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The effect of external masses on internal black hole geometry

  1. Dec 6, 2011 #1
    Suppose you have the simplest type of black hole - time independent, no angular momentum or charge, Schwarzschild solution, then you modify the situation by adding a stationary mass outside the event horizon (I imagined lowering this slowly on an idealised string).

    The question is, does this affect the geometry inside the black hole as well as outside?

    My original argument that it didn't relied on my belief that it was impossible to communicate with points arbitrarily close to the event horizon in finite time, which I inferred from the fact that to a remote observer anything appears to take infinite time to reach the event horizon. So if you can't communicate with this thin region around the event horizon, surely you cannot affect anything further in, and presumably that would even include the space time geometry for the entire region inside the event horizon.

    Having come to this conclusion, I considered colliding black holes and concluded that, to a remote observer, the geometry from the event horizon down would be indistinguishable from spherically symmetric Schwarzschild right up to the time the event horizons collided. Although this is not what one sees in simulations, this could be because there is inevitable distortion to display things on a plane: the region between the black holes would be very severely warped.

    But perhaps I am wrong, and the geometry of the interior of black holes is affected by masses outside. What do the experts say? And if I am wrong, where does the communication argument fall down? Can quantum effects break it? [My argument probably falls down when the thickness of the region around the event horizon is very tiny, as quantum effects probably matter. Especially when it is less than the Planck distance.]
     
    Last edited: Dec 6, 2011
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  3. Dec 6, 2011 #2

    PeterDonis

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    This is not quite correct. It is correct that, as we consider points closer and closer to the horizon, it takes light signals longer and longer to get back out to the remote observer; that is why things appear to take "infinite time" to actually reach the horizon.

    But that only applies to *outgoing* light signals, not to *ingoing* light signals, or indeed to ingoing objects in general. You can fall into the hole in a finite time by your own clock (the time to fall by your clock is a function of the height at which you start to fall, divided by the mass of the hole). Light can also "fall" into the hole the same way. So there is nothing stopping the causal influence of the second massive object from getting inside the black hole to influence its interior geometry.
     
  4. Dec 6, 2011 #3
    I understand what you mean, although in fact both ingoing and outgoing light appear to take an infinite time to traverse a path to the event horizon (the path time is not dependent on the direction). More precisely, the time take to get to points near the black hole and back increase to infinity as the point approaches the event horizon.

    For the relevance, bear in mind that gravity is a kind of signal limited by the speed of light.

    My assertion was about the gravitational signal that gets to a remote observer from the black hole itself through any observation affected by it. Perhaps I did not express it well, but I was asserting that a remote observer could never see anything that suggested that the geometry of the black hole was anything other than Schwarzschild from the event horizon down, even if there was a heavy object near it.

    But if this is true, surely the argument would apply even to a second black hole, as long as the event horizons were separate?
     
  5. Dec 6, 2011 #4

    PAllen

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    This is not correct for an infalling observer. The infalling observer, even past the event horizon, can still receive light signals from outside. The outside observer never sees them cross the horizon, but the infalling observer actually receives them (and takes the message to their grave at the singularity).
     
  6. Dec 6, 2011 #5

    PeterDonis

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    The "and back" is critical. For something outside the hole to affect the geometry inside the hole, the signal only has to go "down"; it doesn't have to come back "up".

    But you have to also ask the question, how did the heavy object get there in the first place?

    To see what I mean, consider first the formation of a single black hole by the collapse of a massive object. People often ask, how does the gravity get out of the black hole, since the event horizon prevents any signals from getting out? But the "gravity" you feel when you are outside the horizon doesn't come from inside the horizon; it comes from the massive object that originally collapsed to form the hole. As the massive object collapses, it leaves an "imprint" on the spacetime, and the imprint is static at a given radius once the object has collapsed inside that radius. (We're assuming an idealized spherically symmetric collapse; in the general case the imprint won't be exactly static, but it will still be there.) The imprint is what we feel as the "gravity" of the black hole. So if you were sitting stationary at some radius and saw the massive object collapsing past you, you would see the spacetime around you change; in the idealized case of a collapsing spherical shell of matter, you would see the "force of gravity" in your vicinity gradually grow as more of the matter collapsed below you, until finally, when it was all beneath you, you would feel the full "gravity" of the object.

    Similarly, if a second heavy object falls towards the hole, it will have to fall past you, and as it does, you will see the "force of gravity" around you change again due to the heavy object's influence. Once the heavy object is below you, its "gravity" will merge with that of the black hole as far as you are concerned. (It won't be exactly like the spherically symmetric case since the heavy object is coming from a specific direction, but again, the general principle is the same.) But the merging is because of the effect of the heavy object as it passed you, not because its gravity has to "get out" to you again from at or near the event horizon.
     
  7. Dec 6, 2011 #6
    But for anyone outside to see an effect something has to come back (a gravitational signal).

    Yes. Might it be reasonable to say that everything from the event horizon down does not exist to any observer that does not fall in (it is not even finitely far in the future)? All that is observable is outside this. I assumed we started with a Schwarzschild black hole but this is not something that can really be created in finite time, so the problem is theoretical.

    This all makes sense. I was only interested in effects that result from some sort of deformation of the black hole, not from the direct effect of the second mass on the gravitational field.

    Here's a more extreme question. If you have two Schwarzschild black holes (or something indistinguishable formed from some mass a very long time ago - spherically symmetric, no angular momentum and nothing visible except maybe a tad of Hawking radiation :-) ). These two black hole fall towards each other along a common axis (so still no angular momentum). If you enter one of the black holes and fall to its centre, does your proper time of fall to the singularity depend on what point you enter the black hole?

    This is a different way of asking whether the black holes distort each others geometry.

    The original question that led to me opening this discussion was a disagreement about whether an external mass made the event horizon of a Schwarzschild black hole non-spherical. Unfortunately that was not well-enough expressed to be unambiguous, and it seems difficult to turn it into a valid, unambiguous question. Any thoughts?
     
  8. Dec 6, 2011 #7

    PeterDonis

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    No, it doesn't. The gravitational "signal" comes from the "imprint" left on the spacetime where you are as mass falls past you, not from the region of spacetime near the horizon. When you feel the gravity of the hole, you are not feeling the spacetime geometry inside the horizon. So if you don't feel it to begin with, you don't feel it after the second heavy object has fallen in. An observer *inside* the horizon would see changes in the geometry there, but you, as a remote observer far away from the hole, do not.

    You'll find that this is a very touchy question around here. :wink:

    The word "exist", IMO, is very problematical. What happens inside the event horizon can't affect you, causally, when you are outside it. That's the physics. Whether that means the spacetime inside the horizon doesn't "exist for you" depends on what you want the word "exist" to mean. Some people (like me) don't like the idea of "existence" being observer-dependent. Others don't seem to mind it. But you can state all the physics without ever using the word "exist", so it's a matter of personal preference, IMO, whether you want to use that word or not in this connection.

    Not an "eternal" Schwarzschild black hole, no. But there are valid solutions of the Einstein Field Equation that consist of a collapsing region of matter matched to a vacuum Schwarzschild solution to the "future" of the collapsing matter. The classic 1939 Oppenheimer-Snyder paper presented such a solution.

    "Deformations of the hole" would cause emission of gravitational waves, which could be detected at a distance. Kip Thorne's book Black Holes and Time Warps has a good discussion of this.

    By "at what point", do you mean "from what direction in space", or "at what point in time relative to the coalescence of the black holes"?

    In either case I'm not sure there's an easy analytical answer; I think you would need a detailed numerical simulation. There's no analytical solution known for the case of two black holes falling into each other; that problem is studied numerically.

    Again, in general I think numerical simulations are needed to answer this. I suspect the general answer is "yes, but only temporarily". In the book by Thorne I referred to, he talks about what the numerical simulations indicate as far as we know now: a large mass falling into a black hole causes the hole to pulsate, something like a bell ringing, but the pulsations cause emission of gravitational waves, which continues until the hole has settled back down to a spherical state again (with a larger mass). Two holes coalescing form a larger hole which, in the idealized case where the mutual angular momentum is zero, would again cause the larger hole to pulsate and emit gravitational waves until it settled back down to a spherical state. In the more general case where the two holes (or the large mass plus the hole it falls into) have nonzero mutual angular momentum, the final hole will be spinning, so it won't be spherical; but it will still radiate gravitational waves until it has settled down to a stationary spinning state (described by the Kerr geometry instead of the Schwarzschild geometry).
     
  9. Dec 6, 2011 #8

    PAllen

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    I found these for another thread. They are the state of the art of numerical GR. Note, there are no causality violations here. Inside the colliding black holes singularities merge. This has no causal effect outside. However, outside observers see (near) horizon surfaces merge, ring down (vibrate) and eventually settle to larger spherical (unless there is too much net spin - then the result is not spherical).

    From looking at the papers noted in the simulations, you can find that in inspiral situations, up to 5 percent of the total mass equivalent of the two black holes gets radiated as gravitational radiation. You can think of this coming from release of gravitational potential energy and angular momentum during inspiral. It does not somehow come from inside the horizon.

    http://www.black-holes.org/explore2.html
     
  10. Dec 7, 2011 #9
    The problem I am particularly interested in is rather simple as problems in general relativity go. A time-independent solution for a black hole with no angular momentum plus a mass suspended near the black hole (imagine it being attached to a string that keeps it at a constant distance from the black hole). Does that make sense?

    The more specific question was about the time taken to fall from different points on the event horizon of the black hole to the singularity in this situation (this is a meaningful way to ask if the external mass disturbs the internal spherical symmetry of the black hole itself). [Better to look at this problem than the case where there are two black holes, where time dependence is inevitable].

    I have realised that this problem could be transformed to a 2-dimensional problem by reparametrisation, because of the time-independence and the axial symmetry. So even if not solvable analytically, it is very accessible to numerical analysis. I would be very surprised if someone had not done this.
     
    Last edited: Dec 7, 2011
  11. Dec 7, 2011 #10

    PAllen

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    There are exact solutions static solutions for two black holes separated by a 'non physical' strut. If you want a mass in place of one of the black holes, you need an equation of state. The simplest would be perfect fluid, but then the result wouldn't be static. If you try to make the mass sufficiently rigid, it still must have an uneven distribution of at least one of: density or pressure. This alone, I am pretty sure, gurantees there will be no exact solution.

    Note also, a string attached to the mass is inadequate for your intended purpose. The black hole will move (except in the limit of zero mass for you suspended body). You can't abstract from this if you are looking for a full solution. So you need some form of strut instead (if you want an ultimately static solution).

    In short, your problem is not simple at all.

    My belief, if this simulation was run, is that the interior of the black hole solution would, indeed be different than the interior of an isolated black hole, above and beyond any difference resulting from the strut. The key point you are having trouble with is that there is no limit on 'outside horizon' influencing 'inside horizon'. There is only a limit the other way. Another point is that from outside, once stabilized, the perception is of a frozen mass bounded by the horizon; but this state of affairs is only due to absence of nearby influence. The horizon can and does become dynamic as matter approaches. It grows even before incoming matter reaches it.
     
  12. Dec 7, 2011 #11
    I anticipated the black hole moving, and envisaged providing precisely the amount of tension to keep the distance to the black hole constant. This would presumably give a time-independent solution but in an accelerating frame, which might (?) cause problems. If so, it would indeed be necessary to use an idealised supporting rod (or alternatively some idealised thing pulling the black hole against the pull of the small mass).
     
  13. Dec 7, 2011 #12

    PAllen

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    If you try to have them move in tandem, it is not a static solution at all. Further, it will be complex and involve gravitational radiation. You will need an equation of state for the body, plus a representation in the stress energy tensor of whatever is pulling the body to keep distance constant (and then the issue of 'constant distance' can get subtle...)

    I don't see any way you can make this problem simple, let alone amenable to exact solution. On the other hand, it is well within what is tackled with numerical relativity. The website I gave earlier is to a numerical relativity collaborative effort; lots of papers and methodology descriptions are provided. It would be a good place to start if you were ambitious enough to tackle this.
     
  14. Dec 8, 2011 #13

    zonde

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    This seems rather puzzling. Gravitational potential energy comes from mass "falling" in gravitational potential. In this case it is black hole on the whole that "falls" in gravitational potential of other black hole. So from where comes this energy?
     
  15. Dec 8, 2011 #14
    Firstly black holes do have mass. A lot of it, for the ones we know about. When black holes approach each other, they start off with a lot of potential energy. This gets converted to kinetic energy, and a substantial part of the kinetic energy gets radiated as gravitational waves as they merge. If they don't collide immediately, in order for them to go into orbit some kinetic energy must be lost as gravitational waves (this will happen because they are accelerating masses, for one thing) to get an orbit at all. After they are in orbit, a lot of orbital momentum needs to be lost, which means kinetic energy being converted into gravitational waves (again this happens because they are accelerating). Since the orbital speed will be a sizeable fraction of the speed of light before they finally merge, there is a lot of energy to be lost. Even after they merge, gravitational waves are emitted until it settles down into a Kerr solution.
     
    Last edited: Dec 8, 2011
  16. Dec 8, 2011 #15

    zonde

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    Look, this is really simple. Before you start you have two black holes with summary mass M1. At the end you have one black hole with mass M2 and energy in gravitational waves equivalent to M1 - M2.
    So it turns out that black holes have lost part of their mass. And it does not matter what happens in between.

    This might be helpful as well - wikipedia: Mass deficit
     
  17. Dec 8, 2011 #16

    PAllen

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    Yes, that's correct, and equivalent to what I and Elroch said. In GR, there are serious technical problems with formulating energy conservation in non-asymptotically flat spacetimes. However, in asymptotically flat spacetimes, you can define a conserved total energy that changes form as the black holes inspiral.

    One subtle point is that if you consider each BH just before merger, and measure its mass using one of the quasilocal rest mass measures available in GR, it will not be different from when it was far away. Similarly, if a 1 kg mass falls onto a planet, radiating KE away as heat after the collision, when measured locally at it will still be 1 kg. There is a strong relationship between time and energy. Thus, 1 kg in a slower time region will be less than one kg as interpreted by an observer in a faster time region. In the BH merger, an observer at infinity sees each BH having lost a lot of rest mass before merger (and gained KE), while one could loosely say, each BH (treated somehow as an observer) , sees no mass loss in itself.
     
  18. Dec 9, 2011 #17
    Of course it does, the geometry of spacetime depends on the total mass/energy.
     
  19. Dec 9, 2011 #18
    I should mention, the reason for this thread was a discussion elsewhere, where I was arguing that when black holes interact with other objects (specifically, one external mass) the event horizon appears to a remote observer to stay exactly the same as it started (in particular, spherical). It is interesting to see in the simulations PAllen drew attention to, that the apparent event horizon does indeed even stay unchanged even when two black holes collide with each other (until they actually collide, of course), despite the true event horizon (what it would look like if you actually went there) being radically different. Quite an interesting and surprising fact. If this is true for a head-on collision, I see no reason why the same argument should not imply for any interaction of stable classical black holes and other masses.

    eg last third of this simulation.
     
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