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Steel bar bisected by event horizon

  1. Dec 17, 2009 #1

    ACG

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    Hi!

    There's got to be something wrong with the following method to take photographs of the area inside a supermassive black hole, but I can't figure it out.

    1. Lower your spacecraft to just outside the event horizon.
    2. Extend a uniform steel bar with cameras embedded in it through the event horizon. The bar's central of mass remains outside the event horizon.
    3. Take a photograph from the cameras inside the event horizon.
    4. Pull the steel bar back into your ship and take off.

    The question is: if there's a flaw in this plan, it's got to be a situation where the bar breaks somewhere around the event horizon so that the part inside the horizon stays in.

    Except for one thing. What would cause the bar to break? It can't be tides, because tides are minimal outside a supermassive black hole. Yes, in theory the stuff inside the black hole is going to be pulled in. However, this is ignoring the material strength (tensile strength or something like that -- I can't remember what it is) of the bar. You need a certain amount of stress on the bar to break it, and I'd expect that downward force by spacetime tugging the bar in could be overcome by the tensile strength of the bar.

    What am I missing?

    Thanks in advance,

    ACG
     
  2. jcsd
  3. Dec 17, 2009 #2
    Nope. I'm sure others could express it more precisely but the whole point of gravitational collapse into a singularity - the process that creates the black hole - is that no known interaction or phenomenon can resist it at the event horizon, including the electromagnetic force (which tensile strength fundamentally is). Put simply, nothing's strong enough to do it.
     
  4. Dec 17, 2009 #3

    sylas

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    Right there the forces required are going to be enormous. To hover in place or even to remain in a circular orbit at a given distance above the event horizon requires continuous forces that diverge to infinite as you approach the horizon. Those forces are what the tensile force on your rod experiences; they go to infinite as the rod approaches the horizon.

    Cheers -- sylas
     
  5. Dec 17, 2009 #4

    JesseM

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    One way to think of it is in terms of the fact that from the perspective of a local inertial frame near the horizon (see the http://www.einstein-online.info/en/spotlights/equivalence_principle/index.html [Broken] if you're not clear what this means), the event horizon is actually moving outwards at the speed of light, which shows why you can't get back out once you've crossed it. The perspective of an observer hovering at a constant radius above the event horizon is closely analogous to that of an accelerating observer in flat spacetime who sees a Rindler horizon which he's at a constant distance from in Rindler coordinates--from the perspective of an inertial frame, the Rindler horizon is just the surface of a light cone, moving outward at the speed of light (see the second diagram on the Rindler horizon page to see how both the Rindler horizon and the accelerating observers at constant positions in Rindler coordinates would look when their worldlines are plotted in an inertial frame). So what would happen if one of these accelerating observers--who accelerate in such a way that the Rindler horizon never catches up to them, even though the distance between them and the horizon is constantly decreasing from the perspective of an inertial frame--were to dip a bar down so it crossed the Rindler horizon? Well, if you think in terms of the diagram of what's going on from the perspective of an inertial frame, there's obviously no way he could pull it back, since the end that crossed the horizon would have to move faster than light to cross back over the horizon onto the observer's side.

    Also, note that if we compare the Rindler coordinate view of this accelerating observer & Rindler horizon vs. the inertial frame's view of the accelerating observer & Rindler horizon, this is very closely analogous to comparing the Schwarzschild coordinate view of an observer hovering at constant Schwarzschild radius & the event horizon vs. the view of the same pair when seen in Kruskal-Szekeres coordinates--in Kruskal-Szekeres coordinates the observer's worldline looks like a hyperbola and the event horizon is moving outward at the same coordinate speed as light. See my post #4 on this thread for a quick description of how Kruskal-Szekeres coordinates work, and see the bottom half of this page for a discussion of different coordinate systems that can be used to describe a nonrotating black hole.
     
    Last edited by a moderator: May 4, 2017
  6. Dec 17, 2009 #5

    bcrowell

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    There are straightforward ways of seeing that there are limits on the strengths of materials in SR, without even worrying about GR. For example, if a rod was sufficiently rigid, then vibrations would propagate along it at greater than c. Since c is the maximum speed of cause and effect in SR, the conclusion is that a rod that rigid can't exist.
     
  7. Dec 17, 2009 #6
    If one end of the steel bar is pointed toward the black hole as it approaches, eventually the difference in gravitational forces from one end to the other will pull it apart, even before it reaches the horizon.
    Bob S
     
  8. Dec 17, 2009 #7
    Not only will it spaghettify the rod the gravity will become infinite pass the event horizon causing it to accelerate to the singularity near c also as the bar gets closer the event horizon the bar will try to orbit faster than the spacecraft making it break off.
     
  9. Dec 17, 2009 #8

    DaveC426913

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    He addressed this. Around a supermassive BH, the tides are low.



    You're right - it's not the tides. Tides are the gradient (i.e. change) of the gravitational force. The gravitational force just outside the supermassive BH doesn't change much - but that doesn't mean it isn't extremely large.

    The bar breaks for the very simple reason that it is suspended under the ship in a gravity field that is millions of g's.

    For example: the ship is hovering (it is not in orbit) in the gravity field where g is - let's say 1,000,000. The bar is extended out the bottom to a place where g is still 1,000,000. (i.e. let's set tides to zero.) But the steel bar still weighs several million kgs.
     
    Last edited: Dec 17, 2009
  10. Dec 18, 2009 #9

    Ich

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    The "g-Force" at the Horizon actually becomes infinite, even if tides are negligible. There's a difference between free falling and static frames.
     
  11. Dec 18, 2009 #10

    Dale

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    Even in a uniform gravitational field (no tidal forces) there will be tension in a hanging bar. You can see this from a simple free-body diagram, take any section of the bar there are three forces acting on it, the tension from above, the tension from below, and the weight. Since the bar is static the tension from above must equal the sum of the tension from below and the weight. Thus the tension increases as you go up the bar. The tension will be unbounded at the top end as the bottom end gets close to the horizon of a supermassive black hole.

    Tidal forces become important for considering stresses in a free-falling bar, not a stationary bar. In the case of a bar free falling through the event horizon of a supermassive black hole there would be no significant tidal forces and it would fall through without incident.
     
  12. Dec 18, 2009 #11

    xantox

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    It's also the tides, as they also diverge in the limit of the horizon for hovering bodies.
     
    Last edited: Dec 18, 2009
  13. Dec 18, 2009 #12

    Dale

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    No, the tidal force is proportional to M/r³, and since r for the event horizon is proportional to M we get that the tidal force at the event horizon is proportional to 1/M². So the larger the mass the lower the tidal force at the event horizon and for a sufficiently massive black hole the tidal force becomes negligible.
     
  14. Dec 18, 2009 #13

    xantox

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    That is only the case for free-falling bodies crossing the horizon, certainly not for hovering bodies. Since their proper acceleration diverges over a finite distance, the acceleration gradient shall also diverge near the horizon – for black holes of any mass.
     
  15. Dec 18, 2009 #14

    Ich

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    It's a matter of definition. While tidal acceleration is strongly tied to curvature and thus finite, some components may diverge in static coordinates.
    If you say that tides are da/dr, they become infinite for a static observer.
     
  16. Dec 18, 2009 #15

    bcrowell

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    Yeah, I think xantox is correct except for a nonstandard use of terminology. The singularity in the Schwarzschild metric at the event horizon is just a coordinate singularity. I think the standard definition of "tides" in GR is that you measure them using the Weyl tensor, and the Weyl tensor doesn't diverge at the event horizon.
     
  17. Dec 18, 2009 #16

    Dale

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    I agree. I don't know what the gradient of the proper acceleration of a hovering body is usually called, but I don't think that is what people usually mean when they refer to tidal effects in GR.
     
  18. Dec 18, 2009 #17

    xantox

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    If we talk about the tidal tensor then it's the same for the free-falling and the hovering observers. However if we talk about stresses felt by the body, then we have:
    • free falling observer:
      • weight is zero,
      • tidal forces are small enough for high enough radius.
    • hovering observer:
      • weight diverges when approaching the horizon,
      • static tidal forces grow without limit when approaching the horizon.

    Both are significant in the detailed calculation of how the hovering body will break and into how many pieces. Perhaps we can call them "pseudo tides" but they need to be mentioned in some way, as the only mention of small gravitational tides near a supermassive black hole horizon can mislead into thinking the acceleration gradients felt by the hovering body are small.
     
    Last edited: Dec 19, 2009
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