Can Humans Survive Near a Black Hole's Event Horizon?

[nutgeb]

That sounds plausible at first, but actually it's more bizarre than that. Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference. The explanation is that the extreme acceleration of the pole, relative to you, causes extreme Lorentz contraction, and the two effects cancel each other out so that the bottom end almost stops. The top of the pole will slip through your hands before the bottom reaches the horizon, no matter how close you are hovering from the horizon and no matter how long the pole is!

I think I agree with your description.

But what happens from the pole's own perspective? The distance it perceives that it falls to the horizon is far less than its length, and of course its length is not contracted in its own frame. So the pole would predict that most of itself is still in the holder's grip when it crosses the horizon.

This seems like a case of failure of simultaneity in SR, like the ladder/barn paradox.

 

[mjacobsca]

Thanks for the great discussion guys. I've gotten more than my money's worth! My question was more of a thought experiment about what the pole holder would feel when holding onto the pole. I think I expected the answer to be conclusively that the pole would be ripped out of his hands or broken apart, but the concepts of Lorentz contraction and slow-down of time presents intriguing alternatives that I did not think of.

Mike

 

[PeterDonis]

Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference.

This is not correct. It will take infinite *coordinate time*, if you are using Schwarzschild coordinates as your coordinate time, but that is *not* the same as infinite proper time for you hovering close to the hole's horizon.

If you drop the pole (letting it free fall) while hovering at a given r-coordinate, you can calculate, fairly easily, how much proper time by your clock will elapse before you reach a "point of no return" where no amount of force you exert on the pole by tightening your grip will prevent the pole from falling through the horizon and either breaking (if your grip is stronger than the pole's tensile strength) or slipping through your grip never to return (if your grip is weaker). The details of the calculation are on Greg Egan's science pages (link below); the result is that after a proper time (for the hovering observer) of:

ln(2) \frac{a}{c^2},

where a is the (proper) acceleration felt by the hovering observer. Note that Egan does the calculation in terms of s_0, the "distance to the horizon" in the Rindler coordinates he's using--see my comment below on that--but to properly carry the result over to the black hole scenario, it's easier to rewrite things in terms of the acceleration; the relation between the two is:

a = \frac{c^2}{s_0}.

See http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html" for a good treatment of situations of this sort, with *lots* more details worked through than most treatments I've seen on the web. He does the calculation for an accelerating observer in Minkowski spacetime, but for this particular scenario, the result is the same as for an observer hovering (and therefore accelerating) at a constant r-coordinate above a black hole horizon. (He also does the calculation for a rope being paid out from the accelerating spaceship, which corresponds better to the original post in this thread, "lowering" the pole instead of just letting it free fall.)

 

[DrGreg]

Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference.

This is not correct. It will take infinite *coordinate time*, if you are using Schwarzschild coordinates as your coordinate time, but that is *not* the same as infinite proper time for you hovering close to the hole's horizon.

But that doesn't make sense. I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

If you drop the pole (letting it free fall) while hovering at a given r-coordinate, you can calculate, fairly easily, how much proper time by your clock will elapse before you reach a "point of no return" where no amount of force you exert on the pole by tightening your grip will prevent the pole from falling through the horizon and either breaking (if your grip is stronger than the pole's tensile strength) or slipping through your grip never to return (if your grip is weaker). The details of the calculation are on Greg Egan's science pages (link below); the result is that after a proper time (for the hovering observer) of:

ln(2) \frac{a}{c^2},

where a is the (proper) acceleration felt by the hovering observer. Note that Egan does the calculation in terms of s_0, the "distance to the horizon" in the Rindler coordinates he's using--see my comment below on that--but to properly carry the result over to the black hole scenario, it's easier to rewrite things in terms of the acceleration; the relation between the two is:

a = \frac{c^2}{s_0}.

See http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html" for a good treatment of situations of this sort, with *lots* more details worked through than most treatments I've seen on the web. He does the calculation for an accelerating observer in Minkowski spacetime, but for this particular scenario, the result is the same as for an observer hovering (and therefore accelerating) at a constant r-coordinate above a black hole horizon. (He also does the calculation for a rope being paid out from the accelerating spaceship, which corresponds better to the original post in this thread, "lowering" the pole instead of just letting it free fall.)

I agree with all of that, but nothing there actually contradicts the statement I made. The (proper) time that it takes until it's too late for me to stop the bottom of the pole fall through the horizon (which is finite) is not the same thing as the (coordinate) time it takes, in my frame, for it to actually fall through (which is infinite).

But thanks for the link. I'd recommend everyone taking part in this thread read it, as it's highly relevant.

 

[PeterDonis]

I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

You can only *measure* your own proper time, but you can *calculate* the proper time elapsed along any world line you like. And only quantities like proper time that are invariant under changes of coordinates have physical meaning; Schwarzschild coordinate time is *not* such a quantity. You're correct that the Schwarzschild coordinate time it takes for an object to free-fall through the horizon is infinite, but that fact by itself has no physical meaning.

(You could give it a kind of physical meaning by noting that Schwarzschild coordinate time is a good approximation of proper time for observers very far away from the hole, so far away that they do not fall appreciably towards the hole for the duration of your scenario. Those observers would indeed have to wait an infinite amount of time to "see" an object cross the horizon. But that does *not* mean that someone hovering close to the horizon would have to wait an infinite amount of time before the pole in free fall would either slip away from them or break when they gripped it. Nor does it mean that an observer free-falling along with the pole would have to wait an infinite amount of time before crossing the horizon. So whatever physical meaning Schwarzschild coordinate time might have isn't relevant to this scenario.)

You can use Schwarzschild coordinates to *calculate* things like the proper time elapsed along a worldline, but only outside the horizon; at the horizon, Schwarzschild coordinates are singular, so you can't calculate with them there. So if you're trying to deal with objects that are crossing the horizon, you need to find another coordinate system that is nonsingular across the horizon. (In Greg Egan's examples that I linked to, when "translated" to the black hole scenario, the relevant coordinates would be Kruskal coordinates, which are the analogue of Rindler coordinates in a black hole scenario. Kruskal coordinates are nonsingular across the horizon, so they work fine for situations like this one.)

 

[nutgeb]

Dr Greg,

The article is very interesting. The section about Eve letting Adam fall toward the horizon with a rope seems pertinent to our discussion.

It suggests to me that when we let our pole slide through the holder's grip (actually, we need to actively propel the pole at a high rate as indicted in the article), the lower end of a pole of finite length can actually cross the horizon while the upper end is still above our grip. This is true even though the pole holder does not see the lower end of the pole ever cross the horizon.

When we eventually tighten our grip, the pole will break when the tension has sufficient time to travel some distance down the length of the pole.

Do you agree?

The article suggests, as a general proposition, that Adam and Eve can (and will) disagree about whether a finite length of rope can cross the horizon. Again, this seems like a failure of simultaneity to me.

 

[PeterDonis]

The article suggests, as a general proposition, that Adam and Eve can (and will) disagree about whether a finite length of rope can cross the horizon. Again, this seems like a failure of simultaneity to me.

I agree in general with your description of what will happen to the pole. However, as a general proposition, whether or not a finite length of rope can cross the horizon is an invariant; it's a proposition about actual physical events (whether or not a given worldline crosses the horizon), and must come out the same for all observers. If Adam and Eve disagree about whether a finite length of rope can cross the horizon, one of them must be wrong.

Egan does say: "There is no finite q coordinate for any point on the Rindler horizon, which means there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon. In that sense, Eve could claim that Adam never reaches the horizon as far as she's concerned." However, he immediately follows this with: "However, not only is it clear that Adam really does cross the horizon, there is a time \tau_{crit} for Eve after which any signal she sends to Adam will reach him only after he's on the other side." For "Rindler horizon" we can substitute "black hole horizon" and the above still holds good.

In other words, Eve's claim does *not* amount to saying that Adam's worldline does not cross the horizon; she may not be able to *observe* Adam do so, but she can certainly *calculate* that he will do so, and even calculate the time \tau_{crit} after which any signal she sends him will only reach him after he crosses the horizon (so he will be unable to ever get a response back to her).

 

[JesseM]

But that doesn't make sense. I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

Why assume that the outside observer is using Schwarzschild coordinates, rather than some other system like Eddington-Finkelstein coordinates or Kruskal-Szekeres coordinates? Does the problem specify?

 

[atyy]

The Schwarzschild time coordinate is the proper time of the observer at infinity. So there's a coordinate invariant meaning for the observer at infinity if it takes infinite coordinate time for something to happen.

 

[JesseM]

The Schwarzschild time coordinate is the proper time of the observer at infinity. So there's a coordinate invariant meaning for the observer at infinity if it takes infinite coordinate time for something to happen.

I don't think that makes sense--there's a coordinate invariant meaning for anything that happens on the worldline of that observer at infinity, but coordinate times of events at finite radius don't have the same kind of physical meaning (although I think Schwarzschild coordinates are designed to have the property that if a clock hovering at a given radius is running slow by a factor of F relative to Schwarzschild coordinate time, then if it sends light signals to the observer at infinity with every tick, then the observer at infinity will also see the clock running slow by the same factor of F visually). Since all coordinate systems are permissible in general relativity, I'm sure you could design plenty of other coordinate systems which had the property that the time interval between events on the worldline of the observer at infinity would match up with his proper time, yet these coordinate systems would disagree about the time intervals between events at finite radii.

 

[PeterDonis]

Why assume that the outside observer is using Schwarzschild coordinates, rather than some other system like Eddington-Finkelstein coordinates or Kruskal-Szekeres coordinates? Does the problem specify?

A problem can't "specify" what coordinates you have to use to work it out--unless you're taking a class with a particularly evil professor... :-)

You can use any system of coordinates you want; you will still get the same answer when you calculate answers that are invariants, like whether Adam's worldline crosses the horizon. The only caveat, as I said in a previous post, is that if you're dealing with objects that cross the horizon, you have to use coordinates that are nonsingular at the horizon. That rules out Schwarzschild coordinates, but any of the other commonly used systems for black hole spacetimes would work fine.

 

[atyy]

I don't think that makes sense--there's a coordinate invariant meaning for anything that happens on the worldline of that observer at infinity, but coordinate times of events at finite radius don't have the same kind of physical meaning (although I think Schwarzschild coordinates are designed to have the property that if a clock hovering at a given radius is running slow by a factor of F relative to Schwarzschild coordinate time, then if it sends light signals to the observer at infinity with every tick, then the observer at infinity will also see the clock running slow by the same factor of F visually). Since all coordinate systems are permissible in general relativity, I'm sure you could design plenty of other coordinate systems which had the property that the time interval between events on the worldline of the observer at infinity would match up with his proper time, yet these coordinate systems would disagree about the time intervals between events at finite radii.

What I'm suggesting is that DrGreg's use of coordinate-dependent "in his frame" should be just a few steps away from a coordinate-independent "reality" such as what the observer at infinity sees - for example, using the light signal property you mentioned. You can of course calculate in any coordinate system, but the "frame of the observer at infinity" is probably short-hand for something physically meaningful for that observer, such as what he sees.

 

[PeterDonis]

(In Greg Egan's examples that I linked to, when "translated" to the black hole scenario, the relevant coordinates would be Kruskal coordinates, which are the analogue of Rindler coordinates in a black hole scenario. Kruskal coordinates are nonsingular across the horizon, so they work fine for situations like this one.)

Oops, I just realized that I misspoke a bit here. Kruskal coordinates are the analogue of *Minkowski* coordinates in flat spacetime (i.e., the usual "t, x" coordinates in an inertial frame in special relativity, in which Egan's scenarios are set). The analogy is not perfect, of course, since Minkowski coordinates "go to infinity" in all directions, whereas Kruskal coordinates stop at the singularity at r = 0. But for the scenarios we're discussing, that's not relevant.

The analogue in a black hole scenario of Rindler coordinates are actually *Schwarzschild* coordinates, but only for the portion of the spacetime exterior to the hole (just as Rindler coordinates only cover the "wedge" of Minkowski spacetime to the "right"--i.e., positive x-direction--of the null lines t = x and t = -x).

 

[xantox]

It suggests to me that when we let our pole slide through the holder's grip (actually, we need to actively propel the pole at a high rate as indicted in the article), the lower end of a pole of finite length can actually cross the horizon while the upper end is still above our grip. This is true even though the pole holder does not see the lower end of the pole ever cross the horizon.

I don't see a contradiction in respect to what DrGreg said above. If the pole is free-falling, a clock located on the pole will not measure the same thing as a clock located on the hovering ship. Now when you say "the lower end of a pole crosses the horizon while the upper end is above our grip" there is an implicit attempt to determine the simultaneity of two events. However – let's assume the clock on the ship reads noon and the upper end is still above my grip – the question "where is "now" the lower end?" is generally meaningless, as the specific answer will be arbitrarily dependent on the choice of a coordinate system. Even if we don't use (unphysical) coordinate time but clock time (ie. a parameterization on some observable quantity), it will be still an arbitrary choice. Incidentally, the lower end of the pole will cross the horizon at t = ∞ also according to the physical clock in the ship. Without contradiction with the above, it is possible to compute that on the lower end of the pole's own wristwatch a short time will elapse before it crosses the horizon. But still, this computed quantity cannot be compared to ship's clock readings so as to allow saying from the ship "now the lower end has already crossed the horizon".

 

[PeterDonis]

Now when you say "the lower end of a pole crosses the horizon while the upper end is above our grip" there is an implicit attempt to determine the simultaneity of two events. However – let's assume the clock on the ship reads noon and the upper end is still above my grip – the question "where is "now" the lower end?" is generally meaningless, as the specific answer will be arbitrarily dependent on the choice of a coordinate system.

That's true, but the choice isn't completely arbitrary. All coordinate systems are equal, but for some purposes, some are more equal than others. :-)

In the case of a free-falling pole, after a proper time has elapsed since letting go of the pole (by the hovering observer's clock) of

\tau_{crit} = ln(2) \frac{c^2}{a}

where a is the proper acceleration of the hovering observer, no force that the hovering observer can exert on the pole can stop the lower tip of the pole from falling through the horizon. So it seems reasonable to say that the lower end of the pole "has crossed the horizon" at time \tau_{crit} by the hovering observer's clock.

Also, if the proper length of the pole (i.e., its length in its own rest frame) is at least

L_{crit} = \frac{1}{4} \frac{c^2}{a}

then the upper end of the pole will still be at or above the hovering observer's hand when a proper time \tau_{crit} has elapsed by his clock since letting go. So again, it seems reasonable to say that if the pole's proper length is at least L_{crit}, then the pole will still be "passing through the observer's hands when it crosses the horizon."

I agree that you can't *force* someone to accept these definitions, but they tell you useful physical facts about the situation, so the choice to use them isn't completely arbitrary.

 

[PeterDonis]

Just to beat this topic to death a little more :-), we can also calculate the required length of the pole (for a freely falling pole) and the proper time by the hovering observer's clock, if we require that the upper end of the pole just pass through the observer's hands at the same proper time, in the *pole's* rest frame, that the lower end of the pole is just crossing the horizon.

If we write \tau_{pole} for the latter proper time (in the pole's rest frame, and taking \tau_{pole} = 0 at the moment when the pole is released (at that moment the lower end of the pole is just at the observer's hand), L_{pole} for the corresponding proper length of the pole (again in the pole's rest frame), and \tau_{obs} for the proper time in the *observer* frame at the moment the upper end of the pole passes through the observer's hand, then we have (again writing a for the proper acceleration experienced by the hovering observer):

\tau_{pole} = \frac{c^2}{a}

L_{pole} = 0.4143 \frac{c^2}{a}

\tau_{obs} = 0.8814 \frac{c^2}{a}

The above describes another reasonable candidate for the "required length of the pole so that the upper end passes through the observer's hand *at the same time* as the lower end crosses the horizon"--the difference, of course, being that here "at the same time" applies in the pole's rest frame.

Since \tau_{obs} above is greater than \tau_{crit} that we calculated before, it is already too late at this point for the hovering observer to pull back the pole in one piece. In fact, it is fairly easy to see that the maximum length of the pole that can be "saved" at this point is

L_{saved} = \frac{1}{2} L_{pole}

or just half the total length of the pole; the rest of it is doomed at this point to drop below the horizon before any force exerted at the upper end can reach it.

 

[xantox]

So it seems reasonable to say that the lower end of the pole "has crossed the horizon" at time \tau_{crit} by the hovering observer's clock.

This sounds like a metaphor – proper times over different worldlines in curved space-time cannot be compared directly, each worldine has its own private metric structure. It is just enough to say that those signals emitted when e.g. the ship clock reads 12:01 cannot reach the lower end before it will cross the horizon, without the part "the lower end of the pole has crossed the horizon at 12:01". Geometrically, the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius (or such surface could not be said to be an "event horizon").

 

[PeterDonis]

proper times over different worldlines in curved space-time cannot be compared directly, each worldine has its own private metric structure.

If you're going to take this position, then this

the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius.

is just a "metaphor" too. Since the event of the pole reaching the horizon is on a different worldline than that of the hovering observer, by your own statement above, you can't compare their proper times directly, so there is no meaning other than "metaphorical" to the statement that the pole crossing the horizon "cannot happen *after* any finite time-like interval crossed by a hovering ship"--because "after" is a comparison of proper times.

All of this is OK with me; I'm willing to admit that statements about comparing proper times along different worldlines are "metaphorical", including the ones I've made in previous posts. (It's different if the worldlines can later converge, as in the twin paradox, so that observers can actually compare their clock readings, but that isn't the case here.)

 

[xantox]

If you're going to take this position, then "the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius" is just a "metaphor" too. Since the event of the pole reaching the horizon is on a different worldline than that of the hovering observer, by your own statement above, you can't compare their proper times directly,

Here coordinate times are implicitly compared, as events are on different worldlines. The clock on the ship may be used to synchronize a shell coordinate system (locally flat) where each tick of the clock is a tick of shell time – a rescaling of Schwarzschild time. EDIT: The time coordinate in the limit of the pole horizon crossing event tends to infinity. I believe DrGreg was meaning this in comment 34. We clearly agree that this implicit choice of coordinates is arbitrary.

 

[PeterDonis]

Here coordinate times are implicitly compared, as events are on different worldlines. The clock on the ship may be used to synchronize a shell coordinate system (locally flat) where each tick of the clock is a tick of shell time – a rescaling of Schwarzschild time. The time coordinate of the pole horizon crossing event is tshell = ∞. I believe DrGreg was meaning this in comment 34. We clearly agree that this implicit choice of coordinates is arbitrary.

Yes, we do, but remember also that some choices of coordinates will not cover the entire manifold, so you have to be very careful talking about the values of those coordinates for events outside the portion of the manifold they cover. For example, the "ship coordinates" you're using do not cover the entire manifold, in either version (Rindler coordinates in Minkowski spacetime or Schwarzschild coordinates in Schwarzschild spacetime). Those coordinates become singular at the horizon: not just because the time coordinate goes to infinity, but because the metric, written in those coordinates, has a "divide by zero error" (i.e., an infinite value in the denominator of one of the metric coefficients).

So to talk about events either on or inside the horizon, you have to switch to a different coordinate system that remains nonsingular on and inside the horizon (for example, Minkowski coordinates in Minkowski spacetime or Kruskal coordinates in Schwarzschild spacetime). You can still say, if you like, that events like the pole crossing the horizon happen at "t = infinity" in the ship's frame, but the deductions you can draw from that statement, by itself, are *extremely* limited because of the coordinate singularity.

 

[xantox]

Yes, we do, but remember also that some choices of coordinates will not cover the entire manifold, so you have to be very careful talking about the values of those coordinates for events outside the portion of the manifold they cover. For example, the "ship coordinates" you're using do not cover the entire manifold, in either version (Rindler coordinates in Minkowski spacetime or Schwarzschild coordinates in Schwarzschild spacetime).

Sure, those coordinates only allow to map the limit of the horizon crossing and not the horizon crossing itself.

 

[stevebd1]

On the subject of hovering near the event horizon, the force required would be-

|F|= \frac{Gm}{r^2} \left(\sqrt{1-\frac{2M}{r}\right)^{-1/2}

where M= Gm/c^2

As hovering in a gravitational field is synonymous with accelerating in flat space, this means regardless of if you’d found a BH where the tidal forces won’t kill you (for BH >4.6e+4 sol, tidal forces at EH are <1g), if you were to ‘become stationary’ anywhere near the EH, the simple force of gravity (or g-force) would.

For example, an intermediate static BH with a mass of 6e+5 sol and a radius (2M) of 1.77e+6 km, the closest coordinate radius you could hover at comfortably (g=Earth g) would be at r=2.85e+9 km (~3200M), nearly 3 billion km from the black hole! If we consider 10g being the maximum a human could tolerate for a period of time, then you could hover at r=9e+8 km (~1000M). Some military electrical equipment can withstand 15,000g, say we have electrical equipment that can withstand 100,000g, the closest this equipment would get is approx. r=9.45e+6 km (~11M).

This is backed up to some extent by the following web page-
http://www.aei.mpg.de/einsteinOnline/en/spotlights/descent_bh/index.html