# Relativity at event horizon

1. Feb 1, 2014

### dyb

When I look at a minkowski diagram for a black hole I can see that time goes to infinity for the outside observer while the infalling object approaches the black hole.

That means that for an outside observer, it takes an infinite amount of time until the infalling object reaches the event horizon.

Does that mean that the infalling object actually appears to decelerate to an outside observer as it falls into the black hole rather than accelerating? Really? So there's a huge mass and the object actually gets slower until it stops?

2. Feb 1, 2014

### waitedavid137

Actually I believe you are looking at a mapping of Schwarzschild coordinate time and radial coordinate onto Kruskal-Szekeres coordinate axis, and I think you mean the Schwarzschild time goes to infinity as the infalling approaches the horizon, not the physical singularity whose world line is a curve above that 45 degree line.
If you are a remote observer using Shwarzschild coordinates then yes, that is what you interpret as happening to the thing infalling, but that isn't the only choice of coordinates that are natural enough for you to use as a remote observer. So for example, if you are a remote observer using Kerr-Schild coordinates, then you will interpret the situation as that the thing infalling accelerated all the way to the singularity, but that "outgoing" light like information has a remote coordinate speed that drops to zero nearing the horizon and as such even though you reckon that it reaches the horizon in finite Kerr-Schild time which you are using as your time standard, you realise that you will never actually see any information from an event at which it crosses because that information never escapes. In the end you get equivalent results that you never see information from the horizon or below, but what coordinates you use to describe and interpret the situation are up to you.

3. Feb 1, 2014

### dyb

I see, thanks!

So is that right: If the object falling into the black hole encoded some information on a light beam and sent it to the outside observer, that light beam would get red-shifted while travelling outward. Whatever information were modulated onto that beam, it would reach the observer with an increasingly lower rate.

So it's like I take an audio cassette tape and I stretch one end further and further apart up to infinity while leaving the other end intact. When I now play the music from the tape, the music would get progressively slower and slower, and the pitch would get lower and lower. It would take me an infinite amount of time to listen to the whole tape, although the information encoded on the tape is actually finite.

4. Feb 1, 2014

### Staff: Mentor

That's how it *appears* to the outside observer, yes. But this is best understood as an optical illusion; as the infalling object gets closer and closer to the horizon, outgoing light emitted by the object takes longer and longer to reach the outside observer, because of the way the spacetime is curved. And outgoing light emitted by the infalling object *at* the horizon never reaches the outside observer at all; it stays at the horizon forever. (That's what the horizon *is*: an outgoing lightlike surface that stays at the same radius forever.)

To the infalling object, OTOH, it only takes a finite time to reach the horizon (and a not much longer finite time after that to reach the singularity at the center of the hole).

5. Feb 1, 2014

### dyb

Is it really just an optical illusion? Because if the person falling into the black hole decided to return to the distant observer before reaching the event horizon, and they would compare clocks, it would turn out that his clock actually did run slower.

6. Feb 1, 2014

### PAllen

But let's break this down. Depending on infall trajectory, someone falling in and past the horizon sees outside clocks running at reasonably normal rate all the way to the singularity (yes, infaller sees outside just fine). There would be (asymptotically) a last outside time seen by the infaller before the singularity. Instead, suppose they decide to escape just before reaching the horizon. They must apply enormous acceleration to do this (proper acceleration - enormous thrust). It is only as this thrust is applied that they see distant clocks extremely speeding up, such that a billion years on a distant clock goes by for a second of their escape. From their point of view, it is only the acceleration (which is invariant) that produces the differential aging (also invariant when they reach the distant clock and compare them).

Last edited: Feb 1, 2014
7. Feb 1, 2014

### Staff: Mentor

The optical illusion is that the person falling in goes slower and slower as he gets closer to the horizon. Local measurements will show something very different. For example, suppose there are a family of "hovering" observers, each one firing rockets to stay at a constant altitude above the horizon. As the person falling in passes each "hovering" observer, that observer will see the person falling inward faster and faster the closer the "hovering" observer is to the horizon.

Of course if the person stops and turns around, that changes things; but even then, the "hovering" observers nearby when the person turns around will not see him do so "in slow motion", as the distant observer does. They will see him fire rockets normally and turn around and head back outward normally.

When the person gets back to the distant observer, it's true that less time will have elapsed on his clock, and that comparison is invariant--all observers will agree on it. But the reason that comparison is invariant is that it's made at a single event: the person and the distant observer are co-located when they compare elapsed times on their clocks.

8. Feb 1, 2014

### waitedavid137

I wasn't referring to red shift, but yes, the signal becomes infinitely red shifted as it comes from a source approaching the horizon. What I mean is that the picture you're usually given is given you in terms of Schwarzschild coordinates. In those coordinates the line element is
$ds^{2}=\left(1-\frac{2GM}{rc^2 }\right)dct^2 - \frac{dr^2 }{\left(1-\frac{2GM}{rc^2 }\right)} - r^2 d\Omega ^2$
The Kerr-Schild coordinate expression of this is
$ds^{2}=dct_{ks} ^2 - dr^2 - r^2 d\Omega ^2 -\frac{2GM}{rc^2 }\left(dr + dct_{ks}\right)^2$
What you interpret as happening to the infalling test particle depends on which way or whatever other way you choose for your coordinates to behave approaching the hole. Using Schwarzschild coordinates you are correct to interpret the situation as you described it. Its not an illusion, but it is merely how the situation is described by that coordinate system. What I was saying is that if you use these Kerr-Schild coordinates for the behavior of your remote observer coordinates as they approach the hole that the remote observer will then interpret the result as that the test particle fell all the way to the singularity in finite time, but he won't receive information about the event because the maximum speed of information travel "outward" drops to zero near the horizon. To see this just consider a light like path of $ds = 0$ and consider radial motion for the signal $d\Omega = 0$ and solve for the remote coordinate speed of the signal $\frac{dr}{dt_{ks}}$. What you interpret as going on remote from you actually depends on how you choose the remote behavior of your coordinates with respect to which you express whats going on, but in the end they tell you the same thing, that the remote observer will never actually receive information about whats going on within the horizon.

9. Feb 1, 2014

### dyb

I understand that depending on the coordinate the interpretation might be different, although in a sense they give the same ultimate conclusion.

What happens if the black hole evaporated shortly after the object fell into the black hole (from the falling object's time)? Is that an event that would never be observed by the outside observer because for him, an infinite amount of time would need to pass first before observing the evaporation? It seems the answer to this question would need to be the same for both interpretations.

On the contrary, let's assume an outside observer sees an object fall close to the black hole and apparently getting stuck and almost completely stop near the horizon after a billion years. Let's assume the black hole evaporates after such a long time. A billion years has passed for the outside observer, but almost no time has passed for the object because his time is stopped. If we assume that all black holes evaporate before an infinite time passes for an outside observer, wouldn't the test particle always experience the evaporation before he can ever pass the horizon?

Last edited: Feb 1, 2014
10. Feb 1, 2014

### Staff: Mentor

No. If the black hole evaporates, that changes things; the outside observer will see the infalling object reach the horizon at the same time he sees the hole's final evaporation. (The outside observer still won't be able to see inside the hole's horizon, however.)

No, that's not correct. In this scenario, the object falls through the hole's horizon long before the hole evaporates. The light rays emitted outward from the object just before it reaches the horizon will take a long time to get to the outside observer, but he will see them before he sees the hole's final evaporation.

So what the outside observer will see is: the object falling more and more slowly, but instead of asymptotically approaching the horizon and never reaching it (as would be the case for an eternal black hole that never evaporates), he sees the object reach the horizon at the same instant the hole evaporates. However, this is, again, best seen as an "optical illusion"; what the outside observer is seeing, when he sees the object reach the horizon at the same instant the hole finally evaporates, is light from events all along the hole's horizon overlapping each other; all those light rays end up on the same outgoing lightlike path to the outside observer, because of the way the spacetime is curved.

No. See above.

11. Feb 1, 2014

### pervect

Staff Emeritus
This is right so far, but now consider that someone outside the black hole has encoded some information (music , say) on a light beam and is transmitting it to the infalling observer, the reverse of what you specified.

We need to specify a bit more here - we'll specify that the infalling observer is falling from at rest from infinity If you are worried about the length of time it takes to fall from at rest at infinity, you can say instead that the infalling observer has just enough velocity to reach infinity if you run time backwards.

When you factor gravity and the motion of the infalling observer both together (and you can't separate them in any coordinate independent way), you find that with the specified initial conditions (of a fall from infinity) that the infalling observer receives at the event horizon a signal that's RED-shfited by a finite amount (2:1). There's an entry in my blog that goes into some detailed, but messy, calculations for the general case, the case of "at rest from infinity" is particularly simple, and if you search you can see that there's general agreement from various PF posters who have worked this problem by various means in various coordinate systems.

The problem with the idea that "time stops" is that one gets the incorrect notion that the infalling observer "sees" infinite blueshift from the outside observer. This never happens, though by starting from at rest sufficiently close to the black hole (rather than at rest at infinity) you can get a finite blueshift.

What really tends to drive the point about event horizons home though is to consider a totally different scneario, the case of the Rindler observer, with a constant acceleration.

This has been posted about before, I don't have any specific references to post at this time, but you can find them if you dig.

It turns out that the Rindler observer sees an event horizon that acts like the event horizon of a very massive black hole (one without any appreciable curvature at the horizon). In this case the interpretation becomes a bit more clear, because if the Rindler observer was watching the Earth fall through his "Rindler Horizon", it's obviously silly to say that time "stops" on Earth while it falls through the horizon.

12. Feb 2, 2014

### dyb

Thanks everyone I now have a better understanding of relativity on the event horizon!