Detecting matter falling into a Black Hole

[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>R_S, is constant.

\Delta t_o=\frac{\partial t}{\partial t'}\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>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.

 

[Logan5]

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.

I tend to agree with this sentence, and it is the reason why I believe the answer is 3., not 2. Because, in this case, how can a "region of spacetime that can never be in the past light cone of any event outside it" can have a causal influence on a signal dected remotely, and give "evidences" that an object crosses it ?

 

[Logan5]

What you are missing here is that the "Minkowski time" of an event is just a number; it has no physical significance in itself.

I think the Minkowski time does have a physical significance : it is the proper and physical time of observers hovering at fixed distance of the BH (isn't it ?). I have never understood statements which states that it is "just a number" (or, in this case, all times coordinates are "just a number", but what is the point to say such a thing ?).

But I agree that this physical significance does not help a lot our problem. I think the horizon crossing problem is undecidable, and we can and will never know if an object as crossed the horizon, or is approaching asymtotically it.

 

[Logan5]

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 ...

If universe/time cease to exists while the object approaches horizon, I can perfectly understand and accept this. There is by the way a big simultaneity challenge here : we have seen that simultaneity is relative, even conventional, in relativity. But if universe/time cease to exists "at once", is it not an ultimate and absolute "simultaneity" top ? I think we do not have the physics to think about this problem.

There is also another possibility : the BH evaporates before object reaches his horizon : in this case, it will happen nothing at 3 PM, and there is no conceptual difficulties.

 

[PAllen]

I think the Minkowski time does have a physical significance : it is the proper and physical time of observers hovering at fixed distance of the BH (isn't it ?). I have never understood statements which states that it is "just a number" (or, in this case, all times coordinates are "just a number", but what is the point to say such a thing ?).

No it isn't. Even for non-inertial motion in SR, you have the issue that different procedures for synchronizing separated clocks, that all are equivalent for an inertial frame, lead to different results. How do you decide which physical procedure is valid? In GR, there is no such thing as a global inertial frame at all, so there is no observer at all with a well defined global simultaneity. All are pure conventions, with different methods leading to different answer.

An example, physically motivated simultaneity convention that ascribes a finite time horizon crossing and interior events for external observers is as follows:

For a given external observer (e.g. earth), there is the moment a signal can be sent that will be received by an infaller exactly at horizon crossing (call this THm). There is another moment, a little later, when a signal would be received by an infaller just before they reached a singullarity (call this TSm). Then, these signal reception events have ceased to be in Earth's past light cone any time after THm and TSm, respectively. They are thus spacelike separation, which means 'plausibly simultaneous' in GR or SR. So, simply add to these times a reasonable light delay time (a suggested computation for this is given next), call them d1 and d2. Then you state, with perfect meaningfulness for earth, that an infaller crossed the horizon at THm+d1, and reached the singularity at TSm+d2.

One way of computing a reasonable d1 and d2 is to measure parallax distance to the BH, and its apparent angular radius (adjusting for lensing), and use this distance / c as the delay time.

 

[PAllen]

If universe/time cease to exists while the object approaches horizon, I can perfectly understand and accept this. There is by the way a big simultaneity challenge here : we have seen that simultaneity is relative, even conventional, in relativity. But if universe/time cease to exists "at once", is it not an ultimate and absolute "simultaneity" top ? I think we do not have the physics to think about this problem.

There is only a simultaneity challenge if you insist that distant simultaneity has well defined meaning in GR. I would claim this is simply a false claim in GR. If you use causal structure (causal past, future, and neither (spacelike)), then infaller and distant observer completely agree on the relationships between events. In particular, they agree on which events can be declared 'possibly now at a distance'.

There is also another possibility : the BH evaporates before object reaches his horizon : in this case, it will happen nothing at 3 PM, and there is no conceptual difficulties.

Look up other threads on this. It is false. A horizon and horizon crossings occur long before evaporation. Further, for an evaporating BH, horizon crossing events (but not events interior to the horizon) are given finite times by the Einstein clock synch (radar convention). This is because all the light trapped on the horizon is released when the BH finally evaporates. A consequence is that all of the following have finite radar times measured by a distant observer:

T1 - time horizon forms during collapse of a body
T2 - time of a later infalling body crossing the horizon
T3 - time of evaporation

Per radar time, not only are all these times finite, but T1 < T2 < T3, as expected.

 

[PeterDonis]

I think the Minkowski time does have a physical significance : it is the proper and physical time of observers hovering at fixed distance of the BH (isn't it ?).

That isn't "Minkowski time"; it's proper time along the hovering observer's worldline. That has physical significance, yes, but only on that worldline. Events near or on or inside the horizon are certainly not on the worldline of an observer hovering at a fixed distance from the hole far away. So any "Minkowski time" you assign to those events is just a number, with no physical significance, as I said.

If universe/time cease to exists while the object approaches horizon, I can perfectly understand and accept this.

I don't think you've really thought through what this means. Consider the analogue in flat spacetime: you and I are in an accelerating rocket. At some instant you drop out of the rocket in your space suit and start free-falling. To me, in the accelerating rocket, I simply see your light signals redshift more and more until I can't detect them. But to you, for a finite time by your clock, you free-fall; then spacetime ceases to exist and you just vanish. Not "turn into something else", but "vanish"--all the matter and energy in your body suddenly isn't there any more. Conservation of energy is violated. Does that really seem physically reasonable?

There is also another possibility : the BH evaporates before object reaches his horizon

It is, of course, logically possible for an object to start far enough away from the hole that it doesn't reach it until after it's evaporated--for a hole with the mass of the Sun, you would have to start some $10^{67}$ light-years away, more or less. (I say "logically possible" because of course such an infall would not happen in our actual universe.) But it is not possible to restrict all possible infalls to this case. That is, for any size black hole, there is some finite distance above the horizon from which an object, if it starts falling from that distance or less, will reach the horizon before the hole evaporates. (And, as I've just noted, that distance is extremely large for a hole of stellar mass or larger.) Here "before" means "there is a region of spacetime bounded by the horizon that cannot send light signals to infinity, and the worldline of the infalling object crosses the horizon and enters this region"; that is, the statement is about the geometry of the spacetime and how the object's worldline fits into it, and is independent of your choice of coordinates.

 

[Logan5]

No it isn't. Even for non-inertial motion in SR, you have the issue that different procedures for synchronizing separated clocks, that all are equivalent for an inertial frame, lead to different results. How do you decide which physical procedure is valid?

I let Wheeler/Thorne/Misner, in "Gravitation", p. 596 decides. Let me quote this

One "machine" design which constructs (mentally) such a t coordinate, and in the process measures it, is the following. Observers using radar sets arrange to move along the coordinate lines r, phi, theta = const. They do this by adjusting their velocities until each finds that the radar echos from his neighbors, or from "benchmark" reference points in the asymptotically flat space, require the same round-trip time at each repetition. Equivalently, each returning echo must show zero doppler shift; it must return with the same frequency at which it was sent out. Next a master clock is set up near spatial infinity (far from the star). It is constructed to measure proper time-which, for it, is Minkowski time "at infinity"-and to emit a standard one Hertz signal. Each observer adjusts the rate ofhis "coordinate clock" to beat in time with the signals he receives from the master clock. To set the zero ofhis "coordinate clock," now that its rate is correct, he synchronizes with the master clock, taking account ofthe coordinate time Delta t required for radar signals to travel from the master to him. [...] Each observer moving along a coordinate line (r, theta, phi = const.) now has a clock that measures coordinate time t in his neighborhood. The above discussion identifies the Schwarzschild coordinates of equation (23.7)

I agree this is a convention (very close to Einstein synchronisation of clocks in SR), but this convention gives a physical meaning to Minkowski (or Schwarzschild) time. Note I don't write "gives THE physical meaning", but "A physical meaning", which has some validity since determined by valuable scientists, very close from an idea of sync from another valuable scientist : Einstein.

 

[PeterDonis]

this convention gives a physical meaning to Minkowski (or Schwarzschild) time.

Yes, but only for events outside the horizon, because the observers required to realize it physically can only exist outside the horizon, because only outside the horizon are the coordinate lines r, phi, theta = const timelike. So you can't use this method to give a physical meaning to Schwarzschild time on the horizon, or inside it.

Also, this method only works for the case of a black hole that has the same mass forever; as soon as you allow the hole to gain mass, there are no longer any observers at all who meet the physical requirements described--that is, there are no possible worldlines those observers can follow which give a constant, unchanging round-trip light travel time between any two of them. So again, you can't use this method to give physical meaning to Schwarzschild time for the case of a hole that gains mass by having matter fall into it.

 

[Logan5]

That isn't "Minkowski time"; it's proper time along the hovering observer's worldline. That has physical significance, yes, but only on that worldline. Events near or on or inside the horizon are certainly not on the worldline of an observer hovering at a fixed distance from the hole far away. So any "Minkowski time" you assign to those events is just a number, with no physical significance, as I said.

Please see my #67 post. What I meant is that Wheeler/Thorne/Misner meant. They construct a physically significant machine including elements which stay at (r,theta,phi) = const (so, hovering). The "proper time along the hovering observer's worldline" is the Minkowski (Schwarzschild) time according to them. They don't forbid this machine to be at r "near" the EH (but they forbid it on, or inside of course).

But it is not possible to restrict all possible infalls to this case.

From a certain point of view, this is possible. Assume the POV (and this the POV I deseperately seek counter examples or falsification, by thought or actual experience) where the object enter the horizon at Schw. time = infinite (as said by formulas). Whatever large is the live duration of BH, say 10^67 year, the object whill encounter a W/T/M clock which indicates (at the same place than the falling object) 10^67 before (spatially) the object reaches the horizon, because 10^67 < infinity. And is not the life time of the BH defined and counted in Minkowski time ?

 

[PAllen]

I let Wheeler/Thorne/Misner, in "Gravitation", p. 596 decides. Let me quote this
I agree this is a convention (very close to Einstein synchronisation of clocks in SR), but this convention gives a physical meaning to Minkowski (or Schwarzschild) time. Note I don't write "gives THE physical meaning", but "A physical meaning", which has some validity since determined by valuable scientists, very close from an idea of sync from another valuable scientist : Einstein.

What makes a convention like this have weight is uniqueness or preference. For the following special cases you have the feature several plausible physical conventions all produce the same result:

1) Inertial observers in flat spacetime
2) Uniformly accelerating observers who maintain constant apparent distance from each other
3) The stationary observers in a stationary GR solution (this is specifically the case MTW is discussing).

For any of the following cases, there is no equivalence between equally plausible conventions, thus no preference at all:

1) observers with varying acceleration or any rotation in flat spacetime (e.g. a spinning rocket)
2) Any GR solution in which anything at all moves (other than test bodies assumed to be of insignificant mass) or changes

 

[PAllen]

From a certain point of view, this is possible. Assume the POV (and this the POV I deseperately seek counter examples or falsification, by thought or actual experience) where the object enter the horizon at Schw. time = infinite (as said by formulas). Whatever large is the live duration of BH, say 10^67 year, the object whill encounter a W/T/M clock which indicates (at the same place than the falling object) 10^67 before (spatially) the object reaches the horizon, because 10^67 < infinity. And is not the life time of the BH defined and counted in Minkowski time ?

No, an evaporating BH is not a static solution (by definition). The SC metric is inapplicable, and the described MTW procedure cannot be accomplished at all (approximately, maybe, but not exactly - and in the long run, it is the tiny differences that matter).

For an evaporating hole, in fact, a procedure similar to this (using exchange of radar signals based on a chosen outsider oberver, e.g. Earth - because achieving the symmetry of the MTW procedure is impossible for an evaporating BH) give the result I gave in my prior post #65.

 

[pervect]

Some very general comments. Most threads on evaporating black holes turn into long arguments, due to the complexity of the problem. Additionally,and unfortunately, many of the arguments we see here on PF are not very rigorous. There are many important things can be learned by studying the much simpler case of static black holes. I would strongly recommend, especially to people who do not have a strong physics or mathematical background in special relativity, tensors, and differential geometry to consider the case of non-evaporating black holes first and leave the case of the evaporating black hole for later after they understand the non-evaporating case.

There are many important lessons that could be learned and learned much more easily by studying the simple case of the non-evaporating black hole first. My personal impression is that many people with preconceived notions are attracted to the harder problem of the evaporating black hole just because the issues are less clear, leaving more freedom for them to argue for their preconceived notions. Unfortunately, arguing for ones preconceived notions is not the best way to learn new things :(.

A few of the key points that I think that any mental model of black holes (simple non-evaproatingones) should explain and predict:

1) That the proper time to fall through the event horizon is finite. This generally tends to make common lay interpretations that suggest that crossing the event horizon 'doesn't happen' very unattractive (or perhaps even wrong), since it's difficult to explain clearly how things that "don't happen" do happen at a finite time for some observers.

2) That there is no infinite blueshift of light for an infalling observer while falling through the event horizon of a black hole. To calrify, the doppler shift being observed is observed and recorded from a non-falling observer "at infinity", who is shining a fixed-frequency light or laser beam in a radial "downward" direction at the falling observer. This tends to make interpretations that "time stops" at the event horizon rather unattractive, due to the fact that such interpretations can't explain why the free-falling observer don't observe infinite blueshifts.

A further clarification of point 2. It is correct to say that the time dilation for a hovering observer becomes arbitrarily large as one approaches the horizon. But because there is no "absolute time", one cannot correctly conclude from this that time slows arbitrarily for any observer, the observer falling through the event horizon being a direct counterexample. The issue of "absolute time" also has a long history of misunderstanding, but it would be too long of a digression to get into this more without some indication of interest.

3) An additional strong point, but one that would take a while to explain more fully, is the existence of event horizons for accelerating observers that are formally similar to the event horizons we see in black holes. The name for these horizons is "Rindler Horizons". It turns out that many of the typical interpretations of event horizons fail to explain clearly what happens with the horizon of an accelerating observer. For instance, it's reasonably obvious that the Earth exists and will continue to exist, regardless of whether someone accelerates away from it at 1G for slightly over a year. However, the Earth will fall beyond the accelerating observers event horizon , and the accelerating observer will never "see" this happen on any signal unless they stop accelerating.

If textbook, web, or literature references are needed for the above points, do ask. I believe that a good interpretation of the physics should explain all three of the points above, points that can be verified in the literature.

 

[PeterDonis]

They construct a physically significant machine including elements which stay at (r,theta,phi) = const (so, hovering). The "proper time along the hovering observer's worldline" is the Minkowski (Schwarzschild) time according to them.

No, the proper time along the hovering observer's worldline is not the same as the Schwarzschild coordinate time. For a given hovering observer, the ratio of the two is constant; but they are not the same. (And, as I said, all of this only applies in the idealized case where the hole has the same mass forever.)

From a certain point of view, this is possible.

No, it is NOT. Pardon me for the capitals, but you have already been told that this "POV" is not correct, so please do not keep repeating it.

is not the life time of the BH defined and counted in Minkowski time ?

No. That has already been stated in this thread. Again, please do not keep repeating things that you have already been told are wrong.

As I hinted in post #66, you need to stop thinking in terms of "Minkowski time" or indeed in terms of any coordinates at all. You need to think in terms of spacetime geometry. The spacetime of an evaporating black hole has a certain geometry. The question of whether or not that geometry includes a region that cannot send light signals to infinity is a straightforward question of geometry: the answer is yes, it does. The question of whether infalling timelike geodesic worldlines can enter that region is also a straightforward question of geometry: the answer is yes, they can. These facts of spacetime geometry are sufficient to show that objects can fall into the hole before it evaporates, regardless of any choice of coordinates or any kind of "time" you might want to use. You need to get the facts of spacetime geometry straight first, before even thinking about what coordinates you might use to label events in that geometry.

I realize this spacetime geometry is difficult to visualize, but for the idealized, perfectly spherically symmetric case (which, note, is not quite the case we've been discussing--see below), there is a helpful kind of diagram known as a Penrose diagram, which can be very useful in visualizing the causal structure of spacetimes. The Penrose diagram for an evaporating black hole is shown on this Wikipedia page:

http://en.wikipedia.org/wiki/Black_hole_information_paradox#Hawking_radiation

The black hole region is the triangle at the upper left, with the jagged line on the top; the 45 degree line forming the lower right boundary of this triangle is the horizon. Timelike curves are any curves closer to vertical than 45 degrees in this diagram, and it is easy to see that there are such curves that go from the exterior region (the rest of the diagram besides the black hole) into the black hole region.

Note that this diagram can also represent the case of a hole that forms by a spherically symmetric collapse of matter, then gains further mass by a spherically symmetric process (for example, a spherically symmetric shell of matter might fall in), and eventually evaporates. The same general causal structure still applies. As has been mentioned, we do not have exact solutions for the more general case of processes that are not spherically symmetric, but numerical solutions indicate that these more general cases still have the same general causal structure. So from the standpoint of answering the questions we've been discussing in this thread, this Penrose diagram is applicable.

 

[stedwards]

Getting back to the original question...

Consider a test particle falling radially from rest at asymptotically infinite r. To put some numbers to things, the observer, O is at 10Rs; 10 times the schwarzschild radius. The test particle passes O at some velocity less than c.

At a radius greater than about 4.5Rs, luminal signals from the infalling test particle will not arrive at 10Rs before time "ends" (t>=infinity). There is no need to consider the physics at or near Rs.

It is mute whether the clock measuring t is coordinate time, or the proper time at O. They are in finite ratio.

 

[PeterDonis]

At a radius greater than about 4.4Rs, luminal signals from the infalling test particle will not arrive at 10Rs before time "ends" (t>=infinity).

Please show your work. This answer does not look correct.

 

[PAllen]

Getting back to the original question...

Consider a test particle falling radially from rest at asymptotically infinite r. To put some numbers to things, the observer, O is at 10Rs; 10 times the schwarzschild radius. The test particle passes O at some velocity less than c.

At a radius greater than about 4.5Rs, luminal signals from the infalling test particle will not arrive at 10Rs before time "ends" (t>=infinity). There is no need to consider the physics at or near Rs.

It is mute whether the clock measuring t is coordinate time, or the proper time at O. They are in finite ratio.

This is false. Note that as the infalling test particle reaches 4.5 Rs and emits light, the light emits radially outward propagates on the same null geodesic as emitted by a hovering observer at 4.5 Rs. The only difference (for the same emission process) is frequency - not spacetime path. Your claim then is equivalent to the claim that a hovering observer at 4.5 Rs cannot send a signal to a hovering observer at 10 Rs. This is patently absurd.

Anyway, light emitted from anywhere outside of Rs, irrespective of the 4-velocity of the emitter, can reach infinity as long as emitted in the outward radial direction.

 

[stedwards]

This is false. Note that as the infalling test particle reaches 4.5 Rs and emits light, the light emits radially outward propagates on the same null geodesic as emitted by a hovering observer at 4.5 Rs. The only difference (for the same emission process) is frequency - not spacetime path. Your claim then is equivalent to the claim that a hovering observer at 4.5 Rs cannot send a signal to a hovering observer at 10 Rs. This is patently absurd.

Interesting. Please show your work.

 

[stedwards]

Please show your work. This answer does not look correct.

No reason for that, as PAllen so politely put it.

 

[PAllen]

Interesting. Please show your work.

No need in this case. If 4.5 Rs couldn't send a message to 10 Rs, it would mean they were separated by a horizon which is well known to be false. Formal reasoning from known mathematical facts is valid - it is known as a proof.

[edit: In any case, it is really quite easy to compute. I will show the proper time along an r = 10R static world line (using R for SC radius, c=1) elapsed sending a signal ro r=4.5R and back. Doing this way, I compute an invariant rather than a coordinate dependent quantity.

A radial null geodesic is defined by dt/dr = 1/(1 - R/r). The coordinate time for the round trip described above is simply twice the integral of this from 4.5R to 10R. Multiply that by √.9 to get proper time along r=10R world line. The result is:

2(√.9) R (log (9/3.5) + 5.5)hardly 'the end of time'.]

You may complain about lack of politeness, but you are insisting on making emphatic statements in contradiction to axioms (e.g. pretending it is valid to use SC metric formulas for a non-static, non-spherically symmetric scenario).

 

[PeterDonis]

No reason for that, as PAllen so politely put it.

So you agree that PAllen is correct and that your claim in post #74 was false?

 

[yoron]

There is one thing I wonder about. What Peter wrote and Logan commented on. " 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."

I started to think about the gravitational 'clock ringing', aka those gravity waves created by a object passing a event horizon. From where would they emanate? Assuming them to have a finite propagation 'c'. Outside the event horizon, as defined by? Or at/inside? It's not that I think that Peter is wrong in his formulation, it makes a lot of sense to me.
=

Also, as I suddenly started to think of it as 'gravitons'.

 

[PeterDonis]

I started to think about the gravitational 'clock ringing', aka those gravity waves created by a object passing a event horizon. From where would they emanate?

Gravitational waves do not emanate from a point. The GWs emitted by a black hole as an object falls in have wavelengths of the same order as the radius of the horizon, so to the extent they "emanate" from anywhere, they emanate from the entire hole--more precisely, from just outside the horizon, but the entire horizon, not just one particular point.

Also, bear in mind that GWs are made of spacetime curvature; they are not something over and above spacetime curvature. So another way of looking at GW emission by a black hole is that the curvature of spacetime has a "wavy" region in it, which starts, more or less, when the object falls into the hole, and gradually propagates out to infinity. On this view, the GWs are not "emitted" from anywhere; they're just part of the overall spacetime geometry.

 

[yoron]

Yes, gravity is one of those things that is assumed to communicate everywhere, in this case the interaction between a event horizon and some in-falling object. And if I think of it as something able to coexist in SpaceTime as defined by gravity, as that to me is what defines its existence even though it's a 'singularity', then it becomes a 'whole description' of this SpaceTime (what I think of as a container model). But if I think of it as gravitons then? Wouldn't that give those rather unique qualities not seen in any other 'forces'?
=

And thanks for the answer Peter :)

 

[PeterDonis]

gravity is one of those things that is assumed to communicate everywhere

I'm not sure what you mean by this. Gravity propagates at the speed of light, just like other forces.

if I think of it as gravitons then? Wouldn't that give those rather unique qualities not seen in any other 'forces'?

Gravitons are spin-2; the other known force carrier particles are spin-1. That's the only difference.

(Also bear in mind that gravitons have not been detected and we don't anticipate detecting them any time soon. They are purely hypothetical, based on the assumption that gravity should have quantum aspects like the other forces do.)

 

[yoron]

Sorry Peter, was thinking of that a black hole, no matter a event horizon, still exist as gravity outside that event horizon, as well as what I've heard about branes. https://edge.org/conversation/theories-of-the-brane-lisa-randall And yes, your last comment makes also a lot of sense to me. Me writing 'communicate' may be a bad choice of word though.

 

[PeterDonis]

was thinking of that a black hole, no matter a event horizon, still exist as gravity outside that event horizon

Ah, ok. Note, though, that a charged black hole also has an electric field outside its horizon, even though there is no charge outside the horizon. So gravity is not the only interaction that has that property.

The key distinguishing feature of gravity as an interaction is that an object's motion under gravity is independent of any of its properties--all objects "fall" the same way in a gravitational field. For any other interaction, different objects respond to it in different ways.