Rod falling through the event horizon of a black hole

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  • #1
timmdeeg
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Having discussed recently
https://www.physicsforums.com/threads/rod-falling-radially-towards-the-center-of-a-mass.871169/
I'm now puzzled by the question, what happens to the rod during his radial fall through the event horizon and what would the hypothetical observer at ##r=2M## measure.

Let's consider ##M## large enough to prevent any damage of the rod. Remembering said discussion, there is a "force-free" point somewhere on the rod, also called "locus of no acceleration". I think an observer at a constant radius ##r## will measure a radial velocity ##-\sqrt{2M/r}## of the force-free point only but of the low end a lower und for the top end a higher velocity compared to that. But if correct, would this be valid for the event horizon too? It seems implausible however, because it seems to imply that the low end would fall through the event horizon with ##v<c## and the top with ##v>c##.
Any help is appreciated.
 

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  • #2
Nugatory
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what would the hypothetical observer at ##r=2M## measure....
You have to specify that observer a bit more carefully. He's not hovering at ##r=2M## because that's the event horizon; he's passing through it just like the rod (although he need not be in free fall),

I think an observer at a constant radius ##r##....
OK, there's the problem - "constant radius ##r##" is hovering, and there is no hovering at the event horizon.
 
  • #3
timmdeeg
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You have to specify that observer a bit more carefully. He's not hovering at ##r=2M## because that's the event horizon; he's passing through it just like the rod (although he need not be in free fall),
Yes, nothing with rest mass can hover at ##r=2M##. But I think it follows from ##-\sqrt{2M/r}## that an object in free fall moves locally with the speed of light at radius ##r=2M##. Do you agree with that? If yes, then the problem seems to be still that with the exception of the force-free point somewhere on the rod all other points aren't in free fall, right?
 
  • #4
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But I think it follows from ##-\sqrt{2M/r}## that an object in free fall moves locally with the speed of light at radius ##r=2M##. Do you agree with that?
No. Consider the behavior of flashes of light emitted from the free-falling particle in the two radial directions to visualize what's going on without falling back on any particular coordinate system.

And you have to be extremely cautious about using any results expressed in terms of Schwarzschild coordinates at the event horizon, because they're singular there.
 
  • #5
pervect
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Yes, nothing with rest mass can hover at ##r=2M##. But I think it follows from ##-\sqrt{2M/r}## that an object in free fall moves locally with the speed of light at radius ##r=2M##. Do you agree with that? If yes, then the problem seems to be still that with the exception of the force-free point somewhere on the rod all other points aren't in free fall, right?
I would put things differently. I would say that the event horizon is "moving at the speed of light", i.e. we imagine non an "observer", but an object with zero rest mass, hovering at r=2M. Because that object has zero rest mass, it's moving at the speed of light. It's just because it's a massless object that it doesn't have a point of view, the "point of view" of a massless object would be a frame of reference where said massless object was at rest. But there can be no such frame, a massless object always moves at "c" - more technically we say that it follows a lightlike (or null) worldline. There is a short FAQ on this topic that says much the same thing. See https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/.

Thus the event horizon is best regarded as a light-like, or null surface, so there is and can be no "observer" at the event horizon. The only "observers" here are those that are falling through the event horizon - we say those observers have timelike worldlines. We CAN say that the relative velocity between the zero rest mass particle trapped on the horizon and the infalling observer is equal to "c". The trapped particle on the event horizon is actually just an expository point, but I think it's helpful - we can say that the relative velocity between the event horizon and the infalling observer is equal to "c" without imagining this "trapped massless particle" if we wish to. What tends to confuse the issue is the seemingly innocuous idea that we can have an "observer" at the event horizon. But because of the issues raised in the FAQ about the "point of view of a photon", the only valid observer is the one that is falling through the horizon.
 
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  • #6
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but of the low end a lower und for the top end a higher velocity compared to that
Measured how? Why do you expect those differences, and do you expect them to hold whole a part of the rod is inside?
 
  • #7
timmdeeg
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We CAN say that the relative velocity between the zero rest mass particle trapped on the horizon and the infalling observer is equal to "c". The trapped particle on the event horizon is actually just an expository point, but I think it's helpful - we can say that the relative velocity between the event horizon and the infalling observer is equal to "c" without imagining this "trapped massless particle" if we wish to.
Yes this helpful, thanks. From this point of view, although the term ##-\sqrt{2M/R}## yields ##c## too, it is not really applicable, because a massless observer hovering at the event horizon makes no sense.
Then it seems that the relative velocity between the event horizon and all parts of the rod is ##c##, correct?
 
  • #8
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While they cross the event horizon? Yes.
 
  • #9
timmdeeg
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While they cross the event horizon? Yes.
Ok.
Now, according to the conclusions in the link shown in #1, bottom and top of the rod are not in free fall, i.e. they are not force-free and thus are not following a geodesic. Is this true also for the moment, when they cross the event horizon?
Another question. Consider the moment when the bottom is falling through the event horizon with the relative velocity ##c##. Which velocity has the top of the rod at a radius ##r>2M## in this moment relative to a shell observer at the same radius ##r##? The relative velocity between a particle in free fall and this observer is given by ##-\sqrt{2M/r}##. Do you agree that top is falling faster than this particle?
 
  • #10
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For the moment where they cross the event horizon, no external force is relevant. The event horizon moves at the speed of light, the history of those points does not matter, they will always cross it at the speed of light.
 
  • #11
Orodruin
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Consider the moment when the bottom is falling through the event horizon with the relative velocity c.
You keep saying this but it really is not a good description of things.

The key distinction here is that it is not the parts of the rod that are "moving with the speed of light", it is the event horizon that is moving at the speed of light in any locally inertial frame. This includes the local instantaneous rest frame of any part of the rod when the event horizon passes by.

In any local inertial frame, all parts of the rod will be moving at speeds lower than c.
 
  • #12
PeterDonis
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I think it follows from ##-\sqrt{2M/r}## that an object in free fall moves locally with the speed of light at radius ##r=2M##.
No, it doesn't. It follows that the horizon is moving at the speed of light in the local inertial frame of an object falling past it.

with the exception of the force-free point somewhere on the rod all other points aren't in free fall, right?
If the rod is being held together by inertial forces, then yes. But that doesn't change the fact that the horizon will be moving at the speed of light as it passes each point on the rod.

More precisely: each point on the rod has its own worldline, and none of these worldlines intersect. Each worldline crosses the horizon at a distinct event. If we set up a local inertial frame centered on any of these distinct events, the horizon moves at the speed of light in that local inertial frame. The only difference will be that in each of these frames, the worldline of the rod's locus of zero acceleration will be a straight line, while the worldlines of the other points on the rod will not--they will have nonzero proper acceleration so they will be curved. But for any rod made of known materials, the curvature will be very small--so small that the black hole would have to be enormous, large enough to have a Schwarzschild radius of many light-years, for the curvature of the worldlines of other points on the rod to be measureable within a single local inertial frame.
 
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  • #13
timmdeeg
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Thanks to everybody for clarifying that the horizon moves with the speed of light in the local inertial frame of any object passing by. So it seems that Taylor & Wheeler are a bit sloppy saying in "Exploring Black Holes" "there [means the event horizon] the in-falling stone moves with the speed of light as recorded by one observer (equation [24]). Equation [24] is the expression for the radial velocity in shell coordinates. But I understand your point now, it's more accurate to focus on the speed of the horizon in the local inertial frame of the in-falling stone.

More precisely: each point on the rod has its own worldline, and none of these worldlines intersect. Each worldline crosses the horizon at a distinct event. If we set up a local inertial frame centered on any of these distinct events, the horizon moves at the speed of light in that local inertial frame. The only difference will be that in each of these frames, the worldline of the rod's locus of zero acceleration will be a straight line, while the worldlines of the other points on the rod will not--they will have nonzero proper acceleration so they will be curved.
Ah I understand, this helps a lot, thanks.
 
  • #14
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Yes this helpful, thanks. From this point of view, although the term ##-\sqrt{2M/R}## yields ##c## too, it is not really applicable, because a massless observer hovering at the event horizon makes no sense.
Then it seems that the relative velocity between the event horizon and all parts of the rod is ##c##, correct?
I'm not sure why a massless particle hovering at the event horizon "makes no sense" to you. The closest statement that I can make to that that I'd agree with is that "Schwarachild coordinates make no sense at the event horizon, because they become singular there". That I'd agree with. To me, saying that a particle at the event horizon follows a lightlike worldline makes perfect sense, and is the easiest reasonably non-technical way of describing what happens. At this point I'm leery of attempting a any more technical explanation, as I don't know your background. I can only suspect that what's happening is that you're trying to make sense of the geometry of the black hole by using Schwarzschild coordinates, becoming confused by the fact that said coordinates are singular, then attributing the resulting confusion to the geometry itself, rather than to the choice of cooridnates.

As far as velocities go, from a purist point of view, velocity in GR is defined only for two observers at the same point in space-time. See the following quote from "The Meaning of Einstein's equation", https://arxiv.org/abs/gr-qc/0103044.

In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime — that is, at the same place at the same instant. The reason is that in general relativity, we take very seriously the notion that a vector is a little arrow sitting at a particular point in spac etime.

To compare vectors at different points of spacetime, we must carr y one over to the other. The process of carrying a vector along a path without turning or stretching it is called ‘parallel transport’. When spacetime is curved, the result of parallel transport from one point to another depends on the path taken! In fact, this is the very definition of what it means for spacetime to be curved.

Thus it is ambiguous to ask whether two particles have the same velocity vector unless they are at the same point of spacetime.
I'm afraid this might be a little too technical as well, but I hope it is of some help.

It might be worth considering things from another point of view. Consider an array of static observers, holding a constant distance away from the event horizon. Using Baez's formulation, we can find the relative velocity between the static observer and the worldline of some infalling particle at any point. Using this definition, we find that the relative velocity between a static observer and an infalling observer becomes larger and larger, and approaches "c" at the event horizon. At the event horizon, it becomes equal to "c", which implies that one of the worldlines is light-like. As already mentioned though, the light-like worldline is not the infalling observer, the light-lik worldline is the static observer at the event horizon. Said light-like worldline doesn't represent an "observer" in the usual sense, for reasons already mentioned.
 
  • #15
PeterDonis
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I'm not sure why a massless particle hovering at the event horizon "makes no sense" to you.
To be fair, he didn't say a massless "particle", he said a massless "observer". You appear to agree that the word "observer" is not really appropriate.
 
  • #16
Part of the problem is that for all observers outside no objects ever fall into the event horizon. Right? Doesn't it take an infinite time? And relative to those falling in there is no event horizon to cross!
 
  • #17
PeterDonis
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Part of the problem is that for all observers outside no objects ever fall into the event horizon. Right?
Wrong. We have had many, many previous threads on this. See, for example, here:

https://www.physicsforums.com/threa...ses-the-horizon-question.810190/#post-5086407

Doesn't it take an infinite time?
No.

And relative to those falling in there is no event horizon to cross!
No. The event horizon is an invariant global feature of the spacetime.
 
  • #18
timmdeeg
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I'm not sure why a massless particle hovering at the event horizon "makes no sense" to you. The closest statement that I can make to that that I'd agree with is that "Schwarachild coordinates make no sense at the event horizon, because they become singular there".
Ok, I think that clarifies why ##-\sqrt{2M/r}## makes no sense at the event horizon. Perhaps one can say it's the limiting case. So it seems that this formula is truly applicable only as long as the shell observer is time-like.

Thanks for mentioning John Baez's article "The Meaning of Einstein's equation".
Saying "In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime — that is, at the same place at the same instant." there is no reason of not interpreting one of the two particles as light-like, correct?
One can easily drop the phrase "as measured by the massless shell observer" and replace it by "massless particle", but to me this means just wording and not physics. We are talking about relative velocity. Therefore it should make no difference, if I consider the velocity of an infalling time-like object relative to the light-like horizon or the velocity of the light-like horizon relative to an infalling time-like object, as some around here have pointed out.

Now, after overthinking both, horizon and object are "at the same place at the same instant." place sounds a bit strange however, because of the singular coordinates at the horizon. Hm, I'm getting confused, it may well be that Baez's particles should be time-like both!?

Consider an array of static observers, holding a constant distance away from the event horizon. Using Baez's formulation, we can find the relative velocity between the static observer and the worldline of some infalling particle at any point. Using this definition, we find that the relative velocity between a static observer and an infalling observer becomes larger and larger, and approaches "c" at the event horizon. At the event horizon, it becomes equal to "c", which implies that one of the worldlines is light-like.
Ok, thanks. So, if I understand you correctly ##c## at the horizon can be understood as the limiting case. I think from this point of view there seems no necessity to imagine a massless particle at the event horizon.
 
  • #19
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there is no reason of not interpreting one of the two particles as light-like, correct?
Yes there is, if a particle is light-like or not is completely determined and you cannot go around "interpreting" a particle as something it is not.

One can easily drop the phrase "as measured by the massless shell observer" and replace it by "massless particle", but to me this means just wording and not physics.
No, you cannot do this. You cannot describe something from the rest frame of a massless particle because a massless particle has no rest frame.
 
  • #20
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Now, after overthinking both, horizon and object are "at the same place at the same instant." place sounds a bit strange however, because of the singular coordinates at the horizon
Saying that two things are at the same place at the same time is saying that two worldlines (or in this case, a worldline and a lightlike surface) intersect. The presence of a coordinate singularity at the point of intersection is irrelevant; the singularity will go away if you use different coordinates.
Hm, I'm getting confused, it may well be that Baez's particles should be time-like both!?
There's no problem with the intersection of a timelike and a lightlike worldline, or the intersection of two lightlike worldlines, so no need for a requirement that both particles be following timelike worldlines.
 
  • #21
timmdeeg
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Orodruin and Nugatory, thanks for helping.
Saying that two things are at the same place at the same time is saying that two worldlines (or in this case, a worldline and a lightlike surface) intersect. The presence of a coordinate singularity at the point of intersection is irrelevant; the singularity will go away if you use different coordinates.
There's no problem with the intersection of a timelike and a lightlike worldline, or the intersection of two lightlike worldlines, so no need for a requirement that both particles be following timelike worldlines.
Yes, good point, thanks.
 
  • #22
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I have a similar question i was hoping someone could help me with. Why is the event horizon of a black hole so cold? I thought when matter is compressed, it causes it to heat up do to friction
 
  • #23
timmdeeg
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Matter is compressed in the center of a black hole, not at the event horizon. There it is in free fall and feels tidal forces, more or less depending on the mass of the black hole.
 
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  • #24
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I have a similar question i was hoping someone could help me with. Why is the event horizon of a black hole so cold? I thought when matter is compressed, it causes it to heat up do to friction
But how would the heat escape? Infra red is just a kind of EM radiation, and it can't escape.

A small amount of radiation (Hawking radiation) does escape due to quantum mechanical processes that I don't understand myself, but this is tiny.
 
  • #25
PeterDonis
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Matter is compressed in the center of a black hole
No, it isn't. A black hole is a vacuum solution.

Why is the event horizon of a black hole so cold?
If you mean in the classical sense, i.e., the temperature that would be seen by someone free-falling into the hole, it's because there is vacuum there. If you are talking about the Hawking temperature of the hole, that brings in other issues--see below.

I thought when matter is compressed, it causes it to heat up do to friction
The matter that originally collapsed to form the hole will have been compressed while it was collapsing, yes. But it isn't there at the event horizon after the hole is formed. It will have continued on inward to form a singularity at the center, at which point it will have disappeared, so that the black hole that remains behind is vacuum everywhere.

At least, this is what is predicted to happen assuming that the classical laws of GR continue to be valid all the way down. Most physicists actually don't believe that; most physicists think that quantum gravity effects will become important before the singularity is reached, so that we will need a theory of quantum gravity to tell us what happens there. That then brings up a second question: will those quantum gravity effects also be important at the horizon? Or will physics at the horizon still be described well enough by classical GR?

There are two positions on this issue, both of which probably contain some truth, but both of which also raise further issues. They are:

(1) We already know that, at or near a black hole horizon of any size, there is at least one quantum effect that produces a significant prediction: Hawking radiation. One way of looking at the prediction of Hawking radiation is that it is because, when quantum field theory is taken into account, it is impossible to have a true "vacuum" everywhere in a black hole spacetime, because "vacuum" is not an invariant term--more precisely, which state of the quantum field is a "vacuum" state depends on the state of motion of the observer. So an observer free-falling into a black hole could see a particular state of the quantum field as vacuum (one of the simplest such states is called the "Hartle-Hawking vacuum state" in the literature) while an observer who is accelerating in order to "hover" at a constant height above the hole's horizon could see the same quantum field state as containing particles--radiation being emitted outward. That radiation is Hawking radiation. This quantum correction in itself is small, but it gives rise to the possibility that there could be other, larger ones--even ones that would make the temperature at the horizon seem "hot" to all observers, not just accelerated ones.

(2) For any black hole of macroscopic size (such as the ones formed by the collapse of stars or the ones at the centers of galaxies), the spacetime curvature at the horizon is way too small for quantum effects to be significant. So classical GR should hold at the horizon for these types of holes. Even if there are quantum effects like Hawking radiation that make it problematic to say the state at the horizon is an exact "vacuum", those corrections are very small and should not affect the global geometry of spacetime significantly. The Hawking temperature, for example, for any macroscopic hole is many orders of magnitude smaller than the temperature of the CMBR, so things at the horizon of such a hole would be indistinguishable to us from anywhere else in outer space; it certainly would not seem "hot" there.

Whatever eventually turns out to be right, it seems pretty safe to say that there is no effect at a black hole horizon corresponding to the matter that originally formed the hole being "compressed" there and heating up to the point where the temperature at the horizon would be high just for that reason.
 

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