# I Rod falling through the event horizon of a black hole

1. Nov 5, 2016

### timmdeeg

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.

2. Nov 5, 2016

### Staff: Mentor

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),

OK, there's the problem - "constant radius $r$" is hovering, and there is no hovering at the event horizon.

3. Nov 5, 2016

### timmdeeg

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. Nov 5, 2016

### Staff: Mentor

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. Nov 5, 2016

### pervect

Staff Emeritus
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.

6. Nov 5, 2016

### Staff: Mentor

Measured how? Why do you expect those differences, and do you expect them to hold whole a part of the rod is inside?

7. Nov 5, 2016

### timmdeeg

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. Nov 5, 2016

### Staff: Mentor

While they cross the event horizon? Yes.

9. Nov 5, 2016

### timmdeeg

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. Nov 5, 2016

### Staff: Mentor

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. Nov 5, 2016

### Orodruin

Staff Emeritus
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. Nov 5, 2016

### Staff: Mentor

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.

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.

13. Nov 6, 2016

### timmdeeg

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.

Ah I understand, this helps a lot, thanks.

14. Nov 6, 2016

### pervect

Staff Emeritus
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.

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. Nov 6, 2016

### Staff: Mentor

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. Nov 6, 2016

### Justintruth

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. Nov 6, 2016

### Staff: Mentor

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

No.

No. The event horizon is an invariant global feature of the spacetime.

18. Nov 7, 2016

### timmdeeg

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!?

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. Nov 7, 2016

### Orodruin

Staff Emeritus
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.

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. Nov 7, 2016

### Staff: Mentor

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.

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