignity and answer your #16 there here:
 "What is the differential comparison technique you have in mind? Perhaps something like sending a sending a light pulse from one side of the sphere to the other, sending a light pulse out to infinity at the start and a second pulse out to infinity at the end, and measuring the time between receiving the two pulses at infinity?"
Infinity takes a long time, but you more or less get that there are means to eliminate 'difficult' factors if present. Really in my example it's no more than the before/after thing - measure f, D, before shell present, and then the same with shell. Simple (in principle). But this is not really the issue.

 Quote by PeterDonis No--what I apparently have to convince you of is that, in a non-vacuum region, the tangential measure *does* depend on the potential. Obviously it has to, for the rest of what I said to work. To see why it does, consider why it *doesn't* in the exterior vacuum region: it's because the "direction of gravity" is purely radial--the only "gravity" that is "pulling" on you at a given point is the inward radial "gravity" of the distant massive object. In a non-vacuum region, that's no longer true; there is non-zero stress-energy surrounding any given point in the non-vacuum region, on all sides, so there is "gravity" pulling on any given point on all sides, not just a purely radial "acceleration" due to the distant gravity source as there is in the exterior vacuum region. So the effects of gravity in the non-vacuum region change *all* of the spatial components of the metric, not just the radial one...
Nice try, but sadly still not buying that as is. Still opting that the nice transition is purely mathematical artifact of a force-fit coordinate scheme. Here's the crux of the problem imo. We agree tangent spatials by SC's are invariant everywhere exterior to rb. There is in that region the potential (1-rs/r)1/2, plus it's spatial derivatives to all orders, just as there is within the matter region of shell wall. Only essential difference I see is the relative size distribution of potential and derivatives. This gets down then to the *fundamental character* of the relation between potential and various metric components. It makes perfectly good sense that there is a fairly abrupt transition in the 'g' field from maximum at rb, to zero at r <= ra, and likewise for tidal terms - they explicitly are potential derivative *in nature* and must cease in the equi-potential interior. Where is there any analogous physical basis, in the tangent metric component case, for *total* exterior indifference to potential *and all it's derivatives*, yet a relatively steep dependence just within the matter region? 'Pulling in all directions' (I realize this is just you 'dumbing it down' for my benefit) just won't near cut it as explanation, as I have outlined above.

What 'essence' or geometric 'object' can be entirely absent >= rb, yet there strongly for rb>r>ra, so as to explain it? And what's more it has to be shown to be cumulative in effect, and not a mere 'blip' that leaves no trace on exit past r<ra, so to speak. Tall order indeed! Sole uniquely present identity I can think of might be divergence, but that seems most unlikely a solution, and in itself creates another issue. Namely, if divergence is truly absent exterior to rb, this gives the lie to those claiming that in GR 'gravity truly gravitates'. What say you sir?
[One final comment: in #222 you mentioned agreement between yourself and DrGreg's finding in http://www.physicsforums.com/showpos...5&postcount=10, but I read him there as saying interior length are as at infinity, once the metric is applied. A misunderstanding?]

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 Quote by Q-reeus What 'essence' or geometric 'object' can be entirely absent >= rb, yet there strongly for rb>r>ra, so as to explain it? And what's more it has to be shown to be cumulative in effect, and not a mere 'blip' that leaves no trace on exit past r
Um, a non-zero stress-energy tensor, which means a non-zero Einstein tensor, which is the primary geometric object in the Einstein Field Equation? That is what does the work in the non-vacuum region, rb>r>ra. For r<ra, the vacuum Einstein Field Equation is enough to ensure that the "potential" is constant at its value at ra.

 Quote by Q-reeus One final comment: in #222 you mentioned agreement between yourself and DrGreg's finding in http://www.physicsforums.com/showpos...5&postcount=10, but I read him there as saying interior length are as at infinity, once the metric is applied. A misunderstanding?
No, just a difference in terminology. What he means by "once the metric is applied" is "from the viewpoint of an observer in the interior vacuum region". Such an observer can't tell that he is not in the flat spacetime region at infinity by purely local measurements; locally the two regions look the same. Only by global observations can the two regions be distinguished.

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 Quote by DrGreg In this context, infinity is a limit where there are no t coordinates beyond that limit. A lack of coordinates, in a particular choice of coordinates, does not imply non-existence. It just means those coordinates don't cover the whole of spacetime.
Infinity is imaginary (not real) limit where there are no real values beyond this limit no matter what choices you make.
Look if you say that function asymptotically approaches value a as it's argument approaches infinity then it means a is the limit no matter what you do with the argument.

Maybe there is some confusion with my argument that I can still clear up.
I can explain my argument in two steps rather than one:
1) in Schwarzschild metric interior of black hole is completely disconnected from exterior because there is no future beyond infinite future and there is no past before infinite past (where you could hope to connect interior with exterior).
2) there can be any number of spacetime patches that are completely disconnected from our spacetime. There can be even any number of universes that are completely disconnected from our universe. As they do not affect our reality in any way it can be stated that they are not real or alternatively they do not exist.

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 Quote by zonde As they do not affect our reality in any way it can be stated that they are not real or alternatively they do not exist.
Even in flat spacetime, the "inside" of a Rindler horizon does not "affect the reality" of a Rindler observer (and it is "beyond" T=∞ in Rindler coordinates), so does the "inside" of a Rindler horizon exist?

(See post #75, the "inside" is the blue region, the observer is the black line, the red and green lines specify Rindler coordinates.)

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 Quote by zonde 1) in Schwarzschild metric interior of black hole is completely disconnected from exterior because there is no future beyond infinite future and there is no past before infinite past (where you could hope to connect interior with exterior).
This is not correct. The *geometry* of the interior is not disconnected from the geometry of the exterior. They are connected, as can be easily seen by analyzing covariant or invariant objects like geodesics, curvature tensors, etc.

It is true that the interior Schwarzschild *coordinate patch* is disconnected from the exterior Schwarzschild coordinate patch; that is what is meant by statements about the "infinite future" and whether anything is "beyond" it. But that statement does not support your argument, because it only applies to a particular coordinate system; it is not a statement about the underlying geometry, which is what is important for the physics.

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 Quote by Q-reeus Infinity takes a long time, but you more or less get that there are means to eliminate 'difficult' factors if present. Really in my example it's no more than the before/after thing - measure f, D, before shell present, and then the same with shell.
Yes, I get the idea, but different measurement techniques will give different answers which is why a complete description of the measurement technique becomes important. Once you have done that then the result of that measurement technique is guaranteed to be coordinate independent.

 Quote by Q-reeus But this is not really the issue.
OK, so I am not sure what you still think the issue really is. You now understand that the anisotropy was merely an artifact of the coordinates, and I thought that the dissapearance of that anisotropy was what was bothering you.

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Quote by DaleSpam
 You can't have two different "infinite futures". What you can do is you can construct two different dimensions and claim that they both can be considered as going in direction of "future" under different viewpoints. But under single viewpoint there will be only one "infinite future".
I am not certain that I understand what you are saying here. Let me rephrase it the way I would say it: A spacetime with multiple dimensions may have different regions considered to be the infinite future, but the worldline of a single observer will only have one infinite future.

If this is what you meant then I agree, otherwise could you clarify your meaning?
Yes, this needs clarification.

So I am stating that the same "infinite future" applies to set of observers that have parallel time dimension i.e. it applies to global coordinate system as a whole.
I suppose that you can come up with some nasty example where I would have hard time defining "observers with parallel time dimension" but as we are talking about spherically symmetric coordinate systems centered on black hole I can always come up with Euclidean coordinate system after factoring out time dilation (and change in radial length unit if something like that shows up).

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 Quote by DrGreg Even in flat spacetime, the "inside" of a Rindler horizon does not "affect the reality" of a Rindler observer (and it is "beyond" T=∞ in Rindler coordinates), so does the "inside" of a Rindler horizon exist? (See post #75, the "inside" is the blue region, the observer is the black line, the red and green lines specify Rindler coordinates.)
"Inside" of a Rindler horizon does not exist for Rindler observer.

But I have one question about Rindler coordinates.
It seems to me that time dimension can not be arbitrarily extended for Rindler observer, is it right?
There is certain point ahead of Rindler observer (in flat coordinates) where Rindler observer reaches speed of light and time stops for him.

And because of this it's hard for me to associate real observers with Rindler observer.

 Quote by DaleSpam OK, so I am not sure what you still think the issue really is. You now understand that the anisotropy was merely an artifact of the coordinates, and I thought that the dissapearance of that anisotropy was what was bothering you.
Main issue is that which I posted in #240. Realizing SC and ISC are two sides of the same coin took much out of the need out of that other thread, although I still wanted to see whether my expectations of anisotropic spacial distortions for r>rb were confirmed by others calcs. Issue now, following #241, is to nail down just what property/operation of ET actually yields tangential contraction. I recall now yuiop did an analysis getting the opposite - zero tangential contraction, and a radial component that jumped back to potential free value. Was based on some work by Gron I think.

 Quote by PeterDonis Um, a non-zero stress-energy tensor, which means a non-zero Einstein tensor, which is the primary geometric object in the Einstein Field Equation? That is what does the work in the non-vacuum region, rb>r>ra. For r
Well it would be nice to expand on that a bit. Best I could make of Einstein tensor is that it is divergenceless - so much for surmising about divergence as conceivable factor. Given the shell spherical symmetry and static state, can the operation of said tensor within shell wall be expressed entirely in terms of potential and gradients thereof, preferably in polar form? I think all but the T00 term is operative as source on rhs, yes? So there should when it's all broken down, only be potential term and derivatives at work? So the specialness of matter region re tangent contraction should be explicable just in those terms. I think.
 No, just a difference in terminology. What he means by "once the metric is applied" is "from the viewpoint of an observer in the interior vacuum region". Such an observer can't tell that he is not in the flat spacetime region at infinity by purely local measurements; locally the two regions look the same. Only by global observations can the two regions be distinguished.
Thanks for clearing that up.

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 Quote by zonde "Inside" of a Rindler horizon does not exist for Rindler observer.
But do you think that "inside" of a Rindler horizon exists for a Minkowski observer? If yes, then your interpretation of "existence" is observer dependent?

 Quote by zonde But I have one question about Rindler coordinates. It seems to me that time dimension can not be arbitrarily extended for Rindler observer, is it right?
The Rindler time coordinate approaches ∞ as you approach the Rindler horizon, and coordinates can't go beyond ∞, so you are right. (The Rindler time coordinate equals the Rindler observer's proper time along his own worldline, and locally represents Einstein-simultaneity for any other observer at rest relative to the Rindler observer.) But you can do the thing that happens with Schwarzschild coordinates -- you can set up a separate "interior Rindler" coordinate system that covers the inside of the horizon. But neither of the two separate exterior and interior Rindler coordinate systems include the horizon itself (where there is a coordinate singularity in each system).

 Quote by zonde There is certain point ahead of Rindler observer (in flat coordinates) where Rindler observer reaches speed of light and time stops for him.
Not true. The Rindler observer gets ever closer to the speed of light as measured by any inertial observer but never actually gets there. The Rindler observer always measures the local speed of light relative to himself to be c, so from his point of view he never gets any closer.

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 Quote by zonde So I am stating that the same "infinite future" applies to set of observers that have parallel time dimension
OK, I think that the concept you are trying to describe here is a congruence, which is essentially a family of worldlines. For example, you could associate one worldline with each spatial location in Schwarzschild coordinates. This would give a family of timelike curves which could each represent a stationary observer relative to the central mass. Then this set of observers would all share the same "infinite future" region.

The existence of one congruence which share the same "infinite future" does not in any logical way forbid the existence of another set of congruences which share a different "infinite future". Your line of reasoning seems to be that there is a timelike congruence which ends up in the usual "infinite future" therefore all timelike congruences must end up in the same "infinite future". This is not sound logic.

 Quote by zonde i.e. it applies to global coordinate system as a whole.
Many spacetimes don't admit a single global coordinate system that covers the whole manifold. The Schwarzschild coordinates certainly don't cover the whole manifold.

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 Quote by Q-reeus Well it would be nice to expand on that a bit. Best I could make of Einstein tensor is that it is divergenceless - so much for surmising about divergence as conceivable factor. Given the shell spherical symmetry and static state, can the operation of said tensor within shell wall be expressed entirely in terms of potential and gradients thereof, preferably in polar form? I think all but the T00 term is operative as source on rhs, yes? So there should when it's all broken down, only be potential term and derivatives at work? So the specialness of matter region re tangent contraction should be explicable just in those terms. I think.
Some clarifications:

* The entire stress-energy tensor is the "source" on the RHS of the Einstein Field Equation (EFE). The LHS is the Einstein tensor, which is defined in terms of the Ricci tensor, which is a contraction of the Riemann curvature tensor.

* For the non-vacuum region in question, a reasonable stress-energy tensor would be a "perfect fluid" with non-zero pressure:

$$T_{00} = \rho$$

$$T_{11} = T_{22} = T_{33} = p$$

where $\rho$ is the energy density and $p$ is the pressure. The off-diagonal components are all zero. For the spacetime to be static and spherically symmetric, both $\rho$ and $p$ must be functions of the radial coordinate r only. Also, $\rho$ should be positive everywhere in the non-vacuum region, but $p$ will not be; a negative pressure is a tension, and in order for the non-vacuum region to be stable, there will have to be tension somewhere, to keep it from falling apart (a positive pressure everywhere would be possible only if the non-vacuum region went all the way down to the center, r = 0, without any hollow interior).

* The fact that the stress-energy tensor (and therefore the Einstein tensor) is divergenceless is one way of expressing local conservation of energy; the divergence of the stress-energy tensor, over some small piece of spacetime, is just the net energy going in or out of that piece of spacetime. The divergence being zero just means energy in equals energy out; i.e., energy is conserved. So you're right that this probably doesn't help much for this particular problem.

* You can still define a "potential" in the non-vacuum region, because you can do that for any spherically symmetric geometry. And you can still view the gradient of this potential as being the "acceleration due to gravity". I *think* that the effects on the metric, including its tangential part, in the non-vacuum region can be expressed in terms of the potential, but I'm not certain; the metric coefficients may be more complicated than that in the non-vacuum region. The only metric coefficient that I'm pretty well certain can be expressed entirely in terms of the potential is $g_{00}$, the timelike coefficient.

 Quote by PeterDonis * For the non-vacuum region in question, a reasonable stress-energy tensor would be a "perfect fluid" with non-zero pressure: $$T_{00} = \rho$$ $$T_{11} = T_{22} = T_{33} = p$$ where $\rho$ is the energy density and $p$ is the pressure.
Could you demonstrate that such a 'blob' could turn into a black hole?

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