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Black hole horizon puzzle

  1. Feb 18, 2013 #1

    PeterDonis

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    This is a puzzle based on a purported "refutation of GR" on this webpage, which is also the subject of a subthread in this Hacker News discussion. (Btw, the Hacker News discussion contains my answer to the puzzle if you scroll down enough, so you're on your honor not to look at it until you've tried to answer the puzzle yourself. :wink:) I emphasize that I am not saying this puzzle actually does "refute" GR; of course it doesn't. But it does pose an interesting question about how inertial frames centered on the black hole's horizon work, and I would like to see how people respond to it.

    Here's the puzzle: an astronaut free-falls radially into a black hole. Just before he reaches the horizon, he launches a probe radially outward at just over the escape velocity at the launch altitude (which will be a bit less than the speed of light because the launch point is above the horizon). Just after he crosses the horizon, he launches a second probe radially outward at a slightly larger speed, relative to him, than the first probe. For concreteness, suppose the first probe is launched at a speed [itex]( 1 - \epsilon ) c[/itex] relative to the astronaut, and the second probe is launched at a speed [itex]( 1 - 1/2 \epsilon ) c[/itex] relative to the astronaut, where [itex]\epsilon << 1[/itex].

    Now, it seems obvious that these two probes should separate, because the one launched above the horizon will increase its radius with time (since it's moving at greater than escape velocity), while the one below the horizon will decrease its radius with time (since every object below the horizon does). However, if we look at this scenario in a local inertial frame whose origin is the event of the astronaut crossing the horizon, and in which the astronaut is at rest, we should see the two probes moving closer together! This is because the second probe is moving at closer to the speed of light, relative to the astronaut, than the first probe, so in a local inertial frame in which the astronaut is at rest, the second probe should catch up to the first; with the given speeds above, it should catch up at a rate [itex]1/2 \epsilon c[/itex]. What gives?

    Btw, I should also emphasize that "tidal gravity" is *not* the answer. Assume the hole's mass is large enough that tidal gravity is negligible; that is, assume that we can set up a local inertial frame of sufficient size to realize the given scenario. (It's fairly easy to show mathematically that this is possible for a hole with large enough mass.)

    Comments welcome. :wink:
     
    Last edited: Feb 18, 2013
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  3. Feb 18, 2013 #2

    atyy

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    My guess: not every object below the event horizon always decreases its radius regardless of its initial speed and direction.

    "We should see" presumably refers to the free falling astronaut.
     
  4. Feb 18, 2013 #3
    I'm just an undergrad who's not yet very familiar with GR, but wouldn't the observer outside the event horizon never see the second probe get launched? I'm sure that's not the explanation here, but is that not correct?
     
  5. Feb 18, 2013 #4

    George Jones

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    No, this isn't true. Any tangent space, even inside the event horizon, is a Minkowski vector space. In what follows, I use a metric signature of (1, -1, -1, -1).

    In any Minkowski vector space, the Minkowski inner product of any two future-directed timelike vectors is positive. Inside the event horizon, [itex]-\partial / \partial r[/itex] is a future-directed timelike vector. The tangent vector [itex]u[/itex] to any observer worldline is future-directed timelike, and has, for constant theta and phi, coordinate representation

    [tex]
    u= \frac{dt}{d\tau} \frac{\partial}{\partial t} + \frac{dr}{d\tau} \frac{\partial}{\partial r}.
    [/tex]

    Consequently,

    [tex]
    0 < g \left(u, - \frac{\partial}{\partial r} \right) = - \left( 1 - \frac{2M}{r} \right)^{-1} \frac{dr}{d\tau},
    [/tex]

    so [itex] dr/d\tau[/itex] must be negative, i.e., [itex] r [/itex] must decrease.
     
  6. Feb 18, 2013 #5

    PeterDonis

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    It is correct that an observer outside the horizon won't see the second probe get launched (because light from that event will never get outside the horizon). It's also correct that that is not the explanation of the puzzle.
     
  7. Feb 18, 2013 #6

    PeterDonis

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    Yes; more precisely, it means "with respect to the free-falling astronaut's local inertial frame whose origin is the event at which he crosses the horizon".
     
  8. Feb 18, 2013 #7
    Is it that the length contraction and time dilation of the object inside the event horizon (if that's possible) would be greater than the length contraction and time dilation of the object outside the horizon so the observer would see the two objects (if it were possible to see an object reaching the horizon) seperate?
     
  9. Feb 18, 2013 #8
    Hi PeterDonis,
    Thanks for the puzzle, so i won't read the thread you link at, and it will probably be obvious since there I will probably won't get it.
    My take on it is that, there is no real 'contradiction'.
    I suppose that the ship sending a probe after passing the horizon will have no problem seeing it catching the other one, because, it will happen in his own frame of reference, according to his own time.
    As far as an outsider observer is concerned, it will take forever for the ship to even get into the horizon, and there is no way this outside observer would know what happens 'after infinity'

    Am i on the right track ?
     
  10. Feb 18, 2013 #9

    Dale

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    Interesting puzzle. I don't know the answer, but I am willing to take a guess.

    I suspect that it has to do with simultaneity. Events on the two worldlines are spacelike separated, so the order depends on your simultaneity convention. Under the infalling convention they are getting closer and under the outgoing convention they are getting further.
     
  11. Feb 18, 2013 #10

    PeterDonis

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    Length contraction and time dilation aren't involved. Remember that we are viewing things from within a single local inertial frame, and all speeds, distances, and times are relative to that inertial frame. And relative to that inertial frame, the second probe (the one launched inside the horizon) is moving outward faster than the first (the one launched outside the horizon), so the two probes will move closer together. In principle this could be verified by light signals sent from each probe to the infalling astronaut (or by the second probe measuring the Doppler shift of light signals from the first--see my response to DaleSpam below).

    That's right; there isn't.

    Exactly; this is basically what I was saying in response to SpinAnalyser above.

    That's true, but it's not relevant. We're not talking about an outside observer; we're talking about the infalling observer who launches the two probes, and can verify by measurement that within his own local inertial frame, the two probes move closer to each other, even though with respect to the global radial coordinate r, they are separating. All of the light signals the infalling observer receives that carry this information reach him below the horizon, so the fact that an observer outside the horizon will never see them is irrelevant.

    (Also see my response to DaleSpam below; there is another invariant sense in which the two probes are moving closer together.)

    This is kind of on the right track. Under the simultaneity convention of the local inertial frame, the two worldlines are converging. There is another simultaneity convention according to which they are diverging, but it's not quite an "outgoing" one.

    However, I can sharpen the puzzle by giving an invariant (i.e., independent of simultaneity convention) sense in which the two worldlines are converging: if the second probe measures the Doppler shift of light signals sent radially inward by the first probe, it will measure a Doppler blueshift (at least, it will while it is within the confines of the given local inertial frame).

    The Doppler shift observation brings up another point. The "simultaneity" explanation by itself still leaves something fishy: in flat spacetime, it's impossible to reverse the sign of the relative velocity of two objects by a Lorentz transformation (because doing so would amount to changing a direct observable, the Doppler shift, by a Lorentz transformation). This means it's impossible to do so by changing simultaneity conventions. Yet somehow we seem to be saying that we *can* reverse the sign of the relative velocity of the two probes by changing simultaneity conventions. How is that possible?

    One other hint that may help: what do curves of constant radial coordinate r look like in the local inertial frame?
     
  12. Feb 18, 2013 #11

    Bill_K

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    Hyperbolas?
     
  13. Feb 18, 2013 #12

    Dale

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    Of course. It is the Schwarzschild simultaneity convention. In the limit of a supermassive black hole the flat spacetime at the horizon would use a simultaneity convention more like Rindler simultaneity, not Minkowski simultaneity.
     
  14. Feb 18, 2013 #13

    Ben Niehoff

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    Let me see if I can answer it without actually doing a calculation...

    First of all, the fact of whether or not two worldlines intersect is absolute. The intersection point is a single event, and therefore invariant under all coordinate transformations. We know an outside observer cannot observe the worldlines intersecting (in fact, if you draw them in an Eddington-Finkelstein diagram, they diverge).

    Nonetheless, the infalling observer must see that they "appear" to converge. To discuss appearances, one must trace light paths. Peter also brings up the fact that the probes could measure Doppler shift between them and transmit that information down to the infalling observer; and the probes must measure blueshift. Hence, something must cause blueshift between the probes, and the probes must appear to infalling observer that they are approaching one another. But they must never actually collide, because we know that does not happen.

    The way out of the paradox is that the infalling observer must reach the singularity before the probes appear to collide. In fact, one thing we know about appearances from the perspective of infalling observers is that the black disc ahead of them appears to grow and surround them in a black sphere, while all the light of the universe gets squeezed into an ever-smaller light disc behind them. At the moment the observer reaches the singularity, the entire visible universe collapses to a single point behind them. My guess is that this is the exact moment when the probes finally "collide". I suspect that doing some ray-tracing on an Eddington-Finkelstein diagram will reveal this is the case.

    Now we just need to explain the blueshift that occurs before this moment. For the second probe (inside the black hole), this is easy, because all exterior light is blueshifted as it falls toward a black hole.

    What is difficult to explain is that light from the interior probe never reaches the exterior probe, even if we have a supermassive black hole so that spacetime at the event horizon is arbitrarily flat. To prove this may need a precise calculation, taking the limit as ##\epsilon \to 0##. The outgoing light from the interior travels at c, of course, and is emitted arbitrarily close to the horizon; but meanwhile as we get arbitrarily close to the horizon, the exterior probe travels arbitrarily closer to c as well. I suspect these competing effects result in a projected "collision time" between the outgoing light and the exterior probe that always manages to fall outside the region that can be considered approximately flat, thereby invalidating the inertial frame approximation.
     
  15. Feb 18, 2013 #14
    Alright PeterDonis thanks for confirming I didn't get it right :)
    I saw DaleSpam answer before yours, but I thought we were saying the same thing, except I was going to infinity with the same argument.
    So since my making a difference outside/inside the horizon is irrelevant, I look again with more attention at your puzzle, and, then I see I made the assumption that the contradiction was about being inside or outside the horizon, but, you talk about the reference frame of an inertial frame being exactly at the horizon (? is this correct ?)
    If that is correct, well (I'm just asking to make sure I'm not chasing a ghost), I'd argue that being on the edge of the horizon is not even possible, it must be an unstable orbit (just guessing) and you would have to go exactly at the speed of light in a very precise direction, otherwise you would go in, or out.
    Supposing you do stay in this stable orbit, you are still, instead of needing to see something 'after infinity', only able to see it exactly 'at infinity' (of course since you travel at the speed of light, time does not tick, so I don't know what infinite time could be like for you)
    Cheers...
     
  16. Feb 18, 2013 #15
    Oh I see, it's already been taken into account and is part of the observers frame.

    Reach the singularity? How? It's a singularity. It exists in an infinitely small portion of space-time so how could anything possibly reach the singularity before it's gone, or are they infinitely small in time but not space? That doesn't seem right.
     
  17. Feb 18, 2013 #16

    PeterDonis

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    This wasn't the solution I had in mind, and I don't think it will work as a solution to the puzzle. For one thing, the optical effects you are talking about include the angular coordinates, but the puzzle as I posed it involves only radial motion; everything takes place at the same constant values of theta, phi. (See the end of this post for another point about whether anything needs to reach the singularity.)

    Also, bear in mind that the puzzle is set in a single local inertial frame; in such a frame, effects due to spacetime curvature are not detectable. The optical effects you are talking about are due to spacetime curvature. (Bear in mind also that the puzzle does not say the two probes collide in the local inertial frame; it only says they approach each other, even though the first probe is increasing its radial coordinate and the second is decreasing it.)

    This is also due to spacetime curvature, so it won't work as a solution. Besides, consider the following addition to the puzzle: in the local inertial frame I defined, the infalling astronaut could launch a third probe at exactly the same event as he launches the second one, but with an outward velocity of only [itex]( 1 - 2 \epsilon ) c[/itex] relative to him. This third probe would see a Doppler redshift from the first probe (the one launched outside the horizon). The infalling astronaut himself, of course, will also see a Doppler redshift from the first probe, at least within the local inertial frame (since the probe is moving away from him).

    It's true that the projected "collision time" between the two probes (or between outgoing light emitted by the second probe and the first probe) is outside the range covered by the local inertial frame, yes. It's fairly simple to show that this must be the case, using the relative velocities I gave. That's part of the solution, but it still doesn't explain how to reconcile the fact that the probes are approaching each other with the behavior of their respective radial coordinates. (Also, proving this fact about the projected "collision time" does not involve any requirement that anything reach the singularity before the projected "collision time".)
     
  18. Feb 18, 2013 #17

    PeterDonis

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    Yes, but remember that we're talking about an inertial frame--a small patch of *spacetime* described by ordinary Minkowski coordinates. The time axis of this frame is the infalling astronaut's worldline (or the small segment of it that lies within the patch of spacetime we're talking about), and the origin of the frame (time zero, spatial coordinate zero) is the event at which the astronaut crosses the horizon (as I've said, we're dealing with purely radial motion, so we can ignore the other two spatial coordinates--the positive x direction in this frame points radially outward). So this is obviously not a frame that is stationary at a constant altitude; it is an infalling frame (objects at rest in it are falling into the hole). The frame is only "at the horizon" for an instant, the instant "time = zero" in the frame. The horizon itself, in this frame, is a 45-degree line going up and to the right, passing through the origin (since the origin is the event where the astronaut crosses the horizon).
     
  19. Feb 18, 2013 #18

    Ben Niehoff

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    This is a red herring. The answer to the puzzle must lie in curvature effects, because if we could truly ignore all curvature effects, then the axioms of the local inertial frame would demand that the probes collide.

    So the answer depends on somehow showing that the local inertial frame approximation is not valid at the projected point of collision.

    Fine, but for this hypothetical third probe, there are competing effects: the probe's negative relative velocity compared to the first probe, and the blueshift due to infalling radiation from the first probe. Answering the question of which effect wins out requires calculation, but it should not be surprising that the answer meets the expectations of the local inertial frame where that approximation is valid.

    I wouldn't put too much stock in the r coordinate, especially near the horizon. On the interior, the r coordinate is timelike. At the horizon itself, the standard Schwarzschild coordinates are degenerate, so they won't correspond in any useful way to the axes of a local inertial frame.

    Frankly, I think to resolve the paradox it is enough to show that the projected "collision event" happens outside the region where the flat space approximation is valid. Therefore the question of whether a collision actually occurs is subject to curvature effects. That alone addresses the conceptual errors made by whoever came up with this paradox.

    To address the issue of blueshift is just a matter of doing a straightforward calculation (which I am too lazy to do at the moment).

    Edit: I should also add that "The observer sees the difference in r coordinate between Probe A and Probe B decreasing" is not a physical statement at all, and certainly does not mean the same thing as "The observer sees the distance between Probe A and Probe B decreasing" (where "distance" can be defined as "along curves of simultaneity"). As for what the observer actually does see, I should expect that initially, the distance between the probes is decreasing, but it should actually reach a turnaround point and start increasing. After all, the observer has tossed one ball up at escape velocity, and then a second ball at less than escape velocity. The second ball must return.
     
    Last edited: Feb 18, 2013
  20. Feb 18, 2013 #19

    PeterDonis

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    Ah, ok, I wasn't being specific enough. Yes, you're right, explaining why the probes don't collide does require bringing in curvature. But explaining why the two probes are moving closer to each other in the local inertial frame despite the fact that their r coordinates are diverging does not.

    Not in the local inertial frame. We can set up the numbers so that the second and third probes both receive ingoing light signals from the first probe while they are still well within the local inertial frame--i.e., well before curvature effects become detectable. So curvature effects can't explain why the third probe starts out seeing a Doppler redshift while the second sees a Doppler blueshift. The Doppler shifts the probes see immediately upon being launched ("immediately" meaning "for a short enough time that curvature effects are negligible") can only be due to their different velocities relative to the local inertial frame (meaning, relative to the infalling astronaut that launches them).

    Yes, I would expect a calculation using global coordinates to give the same answer; but as I said above, we can set things up so any curvature terms that arise in the global coordinates are negligible at the order of approximation used.

    Yes, exactly; that's one of the key points I was trying to get at. Viewing the r coordinate as an ordinary "spatial coordinate" has issues when you're dealing with events on or inside the horizon. But that doesn't mean we can just discard it.

    For one thing, the statement "the r coordinate is timelike" is coordinate-dependent. [itex]\partial / \partial r[/itex] is timelike inside the horizon in the Schwarzschild chart, but it is spacelike everywhere in the Painleve chart (and the ingoing Eddington-Finkelstein chart). So if we use, for example, the Painleve chart, we can construct a (curvilinear) chart covering the same patch of spacetime as the local inertial frame, in which the spatial distance between the first and second probes is strictly increasing with time (since spatial distance corresponds exactly to the r coordinate in this chart).

    For another thing, the r coordinate does have a physical meaning. See further comments below.

    I have to disagree here. The r coordinate has a physical meaning: the surface area of the 2-sphere labeled by r is 4 pi r^2. That's an invariant physical statement. The fact that the surface areas of the 2-spheres seen by the second and third probes are strictly decreasing, while the surface areas of the 2-spheres seen by the first probe are strictly increasing, is also an invariant physical statement.

    Perhaps that's an even sharper way of stating the puzzle: the second probe starts out seeing a Doppler blueshift from the first, even though the surface areas of the 2-spheres seen by the two probes are changing in opposite directions from the start.

    Yes, I agree that's what will happen, and it's part of the resolution of the puzzle: as you said above, we can't extend the predicted trajectories in the local inertial frame indefinitely because of curvature effects.
     
  21. Feb 18, 2013 #20

    Ben Niehoff

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    In the LIF, probe B is closing distance to probe A, and probe C is receding from probe A, so I don't see what the great mystery is here.

    The fact that other coordinate charts have coordinates named 'r' does not change the fact that I was talking about the coordinate named 'r' in the Schwarzschild chart.

    In the Schwarzschild chart. Shall I bring up all the other charts that use a coordinate named 'r' to debunk this notion?

    Not a very useful one, for two reasons: 1. The area of a sphere around the black hole is not a local observable, and 2. In the region where curvature effects are not negligible, the area of a spherical shell around the black hole has nothing to do with the distance between spherical shells (well, there is a formula that relates them, but it is not the naive formula).

    Of course, the distance between two probes is also not a local observable, except in the region where the LIF approximation holds. Outside that region, one has to use light signals. Or at least curves of simultaneity, to discuss what one might hypothetically conclude if one could receive light signals.

    It might actually be easiest to look at a Penrose diagram. Since a Penrose diagram is conformal, it is easy to check that the initial velocities are indeed converging, but must turn and diverge at some point.
     
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