I Measuring Light Reflection in a Black Hole

[Mike S.]

TL;DR Summary

A statite hovering at two Schwarzschild radii from a black hole drops a probe into the hole, and shines a clock signal inward after it. The probe knows its position relative to the event horizon. What is the last time on the clock received by the probe while outside the event horizon?

To keep things "simple", the black hole is 1E30 kilograms. The statite (stationary satellite, blue) hovers above the hole at a fixed location (twice the Schwarzschild radius from the singularity) by tremendous acceleration. The statite drops a probe (green) that begins to fall toward the hole at t=0 s. The statite beams a signal (yellow) with a different time stamp for each second. The probe can figure out how close it is to the event horizon, and beams that information back to the statite, accompanied by a mirror reflection of the last time stamp it received.

1) What is the last timestamp received by the probe while it is outside the event horizon? (Are there any timestamps not received?)
2) When is the probe's transmission of its last timestamp received by the statite?

Please feel free to explain very simply how you set up a problem like this.

 

[Ibix]

What's your thinking so far?

 

[PAllen]

Qualitatively, the probe will receive time stamps from the statite at and inside the horizon, until it reaches a singularity. Let’s just say, for example, it receives a time stamp of 3:00 at the horizon, and approaching 3:15, as it approaches the singularity. Any signal from the statite after 3:15 will not be received by the probe, due to its presumed destruction (mathematically, geodesic incompleteness). Meanwhile, the last reflected signal from the probe received by the statite will be asymptotically approaching 3:00, and this will be received at asymptotically infinite time later on the statite. No reflections of time stamps received by the probe with values at or later than 3:00 will ever make it back to the statite

 

[Mike S.]

What's your thinking so far?

1. The way I understand it, photons never truly stop coming from an object that falls toward a black hole - though detecting one rapidly becomes a very rare and very low energy event.

2. In concept a photon from the infalling object right next to the event horizon could come out in 10,000 years.

3. If the photon was reflected from the infalling object, then (a) the photon fell toward the hole *after* the infalling object, because it is moving at the speed of light...

4. And (b) the photon must have taken just as long to fall *toward* the hole as to come *away* from it. I'm thinking CPT symmetry: change the polarization of the photon, the phase of the photon (+ and - swap places for a half wavelength shift), and then that photon moving backward in time, in the same gravitational field, takes just as long to come out in reversed time as the original photon would in forward time.

5. So I am thinking that *all* time signals will reach the infalling object, which is moving very rapidly through very curved space, never reaching the horizon. They strike it as (near) "infinitely blueshifted radiation" that is said to form a "firewall" at the event horizon. But if the probe is not destroyed, the time signals bounce back, arriving at the station that sent them (in this example) 20,000 years after they were first sent (and indeed, at the original frequency)

6. I'm not sure, but I'm tempted to conclude that no matter what, bouncing a light beam from source A to mirror B and back to detector A' would establish the midpoint between A and A' as "simultaneous", in that frame of reference, with B. Which would establish that the infalling probe does not pass the event horizon within any reasonable amount of time - as seen by an outside observer. So I think there (1) is no last timestamp, and (2) it will take some time much longer than the time that the probe has been left to fall, before light catches up to the probe and reflects from it.

 

[Ibix]

1. The way I understand it, photons never truly stop coming from an object that falls toward a black hole - though detecting one rapidly becomes a very rare and very low energy event.

That's not correct - you seem to be mixing classical and quantum concepts. Classically, the radiation continues forever, fading and dimming but never quite disappearing. But in a quantum universe there is a last photon, although you'll never be certain it has arrived unless you dive into the hole. There could always be one more.

It's probably best to forget about photons. They're complicated and their behaviour in curved spacetime is even more so. I'm going to refer to light pulses instead, and read your "photon" to be a light pulse.

2. In concept a photon from the infalling object right next to the event horizon could come out in 10,000 years.

Yes.

3. If the photon was reflected from the infalling object, then (a) the photon fell toward the hole *after* the infalling object, because it is moving at the speed of light...

Yes

4. And (b) the photon must have taken just as long to fall *toward* the hole as to come *away* from it. I'm thinking CPT symmetry: change the polarization of the photon, the phase of the photon (+ and - swap places for a half wavelength shift), and then that photon moving backward in time, in the same gravitational field, takes just as long to come out in reversed time as the original photon would in forward time.

This is an unnecessarily complex way of looking at it. The light pulse bounced off the object and came back. Nothing changed about the environment during that process, so the return path is symmetric and it takes the same coordinate time to return as it did to fall in.

5. So I am thinking that *all* time signals will reach the infalling object

No. There is a last signal that the infalling object receives before it crosses the horizon, which will be very redshifted. There is a last signal it receives before striking the singularity, again, very redshifted.

which is moving very rapidly through very curved space, never reaching the horizon.

The object passes through the horizon in finite time by its clocks. Depending on which coordinate system external observers choose to use, the object either passes through in finite time or never passes through - it's a matter of choice that does not affect any observations. (I'm assuming the probe does not have sufficient delta-v to turn around - if it does, external observers obviously conclude that it did not pass the horizon.)

But if the probe is not destroyed, the time signals bounce back, arriving at the station that sent them (in this example) 20,000 years after they were first sent (and indeed, at the original frequency)

Crossing the horizon is a perfectly survivable experience given a sufficiently large black hole. A non-pointlike probe will be shredded before reaching the singularity, but the event horizon is no problem. Signals will continue to return forever, in principle, as discussed.

6. I'm not sure, but I'm tempted to conclude that no matter what, bouncing a light beam from source A to mirror B and back to detector A' would establish the midpoint between A and A' as "simultaneous", in that frame of reference, with B.

"Frame of reference" is not the correct term - you mean "coordinate system". You are here adopting the simultaneity condition used by Schwarzschild, which is perfectly reasonable but cannot assign a time to an object crossing the horizon because the radar pulses never return. That doesn't mean that the probe doesn't cross, just that your method for assigning a time to that event doesn't work.

So I think there (1) is no last timestamp, and (2) it will take some time much longer than the time that the probe has been left to fall, before light catches up to the probe and reflects from it.

There is a last timestamp. The stamps return ever more spread out in time. Assuming you are sending timestamps at discrete intervals, there will be a last one and you can calculate when it will return, although that may be a long time. The probe will continue to receive time stamps after that, but will no longer be able to return them to you.

 

[PAllen]

I don’t think the signals received by the probe would be very redshifted under the assumption that the probe is dropped by the statite, and signals are sent by the statite. The only case I know exactly is for a stationary platform very far from the BH ,dropping a probe, and sending signals. In this case, the probe sees a redshift of exactly 2 as it crosses the horizon.

 

[Mike S.]

The object passes through the horizon in finite time by its clocks. Depending on which coordinate system external observers choose to use, the object either passes through in finite time or never passes through - it's a matter of choice that does not affect any observations.

I've heard this many times, but I can't see the sense in it. The black hole *will* evaporate, even if it takes trillions of years. If we are even conceptually able to send signals back and forth to the infalling object over all those trillions of years, during which it never passes the horizon, then I would think eventually it should make itself apparent again in normal space, having still never crossed the horizon.

 

[PeterDonis]

I've heard this many times, but I can't see the sense in it.

We have had a number of previous threads on this. The short answer is that yes, objects can fall into a black hole that is eventually going to evaporate. I'll see if I can find references to appropriate prior threads. I suspect I'm going to have to write an Insights article on this question at some point.

If we are even conceptually able to send signals back and forth to the infalling object over all those trillions of years, during which it never passes the horizon

We aren't. We can receive signals from the infalling object for an arbitrarily long time (depending on how close to the horizon it gets before it sends the last signal we detect), but we cannot send signals to the infalling object for an arbitrarily long time, and the only reason we can receive them for an arbitrarily long time is that the time delay for a signal to climb back out to us from the infalling object increases without bound as the infalling object approaches the horizon.

More precisely: if we put "timestamps" in the light signals we send inward, showing the time on our clock when the signal was sent (call this timestamp A), and the infalling object adds its own "timestamp" when it sends back the return signal (call this timestamp B), then, no matter how long we wait, the values for timestamp A that we see in light signals coming back out to us will never exceed some finite value $T_A$, and the values for timestamp B that we see in light signals coming back out to us will never exceed some finite value $T_B$, which will be later than $T_A$ by a fixed offset that can be calculated from the ratio $R / R_S$, where $R$ is the radius from which the infalling object starts falling (and at which we are hovering, waiting for return signals), and $R_S$ is the Schwarzschild radius of the hole.

That means that any light signal we send inward after our clock reads $T_A$ will never be reflected back to us, no matter how long we wait; so after our clock reads $T_A$ we are no longer able to send light signals "back and forth" with the infalling object. And after the infalling object's clock reads $T_B$, the infalling object will be below the horizon and unable to send light signals to anywhere outside it.

 

[PAllen]

I've heard this many times, but I can't see the sense in it. The black hole *will* evaporate, even if it takes trillions of years. If we are even conceptually able to send signals back and forth to the infalling object over all those trillions of years, during which it never passes the horizon, then I would think eventually it should make itself apparent again in normal space, having still never crossed the horizon.

No, not back and forth. The statite (stationary platform) can send signal received by the free faller inside the horizon, but no return signal is possible. I described this precisely in my post #3. Evaporation happens on a much longer time scale. In fact, for hundreds of trillions of years, a stellar BH is growing faster by absorption of CMB radiation than it is losing mass by evaporation. As I mentioned in post #3, there is a specific signal sent by the statite that will reach the probe on its approach to the singularity (or whatever is really there instead). After this time for the platform, no other signals reach the probe, because it is presumably destroyed. After this specific time for the platform, none of the history of the probe within the BH horizon is in the causal future of the platform any more. This part of the probe's history has shifted to being spacelike separated from the platform, and its destruction can be legitimately considered to be happening 'now'. The fact that light is still received from the past of the probe is irrelevant to this.

 

[Nugatory]

I've heard this many times, but I can't see the sense in it. The black hole *will* evaporate, even if it takes trillions of years. If we are even conceptually able to send signals back and forth to the infalling object over all those trillions of years, during which it never passes the horizon, then I would think eventually it should make itself apparent again in normal space, having still never crossed the horizon.

The infalling object falls through the horizon, approaches the singularity, and is destroyed long before the conjectured (very plausible, convincingly argued by someone who knew what he was talking about, but still conjectured and lacking any experimental confirmation) evaporation of the black hole. Once inside the horizon all light signals emitted in all directions will end up at the singularity, also very quickly.

When the black hole evaporates, any signals emitted outside the horizon that haven't already made it out will finally be released, as will the trapped light at the horizon. But anything emitted after the infalling object passed through the horizon will have been destroyed at the singularity long ago.

 

[Orodruin]

I don’t think the signals received by the probe would be very redshifted under the assumption that the probe is dropped by the statite, and signals are sent by the statite. The only case I know exactly is for a stationary platform very far from the BH ,dropping a probe, and sending signals. In this case, the probe sees a redshift of exactly 2 as it crosses the horizon.

I put this as an exam problem a couple of years back. The redshift ($z$) is equal to one for the case of the probe falling from infinity. Of course, this means half the frequency.

 

[PAllen]

I put this as an exam problem a couple of years back. The redshift ($z$) is equal to one for the case of the probe falling from infinity. Of course, this means half the frequency.

I think of red shift (outside of cosmological contexts) as factor of frequency or wavelength. So we agree.

 

[Ibix]

If we are even conceptually able to send signals back and forth to the infalling object over all those trillions of years,

But we aren't able to do that.

If we send signals once per second down to the probe then there is a last signal that we get back. It will have been sent a very short time after the probe was released. Later signals will never come back. You can "tune" when that last signal returns by timing it to bounce off the probe arbitrarily close to the horizon crossing, but there is always a time after which no signal will ever return.

The fundamental problem here is that your definition of "time", which you based on radar echoes, does not work when the echoes never return. This does not mean that the probe doesn't cross the horizon, just that your method of assigning a time to that event doesn't work. When the hole evaporates you won't suddenly get multiple trillions of years worth of time stamps coming out at you, because they all fell into the singularity (or whatever's really there). You'll just get a really bright flash of blackbody radiation.

 

[JLowe]

I suspect I'm going to have to write an Insights article on this question at some point.

That would be great, as many of those old threads arent all that helpful.

 

[Mike S.]

@PAllen - your argument *is* valid if there is in fact a last received signal. That's why the question is about how you calculate what the last signal received by the probe is; if someone can show there is (or isn't) a specific cutoff, the qualitative arguments ought to be settled also.

 

[Orodruin]

@PAllen - your argument *is* valid if there is in fact a last received signal.

There is. This can be trivially seen in, for example, Kruskal coordinates. I am 100% sure @PAllen knows this.

 

[Mike S.]

The coordinate system is a big part of the issue. I don't have a deep understanding of Schwarzschild coordinates, but they seem "honest", whereas as I understand it the Kruskal coordinates describe the finite proper time of a hypothetical object falling into a black hole that is fixed and immutable for all eternity. Here, however, we're not looking at proper time, but at the time it takes relative to the frame of relatively "ordinary" space.

Another way to put this is based on the factoid (accurate?) that things fall through the event horizon at the speed of light. How long does it take a massive object, in any frame (other than its own light-speed frame), at any acceleration, to reach the speed of light?

 

[PeterDonis]

The coordinate system is a big part of the issue.

It can be, but not for the reasons you give.

None of the actual physics is coordinate-dependent. The physics is contained in invariants, and you can calculate invariants in any coordinate chart you like, provided that the chart covers the necessary region of spacetime. But not all coordinate charts do, and in fact the chart you seem to trust the most is the one with the biggest issue in this respect for the problem under discussion. See below.

I don't have a deep understanding of Schwarzschild coordinates, but they seem "honest"

Unfortunately, they have a huge problem for this discussion: they are singular at the horizon.

whereas as I understand it the [https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates](Kruskal coordinates) describe the finite proper time of a hypothetical object falling into a black hole that is fixed and immutable for all eternity.

No. The chart whose coordinate time is the same as the proper time of an object free-falling into an "eternal" black hole from rest at infinity is the Painleve chart.

Kruskal coordinates do describe an "eternal" black hole, but so do Schwarzschild coordinates. If you are concerned about how your coordinates describe a black hole that eventually evaporates, you are going to have that issue no matter what chart you use; but Schwarzschild coordinates are still the worst choice you can make, for the reason given above.

Here, however, we're not looking at proper time

Yes, we are.

but at the time it takes relative to the frame of relatively "ordinary" space.

There is no such thing. Simultaneity is a convention, not a physical thing. There is no physically meaningful concept of "the time it takes" for something to happen "relative to the frame" of someone else that is spatially distant. So if you are depending on any such concept in your reasoning, your reasoning is wrong.

Another way to put this is based on the factoid (accurate?) that things fall through the event horizon at the speed of light.

No, this "factoid" is not accurate. We have had many past threads on this as well. The thing that moves at the speed of light is the horizon itself: it is an outgoing lightlike surface. Massive objects falling through the horizon follow timelike worldlines, just like they do everywhere else.

 

[Mike S.]

@PeterDonis - My question as asked only addresses the space outside the event horizon, so I'm hoping someone can confidently work it out with Schwarzschild coordinates, or at least, that the other coordinate systems won't produce wrong answers.

Your objection to "relative to the frame" surprises me. When learning special relativity, there are few things so common as calculating the time it takes a particle to decay or a twin to age.

The thing that moves at the speed of light is the horizon itself: it is an outgoing lightlike surface.

Is it accurate to say that the event horizon of a black hole is stationary, or that its velocity is zero, relative to the outside observer on the statite?

 

[PeterDonis]

My question as asked only addresses the space outside the event horizon

Yes, but you are also relying on Schwarzschild coordinates as a "frame". They're not, at least not in the sense you are using the term.

Your objection to "relative to the frame" surprises me. When learning special relativity, there are few things so common as calculating the time it takes a particle to decay or a twin to age.

Yes. Are you aware that you can do such calculations without using any "frame" at all, inertial or otherwise? That, for example, the "twin paradox" can be solved without any use of a frame?

If you are not aware of these things, I strongly suggest spending some time with a textbook that emphasizes invariants over "frames". I first got this viewpoint from the classic textbook on GR by Misner, Thorne, and Wheeler, but that book is a heavy lift. Taylor & Wheeler's Spacetime Physics used to emphasize this at least somewhat in its old edition (which was the book I first learned SR from), but I don't know if it still does in its new edition.

As I said before, you can, of course, calculate invariants in any coordinates you like, as long as they cover the relevant region of spacetime. If you are indeed only concerned about what happens outside the horizon, you can indeed calculate all the required invariants in Schwarzschild coordinates. But you are going to have to suppress plenty of intuitions while doing so, because if you take those coordinates to have the sort of physical meaning that presentations of SR often ascribe (wrongly, IMO) to inertial frames, you are going to be led down some very tempting but wrong paths of reasoning. Your previous posts in this thread have already shown some examples of this.

Is it accurate to say that the event horizon of a black hole is stationary

Not with the meaning of "stationary" that you appear to be using. The horizon is a null surface, which is composed of light rays (radially outgoing light rays, in this case), and light rays can never be stationary in this sense. To put it another way, the horizon is not a "place": any "place" must be described by a timelike curve (or a set of them), not a null curve (or a set of them).

There is a meaning of "stationary" that (almost) applies to the horizon, but it doesn't have the implications that the ordinary meaning of "stationary" does. This meaning is technical: a curve can be said to be "stationary" if it is an integral curve of a timelike Killing vector field. (I won't explain that further here, but it should be easy to find a definition of this term online, or in any number of GR textbooks.) The worldlines of observers that are "hovering" at constant altitude above the black hole, and have no tangential velocity (i.e., they are not in orbits about the hole, just "hovering" radially) are stationary in this sense. The horizon is composed of integral curves of the same Killing vector field as those worldlines, but they are null, not timelike, so they are not stationary in this sense, but they "almost" are since they are the limits of such timelike curves as the radial coordinate $r$ goes to $r_s$, the Schwarzschild radius of the hole.

or that its velocity is zero, relative to the outside observer on the statite?

No. In a curved spacetime, there is no valid concept of "relative velocity" between things that are not co-located. The only observers who can assign any meaningful "velocity" relative to them to the horizon are observers who are falling through the horizon, and they all assign to it a relative velocity of $c$, radially outgoing.

 

[PeterDonis]

@Mike S. you might find this series of Insights articles on the Schwarzschild geometry helpful:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

 

[Orodruin]

My question as asked only addresses the space outside the event horizon

No, it does not as it specifically concerns an event on the event horizon.

Your objection to "relative to the frame" surprises me. When learning special relativity, there are few things so common as calculating the time it takes a particle to decay or a twin to age.

Which all depend on a simultaneity convention (albeit a relatively reasonable one), which is the source of most - if not all - relativity ”paradoxes”.

Is it accurate to say that the event horizon of a black hole is stationary, or that its velocity is zero, relative to the outside observer on the statite?

No.

 

[Ibix]

You can kind of cheat and use Schwarzschild coordinates, I think. Solve for $r(\tau)$ and $t(\tau)$ for the probe and light ray ($\lambda$ not $\tau$ for the latter) and eliminate $t$ to get an expression for the proper time and $r$ coordinate that the probe receives a light pulse emitted some chosen time after launch. Since $r$ and $\tau$ are well defined through the horizon you ought to be able to find the limiting signal emission time.

The cheating bit is that I don't know if you can prove that the equations of motion we'd be using are well defined at the horizon without using something other than Schwarzschild coordinates. So OP has to trust coordinates he doesn't trust even this way.

 

[Orodruin]

You can kind of cheat and use Schwarzschild coordinates, I think. Solve for $r(\tau)$ and $t(\tau)$ for the probe and light ray ($\lambda$ not $\tau$ for the latter) and eliminate $t$ to get an expression for the proper time and $r$ coordinate that the probe receives a light pulse emitted some chosen time after launch. Since $r$ and $\tau$ are well defined through the horizon you ought to be able to find the limiting signal emission time.

The cheating bit is that I don't know if you can prove that the equations of motion we'd be using are well defined at the horizon without using something other than Schwarzschild coordinates. So OP has to trust coordinates he doesn't trust even this way.

Up to the horizon you can use Schwarzschild coordinates without a problem snd for the horizon crossing the limit is perfectly fine. If I recall correctly, the coordinate singularities of the horizon also cancel out in the computation and give the correct result.

 

[Mike S.]

I don't have the full calculation at hand, but if it's meaningless to say that an object hovers at a distance D from an event horizon, and wrong to assume that just because horizon passes an object at the speed of light, the object passes the horizon at the speed of light, I should check to see if this math is also irrelevant.

According to https://cosmo.nyu.edu/yacine/teaching/GR_2019/lectures/lecture22.pdf , for light
dr/dt = -(1-2M/r)
t = ± [r + 2M ln(r/2M − 1)] + constant

And according to https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/ (Peter Donis), for matter at rest
dr/dt = -(1-2M/r) sqrt(2M/r)

Now, I ought to remember how to get that second differential equation sorted out quickly, but I'm afraid it's been a while. Is this going to yield an accepted result where the calculated difference in t actually has something to do with the elapsed time at the statite or anywhere else in the universe?

 

[Nugatory]

I don't have the full calculation at hand, but if it's meaningless to say that an object hovers at a distance D from an event horizon...

The "meaningless" (I'd prefer to say "hopelessly ambiguous", but that's just me) statement "an object hovers at a distance D from an event horizon" can be replaced with the more precise statement "the object's Schwarzschild $r$ coordinate is $D$".

 

[PeterDonis]

if it's meaningless to say that an object hovers at a distance D from an event horizon

It's not. Who said it was?

What you need to be careful of is interpreting the radial coordinate $r$ as being the same as the radial distance $D$. It isn't. But there is a well-defined "distance to the horizon" for a hovering observer (although there are some technical points involved in calculating it).

The "meaningless" (I'd prefer to say "hopelessly ambiguous", but that's just me) statement "an object hovers at a distance D from an event horizon"

It isn't either meaningless or hopelessly ambiguous. It just needs some care in its definition and calculation.

I would say the real issues come if one tries to define a "distance" to the singularity.

 

[PeterDonis]

wrong to assume that just because horizon passes an object at the speed of light, the object passes the horizon at the speed of light

Who said that was wrong? Relative velocities are symmetric. If a light ray passes you anywhere, your speed relative to it is the same as its speed relative to you, namely $c$.

What is wrong is to assume that having a light ray pass you turns you into a light ray.

 

[PeterDonis]

I should check to see if this math is also irrelevant.

The equation from the Insights article is specifically for an object that free-falls inward from rest at infinity. It is not valid for an object that free-falls inward from rest starting at a finite value of $r$ (such as that of the statite). It can be a reasonable approximation if you make the starting value of $r$ very, very large compared to the hole's Schwarzschild radius.

Is this going to yield an accepted result where the calculated difference in t actually has something to do with the elapsed time at the statite or anywhere else in the universe?

Yes, as long as you limit yourself to the portion of the infalling object's worldline outside the horizon, you can calculate the coordinate time it takes for the infalling object to fall from some given value of $r$, such as that of the statite, to some other value of $r$. And you can then calculate the additional coordinate time $t$ it will take for an outgoing light signal emitted by the infalling object from that value of $r$ to get back out to the value of $r$ where the statite is hovering. And since the statite is hovering, its proper time has a fixed relationship to coordinate time $t$, so you can equate an elapsed coordinate time to an elapsed proper time for the statite. Since this final elapsed time is between two events on the statite's worldline (the infalling object starting its fall and the outgoing light signal arriving), it corresponds to an actual elapsed time on the statite's clock between those two events.

 

[Orodruin]

But there is a well-defined "distance to the horizon" for a hovering observer (although there are some technical points involved in calculating it).

This should depend on the simultaneity convention used. For a given convention that would be well defined but I’d still not call that physically relevant.

so you can equate an elapsed coordinate time to an elapsed proper time for the statite.

... according to the fixed proportionality. The coordinate time itself is not equal to the proper time of the statite and is being used as a mathematical tool. In and of itself it has no physical relevance.

 

[PeterDonis]

This should depend on the simultaneity convention used.

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

 

[PeterDonis]

The coordinate time itself is not equal to the proper time of the statite

Yes, agreed. But since in the particular case under consideration there is a fixed proportionality between coordinate time and the statite's proper time, and since the events at each endpoint of the relevant interval are on the statite's worldline, calculating in terms of coordinate time can be used to obtain a physically relevant answer in terms of the statite's proper time.

 

[Orodruin]

calculating in terms of coordinate time can be used to obtain a physically relevant answer in terms of the statite's proper time.

That is true for any coordinate system. It is just that the Schwarzschild coordinates give a particularly simple relation because of using the hypersurfaces orthogonal to the timelike KVF as coordinate surfaces.

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

True, but it could be discussed how meaningful this is. For example, the event that you are computing the distance to is the same for all of those hypersurfaces (the origin in a Kruskal diagram, which is the r = R event in of all of them) similar to the distance to the origin for a Rindler observer in SR.

 

[cianfa72]

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

Just to check my understanding: consider two hovering observers. One is the given observer hovering at fixed $(R_1, \theta, \phi)$ Schwarzschild coordinates and the other is hovering at the event horizon with the same angular coordinates $(R_0, \theta, \phi)$.

Now since the worldlines of both hovering observers are integral curves of the timelike KVF if we pick the foliation above (namely the family of spacelike hypersurfaces that is everywhere orthogonal to the timelike KVF) the distance between them in any of those spacelike hypersurfaces is always the same.

 

[Orodruin]

the other is hovering at the event horizon with the same angular coordinates (R0,θ,ϕ).

It is impossible to hover at the event horizon. It is a lightlike surface.

The statement is however true for any stationary observer above the event horizon since the KVF by definition is a symmetry of the spacetime.

 

[cianfa72]

It is impossible to hover at the event horizon. It is a lightlike surface.

Ok, the point is that along the hovering observer worldline the spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ have always the same 3D geometry. So on each of these spacelike hypersurfaces the distance between points at given coordinate $(R_1, \theta, \phi)$ and $(R_0, \theta, \phi)$ is always the same (the event horizon radial coordiante is always $R_0$).

 

[Orodruin]

Ok, the point is that along the hovering observer worldline the spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ have always the same 3D geometry. So on each of these spacelike hypersurfaces the distance between points at given coordinate $(R_1, \theta, \phi)$ and $(R_0, \theta, \phi)$ is always the same (the event horizon radial coordiante is always $R_0$).

You have to be much more careful here. The point on the horizon that you are referring to is the same event for all of those hypersurfaces. This means that it is also not in any way or form a distance to ”where the dropped object crosses the horizon”. That event is in the causal future (actually on the light cone) of the event on the hypersurfaces.

 

[cianfa72]

You have to be much more careful here. The point on the horizon that you are referring to is the same event for all of those hypersurfaces.

Sorry I've some difficulty to grasp it. The exterior of BH is foliated by spacelike hypersurfaces of constant Schwarzschild coordinate time $t$.

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

 

[Orodruin]

Sorry I've some difficulty to grasp it. The exterior of BH is foliated by spacelike hypersurfaces of constant Schwarzschild coordinate time $t$.

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

The exterior does not include the event horizon. All of those hypersurfaces, when extended to the event horizon, include the same event on the horizon, much like all values of $\theta$ would map to the origin when $r=0$ in polar coordinates. This event is not the same event as where any observer would pass the event horizon.

Edit: This is of course related to the fact that no observer can pass the horizon in finite coordinate time in Schwarzschild coordinates.

 

[Ibix]

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

Have a look at a Kruskal diagram. Lines of constant Schwarzschild $t$ (the hypersurfaces you are talking about) are straight lines through the origin. So is the event horizon. This is the point @Orodruin is making - that all those surfaces meet at one event, the origin of the Kruskal diagram, which is in the causal past of any massive object's horizon crossing.

 

[cianfa72]

Have a look at a Kruskal diagram. Lines of constant Schwarzschild $t$ (the hypersurfaces you are talking about) are straight lines through the origin. So is the event horizon.

Ah ok, so the event horizon in Kruskal diagram is actually a straight line of 45 degree through the origin of the diagram. For instance in region I all the events that belong to the event horizon have Schwarzschild coordinates $(r=1,t=\infty)$ and varying $\theta, \phi$. In this sense - as @Orodruin pointed out with the example of polar coordinates - for a given couple $\theta, \phi$ there is an entire straight line of events all mapped to the same $(r=1,t=\infty)$.

that all those surfaces meet at one event, the origin of the Kruskal diagram, which is in the causal past of any massive object's horizon crossing.

That means that on all spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ the event with coordinate $r=1$ is actually the same event (i.e. the event mapped in the origin of the Kruskal diagram). It is an event that belong to the event horizon.

 

[PeterDonis]

the event that you are computing the distance to is the same for all of those hypersurfaces

Hm, yes, this is an interesting point, because in a more realistic spacetime containing a black hole that forms by gravitational collapse (the Oppenheimer-Snyder model would be the idealized case of this), the event you describe is not there (because that portion of the Kruskal diagram is not present, it's replaced by the region occupied by the collapsing matter). I have seen the "distance to the horizon" formula (and the integral that is used to derive it) discussed in plenty of GR textbooks, but I have never seen any discussion of how it is to be interpreted in a more realistic model.

 

[Orodruin]

Hm, yes, this is an interesting point, because in a more realistic spacetime containing a black hole that forms by gravitational collapse (the Oppenheimer-Snyder model would be the idealized case of this), the event you describe is not there (because that portion of the Kruskal diagram is not present, it's replaced by the region occupied by the collapsing matter). I have seen the "distance to the horizon" formula (and the integral that is used to derive it) discussed in plenty of GR textbooks, but I have never seen any discussion of how it is to be interpreted in a more realistic model.

Indeed, I think this just goes to show that ”distances” are generally based on conventions and are not really physical. In a stationary spacetime there is a natural definition of distances due to the timelike KVF, but Schwarzschild spacetime is only stationary outside the event horizon.

 

[PeterDonis]

Schwarzschild spacetime is only stationary outside the event horizon.

And also, in the Oppenheimer-Snyder model, the region containing the collapsing matter (a portion of which is outside the horizon) is not stationary.

 

[DrGreg]

Here is an image I created and posted (at https://www.physicsforums.com/threads/oppenheimer-snyder-model-of-star-collapse.651362/post-4164435) nearly 10 years ago.

It's a Kruskal diagram of a black hole formed by collapse of matter. The white area in the bottom left is where the matter is collapsing, and not covered by these coordinates. In an eternal black+white hole, the pattern would continue in that area.

The straight pink lines represent the foliation into spacelike hypersurfaces of constant Schwarzschild coordinate time. If extrapolated (in an eternal black+white hole) they would all meet at the same event, but each one hits the surface of the collapsing matter (the grey line).

(The coloured gridlines were calculated and plotted using Matlab, but the grey line was just sketched and is an intelligent guess.)

So this reveals something I'd never consciously realized before: that, using the Schwartzschild definition of simultaneity, not only does no infalling person ever cross the horizon in finite Schwartzschild time, the horizon never even comes into existence in finite Schwartzschild time. (The Schwartzchild model does not take account of outward Hawking radiation, or any inward radiation or matter after the initial collapse.)

The "distance to the horizon" being discussed in previous posts is measured along the straight pink lines (via the metric of course), but the none of these lines actually reaches the horizon (unless it's an eternal black+white hole).

 

[PAllen]

There is one geometric definition of distance from an event to a body that I’ve seen used in GR manifolds. It is the geodesic interval along a spacelike geodesic 4-orthogonal to the world tube boundary that reaches the given event. There can possibly be more that one such geodesic, but then they are all physically plausible distances to different surface features.

In the case of the horizon, there are no such geodesics at all. This is consistent with the observation that spacelike geodesics between some event and a lightlike geodesic in SR can range from (0,infinity) in length. (Whereas, between an event a timelike geodesic, there is a maximum such geodesic length, which is also the orthgonal one).

Thus I claim that no meaningful definition of distance to a horizon is possible at all.

 

[PeterDonis]

the horizon never even comes into existence in finite Schwartzschild time

Yes, that's correct. Schwarzschild coordinates cannot cover the horizon at all. Even the "interior" Schwarzschild coordinate patch, which covers the region inside the horizon, does not cover the horizon itself.

 

[PeroK]

I don't have the full calculation at hand, but if it's meaningless to say that an object hovers at a distance D from an event horizon, and wrong to assume that just because horizon passes an object at the speed of light, the object passes the horizon at the speed of light, I should check to see if this math is also irrelevant.

According to https://cosmo.nyu.edu/yacine/teaching/GR_2019/lectures/lecture22.pdf , for light
dr/dt = -(1-2M/r)
t = ± [r + 2M ln(r/2M − 1)] + constant

And according to https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/ (Peter Donis), for matter at rest
dr/dt = -(1-2M/r) sqrt(2M/r)

Now, I ought to remember how to get that second differential equation sorted out quickly, but I'm afraid it's been a while. Is this going to yield an accepted result where the calculated difference in t actually has something to do with the elapsed time at the statite or anywhere else in the universe?

Using Schwarzschild coordinates, it's possible to prove that we can wait only a maximum finite time to send light after an infalling object in order for the light to overtake the object before the event horizon.

Starting from rest at initial radius $r = R$, we can generate the equation for the infalling object:
$$\frac{dr}{dt} = -\frac{1}{\sqrt{1 - \frac{2M}{R}}}\big ( 1 - \frac{2M}{r} \big ) \sqrt{\frac{2M}{r} - \frac{2M}{R}}$$This can be integrated to give:
$$t = 2M\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + F(r)$$Where $F(r)$ is some additional finite function of $r$. I.e. finite as $r \to 2M$.

Then, we have the relatioship between $r$ and $T$, which is the coordinate time of a inbound light pulse stating from $r = R$:
$$T = -2M\ln\big (\frac{r}{2M} - 1\big ) - r + T_0$$Now, we for a given $r$, we can calculate the difference between the time, $t$, it takes the infalling object to reach $r$ and the time it takes a light pulse. This is:
$$t - T = 2M\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + F(r) + 2M\ln\big (\frac{r}{2M} - 1\big ) +r - T_0$$What we need to show is that tends to some finite limit as $r \to 2M$. Everything is finite, except possibly the combined log term:
$$\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + \ln\big (\frac{r}{2M} - 1\big )$$To see that this is finite, we re-express the term inside the first log as:
$$\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} = \frac{\frac{r}{2M} -\frac{2r}{R} + 1 + 2\sqrt{\frac{r}{2M} -\frac{r}{R}}\sqrt{1 - \frac{r}{R}}}{\frac{r}{2M} -1} $$And, when we combine the logs the denominator will cancel with the second log term, leaving a finite limit as $r \rightarrow 2M$.

In other words, the difference in time between a light pulse and an infalling object tends to some finite limit as $r \to 2M$, Which means that we have a finite time we can wait before sending a light signal that will reach the object before it gets to the event horizon.

 

[PeterDonis]

In the case of the horizon, there are no such geodesics at all.

I would say that a horizon does not have the kind of "world tube boundary" that is required. In particular, if the horizon is an integral curve of a KVF, since the KVF is null on the horizon, it is orthogonal to itself there, so the concept of "spacelike surface orthogonal to the KVF" is no longer valid.

In the case of the distance formula in Schwarzschild spacetime that I was referring to, that is standardly interpreted as the limit as $r \to 2M$ of the formula for the distance between stationary observers at two different values of $r$ that is valid outside the horizon. That limit is valid mathematically, but I'm no longer sure it has the physical meaning that is usually attributed to it. See further comments below.

no meaningful definition of distance to a horizon is possible at all.

In the light of the discussion we have been having about what portion of spacetime the spacelike hypersurfaces orthogonal to the timelike KVF in Schwarzschild spacetime actually cover, I now agree with this. There is a mathematical formula that will give a "distance to the horizon" in one of these hypersurfaces, but the spacelike geodesic segment it describes either doesn't exist as a complete segment in the spacetime (because a portion of it would be inside the collapsing matter region and the formula is not valid there), or goes to an event that is the intersection of multiple spacelike geodesics from different points on the timelike worldline of the hovering observer (the same issue that arises with the Rindler horizon in SR).

 

[PAllen]

@PAllen - your argument *is* valid if there is in fact a last received signal. That's why the question is about how you calculate what the last signal received by the probe is; if someone can show there is (or isn't) a specific cutoff, the qualitative arguments ought to be settled also.

I am going to take this up in two different posts. First, pictorially, using Kruskal coordinates, then quantitatively, with the computation made much easier by using the generalized Lemaitre coordinates derived in this paper:

https://arxiv.org/abs/1911.05988

around page 8. (This and another paper: https://arxiv.org/abs/1211.4337, are very interesting to see many generalizations of Gullestrand-Panlieve coordinates and their relationships to Lemaitre type coordinates).

So pictorially, we have:

https://www.physicsforums.com/attac...296920/?hash=0ada85ddb1326ac3dba60e793c42457e

In a later post, I will calculate the proper time elapsed for the statite between A and B and A and C. I will also calculate the proper time for the probe between A and its receipt of signal B, and between A and its receipt of signal C.

To address Hawking radiation, once the computations are done, you will see that events B and C, as well as their whole signal trajectories are in an era when the BH is growing due to absorption of CMB radiation. Only many trillions of years later, when expansion has reduced the CMB temperature to below the Hawking temperature for a BH, will the BH start losing mass. This is kilometers to the upper right in the attached diagram.

On a technical note, the true mathematical model applicable would be some fusion of an ingoing vaidya metric (for the BH growth phase), and outgoing vaidya metric for the shrinking phase, both modified to join with an FLRW metric. This has probably never been done exactly, but the differences in consequences for the region under consideration would only affect computations well beyond 10 significant digits.