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How do black holes gain mass?

  1. Nov 18, 2009 #1
    This question was asked by a friend recently, and I couldn't give an answer!

    To an observer on the Earth, when anything falls towards a black hole, it never actually crosses the event horizon. Therefore, from our observational point of view, how do black holes gain mass? As time goes by, all we would see would be things piling up on the event horizon.

    Is this something to do with the coordinate system used? Or does matter count towards the black hole's mass before crossing the event horizon?

    Cheers.

    Liam
     
  2. jcsd
  3. Nov 18, 2009 #2
    No, the event horizon size is determined by the mass inside, not outside. Perhaps once the matter gets within a Plank length of the horizon it can tunnel inside? That's pure speculation however. Honestly, based on the classical model, I don't think an observer outside the black hole would ever see it grow in size.
     
  4. Nov 18, 2009 #3
    You are correct that from the point of view of an outside observer they would see material approaching the event horizon but would not see it 'fall in' as such.

    Imagine two observers, Alice and Bob. Suppose that Alice falls into the black hole and Bob is stationary at infinity with respect to the black hole. Then suppose that Alice sends light pulses to Bob at a constant rate (e.g. 1 per second) with respect to the time she measures, then as she falls closer and closer to the black hole Bob would measure a decrease in the rate of the signals received from Alice. In other words, what Alice sees as a second may correspond to 5 seconds to Bob, exactly how much the signals vary from each other depends on how close Alice is to the Schwarzschild radius - The signal rate would approach infinity as she came closer and closer to the event horizon.

    Although Bob would not see Alice cross the Schwarzschild radius, that does not mean that Alice doesn't cross it. From Alice's point of view she would cross the event horizon as you would expect, its just that Bob wouldn't see it happen. As Alice approaches the black hole there would also be gravitational redshift to take into account, so the signals Bob sees from Alice would get increasingly redder as she approached the event horizon. Likewise, if Alice was to look at Bob, the light she would get from him would be increasingly bluer (redshift in reverse - blueshift).

    Once Alice falls past the Schwarzschild radius, Bob won't get any new signals from Alice, although he may get redshifted signals that Alice sent to him before she 'crossed over'.

    If Alice's spacecraft fell into the black hole, Bob could (in theory) measure the increase in mass to the black hole by measuring the Schwarzschild radius of the black hole, but since black holes have alot of mass anyway, unless Alice's spacecraft is similar in mass to a red giant or something it wouldn't have much of an effect.

    btw, although I don't think I explicitly said it above, I've assumed that there is little/no rotation with respect to the central mass which gives rise to the black hole. If there is rotation, then there are other interesting effects e.g. from Bob's point of view, the Schwarzschild radius would be larger than if there was no rotation - Think about the equivalence of matter and energy: increase in rotational energy equates to an increase in mass -> larger Schwarzschild Radius.

    With respect to quantum tunnelling at the Schwarzschild radius, thats related to Hawking radiation - http://en.wikipedia.org/wiki/Hawking_radiation
     
    Last edited: Nov 18, 2009
  5. Nov 18, 2009 #4
    To a certain extent it does. If the sun was replaced by a black hole of the same mass as the sun, then the Earth and other planets would continue to orbit in exactly the same way. If an artificial hollow sphere was made out of some very strong material such that the inner radius was just marginally larger than its Schwarzschild radius it would be very difficult to distinguish such an object from a true black hole unless you got very close to it.
     
  6. Nov 18, 2009 #5
    Wouldn't the lack of light pulses from Alice imply she fell in? And hence Bob would know, using his knowledge of GR, the exact time (finite, not infinite) Alice crossed the event horizon based on what you are saying? How would the time between signals "grow to infinity" and yet still have Bob know at some finite time when Alice crossed?

    Now think about this... Let's suppose for visual simplicity that Alice was on a b-line path to the center of the BH. She is just free falling to the center in a straight path. As she approaches the Event Horizon, her speed will rapidly increase, approaching the speed of light. At the event horizon, she would be at that speed, but we already know that would give her infinite energy (since she wasn't light to begin with), time would slow to a crawl, etc. These are all effects only seen from Bob, in his reference frame, but all imply that from his frame Alice takes infinite time to reach the horizon. It's not just some "optical illusion", but from his frame it literally never happens (or, happens at infinite time).
     
  7. Nov 18, 2009 #6

    JesseM

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    If Alice emitted light continuously rather than in a quantum manner (via photons), and if Bob could detect light from her even when it was redshifted to arbitrarily large wavelengths, then in GR Bob would never stop seeing light from her, getting closer and closer to the horizon.
    No, the idea of energy approaching infinity as you approach the speed of light only works in inertial frames, and it's impossible for any coordinate system covering a finite-sized region of curved spacetime to qualify as "inertial", all finite-sized coordinate systems in GR are non-inertial ones.
    The phrase "Bob's frame" wouldn't have a single well-defined meaning in GR the way it would in SR if you were talking about inertial frames. You could come up with an infinite number of distinct non-inertial coordinate systems in which Bob was at rest, which could have different definitions of simultaneity and thus would say different things about the coordinate time of events far away from Bob. For example, if Bob is at constant Schwarzschild radius, then he'd be at rest in both Schwarzschild coordinates and Eddington-Finkelstein coordinates, but in Eddington-Finkelstein coordinates objects falling into a black hole can reach the event horizon in finite coordinate time, while in Schwarzschild coordinates they can't (see the bottom half of this page for a rundown of some different type of coordinate systems that can be used in a nonrotating black hole spacetime).
     
  8. Nov 18, 2009 #7
    The example given for Bob and Alice placed him infinitely far away (that is to say, a place where there is no curvature and inertial frame has meaning). Is that really a physical example? No. But the Universe is closed / compacted, so this whole issue is really all an approximation anyway. The entire problem is based on an assumption of an infinite sized universe that is flat when there is no matter. If we live in that universe for the purposes of this analysis, than Bob (sitting at infinity in an inertial frame) sees Alice take an infinite amount of time to cross the event horizon.
     
  9. Nov 18, 2009 #8

    tiny-tim

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    Hi Liam! :wink:

    I agree with kev :smile:

    the gravitational flux through a surface depends on the amount of mass inside it, not on where exactly it is …

    the effect at a distance will be the same whether the mass is inside the event horizon or just near it. :smile:
     
  10. Nov 18, 2009 #9

    JesseM

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    But if your coordinate system doesn't extend all the way to event horizon, you can't use that coordinate system to assign a time coordinate to the event of Alice crossing the horizon.
    Again, you can't assign a coordinate time to Alice crossing the horizon (or even getting close) without using a non-inertial coordinate system--a coordinate system only assigns coordinates to events within the region of spacetime covered by that coordinate system, by definition. If you're just talking about when Bob sees Alice reaching different radii, then this is a local question that has nothing to do with your choice of coordinate system, all coordinate systems agree in their predictions about local events like what time Bob's clock reads when a certain light signal reaches him.
     
  11. Nov 18, 2009 #10
    That is true, but I think it's important to note that the black hole radius isn't effected by that. It's only effected by the mass inside the event horizon.
     
  12. Nov 18, 2009 #11

    tiny-tim

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    Who cares about the radius? :confused:
     
  13. Nov 18, 2009 #12

    jambaugh

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    The proposition that an infalling body never crosses the event horizon as seen by an outside observer is an idealization assuming a fixed mass black hole and a relatively massless in-falling body.

    If you add the perturbative effect of the body's mass then at some finite time the event horizon near the mass will "rise up to meet it", i.e. a new event horizon will form due to the added mass and encompassing it. Then the EV of the black hole will vibrate, radiating gravity waves until it settles down to its new larger spherical shape.

    You can idealize this scenario further by letting the in-falling added mass be a point mass (hence another tiny black hole) and read up on the research done on merging black-holes. Outside the extent of the infalling mass distribution the behavior of the black hole's event horizon will be the same as if the infalling mass were a micro BH.
     
    Last edited: Nov 18, 2009
  14. Nov 18, 2009 #13

    pervect

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    You might think about "how do you measure the mass of a black hole"? Typically, you'd just look at the orbital period of a body around it. This body will be well outside the hole. So the argument takes on a distinctly philosophical tone - it's not about things that can be measured.

    That said, there are philosophical problems with the idea that black holes don't exist. This relates to the philosophical view one takes of the Rindler horizon that an accelerated observer sees. If one uses the same logic for the Rindler horizon that one applies to the black hole, one gets misleading at best and simply wrong at worst predictions. For instance, one might conclude that a signal sent from the accelerated observer "can never reach the Earth" because it "can never cross the Rindler horizon". But people on Earth get the signal just fine, even though the people on the rocket never see the signal get past said horizon.
     
  15. Nov 18, 2009 #14
    Oh absolutely this is all based on a thought experiment and an ideal case. In reality, as the body gets within a Plank length of the event horizon, particles can begin to tunnel in, and it gets sucked in (finite time). The full reality of this is more complicated then just the application of GR to point masses. To me, QM saves the day here, because it helps put caps on not only the infinitely small but also the infinitely big.
     
  16. Nov 18, 2009 #15
    It's related directly to the last question of the original post.
     
  17. Nov 18, 2009 #16

    tiny-tim

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    uhh? how? :confused:
     
  18. Nov 18, 2009 #17
    Agree.
     
  19. Nov 19, 2009 #18

    jambaugh

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    No. It has nothing to do with Plank lengths and tunneling (quantum phenomena) but rather what the classical theory predicts. It is not a matter of "QM saving the day" it is a matter of correctly understanding the predictions of classical GR.

    One cannot blithely invoke "tunneling" and "Plank length". The Plank scale is a scale at which one cannot resolve positions further due to gravitational issues just as the Compton wavelength is the scale at which one cannot resolve positions further due to electron-positron pair generation.

    Invoking the Plank scale here is being redundant. The signal presupposed by the Plank scale arguments is another "in-falling stress-energy source" perturbing the Schwarzschild solution.

    Secondly the Plank scale argument presupposes all three spatial coordinates are being resolved simultaneously thereby localizing all of the energy of the measuring signal to within a Schwarzschild radius. There is nothing in the Plank argument preventing a higher resolution of one coordinate (i.e. radial coordinate of an infalling object) while leaving others broadly resolved (e.g. plane-wave one photon at a time "radar" at higher than Plank frequency).

    The Schwarzschild solution is a vacuum (zero stress-energy density) solution which of course is technically invalid when you speak of an in-falling mass (or stress-energy of some massless field) since you no longer have a vacuum. At issue here is how that solution and hence the stationary event horizon of the solution is perturbed by the addition of this mass.

    In order to intuitively grasp what is going on it is useful to remind ourselves that there are a continuum of event horizons out there and we needn't invoke black holes. The future light-cone of any event point defines an event horizon, as does any space-like surface. Finkelstein (my thesis advisor) worked out a nice way to understand gravity in terms of fields of perturbed light-cones (see Eddington-Finlekstein coordinates) which again are themselves event horizons. Mass bends the light-cones of surrounding event-points in towards the mass' world-line. Concentrate mass enough and the outer edges of these in-turning light-cones become radially parallel with the mass' worldline thereby defining a stationary event horizon. You then have a black hole.

    The event horizon of a BH is a sphere tangent to all (future) light-cones of all events on that surface (which thence point inward). Slightly outside the Schwarzschild radius (Sr) a small mass will bend light-cones towards it from both directions. Between the mass and the Sr the cones bend away from the BH and so the r=Sr surface ceases to be an event horizon. Beyond the mass the cones bend further toward the BH and so a surface r = Sr + "a bit" becomes a horizon with the mass interior to it.

    More generally consider the effect of a mass relatively near but not too near the event horizon of a BH. One might, upon first intuition, suppose the event horizon would bulge outward due to the tidal effects. In fact the event horizon will dimple inward slightly (more flatten then dimple it will remain convex). One must remember that the EH is not a physical surface but rather a mathematically defined surface (said mathematics having major physical consequences). As I mentioned we are surrounded by event horizons. So as the mass moves closer the EH's which are outwardly expanding light-cones enclosing both mass and BH's horizon will expand more slowly until one of them (distorted by the gravity of the mass + BH) will form the new quasi-stationary EH of the enlarged black hole.

    The new horizon doesn't materialize out of nowhere and it doesn't really as I first suggested (and quickly qualified) expand from the old EH. Rather one of the otherwise dynamic EH's is overlaid like a veil becoming the new horizon. I say "quasi-stationary" because there will be a "pebble in a pond" dynamic undulation emitting gravity waves as the quadrapole and higher order mass moments of the system settle down.
     
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