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

[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.

 

[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.

 

[Dale]

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.

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.

Even worse, you would die immediately.

Not for a supermassive black hole.

 

[A.T.]

it would be impossible to pull back your hand etc.

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.

 

[phinds]

Even worse, you would die immediately.
So you certainly would notice when you have crossed the event horizon.

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.

 

[PeterDonis]

but inside the event horizon everything moves only in one direction, towards the singularity r=0.

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 probably has to do with what people loosely say as "space and time switch role beyond 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:

the singularity r=0 is in the future

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

 

[pervect]

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.

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.

 

[anorlunda]

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?

 

[PAllen]

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?

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.

 

[Nugatory]

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?

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 Schwarzschild 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.

 

[yenchin]

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

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!

 

[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?)

 

[PeterDonis]

If you sit in a rocket very near to event horizon of a supermassive black hole, how would the horizon look like?

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.)

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?

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.

 

[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.

 

[PeterDonis]

With regards to the question about a portion of a rope being suspended within a Schwarzschild black hole, the answer should be no.

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.

 

[WannabeNewton]

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.

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

 

[jartsa]

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.

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.

 

[PeterDonis]

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)

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

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

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

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.

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.

for the falling person there's not enough time to notice anything special.

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.

 

[jartsa]

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

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.

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.

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.

 

[Dale]

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

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

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.

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.

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.

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.

 

[PeterDonis]

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.

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.)

 

[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?

 

[PeterDonis]

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.

Correct.

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?

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.

What happens with his leg if the observer moves farther away from the EH?

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

 

[WannabeNewton]

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.

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

 

[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.

 

[PeterDonis]

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

Well, if you're modeling an observer and his leg as separate objects, the observer can be at a fixed spatial location while the leg isn't. That's the scenario I thought we were discussing: the observer hovers at a constant r but lowers his leg below the horizon.

 

[WannabeNewton]

That's the scenario I thought we were discussing: the observer hovers at a constant r but lowers his leg below the horizon.

Ah ok, I must have missed that part. I thought we were talking about a situation where the whole guy's body just hovers in place.

 

[PeterDonis]

If it approaches the EH asymptotically, it never crosses it.

No, that's not correct. We've had umpteen threads about this and I don't want to belabor it, but the leg *can* indeed cross the horizon. (And I don't think WannabeNewton was asking about the "asymptotically" bit anyway.)

for the hovering observer his leg will be always outside the EH and won't detach from him

No, that's not correct either. The detaching of the leg will have to take place at some $r$ above the horizon; it can't take place exactly on the horizon (or below it), because the portion of the observer's body that remains attached has to do so while moving slower than light. So it will only take some finite time, by the observer's clock, from the time he starts lowering his leg, for him to see the light from the detaching of his leg. It's true that he will see his leg continue to fall, and it will appear, to him, to approach the horizon asymptotically; but this will be *after* it detaches.

 

[maxverywell]

No, that's not correct. We've had umpteen threads about this and I don't want to belabor it, but the leg *can* indeed cross the horizon. (And I don't think WannabeNewton was asking about the "asymptotically" bit anyway.)

Wait, what I'm saying is that for the hovering observer outside the EH, his freefalling attached leg approaches the Eh asymptotically, so by definition he won't see it crossing the EH. But I agree that the leg crosses the EH, but this is something that an another observer, following the leg, will see.

(I'm assuming that the hovering observer is very close to the EH, approximately at one leg's length)

No, that's not correct either. The detaching of the leg will have to take place at some $r$ above the horizon; it can't take place exactly on the horizon (or below it), because the portion of the observer's body that remains attached has to do so while moving slower than light. So it will only take some finite time, by the observer's clock, from the time he starts lowering his leg, for him to see the light from the detaching of his leg. It's true that he will see his leg continue to fall, and it will appear, to him, to approach the horizon asymptotically; but this will be *after* it detaches.

 

[WannabeNewton]

Wait, what I'm saying is that for the hovering observer outside the EH, his freefalling attached leg approaches the Eh asymptotically, so by definition he won't see it crossing the EH.

Peter already answered this. The detachment will happen at some radial coordinate above the horizon, realistically within a finite time read on the observer's own clock.

 

[maxverywell]

I was wrong. It takes infinite proper time for infalling object (forget legs...) to cross the EH only for an hovering observer far away the EH (actually at infinity) where his proper time is actually the Schwarzschild coordinate t. He won't see the object crossing the EH, but this is an optical illusion -- the light rays from the object will reach him at larger and larger values of t while the leg approaches the EH. But near the EH the coordinate t is not his proper time. But it's still not clear to me if he will see the object crossing the EH in his finite proper time.

 

[PeterDonis]

it's still not clear to me if he will see the object crossing the EH in his finite proper time.

No, he won't. If the observer is hovering at constant $r$, his proper time is Schwarzschild coordinate time times a constant time dilation factor. A finite constant times infinity is still infinity.

 

[WannabeNewton]

But near the EH the coordinate t is not his proper time. But it's still not clear to me if he will see the object crossing the EH in his finite proper time.

Note that if the falling object sends regularly periodic signals to the observer hovering at constant $R$, we take the initial event on the hovering observer's worldline at which the first signal is received, and then calculate the time interval between the initial event and each following event on the hovering observer's worldine at which a signal is received, the time interval will approach infinity as the infalling object gets closer and closer to the horizon. The proper time between events on the hovering observer's worldline is given by $\Delta \tau = (1 - \frac{2M}{R})\Delta t\propto \Delta t$ so the proper time between said events on the hovering observer's worldline will also go to infinity as the infalling object gets closer and closer to the horizon.

 

[jartsa]

The head of person that is standing on a platform near an event horizon will observe gravitational redshifting of signals coming from the feet. Very large redshifting if the platform is very near the EH.

After the person has jumped and has been free falling for some time, the head will say the redshift has almost disappeared.

When the head says "the redshift has almost disappeared", an observer that is hovering nearby will say the head is approaching the feet quite fast, and that is the reason why the head observes almost no Doppler shift of signals emitted by the feet.

The contracting motion of the falling person makes the special situation to feel normal to the falling person. That is the opinion the hovering person.

 

[PeterDonis]

The head of person that is standing on a platform near an event horizon will observe gravitational redshifting of signals coming from the feet. Very large redshifting if the platform is very near the EH.

Not necessarily. The gravitational redshift depends on the person's height relative to the size of the hole; if the hole is large enough the redshift from feet to head will be very small, even if the acceleration required for the person to hover is very large.

However, the more important point is that the redshift is present *because* both the head and the feet are accelerating; if they are freely falling it is absent, regardless of altitude. See below.

After the person has jumped and has been free falling for some time, the head will say the redshift has almost disappeared.

What redshift? The redshift of light signals coming from the feet? That will disappear as soon as both the head and the feet are freely falling; as noted above, the redshift is only there to begin with if both the head and the feet are accelerated. If they are freely falling there is no redshift (assuming that the person's height stays the same).

When the head says "the redshift has almost disappeared", an observer that is hovering nearby will say the head is approaching the feet quite fast

If the acceleration of the hovering observer is large enough, yes, that observer might quite quickly see the free-falling person to be greatly length contracted--although I'm not sure this equates to the hovering observer saying the head is approaching the feet. But that's a side issue.

The main issue is that, as noted above, the redshift disappears as soon as the person starts freely falling--i.e., before he appears length contracted to the hovering observer. So length contraction can't be the explanation for the disappearance of the redshift.

Here is how an observer freely falling in the vicinity of the person interprets the behavior of the redshift: light signals emitted upward from his feet take some time to reach his head. While the person is accelerated, during the time the light is traveling, the head gains speed away from his feet (because the head is accelerated). This causes the redshift.

But as soon as the person starts freely falling, the speed of his head relative to his feet no longer changes while the light is traveling from feet to head. So there is no longer any redshift. Note that, according to the freely falling observer, there is no "gravitational redshift" at all; the only redshift is a straightforward Doppler shift due to a change in velocity.

Now, how does the hovering observer interpret this? According to him, the light does redshift as it rises against the gravity of the hole, whether the person is accelerated or freely falling. But if the person is accelerated, hovering at a constant altitude, the gravitational redshift is the only effect happening, so the light from the feet is observed by the person to be redshifted when it reaches his head.

On the other hand, if the person is freely falling, then during the time the light takes to rise from his feet to his head, he gains just enough downward velocity for the Doppler blueshift due to that velocity to exactly cancel the gravitational redshift due to the change in height. So the person observes no redshift when light from his feet reaches his head.

 

[Ookke]

Let's say that a particle annihilates right after crossing the event horizon of a supermassive black hole. The annihilation produces two gamma ray pulses, one directed towards the singularity and other into the opposite direction.

I would guess that the outwards directed pulse can at least briefly visit above event horizon, since the particle is already quite high in gravitation potential when the annihilation occurs, and tidal forces are not strong at all at EH of a supermassive black hole. It's hard to see what mechanism could grab the gamma ray and pull it back before the EH, which was just crossed moment ago (especially when nothing out of ordinary seems to be going on).

However, I find it quite easy to accept that eventually the outwards directed pulse too will be pulled back into the singularity. It's just that the event horizon doesn't seem to make sense as a sharply defined surface, but rather it could be an approximate concept, like "any stuff below EH doesn't usually visit above it, but even if it does, it will be eventually pulled back into singularity". So the black hole would be totally black, when looked from far away enough, but not necessarily so when looked close above EH.

 

[PeterDonis]

I would guess that the outwards directed pulse can at least briefly visit above event horizon

No, it can't. Both light pulses will continually decrease their $r$ coordinate until they hit the singularity. The "outward" directed pulse will take longer to reach the singularity, but it never gets to any $r$ greater than the one at which it was emitted.

It's hard to see what mechanism could grab the gamma ray and pull it back before the EH, which was just crossed moment ago (especially when nothing out of ordinary seems to be going on).

There isn't any mechanism that needs to "grab" it. Remember that, inside the horizon, curves of constant $r$ are spacelike, not timelike. Any future-directed curve, timelike or null, inside the horizon *has* to have decreasing $r$. So the only "mechanism" that acts on the outgoing gamma ray is spacetime itself.

the event horizon doesn't seem to make sense as a sharply defined surface, but rather it could be an approximate concept

No, it's exact and sharply defined. But it is not a "place". As I posted earlier, the fact that the EH is a surface of constant $r$ does *not* mean it has a constant "position"; for a surface of constant $r$ to be at a constant position, the surface has to be timelike, and the EH, while it is a surface of constant $r$, is not timelike, it's null. (And inside the EH, surfaces of constant $r$ are spacelike, so they're even more emphatically not "places". Inside the EH, surfaces of constant $r$ are best thought of as "instants of time", with the future direction of time being the direction of decreasing $r$.)

 

[maxverywell]

Let's pose the same problem slightly different. Consider two observers A and B that are connected by a rope of constant length L=1m. Initially the two observers are hovering outside the EH at constant distance r=0.5m from it (they are in their spaceship). Then the observer B jumps out of the spaceship and freefalls. What will happen?

From the perspective of the observer B he will cross the EH at finite proper time $\tau_B$ (let's say after a few minutes), then the rope will break and he will continue falling to the singularity.

From the perspective of the observe A, the observer B is approaching the Eh asymptotically, i.e. it takes infinite proper time $\tau_A$ to cross the EH.

Now, if the observer A after his finite proper time (let's say after 1 hour) fires more the engines of the spaceship, to move away from the EH, he will pull the observer B with him. From the perspective of B, his elapsed proper time will be much smaller (probably few seconds, I haven't done the calculations) when the spaceship starts moving away, and at that time he hasn't crossed the EH yet. So the two pictures are consistent because both observers were outside the EH.

The problem when we say that observer B jumps inside the BH is that for the observer A it takes infinite time to see the B reach the EH, so he wont's see the B crossing the EH. So from the perspective of the A, we cannot say that the B jumps inside the black hole. This will never happen. But of course from the point of view of the B, he can jump inside the black hole -- he will cross the EH in finite proper time $\tau_B$. But when this happens, the proper time $\tau_A$ of A will become infinite.

 

[WannabeNewton]

What exactly is your question? By the way, observer B is obviously not in free fall because he is attached to a rope and the rope exerts some tension on him.

 

[jartsa]

Let's pose the same problem slightly different. Consider two observers A and B that are connected by a rope of constant length L=1m. Initially the two observers are hovering outside the EH at constant distance r=0.5m from it (they are in their spaceship). Then the observer B jumps out of the spaceship and freefalls. What will happen?

From the perspective of the observer B he will cross the EH at finite proper time $\tau_B$ (let's say after a few minutes), then the rope will break and he will continue falling to the singularity.

From the perspective of the observe A, the observer B is approaching the Eh asymptotically, i.e. it takes infinite proper time $\tau_A$ to cross the EH.

Now, if the observer A after his finite proper time (let's say after 1 hour) fires more the engines of the spaceship, to move away from the EH, he will pull the observer B with him. From the perspective of B, his elapsed proper time will be much smaller (probably few seconds, I haven't done the calculations) when the spaceship starts moving away, and at that time he hasn't crossed the EH yet. So the two pictures are consistent because both observers were outside the EH.

The problem when we say that observer B jumps inside the BH is that for the observer A it takes infinite time to see the B reach the EH, so he wont's see the B crossing the EH. So from the perspective of the A, we cannot say that the B jumps inside the black hole. This will never happen. But of course from the point of view of the B, he can jump inside the black hole -- he will cross the EH in finite proper time $\tau_B$. But when this happens, the proper time $\tau_A$ of A will become infinite.

Not quite. Safety line stops working after some time of falling. A tug send through the rope from above, propagating at speed of light, approaches the falling person, but never reaches him.

BUT a safety net that hangs arbitrarily close to the event horizon can be constructed.

 

[maxverywell]

What exactly is your question? By the way, observer B is obviously not in free fall because he is attached to a rope and the rope exerts some tension on him.

Initially, when both observers are on the spaceship, they are at the same r., so that the rope is not stretched and he can freefall before his distance from the spaceship becomes equal to the length of the rope.

The question is: is what I wrote correct? :)

Not quite. Safety line stops working after some time of falling. A tug send through the rope from above, propagating at speed of light, approaches the falling person, but never reaches him.

BUT a safety net that hangs arbitrarily close to the event horizon can be constructed.

The tug is propagating at the speed of sound in the rope, isn't it?
But why it would never reach the falling observer? It moves faster than the falling observer.

 

[WannabeNewton]

Initially, when both observers are on the spaceship, they are at the same r., so that the rope is not stretched and he can freefall before his distance from the spaceship becomes equal to the length of the rope.

He is not free falling for all time given the aforementioned setup.

It moves faster than the falling observer.

https://www.physicsforums.com/showpost.php?p=4419049&postcount=13
https://www.physicsforums.com/showpost.php?p=4419061&postcount=14

The question was already answered, just replace the second observer with a brick or a test particle.

 

[PeterDonis]

Now, if the observer A after his finite proper time (let's say after 1 hour) fires more the engines of the spaceship, to move away from the EH, he will pull the observer B with him.

No, he won't. At least, there will be some finite proper time by A's clock after which nothing he does can keep B above the horizon.

To see this, you have to make a careful distinction between *ingoing* and *outgoing* light signals. The reason A sees B approach the horizon asymptotically is that *outgoing* light signals from B, as B approaches the horizon, take longer and longer to get to A.

But if A tries to pull on the rope to keep B from crossing the horizon, that's an *ingoing* signal, not an outgoing one. Now consider the event at which B crosses the horizon. That event has a past light cone, and that past light cone intersects A's worldline at some finite proper time. Any signal or influence emitted by A, whether it's tugging on the rope, sending a radio signal, whatever, must reach B *after* B has crossed the horizon, and therefore cannot prevent B from crossing the horizon.

So there is a quite practical sense in which, after a finite time by A's clock, he can consider B to have crossed the horizon: because after that time, nothing A does can affect B before he crosses the horizon.

 

[PAllen]

I think this set up can be interesting to think about. Consider a ship hovering 10 meters above supermassive BH horizon [note that in the real world, rather than though experiments, even ignoring radiation - even that produce by thrust infalling into the BH - you have a quandary: hovering this close to a supermassive BH, you need proper accelerations so large that matter would be squeezed to greater density than the nucleus of an atom; on the other hand, if you want to be meters from a BH such that the hovering g force is only e.g. a few gees, then the BH is submicroscopic]. So, we have a ship and observers made of unobtainium that can resist the gee forces; and we have super flexible fiber optic cable attached to a weight. We let it drop from the spaceship toward the horizon. It is not clear to me why the cable must snap before the object reaches the horizon in near free fall. Each element of the free falling cable and weight is following a timelike trajectories; distances between elements need not increase, and tidal stresses could be quite modest. So, at least briefly, I don't see any reason the cable couldn't extent through the horizon. You would have the feature that light sent along the cable from the ship just after dropping the weight could reach the weight soon after the weight has crossed the horizon. A return signal would fail, and either the ship would have to keep running cable at near the speed of light, or else the cable would break [ or it would have to have the ability to stretch at near the speed of light].

Am I missing something?

 

[PeterDonis]

It is not clear to me why the cable must snap before the object reaches the horizon in near free fall.

It snaps because the upper end is being tugged on, very hard, by the accelerating spaceship, and any point on the cable at or below the horizon would have to move faster than light to keep up. So the cable must break at some point above the horizon, so that the end above the break can stay with the ship while still remaining on a timelike worldline.

Each element of the free falling cable and weight is following a timelike trajectories

But no timelike trajectory at or below the horizon can remain at a constant $r$.

distances between elements need not increase

Not true for elements at or below the horizon; there's no way for the distance *not* to increase between any such element and any element above the horizon that remains at a constant $r$. See above.

and tidal stresses could be quite modest.

But tidal stresses aren't the key stresses in the problem. The cable is under stress because its upper end is being accelerated, very hard.

 

[PAllen]

It snaps because the upper end is being tugged on, very hard, by the accelerating spaceship, and any point on the cable at or below the horizon would have to move faster than light to keep up. So the cable must break at some point above the horizon, so that the end above the break can stay with the ship while still remaining on a timelike worldline.

No, it's not being tugged at all. The slack is not even used up before the dropped weight crosses the horizon.

But no timelike trajectory at or below the horizon can remain at a constant $r$.

No such trajectory is required by my scenario.

Not true for elements at or below the horizon; there's no way for the distance *not* to increase between any such element and any element above the horizon that remains at a constant $r$. See above.

But elements above the horizon are not maintaining constant r. All parts of the apparatus are in free fall or 'minimal' tension until the slack in the cable is used up, with the weight below the horizon.

But tidal stresses aren't the key stresses in the problem. The cable is under stress because its upper end is being accelerated, very hard.

Not in the scenario I intended.

 

[WannabeNewton]

Hi PAllen! If the rope+weight drops in free fall with ample slack as you describe then I don't see why it couldn't cross the EH glibly as if it were a summer's day in the shire ;) but the problem is if you try to tug back on the rope+weight in order to try to pull it out once it passes the EH. Then the rope will snap at some point above the EH.

 

[PAllen]

Here is a way to clarify my scenario. Imagine said ship dropping pellets in rapid sequence. These pellets, after a short time, form a sequence crossing the horizon. Each pellet can actually send two way signals to its neighbor on either side (except that if one pellet is on one side of the horizon, the next pellet up won't receive the signal until after it has crossed the horizon). Take the limit of this, or imagine a string with slack between each free falling pellet. By all the arguments the crossing the horizon is locally a non-event, I therefore see nothing preventing a free falling cable with slack from crossing the horizon with one end connected to a hovering observer. Breakage will occur when slack is exhausted.

 

[PAllen]

Hi PAllen! If the rope+weight drops in free fall with ample slack as you describe then I don't see why it couldn't cross the EH glibly as if it were a summer's day in the shire ;) but the problem is if you try to tug back on the rope+weight in order to try to pull it out once it passes the EH. Then the rope will snap at some point above the EH.

Well that is obvious. I was not interested in trying to tug anything back, just remain connected for a while across a horizon from a hovering ship. I was just intrigued with the one way signalling possible - until slack used up, rocket could send signals to inside EH weight.

 

[WannabeNewton]

But if the rope/string/cable w\e is the medium being used for the signaling then what kind of signal could you send other than a tug?