Detecting matter falling into a Black Hole

[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.