# Why we won't notice anything special when crossing the horizon?

1. Jun 17, 2013

### maxverywell

Why is it said that we won't notice anything special crossing the event horizon of BH?
I agree that there is no singularity and the curvature there is not infinite etc. but inside the event horizon everything moves only in one direction, towards the singularity r=0. Therefore it's possible to make body moves only in one direction, e.g. it would be impossible to pull back your hand etc.. Even worse, you would die immediately.
So you certainly would notice when you have crossed the event horizon.

I am considering supermassive black holes such that the tidal forces are not noticeable locally far away from the singularity.

Last edited: Jun 17, 2013
2. Jun 17, 2013

### yenchin

This probably has to do with what people loosely say as "space and time switch role beyond horizon".
Consider the Schwarzschild black hole for simplicity. The Schwarzschild r coordinate becomes the time coordinate inside the Schwarzschild radius. Look at the Penrose diagram; the singularity r=0 is in the future, not something the observer can see lying somewhere in space while helplessly falling toward it. The Schwarzschild singularity is unavoidable in the same sense that next Monday is avoidable.

3. Jun 17, 2013

### Staff: Mentor

The reason that we don't notice it is because we are used to it, this happens all the time. Outside the event horizon everything moves only in one direction, towards the future t=∞. It is impossible to reach your hand back into the past.

Consider the event horizon in Rindler coordinates. Every instant that you have been alive you have crossed an event horizon in some set of Rindler coordinates. Every null surface which you can draw in spacetime is a "horizon" where things can cross in only one direction, no matter how hard you accelerate.

Not for a supermassive black hole.

4. Jun 17, 2013

### A.T.

Movement is relative. You cannot move back your hand relative to a non falling observer. But you can move the hand in any direction relative to your falling body.

5. Jun 17, 2013

### phinds

As DaleSpam said, that is NOT necessarily true. For some BHs you would die well BEFORE you reached the EH and for others not until well AFTER passing the EH. It is a function of the tidal forces and for small BHs, the tidal forces are huge well outside the EH and for really large ones the tidal force is trivial until well past the EH.

6. Jun 17, 2013

### Staff: Mentor

This is a sloppy way of putting it. The correct statement is that everything moves only in one *timelike* direction inside the horizon--but as DaleSpam noted, that's also true outside the horizon! The difference inside the horizon is that the timelike direction has to be towards the singularity. But there's no way to detect that locally, which is why you don't notice anything special when you cross the horizon.

This statement is only true in Schwarzschild coordinates--perhaps that's what you mean by "loosely". You state it better further on in your post:

More precisely, the singularity is in the future of any event inside the horizon. That's an invariant statement, independent of coordinates.

7. Jun 17, 2013

### pervect

Staff Emeritus
In the local inertial frame of someone free-falling into a black hole, the event horizon is approaching at the speed of light.

So, drawing a space-time diagram with the event horizon as a "place" is misleading and confusing if you're trying to depict the experience of a free-faller. To understand what's happening from a free-fallers POV, you need to draw the event horizon as a free-faller would experience it, which is as a light-like or null worldline.

You can certainly draw your hand back before and after you reach the horizon (from the viewpoint of your local inertial frame). But in order to get your hand out of the event horizon after passing through it, you'd have to overtake a light signal that has already passed you! Which is obviously not possible in the context of relativity.

In Schwarzschild coordinates, which are singular at the horizon, everything falls into the black hole at the speed of light. This may mislead one into thinking that everythign falls into the black hole at an equal rate. However, this is not true. The velocity of A and B relative to the horizon is always "c", the speed of light. But this doesn't imply that the velocity of A relative to B is zero in their respective local inertial frames.

8. Jun 17, 2013

### Staff: Mentor

If you travel radially inward with your arm extended radially inward in front of you, wouldn't your view of your hand dim as you approached the EH, disappear as your hand enters the EH before you, and remain disappeared inside the EH?

Is it a true statement that every radius interior to an EH is also an EH?

9. Jun 17, 2013

### PAllen

No, and no. Your hand would dim as observed by a static observer, but not as observed by you, the infaller. You are passing the static observers at increasing speed, so your view of your hand is blue shifted relative to theirs, with the result that your hand looks perfectly normal.

The EH is a unique global feature (despite not being locally detectable) of a BH geometry and I have no idea on what basis you might propose that every radius inside is an EH. Inside the horizon, r=const is not lihgtlike, so it cannot conceivably be a horizon.

10. Jun 17, 2013

### Staff: Mentor

No. Track the worldline of your hand, your eye, and the light traveling from one to the other (remember your eye and your hand are both in free-fall - it would be a different story altogether if you were hovering at a constant Schwarzchild r coordinate outside the horizon and lowering your hand through it!) and you'll see that light is always reaching your eye from your hand.

11. Jun 17, 2013

### yenchin

In standard Schwarzschild metric, every constant r surface interior to the event horizon is a trapped surface. The outer boundary of these sets of surfaces is the apparent horizon, which in this case, is the same as the event horizon. However, the notion of apparent horizon is not necessarily coincide with that of event horizon!

12. Jun 17, 2013

### Ookke

If you sit in a rocket very near to event horizon of a supermassive black hole, how would the horizon look like? Is it uniform, black surface or something else?

Let's say that you tie a brick at one end of a rope and throw the brick below horizon, would you be able to get the brick back using the rope? (Old question I found somewhere, is there any opinions about this here?)

13. Jun 17, 2013

### Staff: Mentor

You can't see the horizon from outside the horizon, because nothing can get from the horizon, or inside it, to any place outside it, not even light. The only way to see the horizon is to cross it. When you cross it, you see light emitted by objects that previously crossed the horizon, at the instant they crossed it. Similarly, if you're hovering above the horizon, but close to it, and you look in its direction, you will see light emitted by objects falling towards the horizon, when they were closer to it than you are. (This light will be redshifted as well, more so the closer to the horizon it was emitted.)

No, because, once again, nothing can get from inside the horizon to any place outside it. If you pull on the rope once the brick is below the horizon, the rope will break at some point above the horizon, and all you will get back is the part of the rope that remains after the break.

14. Jun 17, 2013

### WannabeNewton

With regards to the question about a portion of a rope being suspended within a Schwarzschild black hole, the answer should be no. Let's modify the situation a little bit and consider an observer at spatial infinity (where space-time is asymptotically flat) and assume he has a really long rope with a particle suspended (i.e. static) from the bottom end. Some while back, I was solving a problem from Wald's General Relativity textbook on what the force felt by the particle is at the bottom as compared to the force actually exerted by the observer at infinity. Here's the thread on that: https://www.physicsforums.com/showthread.php?t=679255

The point is that the information about the tension as exerted by the observer on his end of the rope must propagate in finite time to reach the end the particle is suspended from. As a result, the magnitude of the force exerted by the observer gets redshifted by the redshift factor i.e. $F_{\infty} = VF$ where $V = (-\xi^{a}\xi_{a})^{1/2}$ ($\xi^{a}$ is the time-like killing vector field). Now assume the observer at infinity lowers the extremely long (but finite) rope into the event horizon at such a slow rate that the points on the rope are approximately static. Then, since the force on the points extremely close to the horizon is given by $F = V^{-1}F_{\infty}$ and $\lim_{r\rightarrow 2GM}V^{-1} = +\infty$, the local force felt on the points of the rope (as well as the particle) which are extremely slowly inching towards the event horizon will blow up to infinity and the rope will in effect be torn apart.

On the other hand if by some means the rope is lowered fast enough so that a portion of the bottom end of the rope (including the particle suspended from it) is lowered into the horizon, the observer will not be able to pull that portion of the rope back out for if he tried to exert a force on the rope in order to pull out the portion of the rope resting below the horizon, then for the same reasons above the force on the points of the rope outside the event horizon will increase without bound in the limit as one reaches the event horizon and the rope will snap at some point near the event horizon.

15. Jun 17, 2013

### Staff: Mentor

Not just should be, is. Your analysis is correct. The same answer can be obtained even more straightforwardly by realizing that the BH horizon behaves, with respect to observers hovering at a constant radius outside it, the same as the Rindler horizon of an accelerated observer in flat spacetime.

16. Jun 17, 2013

### WannabeNewton

Ah yes so if we considered a particle/observer being held static in the Schwarzschild space-time of a black hole so that in a small enough neighborhood the gravitational field is essentially uniform then we can use the equivalence principle to simply describe the local dynamics by the Rindler space-time in which case we see similarly that the proper acceleration diverges as one approaches the Rindler horizon in the usual way, corresponding to static observers placed closer and closer to the event horizon

17. Jun 19, 2013

### jartsa

Let's say there's a short stick hovering near the event horizon, indicating an area where something special might be noticed. (for example that it's impossible to kick your foot up in less then one minute)

An observer falling past the stick will say the length of the stick is much shorter than the rest length of the stick.

The stick will say the length of the falling person is much shorter than the rest length of the person, and the length approaches zero as the person approaches the event horizon.

So my point is that for the falling person there's not enough time to notice anything special.

18. Jun 19, 2013

### Staff: Mentor

Why would this be true? Locally, spacetime looks just like it does anywhere else, and locally time "flows" normally.

Sure, but this is true anywhere in spacetime, not just near the horizon.

You have to be careful here because the length measurement you're talking about can only be done locally. There's no way for the stick to measure the length of the falling person if the stick and the falling person are spatially separated. What you really should say here is that if we have a whole family of sticks, hovering at various altitudes above the horizon, the sticks closer to the horizon will measure the length of the falling person to be shorter than the sticks further away, with the length measured by any given stick going to zero as the altitude of the stick above the horizon goes to zero.

I don't see how this follows from anything else you said. (Not to mention that it seems to indicate the same error as I noted at the beginning of this post--locally nothing special is noticed because locally there *is* nothing special going on.

19. Jun 20, 2013

### jartsa

What I meant was: The foot can not be lifted quickly to the same position where the head is now.

And the original problem was: A foot that is below an event horizon can not be lifted in any time to where a head is, when the head is above the event horizon.

There's seems to be frame jumping going on. First we are in the frame of the falling person, the we are in the frame of a static observer.

Near an event horizon there's an area where a nerve impulse can not travel from a foot to a place where a head is now, in a decent time according to the head. But that area is passed quicly when free falling. (The head was frame jumping when thinking about the situation)

That was the idea.

Now let's extend that idea a bit:

It takes a finite time to reach the singularity. The special effects that require more time than that to be noticed, will not be noticed.

20. Jun 20, 2013

### Staff: Mentor

In the free-faller's frame, it certainly can.

That is true. However, the same thing is true of any arbitrary null surface even in completely flat spacetime. So this does not represent anything special about the event horizon.

I think this is the source of the confusion. You are mixing up frames. In the local free-falling frame there is nothing special about the event horizon. Locally it is simply an ordinary null surface. Nothing you can say about it in the local free fall frame cannot also be said about any arbitrary null surface in an inertial frame in flat spacetime.

You seem to get confused when you think about things from the perspective of the static frames.

Last edited: Jun 20, 2013
21. Jun 20, 2013

### Staff: Mentor

One possible source of your confusion here is that you are thinking of the event horizon as a "place", something that stays at a fixed position. It's not; it's a null surface, and no null surface can be at a fixed position.

(Note: yes, the horizon is at a fixed $r$ coordinate, but that's not the same as being at a fixed position. For a fixed $r$ coordinate to correspond to a fixed position, a curve of constant $r$ must be timelike. That's only true outside the horizon; the horizon itself is a curve of constant $r$, but it's null, not timelike; and inside the horizon, curves of constant $r$ are spacelike. So you can't think of $r$ as "position" when you are talking about events on or inside the horizon, or motion that crosses the horizon.)

22. Jun 20, 2013

### maxverywell

As have been said and from what I have understood, the event horizon is traveling at c with respect to an infalling observer. So there is no problem in pulling (or lifting) your leg back, because its speed, with respect to the infalling observer's body, is always less than c. So your body will cross the EH before you pull back your leg completely.

I wonder what happens when the observer is not infalling, but he is static observer hovering outside the EH at constant r (so his world line is timelike)?
For this observer the EH is not moving and he won't see his leg crossing the EH -- the leg will approach the EH asymptotically. Right? What happens with his leg if the observer moves farther away from the EH?

23. Jun 20, 2013

### Staff: Mentor

Correct.

He wil see his leg approach the EH asymptotically, but that's a minor point compared to the fact that as soon as his leg crosses the EH, either he will have to fall through himself or his leg will be detached from him. Remember that the EH is an outgoing null surface, and inside the EH, curves of constant $r$ are spacelike; so his leg can't stay at a constant $r$, even for an instant, once it reaches the horizon; it *has* to fall inward, to smaller values of $r$.

A question that is often asked in this connection is, what actually pulls the leg off of his body? An answer that is often given is "tidal gravity", but that's not right. Tidal gravity can be made negligible at the horizon by making the hole's mass large enough. Actually, the question as I just posed it gets things backwards: it's not that the leg gets pulled off, it's that the rest of the observer gets pulled away from the leg. Remember that, in order to "hover" at a constant altitude above the horizon, the observer has to accelerate, and the closer he is to the horizon, the harder he has to accelerate. So what's actually happening is that the observer is accelerating very hard to stay at the same altitude above the horizon, and once his leg reaches the horizon, it can't keep up (because to do so, it would have to move at or faster than the speed of light). So the observer's rocket engine (or whatever it is that is exerting the force on him that keeps him at altitude) pulls the observer away from the leg, so hard that the leg's structural strength is exceeded and it breaks off.

The same as above, except his leg getting detached will probably happen sooner (by the observer's clock).

24. Jun 20, 2013

### WannabeNewton

Perhaps I misread something but why would an observer who is at a fixed spatial location be approaching and eventually cross the EH?

25. Jun 20, 2013

### maxverywell

Yeah, I wanted to ask the same. If it approaches the EH asymptotically, it never crosses it.

That's strange. We know that from the point of view of an another observer who is following the leg of the hovering observer, the leg will cross the EH and eventually will detach from the body of the hovering observer. But for the hovering observer his leg will be always outside the EH and won't detach from him, even if he later moves farther away from the EH (bigger values of the r). This two descriptions are completely different but because the two observers will become causally disconnected, there is no problem. I think this is called black hole complementarity.