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

[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?

 

[PeterDonis]

Imagine said ship dropping pellets in rapid sequence.

This is different from the cable scenario because the pellets aren't connected to the ship after they are dropped.

Take the limit of this, or imagine a string with slack between each free falling pellet.

Consider the string between the ship and the first pellet (is there one? If not, again, this scenario is different from the cable scenario). What's the worldline of its upper end (the one at the ship)? It's the same as the ship's worldline, right? Then the string is not in free fall; one end is accelerated. Since there are internal forces in the string, the rest of it won't be in free fall either.

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.

See above; the cable cannot be in free fall because one end is accelerated and there are internal forces in the cable. Our intuitions about "slack" cables don't work well when you're dealing with such large proper accelerations.

Or, alternatively, one could agree that breakage will occur when the slack is exhausted, and just clarify exactly *when* that happens; it must happen at some point on the cable above the horizon, because, once again, the part of the cable above the break has to stay with the ship, and it can only do that while remaining on a timelike worldline if the break is above the horizon. So the slack will be exhausted *before* our naive intuition says it "should" be (when the object at the lower end of the cable makes it taut), because of the large acceleration of the ship.

 

[PAllen]

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?

The cable is a fiberoptic. A light signal. If I am not mistaken, if the ship sends the signal right after dropping the weight, it can catch up with the weight inside the horizon. Further, if my slack is long enough, the weigh can keep getting signals sent from out side the horizon until it hits the singularity. The last of these would be slightly before the intersection of the past light cone of the event of the weight 'reaching' the singularity, with the ship's world line.

 

[PAllen]

Consider the string between the ship and the first pellet (is there one? If not, again, this scenario is different from the cable scenario). What's the worldline of its upper end (the one at the ship)? It's the same as the ship's worldline, right? Then the string is not in free fall; one end is accelerated. Since there are internal forces in the string, the rest of it won't be in free fall either.

If we are allowed to posit that I can drop a weight on some cable and it does not immediately break on leaving the ship, due to slack (idealized materials, but no violation of principle), then this situation near the ship can be maintained. I have to keep pushing cable out as needed to make sure it is never taught. All this is physics local to the rocket. Once outside the ship, each piece is, I claim, close enough to free fall that the difference is inconsequential.

Or, alternatively, one could agree that breakage will occur when the slack is exhausted, and just clarify exactly *when* that happens; it must happen at some point on the cable above the horizon, because, once again, the part of the cable above the break has to stay with the ship, and it can only do that while remaining on a timelike worldline if the break is above the horizon. So the slack will be exhausted *before* our naive intuition says it "should" be (when the object at the lower end of the cable makes it taut), because of the large acceleration of the ship.

Here I agree. My intuition suggests the cable will break at the rocket itself [or close to it, definitely above the horizon] as soon as it runs out of cable to feed.

 

[Dale]

There are several very interesting and related scenarios here:
http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

 

[WannabeNewton]

PAllen I think I am misinterpreting what you are saying because I don't see any problem with ingoing radial null geodesics passing through the EH. In EF coordinates the ingoing radial null geodesics do pass through the EH and could quite possibly intersect the worldline of an observer/particle inside the EH (at least intuitively, the calculations might show otherwise) but what exactly will this ingoing signal get us back at the ship hovering outside the EH?

Thanks for the link DaleSpam, very instructive page!

 

[PeterDonis]

I have to keep pushing cable out as needed to make sure it is never taught.

This is certainly possible, but it seems different from what jartsa was proposing, since he specified that the cable has a constant unstressed length of 1 m.

Once outside the ship, each piece is, I claim, close enough to free fall that the difference is inconsequential.

If the cable is continuously paid out, i.e., if the unstressed length of cable outside the ship is continuously increasing, yes, I think this can be realized. My comments were based on a cable of constant unstressed length.

 

[PAllen]

PAllen I think I am misinterpreting what you are saying because I don't see any problem with ingoing radial null geodesics passing through the EH. In EF coordinates the ingoing radial null geodesics do pass through the EH and could quite possibly intersect the worldline of an observer/particle inside the EH (at least intuitively, the calculations might show otherwise) but what exactly will this ingoing signal get us back at the ship hovering outside the EH?

All of this agrees with what I supposed. As to what you will get back the ship, obviously nothing. I was just playing with the idea (separate from Jartsa's scenario) of maintaining (for a while) some type of material object across the horizon extending to a hovering ship, and the ship sending signals to (but never getting a response) 'someone' at the other end, inside the horizon. Also, noting that any part of the cable can send a backwards signal to a part a little further away and reach it; but only when that piece has itself crossed the horizon. This allows that piece of cable to behave locally like any ordinary fully connected piece of cable.

 

[PAllen]

This is certainly possible, but it seems different from what jartsa was proposing, since he specified that the cable has a constant unstressed length of 1 m.



If the cable is continuously paid out, i.e., if the unstressed length of cable outside the ship is continuously increasing, yes, I think this can be realized. My comments were based on a cable of constant unstressed length.

It is a different scenario. I guess I wasn't clear enough about how there was no relation between my scenario and others. I just wanted to emphasize that, ignoring practical considerations, I saw no reason you could not maintain some form of material connection between an object inside the horizon and hovering ship.

 

[WannabeNewton]

By the way, take a look at the "Free fall" section of the link given by DaleSpam. It deals with a scenario quite similar to the one you described (rope with sufficiently ample slack).

 

[PeterDonis]

It is a different scenario. I guess I wasn't clear enough about how there was no relation between my scenario and others. I just wanted to emphasize that, ignoring practical considerations, I saw no reason you could not maintain some form of material connection between an object inside the horizon and hovering ship.

Ah, ok, I agree this is certainly possible. The only complication I can see that we haven't covered is the question of how fast the cable would need to be paid out; would the speed of cable exiting the ship have to exceed the speed of light at some point before the object at the lower end of the cable hit the singularity? I haven't done any analysis about that, but I think it's a legitimate question.