Detecting matter falling into a Black Hole

[Logan5]

That would only be true (if at all) if the EH were significant locally and has been said repeatedly in this thread, it is not.

But a deformation is not locally detectable (or significant) either. A deformation is a global artefact, which extends in space and time. If we consider a single point, there is no deformation. So this is not incompatible ? GW emission is a global phenomenon not a local one.

And how to interpret all these sources which talks about GW emission when a BH "eats" matter (eat = the matter cross the horizon)

 

[ShayanJ]

Because this source http://kipac.stanford.edu/kipac/black-holes-eating-stars-and-making-waves (and numerous others) seems to say so, and also Jim Graber in this thread. And because crossing EH is a major "hair" (hair are deformation of the horizon, aren't they ?) that should be shaved with a major GW emission, shouldn't it ? And because some theories (like firewall) doesnot agree and leads to the "5" family of answer to this question, which should not be too quickly dismissed ?

Well, eating another star isn't comparable to eating just a small object. The star surely has a size and mass comparable to the black hole and so its passage from the horizon can't be called local. This is much different from what we've been talking about in this thread. There may be many things causing the gravitational waves that are different from what you think. At least I can talk about such a situation.
I should also say that the no-hair conjecture is not yet proved, let alone a hair-shaving theorem!

 

[PeterDonis]

The observer O detects the gravitational waves at his time T3. The question is : what is this time T3 ?

The short answer is (2); jimgraber's post gave the basic explanation of why.

But it is possible (even probable ?) that, at a finite proper time Tau1, while approaching the EH, the external universe ends, or the BH himself evaporates, or all observers has decayed. So for all practical purposes the BH can be a frozen star, no ?

No. We have had a lot of threads in this forum on this; it's a common misconception, but it's still a misconception.

It's important to keep distinct the time according to the distant observer, who always stays far away from the hole, and the proper time of an observer who falls into the hole. They are not the same.

If the distant observer sets up his most "natural" time coordinate (which is just the Schwarzschild time coordinate), and tries to extend it to the horizon, he finds it giving an infinite value of $t$ to any event on the horizon. But the horizon is not just one event; it contains a lot of different events, so a coordinate chart that labels them all with the same value is not valid there. This is usually described as Schwarzschild coordinates having a coordinate singularity at the horizon.

For an observer falling into the hole, however, it only takes a finite amount of his proper time to fall to the horizon. And if he sets up his most natural time coordinate, the one that matches his proper time to fall to the horizon, he finds that this coordinate chart can cover the region at and inside the horizon just fine, all the way down to the singularity at $r = 0$. And according to this time coordinate, the end of the external universe (if there even is one--according to our best current model, our universe will expand forever), or the evaporation of the black hole (if we include quantum effects--but note that for the hole to evaporate, it has to be colder than the CMBR, which even most supermassive holes are not at this point), will be very, very far in the future at the time the infalling observer crosses the horizon.

As for which coordinate is "right", they're both just coordinates, and coordinates in GR have no direct physical meaning. But there is a clear sense in which the infalling observer's coordinates give a better picture of what's happening: those coordinates label distinct events on the horizon with distinct times, and give a clear time ordering to those events, which matches the "natural" time ordering of events on the horizon. So this ordering is invariant, and it says that the evaporation of the hole is, as I said above, far to the future of the event where the infalling observer crosses the horizon.

 

[PeterDonis]

That would only be true (if at all) if the EH were significant locally and has been said repeatedly in this thread, it is not.

First of all, the implication here is not true: if matter falls into a black hole, and the infall process is not perfectly spherically symmetric, then GWs will be emitted. This has been known since the 1970s and has been confirmed by many calculations.

Second, this whole question of whether the EH is "significant locally" is a red herring, because, as Logan5 pointed out, GW emission is itself not a "local" process. To even tell that GWs are being emitted, you have to look on a distance scale comparable to the wavelength; for GWs emitted by a black hole as an object falls in, the wavelength will be on the order of the hole's Schwarzschild radius. From a global perspective, the GW emission process can just as easily be viewed as the horizon (i.e., the boundary of the region that can send light signals to infinity) changing shape from its original spherical shape, to a deformed spherical shape, to a new spherical shape with a larger area.

eating another star isn't comparable to eating just a small object.

True; the amplitude of the GWs will be much larger if the hole eats a star. But in principle they will be there even if the hole eats a small object.

 

[ShayanJ]

True; the amplitude of the GWs will be much larger if the hole eats a star. But in principle they will be there even if the hole eats a small object.

Could you explain why is there any GW at all?

 

[PeterDonis]

Could you explain why is there any GW at all?

Heuristically, it's because the infall process is not perfectly spherically symmetric, so when the infalling object passes through the horizon, the horizon gets deformed. But a stationary black hole horizon must be spherical, so the deformation will get radiated away as GWs. Kip Thorne's Black Holes and Time Warps has a good non-technical discussion. The original discovery of this was by Richard Price in 1972: see here:

http://en.wikipedia.org/wiki/Richard_H._Price

Note that Price's theorem is actually a theorem, even though the no-hair result that it helps to justify is still only a conjecture since it has not been rigorously proven.

 

[Logan5]

The short answer is (2); jimgraber's post gave the basic explanation of why.

I like this answer, but, as I asked to Jim (alas without answer), can you show me a source which gives a formula to compute a finite T3, and how this formula is found ? If your answer is true, it is very precious for me and my understanding.

Also, this answer seems contradictory with other answers in this thread, which denies the possibility to remote detect horizon crossing because "horizon has no signification locally", so no physical phenomenon can be locally emitted when the object cross locally the horizon. Can your position and these positions be reconciliated or are they incompatible ?

 

[Logan5]

No. We have had a lot of threads in this forum on this; it's a common misconception, but it's still a misconception.

I agree, I know that, and it is the reason why I formulated the problem in post 1 (or tried to) in a way which avoids this kind of formulation and time comparison in 2 different referentials. Anyway your summary is very clear and push again in my mind these important concepts.

 

[PeterDonis]

can you show me a source which gives a formula to compute a finite T3

Roughly, it will be $R^{3/2} + R$, where $R$ is the distance of the observer from the hole. I don't have a handy online reference, unfortunately.

 

[PeterDonis]

this answer seems contradictory with other answers in this thread, which denies the possibility to remote detect horizon crossing because "horizon has no signification locally", so no physical phenomenon can be locally emitted when the object cross locally the horizon.

As I noted in a previous post, GW emission is not a "local" phenomenon; you can't pick a particular point on the horizon and say GWs are being emitted from that point. The wavelength of the GWs will be comparable to the Schwarzschild radius of the hole, so the GW emission is really a phenomenon that involves the entire horizon. An observer falling in with the object that adds mass to the hole won't be able to tell that GWs are being emitted as it crosses the horizon. That is only observable on a larger scale.

 

[Logan5]

As I noted in a previous post, GW emission is not a "local" phenomenon; you can't pick a particular point on the horizon and say GWs are being emitted from that point. The wavelength of the GWs will be comparable to the Schwarzschild radius of the hole, so the GW emission is really a phenomenon that involves the entire horizon. An observer falling in with the object that adds mass to the hole won't be able to tell that GWs are being emitted as it crosses the horizon. That is only observable on a larger scale.

I agree. It was the observation I made in post #31.

 

[Logan5]

Roughly, it will be $R^{3/2} + R$, where $R$ is the distance of the observer from the hole. I don't have a handy online reference, unfortunately.

Very unfortunate, indeed ! I have a quest for this formula for years now, but I have not been able to find it, or even to find the theoretical background to justify it. There is something strange here. I will worship your name for ever if you can give me a link or a book title where I can find such things. I am serious

 

[jimgraber]

Logan5, I will try to reply more soon, but it may take a few days. In the meantime, take a look at figure 2 on page 6 of this reference to get a good idea of what I am talking about:

http://arxiv.org/pdf/1106.1021v2.pdfInspiral-merger-ringdown multipolar waveforms of nonspinning black-hole binaries using the effective-one-body formalism
Authors:Yi Pan, Alessandra Buonanno, Michael Boyle, Luisa T. Buchman, Lawrence E. Kidder, Harald P. Pfeiffer, Mark A. Scheelhttp://arxiv.org/abs/1106.1021

best,
a very busy Jim Graber

 

[Logan5]

Logan5, I will try to reply more soon, but it may take a few days. In the meantime, take a look at figure 2 on page 6 of this reference to get a good idea of what I am talking about:

http://arxiv.org/pdf/1106.1021v2.pdfInspiral-merger-ringdown multipolar waveforms of nonspinning black-hole binaries using the effective-one-body formalism
Authors:Yi Pan, Alessandra Buonanno, Michael Boyle, Luisa T. Buchman, Lawrence E. Kidder, Harald P. Pfeiffer, Mark A. Scheelhttp://arxiv.org/abs/1106.1021

best,
a very busy Jim Graber

Thank you very much. I now understand better what you, and probably Peter, meant.

As far as I can understand this document, and your answer in post #22, these elements gives me this information (which is new for me) : GW peak signal does NOT witness horizon crossing, but horizon approaching. BTW, you said in post #22 : "But as it gets close to the horizon, it begins to decrease exponentially and is very soon arbitrarily close to zero, but it never reaches zero"

So my thought experiment is not very good for what I intended. I wanted a remote signal for horizon crossing. But I had no much hope, because I awaited the answer to be 3. and your answer actually implies 3. (or even 4.) not 2. Indeed a peak GW signal reaches the observer at a finite time T3, but this peak signal is not for horizon crossing and the question was all about that.

The conclusion I draw (for now) from this discussion (and other discussions elsewhere also) is that no actual experience, no signal, no thought experiment, can falsify the "frozen star" model. I don't know if this model is true or not, but I understand it, and I don't understand the model where a BH "eats" matter, which cross the horizon within the time the universe or BH lives. And the Newton shell theorem assure that we confound, for all practical and experimental, or even theoretical purposes, a matter shell (which is a BH in the frozen star model) with a singularity. I am eager to hear about any actual or thought experiment which can falsify this model.

If I am not abusing, I have another question to surround the problem. I place the observer, now, on the falling object. Let this observer look behind the fixed clock. Which is the last Minkowski time the observer on the falling object can see on the remote clock, before the signal becomes too weak and before it crosses the horizon ? This gives a minimal bound for T3 (alas not a maximal bound), and I wonder if this time (let name it T'3 < T3) is large or not, and i wonder also how it compares with the peak GW signal time.

 

[PeterDonis]

GW peak signal does NOT witness horizon crossing, but horizon approaching.

This is not really correct. Yes, the peak signal is effectively emitted from the infalling object before it crosses the horizon; but if the infalling object did not cross the horizon, but instead remained above it (suppose, for example, that there was a large rocket attached to it that fired just before it reached the horizon, and kept it "hovering" there), you would see a different pattern of GWs; also, other observations would be different as well (for example, the light from the object would stop redshifting at some finite wavelength, instead of continuing to redshift to arbitrarily long wavelengths). So, while the GWs and other observations do not directly show you the object crossing the horizon, they are certainly evidence of the object crossing the horizon, since all those observations would be different if the object did not cross the horizon.

The conclusion I draw (for now) from this discussion (and other discussions elsewhere also) is that no actual experience, no signal, no thought experiment, can falsify the "frozen star" model.

This is not correct. The "frozen star" model leads to different observations than the black hole model does. See above.

I don't know if this model is true or not, but I understand it, and I don't understand the model where a BH "eats" matter

Picking a model you can understand over a model you can't is only a viable strategy if both models make the same predictions. That is not the case here. See above.

And the Newton shell theorem assure that we confound, for all practical and experimental, or even theoretical purposes, a matter shell (which is a BH in the frozen star model) with a singularity

I don't understand what you mean by this. The shell theorem says that a spherically symmetric matter distribution outside some region of interest does not cause any spacetime curvature (gravity) inside that region. It says nothing about how gravitational collapse looks to someone outside the collapsing region.

There is a theorem called Birkhoff's theorem that says that any vacuum, spherically symmetric region of spacetime must have the Schwarzschild geometry. But that theorem only applies in the case of exact spherical symmetry; we are talking about an asymmetric case, where an object falls into a black hole along one particular radial trajectory.

 

[PeterDonis]

I place the observer, now, on the falling object. Let this observer look behind the fixed clock. Which is the last Minkowski time the observer on the falling object can see on the remote clock, before the signal becomes too weak and before it crosses the horizon ?

The signal seen by the infalling observer, coming from the fixed clock, does not get infinitely weak at the horizon; it is only redshifted by a factor of about 2 (the exact factor depends on the details of the infall). The time seen by the infalling observer on the fixed clock as he crosses the horizon will be roughly the time on his own clock at that instant minus $R$, i.e., minus the radial coordinate of the fixed clock. (Note that I am using units where $c = 1$.)

 

[Logan5]

This is not really correct. Yes, the peak signal is effectively emitted from the infalling object before it crosses the horizon; but if the infalling object did not cross the horizon, but instead remained above it (suppose, for example, that there was a large rocket attached to it that fired just before it reached the horizon, and kept it "hovering" there), you would see a different pattern of GWs; also, other observations would be different as well (for example, the light from the object would stop redshifting at some finite wavelength, instead of continuing to redshift to arbitrarily long wavelengths).

We have a misunderstanding here. The object which fall does not "hover" : it follow a geodesic of course (free fall). If the object was hovering, or does not follow a geodesic, I agree the GW signal should be a different pattern. But this is not the case. In the "frozen star" model, objects who follow a geodesic asymptotically approches the horizon, never stop or hover, and only cross the horizon at Minkowski time = infinity (as given by well known equations or coordinates).

So, while the GWs and other observations do not directly show you the object crossing the horizon, they are certainly evidence of the object crossing the horizon, since all those observations would be different if the object did not cross the horizon.

Nevertheless, this remark can be correct even if I still don't see clearly what are these evidences. If this remark is correct, do you agree that the thought experiment in post #1 can be modified to measure these "evidences" and determine a finite T3, as soon as the signal is significant of a "crossing horizon" event ?

 

[PeterDonis]

The object which fall does not "hover" : it follow a geodesic of course (free fall).

This is correct. But if the object falls into the hole, through the horizon, it is not "frozen" near the horizon. Similarly, if the hole is formed by collapsing matter, it is not a "frozen star"; the collapsing matter is not "frozen". It collapses all the way to form a singularity at $r = 0$.

In the "frozen star" model, objects who follow a geodesic asymptotically approches the horizon, never stop or hover, and only cross the horizon at Minkowski time = infinity (as given by well known equations or coordinates).

If this is what you mean by the "frozen star" model, then it is not a different model from the black hole model; it is just an incomplete version of the black hole model. It only covers the region outside the horizon. This model does not say the horizon is not there, or that nothing can fall through it. It only says that it cannot describe such events.

What you are missing here is that the "Minkowski time" of an event is just a number; it has no physical significance in itself. The proper time elapsed on the distant observer's clock between two events on his worldline has physical significance; but there is no invariant way of saying which event on that distant observer's worldline happens "at the same time" as some event near, or at, or inside the horizon of the black hole. Your "Minkowski time" is simply one convention for saying which events happen "at the same time"; but it is only a convention. And since it assigns the value "infinity" to events on the horizon, it is a convention that simply does not work there. There is no physical meaning to that "infinity" value. The way to fix it is to adopt a different convention that does not have that problem.

Your question about what time T3, by the distant observer's clock, he will observe GWs from an object falling into the hole does have physical meaning; but note that it is a question about events on that observer's worldline. If I tell you a finite value for T3, that does not tell you, in itself, "what time" the object crossed the horizon, according to the distant observer's clock; answering that question still requires adopting a convention, even if you know T3. Even with a finite T3, you could still adopt the convention that assigns the value "infinity" as the "Minkowski time" of events on the horizon, such as the object crossing it. You might not want to, because it seems unreasonable to you; but "unreasonable" is not the same as "mathematically impossible".

If this remark is correct, do you agree that the thought experiment in post #1 can be modified to measure these "evidences" and determine a finite T3, as soon as the signal is significant of a "crossing horizon" event ?

The answer for the finite T3 that jimgraber gave you already does that. The GW pattern he describes is the "evidence" that the object fell through the horizon. If it hadn't, a different GW pattern would have been observed.

 

[Logan5]

The signal seen by the infalling observer, coming from the fixed clock, does not get infinitely weak at the horizon; it is only redshifted by a factor of about 2 (the exact factor depends on the details of the infall). The time seen by the infalling observer on the fixed clock as he crosses the horizon will be roughly the time on his own clock at that instant minus $R$, i.e., minus the radial coordinate of the fixed clock. (Note that I am using units where $c = 1$.)

I never read somebody who stated that as clearly as that (thank you !). And your statement made me reconsider this kruskal szekeres diagram of a fall : http://www.google.fr/imgres?imgurl=http%3A%2F%2Fi.stack.imgur.com%2FYA5TE.png&imgrefurl=http%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F111930%2Fschwarzschild-metric-coordinate-sign-change-in-0-leq-r-leq-2gm&h=898&w=658&tbnid=lAyY3AjZjbaIvM%3A&zoom=1&docid=H-7OyVASRlrGbM&ei=mLpLVY31EeejyAOzooDoDg&tbm=isch&iact=rc&uact=3&dur=251&page=1&start=0&ndsp=1&ved=0CCEQrQMwAA

In a KS diagram, photons conveniently follows a constant 45° path (as in Minkowski diagram), and photon path can be reversed. Indeed, if we follow the past photon cone at tau=33.7M (horizon crossing) until the R=3.5M line (say the distant clock is at this distance), this intersects a "t" (minkowski time) line very near of the "t" line where the object was when it was at 3.5M distance. You should be right and you made me realize this. Thank you again.

Well, this fact gives us nothing about the problem, but is worth to be known.

 

[Logan5]

If I tell you a finite value for T3, that does not tell you, in itself, "what time" the object crossed the horizon, according to the distant observer's clock; answering that question still requires adopting a convention, even if you know T3.

This is the very crux of the problem. In fact, I am not interested by a Minkowski time at which I can tell "the object IS in the BH". And I know (and understand) simultaneity is conventional even in simple SR cases, and even more in extreme GR cases ! I am interested by an evidence which can invalidate frozen star model, and you are near to convince me. I am digesting your other statements I carefully read.

I can't help myself to think, if we have a signal significant of an horizon crossing, then this event is the past of this measure (whatever moment in the past, but in the past), but I think you'll tell me we cannot conclude this. There is still an hard difficulty here.

 

[PeterDonis]

I can't help myself to think, if we have a signal significant of an horizon crossing, then this event is the past of this measure (whatever moment in the past, but in the past), but I think you'll tell me we cannot conclude this.

Yes, you're right, you can't. That's one of the very counterintuitive features of a black hole spacetime; no event at or inside the horizon can ever be in the past (more precisely, in the past light cone) of any event outside the horizon. In fact, this is one way of stating the definition of an event horizon: it's the boundary of the region of spacetime that can never be in the past light cone of any event outside it.

Mathematically, the reason for the presence of the event horizon and the black hole region in the model is clear: that's what the Einstein Field Equation says when you plug in the appropriate stress-energy tensor. Physically, the reason physicists believe the horizon and the black hole region are there, even though they can never be in our past light cone, is simple: where else can the infalling object go? If it stopped falling, or fell more slowly, or otherwise failed to cross the horizon at the event where the mathematical model says it should, we would observe something different.

 

[stedwards]

[...]Mass of object and observer is negligible compared to the mass of the BH.[...]

Hello, Logan.

Exactly, what "negligible" means, in this context might effect the answer--whether the infalling object is small, but finite mass, or the object is zero mass following a time-like geodesic.

If finite, the combined radius of object + black hole becomes
\frac{M_S+m_o}{M_S}R_S
after coordinate time less than infinity.

The problem can be simplified using coordinate time. The ratio between the elapsed coordinate time (the time measured be an observer at asymptotic infinity) and the elapsed local time of the observer stationed at some radius R&gt;R_S, is constant.

\Delta t_o=\frac{\partial t}{\partial t&#039;}\Delta t
\Delta t_o=\sqrt{(1-R_S/R)} \Delta t

I've also found it useful at times to replace the coordinate time.

s=(1=R_S/r)^{-1} t

It is well behaved for r&gt;R_s and s \rightarrow t at asymptotic infinity.

 

[PeterDonis]

If finite, the combined radius of object + black hole becomes

$$\frac{M_S+m_o}{M_S}R_S$$

after coordinate time less than infinity.

Which coordinates are you using? Ordinary Schwarzschild coordinates actually can't be used "as is" in this scenario, because they are specifically constructed for an "eternal" black hole which never gains any mass.

 

[stedwards]

Which coordinates are you using? Ordinary Schwarzschild coordinates actually can't be used "as is" in this scenario, because they are specifically constructed for an "eternal" black hole which never gains any mass.

We ask, at x amount of proper time of the infalling particle is it below an event horizon of a black hole of original radius Rs. The new causal horizon has R=M+m.

If you have a better model, I'd like to see it. To have some chance of a realistic solution the test mass should be non-zero mass. In such case it has nonzero radius. In a realist perturbation the the Schwarzschild metric, Rs is too small. Any question involving the critical radius should involve the mass of the test particle-black hole system at small r>Rs. I'm unaware that this one upgrade would be ultimately insufficient. May questions would remain undecided.

Do you have an extremal action metric solution for radially infalling, finite mass? This would be a degenerate two body problem.

 

[PeterDonis]

If you have a better model, I'd like to see it.

Sure, just use coordinates that aren't singular at the horizon. For example, Painleve or Eddington-Finkelstein coordinates. This allows you to make the mass $M$ of the hole a function of the time coordinate (actually, in the general case it ends up having to be a function of the time and radial coordinates) without causing any problems. An example of this technique is the ingoing Vaidya metric, which describes radiation falling into a black hole for the idealized case of perfect spherical symmetry; it basically amounts to using Eddington-Finkelstein coordinates with a mass $M$ that can increase with time.

Do you happen to know if there exists a extremal action metric solution, of radially infalling, finite mass?

If you mean for the original collapse to a black hole, the Oppenheimer-Snyder model describes that for a perfectly spherically symmetric collapse. But the total mass in this model is constant. If you mean mass radially falling into an existing black hole, I don't think there is an exact solution known; this case is normally solved numerically. (ISTM that there ought to be an analogue of the ingoing Vaidya metric for the case of spherically symmetric matter, instead of radiation, falling into a black hole, but I don't know of one.)

 

[stedwards]

Sure, just use coordinates that aren't singular at the horizon. For example, Painleve or Eddington-Finkelstein coordinates. This allows you to make the mass $M$ of the hole a function of the time coordinate (actually, in the general case it ends up having to be a function of the time and radial coordinates) without causing any problems. An example of this technique is the ingoing Vaidya metric, which describes radiation falling into a black hole for the idealized case of perfect spherical symmetry; it basically amounts to using Eddington-Finkelstein coordinates with a mass $M$ that can increase with time.

Words be words. Do you have an online reference for an infalling particle?

If you mean for the original collapse to a black hole, the Oppenheimer-Snyder model describes that for a perfectly spherically symmetric collapse. But the total mass in this model is constant. If you mean mass radially falling into an existing black hole, I don't think there is an exact solution known; this case is normally solved numerically. (ISTM that there ought to be an analogue of the ingoing Vaidya metric for the case of spherically symmetric matter, instead of radiation, falling into a black hole, but I don't know of one.)

 

[PeterDonis]

Do you have an online reference for an infalling particle?

If you mean an infalling object with non-negligible mass, again, I don't think there is any exact solution known for this case; it is solved numerically. Numerical relativity is a large subject and I am not very familiar with the details; you could try the Wikipedia page to get a brief overview:

http://en.wikipedia.org/wiki/Numerical_relativity

In any case, this is not really relevant to the question I asked about your post #52. If you are using ordinary Schwarzschild coordinates, as I suspect you are, the statement you made in that post that I quoted in post #53 is not correct.

 

[stedwards]

In any case, this is not really relevant to the question I asked about your post #52. If you are using ordinary Schwarzschild coordinates, as I suspect you are, the statement you made in that post that I quoted in post #53 is not correct.

Again, peter, I would like to see more than claims, though am sure you've put in a great deal of thought into your posts. I don't read texts without a critical eye, and this is no less true is this forum. Could you elaborate? I think some mathematics should be in order.

 

[PAllen]

Just to be a bit perverse, while it is true that (without modifying GR), you cannot have a sufficiently massive collapse where the infalling matter (or later infalling matter) stops at some frozen surface outside the horizon (this is precluded by Buchdahl's theorem), if you insist that you cannot stand the horizon and its interior, there is mathematical way to do it without violating the GR field equation. Instead of a frozen star, what you do is more analogous to those infamous Soviet era pictures where purged (killed) leaders were simply edited out of historically significant pictures that might show them in a positive light. You institute a 'manifold purge' (not a freeze), forming the open submanifold that includes no events not in the past of future timelike infinity. Whatever coordinate justification you may use, this is the topological operation you are really doing - simply leaving an open hole in the manifold. The result is a valid manifold, GR equations hold everywhere on it, and gravitational radiation would be the same (to calculate the radiation, you would have to admit the horizon - to compute horizon ring down, but you would treat this as mathematical device, and excise it from your 'real' manifold).

In a mathematically formal way, SR locally holds everywhere - because it is an open set. The close you get to the open 'edge' the more local this statement is, but it is still formally true.

However, the plausibility of the physics is another matter altogether. To highlight the consequence, imagine a lab capsule free falling into a large, old, isolated BH (so no major tidal effects until well inside the horizon). The rate distant clocks are observed to tick at compared to local clocks can be made whatever you you want based on initial conditions (ranging from dropping from fairly near the BH [distant clocks fast/blueshifted], to being thrown toward the BH at some point faster than a free faller from infinity [distant clocks, slow/redshifted as much as you want]). So I will posit initial conditions such that distant clock rates are the same as the lab's on horizon crossing. This means that Doppler shift will also be neutral. The distant universe and distant clocks will be seen to be behaving just like yours. You are bouncing a ball off a wall of the lab. Suddenly as 3 PM (horizon crossing) approaches, with the ball in mid flight, the lab is required to simply cease to exist. There is no other admissible local interpretation. It cannot approach a frozen surface - that is mathematically precluded. There is no local physics to cause a freeze. A freeze implies absence of motion, but still progress of time. However, it is exactly time and existence that must cease at an arbitrary local moment to sustain this 'manifold purge'. The world line of every atom of the lab and its contents must simply cease on approach to 3 PM (a time, not a place). If you find this model palatable, you are welcome to it ...

 

[PAllen]

Again, peter, I would like to see more than claims, though am sure you've put in a great deal of thought into your posts. I don't read texts without a critical eye, and this is no less true is this forum. Could you elaborate? I think some mathematics should be in order.

Not needed. The Schwarzschild solution is a static, spherically symmetric solution. Its uniqueness is given by Birkhoff's theorem. You cannot grow the horizon without violating this defining basis. Solutions for a spherically symmetric horizon that grows by infalling null dust, for example, are known as ingoing Vaidya metrics - they are not static, and are not the SC metric. Google Vaidya metric and you'll get let's of references. [edit: I see Peter already posted a link for this]

For any realistic infall that violates spherically symmetry, you cannot remotely use the SC metric - it is mathematical contradiction - using a static spherically symmetric solution purportedly model a non-static, non-symmetric scenario. There are no exact metrics for this case. As has already been explained, numerical relativity has become adept at modeling such scenarios.