Falling Clock & Black Hole: Measure Time or Disappear?

[Omega0]

TL;DR Summary

How can it be that a binary operator has no influence in the flow of information?

Hi,

This question is so simple - sorry if the answer is also that simple...
It is pretty clear that every matter can cross the event horzion of a black hole. It is said that this process can be even very smooth if the black hole is big enough ("the bigger, the better for you", this is how I would summarize it).
So if a light clock (laser, mirror) is falling into the black hole, there should be - locally - no problem to measure the time and I find nothing suspicious. The problem here is - if I understand it correctly - that the event horizon is a terminator. Nothing going in gets out. If we speak about "going in" or "going out" we obviously mean the transport of information, so it is never a local thing. We need space for it to transport information from A to B. So why shouldn't I can measure that B is gone?
Or with other words, we actually see matter disappearing in black holes, so why should this be not true if you have a tiny laser clock falling? Why shouldn't the clock stop working if B reached the event horizon?

Sorry, this is probably trivial, thanks for your replies in advance - and stay healthy.

 

[.Scott]

The event horizon will not stop the clock from working.

If your watch says 12 noon as you cross the event horizon, you will continue to be able to read your watch for some (short) time thereafter.
However, your 12 noon will correspond to a wide range of time for observers watching you fall from outside the black hole.

It isn't quite fair to say that they will watch you fall until you reach the event horizon, because once you cross the photon sphere, in becomes more and more difficult for information from you to reach the outside world. By 11:59:59, as a matter of practicality, only you will be able to actually read the time from your watch. And as you cross the horizon, anything unique about you will be undetectable.

But if we set aside the impracticality of actually reading your watch from afar, an observer would see your watch approach 12 noon without ever reaching that time.

 

[Omega0]

The event horizon will not stop the clock from working.

If your watch says 12 noon as you cross the event horizon, you will continue to be able to read your watch for some (short) time thereafter.

What is "me"? Something like a solid body with 2 meters length? Or 1 micrometer? Or which length is allowed for the clock?

However, your 12 noon will correspond to a wide range of time for observers watching you fall from outside the black hole.

Yes, but I am not interested in the external observers as I am the observer. I am the person which measures - and I am not local, I am not d/dx or whatever. My clock has a size. Period.

It isn't quite fair to say that they will watch you fall until you reach the event horizon, because once you cross the photon sphere, in becomes more and more difficult for information from you to reach the outside world.

I find it very fair because I have the same "problem" as any other observer outside - I have a volume and it is me watching, nobody else. I am the observer. I am measuring and I don't care about the world outside. I am interested in the local physics, nothing else. The problem here is, in my understanding, that "local" is a basically wrong term in physics, generally. No process can be thought to be locally as we learned from thermodynamics etc.

 

[.Scott]

If the only observer you are interested in is the one falling through the event horizon, then that observer will see the clock move past 12noon as he and the clock cross the horizon.

It is certainly true that your feet will precede your watch through the horizon and that your watch will precede your eyes, but that will not affect your local observations. What will actually be happening is that light will reflect off the face of the watch as it normally does and be directed up towards you eyes. That light will never make it back out of the event horizon, but that won't matter because your eyes will cross the horizon from the outside and catch up to that light. So you will see the watch cross noon - and it will appear to do so normally.

 

[Omega0]

If the only observer you are interested in is the one falling through the event horizon, then that observer will see the clock move past 12noon as he and the clock cross the horizon.

It is certainly true that your feet will precede your watch through the horizon and that your watch will precede your eyes, but that will not affect your local observations. What will actually be happening is that light will reflect off the face of the watch as it normally does and be directed up towards you eyes. That light will never make it back out of the event horizon, but that won't matter because your eyes will cross the horizon from the outside and catch up to that light. So you will see the watch cross noon - and it will appear to do so normally.

Okay, so the light which is after the horizon will make it to reach my eyes?

 

[PeroK]

Okay, so the light which is after the horizon will make it to reach my eyes?

Good question!

I suspect the problem here is trying to analyse things implicitly in Schwarzschild coordinates, where there is a coordinate singularity at the EH. The key may be an analysis in, for example, Eddington-Finkelstein coordinates.

There is no physical singularity at the EH. There shouldn't be a local blip in the working of a clock, or in your feet momentarily disappearing! But, to check this you need coordinates that straddle the EH without a coordinate singularity.

 

[PeroK]

PS: So, here's part of the answer. Take a clock fixed at any Schwarzschild radial coordinate outside the EH (no matter how close to the horizon). You have a singularity at the EH for the time coordinate taken from that clock.

According to any of these "Schwarzschild" clocks, an object never reaches the EH.

Take a clock falling into the black hole. That clock is not a "Schwarzschild" clock and the time coordinate according to that clock has no singularity at the EH. According to that clock it crosses the EH at a finite time. Due to its finite size, technically different parts of the clock cross the EH at different times, but there is no sense in which the part of the clock nearest the EH (that goes in first!) never reaches the EH. Not according to time as measured on the infalling clock itself. According to that clock, the time at which each compoent of the clock crosses the EH is finite and well defined.

 

[Orodruin]

A constructive way of seeing this issue from the perspective of the clock is instead considering the following, which is an identical situation apart from the missing tidal effects: How is a light clock affected if it is passed by a light-like infinitly extended surface? (Ie, a surface moving past the light clock at the speed of light.)

 

[.Scott]

Okay, so the light which is after the horizon will make it to reach my eyes?

Yes.

 

[Omega0]

Good question!

I suspect the problem here is trying to analyse things implicitly in Schwarzschild coordinates, where there is a coordinate singularity at the EH.

Yes, you are right, I thought about Schwarzschild and yes, the singularity at the EH means logical trouble.

The key may be an analysis in, for example, Eddington-Finkelstein coordinates.
There is no physical singulaity at the EH. There shouldn't be a local blip in the working of a clock, or in your feet momentarily disappearing! But, to check this you need coordinates that straddle the EH without a coordinate singularity.

Awesome, thanks a lot! I never heard about Eddington-Finkelstein coordinates. The singularity of the Schwarzschild metric appeared fundamental to me.

 

[Orodruin]

Yes, you are right, I thought about Schwarzschild and yes, the singularity at the EH means logical trouble.

Awesome, thanks a lot! I never heard about Eddington-Finkelstein coordinates. The singularity of the Schwarzschild metric appeared fundamental to me.

The point is that the singularity at the event horizon is only a coordinate singularity, i.e., it is due to the coordinates not behaving well there - a bit like spherical coordinates do not behave very well for $r = 0$ or $\theta = 0$. If you look at calculable invariants, such as the Ricci scalar, they are all finite at the Schwarzschild radius and indeed the singularity of Schwarzschild coordinates can be removed by a more clever choice of coordinates (Eddington-Finkelstein is one such choice, Kruskal-Szekeres coordinates another, which is quite instructive for studying the maximally extended Schwarzschild geometry).

In juxtaposition, the singularity at $r \to 0$ is physical and cannot be removed by coordinate transformations.

 

[jbriggs444]

Okay, so the light which is after the horizon will make it to reach my eyes?

Your [falling] eyes will make it to the light. Your eyes will see the light that was emitted from your toes inside the horizon after the eyes too have passed inside the horizon.

Or, adopting the eye's point of view, after the horizon has swept past the eyes at the speed of light.

 

[Nugatory]

Your eyes will see the light that was emitted from your toes inside the horizon after the eyes too have passed inside the horizon.

Furthermore there is no discontinuity. The light that leaves your toes just before they cross the horizon reaches your eyes just before they cross the horizon; the light that leaves your toes as they cross the horizon reaches your eyes as they cross the horizon; and the light that leaves your toes just after they cross the horizon reaches your eyes just after they cross the horizon. Thus, light from your toes is reaching your eyes for the entire time and you never lose sight of your toes.

@jbriggs444 already knows this of course - teh comment is for others following the thread

 

[Omega0]

due to the coordinates not behaving well there - a bit like spherical coordinates do not behave very well for $r = 0$ or $\theta = 0$.

Physics does not now coordinates. Physics is to measure things. It is only to find out that there happened something between A and B. If you ping from A and there nothing from B this is a strong issue in the measurement process. The basic point I am speaking about is that you need time and space to measure things.
Correct, the coordinate systems we have chosen may collibe but physics does not.

...
In juxtaposition, the singularity at $r \to 0$ is physical and cannot be removed by coordinate transformations.

If so, what is Hawking radiation? It comes from a process like 0 or 1, a binary thing, right? I am no expert about this, not at all, but I thought I understood that there is a 0 or 1, or say -1 and 1, this doesn't matter. So we do have a radiation which comes from the existence of a Schwarzschild radius or so.Correct?
Now we can't think about a falling clock with some dimension where we have a 0 and 1? So it is allowed to separate something inside or outside the EH in quantum field theory but the picture is wrong in a classical sense?

 

[PeterDonis]

some dimension where we have a 0 and 1?

Where are you getting 0 and 1 from? Everything we are discussing is continuous. (That is even true of quantum field theory, btw. Some observables have a discrete spectrum of possible results--but not all--but the underlying quantum fields are continuous, and so is spacetime.)

 

[Dale]

The light that leaves your toes just before the cross the horizon reaches your eyes just before they cross the horizon; the light that leaves your toes as they cross the horizon reaches your eyes as they cross the horizon; and the light that leaves your toes just after they cross horizon reaches your eyes just after they cross the horizon. Thus, light from your toes is reaching your eyes for the entire time and you never lose sight of your toes.

Wow! Nicely said. @Omega0 pay attention to that post.

 

[PeterDonis]

we do have a radiation which comes from the existence of a Schwarzschild radius or so.Correct?

This is actually an open question. Hawking's original derivation of the existence of Hawking radiation used quantum field theory in curved spacetime--Schwarzschild spacetime. His derivation basically said that in this spacetime, a quantum field state that looks like vacuum in the far past looks like a thermal state with radiation coming from the black hole region in the far future. It said nothing about the specific mechanism that produced the radiation.

Since then there has been a lot of speculation and research on this topic, and one of the speculations is that the actual mechanism that produces Hawking radiation might have to do with apparent horizons, not event horizons. In other words, it might have to do with surfaces at which, locally, radially outgoing light does not move outward but stays in the same place, rather than with surfaces that globally are boundaries of regions that can't send light signals to infinity. In Schwarzschild spacetime, the two coincide, so it's impossible to distinguish which one is causing the radiation in Hawking's original derivation.

 

[Omega0]

Furthermore there is no discontinuity. The light that leaves your toes just before the cross the horizon reaches your eyes just before they cross the horizon;

Toes are usually not emmiting light.
Those freaks need something like reflection.

as they cross the horizon reaches your eyes as they cross the horizon; and the light that leaves your toes just after they cross horizon reaches your eyes just after they cross the horizon. Thus, light from your toes is reaching your eyes for the entire time and you never lose sight of your toes.

Awesome, sounds like science before Newton or so, you not even mention that there is something like interaction. Toes are emitting information, damn that's fun. :-D

 

[Ibix]

Awesome, thanks a lot! I never heard about Eddington-Finkelstein coordinates. The singularity of the Schwarzschild metric appeared fundamental to me.

I'd use Kruskal-Szekeres coordinates, because then you can draw a Kruskal diagram (with apologies to @DrGreg, whose diagram I've defaced):

In this diagram, the outside world (or at least a slice of it) is distorted into the right hand wedge (the red region labelled I) and the interior of the hole is the upper wedge (blue, labelled II). The other two regions don't matter here (and don't exist in a realistic black hole). The event horizon is the boundary between I and II, and the singularity is the curve at the top of the diagram.

An important feature of this diagram is that light moving radially inwards moves on a 45° line sloping up to the left, and outward moving light follows a 45° line up to the right - parallel to the event horizon.

I've sketched the paths of two infalling observers in black. The left hand one is your toes, emitting pulses of light (orange arrows). All pulses emitted before your toes cross the horizon are received by the second observer (your eyes) before they cross the horizon. All pulses emitted after your toes cross the horizon are received after your eyes also cross. Given that you can pick a coordinate system where the light rays' worldlines are parallel to the event horizon, how could it be otherwise?

 

[jbriggs444]

Those freaks need something like reflection.

Mirrors on the toes of one's shoes. Classic move.

 

[pervect]

Toes are usually not emmiting light.
Those freaks need something like reflection.

Awesome, sounds like science before Newton or so, you not even mention that there is something like interaction. Toes are emitting information, damn that's fun. :-D

That's an odd objection to make.

It doesn't really matter if your toes are illuminated by an external source or whether they have "glow in the dark" nail polish as far as the physics goes. It's simpler to answer the question about light from the toes without considering exactly how the light was emitted - it simply doesn't matter if the light was emitted by glow-in-the dark nail polish, or whether it was reflected from a lamp.

It's such an odd objection that I wonder if we're being trolled here. Of course, if we are being trolled, it's rather unlikely that we'll get an acknowledgment. However, it's still plausible that we are not being trolled, and there's some other reason you're bringing up such irrelevant stuff - sometiems people do that when they get an answer they don't like, they try to change the question.

 

[Dale]

It's such an odd objection that I wonder if we're being trolled here

Me too. It is a really poor objection to one of the best posts I have seen. Very disappointing