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Properties of event horizon for incoming matter

  1. Oct 3, 2014 #1
    Dear all,
    In one of his lectures,Prof. Susskind mentioned that the event horizon "bulges" forward to meet any incoming radiation or matter; and it is a property of Einstein field equations. I have not come across any such property, and if it exists, shouldn't it belong to the Schwarzchild(or Kerr,etc) metric rather? Can someone mathematically explain that property in detail? I want to know the mathematical description of this property.
    Thanks in advance!
     
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  3. Oct 3, 2014 #2

    Nugatory

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    Whatever it is, it won't be the Kerr or Schwarzchild metrics, as these describe a constant-mass static black holes and apply only when any infalling particles are small enough that the gravitational effects of their mass is negligible.

    I don't know if there's an exact solution to the Einstein field equations for the case of a massive object falling into a black hole.
     
  4. Oct 3, 2014 #3

    PeterDonis

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    I believe there is one for the extremely idealized case of a spherically symmetric thin shell of matter falling into a Schwarzschild black hole. But I don't think there is for any more realistic case; it's all numerical simulations.
     
  5. Oct 3, 2014 #4

    PeterDonis

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    Here's one way of thinking about this: suppose there is a Schwarzschild black hole of mass ##M##, and a thin, spherically symmetric shell of matter is falling towards it such that, once the matter has fallen in, the black hole will have mass ##M + m##. Suppose you are hovering just outside the hole's horizon before the shell of matter falls in, at radius ##r = 2M + \delta##, where ##\delta## is small enough that it puts you at a radius less than ##2 ( M + m )##, i.e., less than the horizon radius will be after the shell has fallen in.

    Now there will be some point along your worldline at which, if you emit a flash of light radially outward, it will reach radius ##r = 2 ( M + m )## at the same instant that the shell of matter falls to that radius; and at that point, the flash of light will be trapped at ##r = 2 ( M + m )## forever, i.e., it will be at the horizon. Then the path of that flash of light marks the movement of the horizon outward to "meet" the incoming shell of matter: i.e., at the instant when you emit that flash of light, you are at the horizon, because it has started moving outward from ##r = 2M## to ##r = 2 ( M + m )##, and it is passing your radius, ##r = 2M + \delta##, at the instant you emit the flash of light.

    The reason things work this way is that the horizon (more precisely, the absolute horizon--see below) is not locally defined; it's globally defined: it's the boundary of the region of spacetime (the black hole) that can't send light signals out to infinity. That means that, to know exactly where the horizon is, you have to know the entire future of the spacetime; you can't tell just from measurements in your local region. That's why the horizon can "move" outward in a way that appears to "anticipate" what is going to happen in the future; it's not a "real" thing that's actually moving, it's just a surface in spacetime that's defined in a particular way.

    Note: what I said above applies to the absolute horizon, which is defined as I did above. But there is another kind of horizon, called an "apparent horizon", which is defined differently: it is a surface at which outgoing light rays don't move outward, but stay at the same radius. (You can think of this as a sphere of light rays emitted outward not expanding, as it would in ordinary flat spacetime, but maintaining the same area.) For a black hole that is stationary, i.e., whose mass is not changing, the apparent horizon and the absolute horizon coincide; but for a black hole gaining mass, as in the example above, they don't. You can see that from the above: you emitted a light flash outward from ##r = 2M + \delta##, and it did move outward, to ##r = 2 ( M + m )##, so when you emitted it, you weren't at an apparent horizon, but you were at the absolute horizon, as I showed above.

    I'll see if I can find an online reference for the idealized model I referred to above. One caution: you can't use the usual Schwarzschild coordinates for a problem like this, so you'll need to be familiar with other charts--Eddington-Finkelstein is probably the best one.
     
  6. Oct 4, 2014 #5
    Love your answer @PeterDonis...The thought experiment is excellent. So now I do have an idea of how it works, though it is still purely visualisation. I thought there would be something like it in my new book (straumann's GR and astrophysics), but couldn't find anything......Thanks a lot though.
    Can you furnish a more mathematically rigorous approach ( I am trying)? And I really appreciate it that you are trying to find an online reference( I googled for hours, but couldn't get a thing).....Thanks again.
    EDIT: Can this, in any case, be related to intrinsic gravitational entropy?
     
  7. Oct 4, 2014 #6

    PeterDonis

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    Poking around in past PF threads, I found this one:

    https://www.physicsforums.com/threads/oppenheimer-snyder-collapse.753312/

    Reading that thread, I realized that I was misremembering what exact solutions exist for things falling into a black hole. The link I gave in the second post in the thread is to an article about the Vaidya null dust, which models a thin shell of radiation (such as EM waves) falling into a black hole; but from what George Jones says in his thread later on, nobody has found a similar solution for a thin shell of ordinary matter falling into a black hole. So I was probably remembering the heuristic features I was describing from the Vaidya model--i.e., to be strictly correct, you should imagine the thin shell I referred to as being radiation, not matter. I would expect the qualitative features of a model with a thin shell of matter falling in to be similar, but if any solutions exist for that, they are apparently numerical only, not exact analytical solutions. Sorry for the confusion.

    Not sure what you mean by "intrinsic gravitational entropy". The entropy of a black hole is its horizon area (more precisely, it's 1/4 of its horizon area in Planck units, so one unit of area is one Planck length squared).
     
  8. Oct 4, 2014 #7
    It means that there is an entropy associated with gravitational fields that is quite separate from any any entropy associated with matter or radiation. In other words this entropy is intrinsic to the gravitational field and will always be present regardless of the arrangement of matter and energy. It does exist, has been mentioned in "Nature of space and time" by Penrose and Hawking.

    The obvious example of this is the entropy associated with a black hole. This entropy depends only on the area of the event horizon and not on whatever matter fell into the black hole.
     
  9. Oct 4, 2014 #8

    Dale

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    I had not heard of that before. That is interesting during the collapse. After the collapsing shell falls past ##r=2(M+m)## but before it falls past ##r=2M## there are (I think) two apparent horizons. Light at the inner horizon is trapped by the original mass, M, and light at the outer horizon is trapped by the new mass, M+m. But light between the inner horizon and the collapsing shell is not trapped and can move outwards.
     
  10. Oct 4, 2014 #9

    PAllen

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    I think light trapped at the horizon before infalling matter approaches gets to move out, thus the inner trapped surface ceases to be ( a trapping surface). However, it is possible there are two trapping surfaces for some period of time - I've read several papers analyzing complex evolutions of BHs that discuss 'innermost trapped surface', suggesting that there can be several before the dynamics settle.
     
  11. Oct 4, 2014 #10

    PeterDonis

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    For a bit after the absolute horizon starts to move outward, yes, this is true. But then that light starts falling inward as the trapping horizon moves outward; see below.

    No, it doesn't; it stays at the same radius for a bit, but then starts falling inward, as above. If you look at the apparent horizon as a 3-surface in spacetime, it actually has a spacelike segment at this point. That is, in the model I gave in my earlier post (and using the time coordinate of that model), it stays at ##r = 2M## after the absolute horizon has started to move outward from that radius (I think this portion of the apparent horizon, as a 3-surface in spacetime, is actually timelike, but I'm not sure). Then, before the absolute horizon reaches ##r = 2 ( M + m )##, the apparent horizon moves outward "faster than light" from ##r = 2M## to ##r = 2 ( M + m )##, such that it reaches the latter radius just as the absolute horizon does (and as the infalling shell of matter or radiation does).

    One caveat: all this is using the particular coordinate chart I used in my earlier post. There are other charts that can show different behavior because of the spacelike segment of the apparent horizon; see below.

    Not in the spacetime we've been discussing and in the chart I've been using, no. Once the apparent horizon has moved outward of a given radius, any light that was trapped at that radius must start falling inward. No light that is once caught at a trapping horizon ever gets to move outward from the radius at which it was caught.

    (The same caveat as I gave above applies here too--see below.)

    Yes, this can happen along the spacelike segment; there are space+time slicings that will show trapped surfaces at multiple radial coordinates "at the same time". (But there are others, such as the one I used above, that don't.)

    Note that in some of these charts, the spacelike segment of the apparent horizon is moving "backwards in time"; that is, in these charts, there is a period of time in which there are three trapping surfaces, an inner one, a middle one, and an outer one. In the region between the inner and middle ones, light is not trapped and can in fact move outward; however, this light will end up somewhere on the spacelike segment of the apparent horizon and will then be trapped (i.e., none of this light will ever actually escape to infinity--it is all inside the absolute horizon).
     
    Last edited: Oct 4, 2014
  12. Oct 8, 2014 #11

    PAllen

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    Do you know how well this shell result generalizes? If it is general, it suggests that a tiny body falling into an existing BH results in all light trapped at the (prior) horizon falling in.
     
  13. Oct 8, 2014 #12

    PeterDonis

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    AFAIK that feature is general, yes. I don't know of any exact solutions describing an object falling in from a particular direction; I think there are only numerical simulations of the detailed behavior. But the increase of the absolute horizon area is general; IIRC that's proved in Hawking & Ellis without any assumptions about symmetry. And the increase in absolute horizon area is what ensures that light trapped at the old horizon falls inward (because it is what ensures that the apparent horizon has to jump outward at some point, since after the hole has settled into its new stationary state the apparent and absolute horizons must again coincide--the difference with a non-symmetric infall is that the "jump" might happen at different times for different points on the horizon, so to speak).

    (Also, I think Price's theorem comes into play as well for a non-symmetric infall, since that is what ensures that any non-symmetry in the horizon gets radiated away as gravitational waves, so that the end state of the hole is symmetric.)
     
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