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Black hole formation with golf balls in GR

  1. Sep 30, 2014 #1
    Start with an existing black hole and an event horizon radius R at time T. Say the black hole is being "fed" an infinite series of golf balls, one after the other, which are all stamped numerically such that the current golf ball external to the event horizon is 1.0 * 10^32.

    See linked img: http://i1373.photobucket.com/albums/ag380/rjbeery/golfball_black_holes_zps339d1899.png

    Now, starting at time T, run the clock backwards to T_past until R_past = R/2. What does the scene look like? Do golf balls with numbers less than 1.0 * 10^32 appear? If they do then there is a time T_crossover such that T_past < T_crossover < T where we could have witnessed the event horizon expand due to matter crossing it. In my understanding of GR, this cannot happen because golf balls external to the event horizon remain theoretically observable (with perfect instrumentation) forever. But in this thought experiment the black hole at time T is made of nothing but golf balls numbered 1 through (1.0*10^32)-1. I find it difficult not to view this as a contradiction, so what is the resolution?

    golfball_black_holes_zps339d1899[1].png
     
  2. jcsd
  3. Sep 30, 2014 #2

    Dale

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    Are you explicitly considering the golf balls to be sufficiently massive to alter the size of the horizon or are you thinking of golf balls as idealized test particles.

    By "see" are you talking about some specific measurable observation or are you talking about the scenario as described by some coordinate system. If the latter, which coordinate system?
     
  4. Sep 30, 2014 #3
    The golf balls are 46 grams each; by "see" I mean record visible evidence of with an infinitely precise EM detector.
     
  5. Sep 30, 2014 #4

    PeterDonis

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    In other words, you are postulating that there was no black hole at all, just empty space, when golf ball #1 was fed in? In that case, how did the hole form in the first place? Golf ball #1 was not a black hole, right?

    They have to be, otherwise his statement that I quoted above would make no sense.
     
  6. Sep 30, 2014 #5

    PeterDonis

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    The latter statement is only true for a black hole whose mass (and therefore horizon area) never changes. Obviously this does not apply to your scenario.
     
  7. Sep 30, 2014 #6
    Yes, the black hole is presumed to have formed by the earlier golf balls. That's why, presuming my math is correct, that I'm starting the experiment at such a high numbered ball.
    Agreed, but I'm seeing a chicken-and-egg problem here. How does the mass of the black hole change in the first place if additional mass is never seen to cross the event horizon? To put it another way, how can we claim that mass would be eternally visible on the outside of the event horizon? To my understanding, this is what the mathematics of GR clearly shows.
     
  8. Sep 30, 2014 #7

    PeterDonis

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    How? You can't form a black hole one golf ball at a time; one golf ball isn't a black hole, neither is two or three or a thousand. If you could form a black hole this way, it would certainly already have happened given how popular golf is. ;)

    We've had umpteen threads on this already, in some of which you have participated IIRC. I don't see that your scenario here adds any new element to the discussion. I guess we need a forum FAQ entry on this topic.
     
  9. Sep 30, 2014 #8
    You can't form a black hole one golf ball at a time? How does a giant mass of golf balls differ from a giant star? Matter is matter, correct? At some point the critical mass is reached and "something" happens...
     
  10. Sep 30, 2014 #9

    PeterDonis

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    Nope. A golf ball is a stable structure that will never undergo gravitational collapse. (See further comments below.)

    A giant mass of golf balls is not the same as one golf ball; you're moving the goalposts.

    That said, if the golf balls are still golf balls, a giant mass of them is still a stable structure that will never undergo gravitational collapse. A giant star is not; it is only able to maintain its structure while it is shining (meaning, while fusion reactions in its core keep it at a high temperature). When it stops shining, it collapses; and if it's massive enough, so that it can't form a different stable configuration at low temperature (a white dwarf or a neutron star), it collapses into a black hole.

    I suppose you could, in principle, collect enough golf balls together to make their self-gravity non-negligible--but long before anything like this could form a black hole, the golf balls would no longer be golf balls, because their internal structure would have been totally destroyed by the self-gravity of all of them taken together.

    You could also, I suppose, subject a single golf ball to some extreme compression process that would force it to collapse inside its own Schwarzschild radius (which would be on the order of ##10^{-26}## cm based on my quick Google search on golf ball masses ;)). But again, it would stop being a golf ball long before you got it compressed that far; it would first have to go through stages of being like white dwarf matter, then like neutron star matter (and possibly through other more exotic states as well).

    None of this invalidates the basic idea of throwing golf balls into a black hole that already exists. But it does invalidate the idea that if we "reverse time" on such a process, we will never see anything more exotic than a succession of golf balls emerging from a smaller and smaller black hole. At some point a much more violent process will have had to take place.
     
  11. Sep 30, 2014 #10
    I feel like I've put emphasis on the wrong point. The structure of the golf balls is not relevant; their accumulated mass is. At some point the mass would be enough such that their gravitational forces would compress them together; perhaps it would first form a star, but at some point traditional GR states that a black hole would form, agreed? Let's call that time T_blackhole.

    Do you agree that GR suggests all golf balls approaching the newly-formed event horizon after T_blackhole should be visible at time T?
     
  12. Sep 30, 2014 #11

    PAllen

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    I will describe what you would see in forward time, and time reversed. In forward time, you would see each golf ball flatten, progressively blacken (via asymptotically infinite red shift). In watching a long sequence of infalling golf balls, you would see the point of blackening (to any specified degree) happen at larger and larger radius.

    The time reversal would simply be the precise inverse of this.

    There is no contradiction at all.

    Any contrary 'understanding' of yours is simply not correct.

    Note, you cannot directly see an event horizon; however, an event horizon can grow, and observable quantities related to it will show the grown.
     
    Last edited: Sep 30, 2014
  13. Sep 30, 2014 #12

    Dale

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    OK, I misunderstood the scenario.

    I don't know enough about the relevant metrics to answer the question. In general, however, it seems silly to assume a chicken and egg scenario rather than just a complicated metric.
     
  14. Sep 30, 2014 #13
    You say "progressively blacken via asymptotically infinite red shift" but then you also use the term "point of blackening". You don't see this as a contradiction? The golf balls residing at R < your "point of blackening" would still be theoretically approaching the asymptotically infinite red shift. Are the golf balls near R = 0 visible or not?
     
  15. Sep 30, 2014 #14

    PAllen

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    Yes they are. There is actually a fair amount of misinformation on this in popular literature (by writers only familiar with ideal SC geometry, rather than collapse and growth of BH's). You can still see an earlier infalling golf ball apparently at a smaller radius, at a higher degree of redhsift than a later infalling golf ball. Each golf ball remains 'visible but ever blacker' as of when it was just outside the (growing) event horizon. Again, no contradiction at all.
     
  16. Sep 30, 2014 #15

    pervect

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    The accumulated mass of the golf balls exists before they cross the line on your diagram that you've labelled as the horizon, but it seems that you ignore it until they cross that line. It's as if to the right of the line they are test particles, when they cross the line you suddenly start thinking of them as gravitating, but you don't consider what their gravity would do when they were on the other side of the line.

    I'm afraid I'm not sure what their gravity would do, I'd need a metric to try and begin to figure it out. It would be natural to figure it out first in a frame where the golfballs were all initially at rest, rather than in a frame where they were moving to the left.

    WIth any finite line, both ends would move towards the middle. This semi-infinite line only has one end though, so I'm not sure what would happen.
     
  17. Sep 30, 2014 #16

    PeterDonis

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    This is basically one of the alternatives I gave in my last post: at some point the self-gravity of the assemblage of golf balls is non-negligible, so they gradually transition from an assemblage of golf balls to a single big sphere of matter, something like a planet. At some much further point, if we keep on adding mass, yes, the mass of the object will become larger than the mass limit on white dwarfs or neutron stars or any other stable structure, and at that point, it will collapse to a black hole.

    Time in what chart? Remember that the Schwarzschild chart does not cover the black hole or its horizon, so you can't describe the process of black hole formation in this chart. You have to use some other chart--for example, Painleve or Eddington-Finkelstein. In those charts, yes, you can assign some particular coordinate time ##T_b## to the formation of the black hole.

    The same question as I asked above applies here (time in what chart?), but I'll assume we're using a chart that meets the requirement I gave above, that you can assign a particular finite time ##T_b## to the formation of the hole.

    Let's first consider a simpler case than yours, where a single object falls into the hole, and the object's mass is negligible--it does not change the hole's mass in any way. (If we wanted to be somewhat rigorous, we could consider this as a limiting case of the infall of an object of mass ##m## as ##m \rightarrow 0##.) In this case, yes, light rays from the object will be visible at any time ##T > T_b## in the region outside the hole's horizon. As ##T## increases, the light visible at that time will have been emitted by the object increasingly close to its crossing the horizon. (Technically, this light will also be increasingly redshifted, so we have to assume that the light is being observed by idealized detectors that can detect light of arbitrarily long wavelength. We also have to assume that the light is continuously emitted, i.e., we are ignoring the quantum nature of light.)

    However, in a chart in which we can assign a finite time ##T_b## to the hole's formation, we can also assign a finite time ##T_c## to the infalling object crossing the horizon. How does that work? Simple: in this chart, light emitted by the object as ##T \rightarrow T_c## gets received by an observer far away at a time ##T \rightarrow \infty##, because of the time delay (in this chart) caused by the hole's gravity. (This is the same general kind of effect that has been measured in the much weaker field of the Sun as the Shapiro time delay of radar signals.) Light emitted at exactly ##T = T_c## by the object never gets to the faraway observer: it stays at the horizon forever. And light emitted by the object at ##T > T_c## hits the singularity, though at a later time than the object itself.

    All of the above assumed that the mass of the hole never changed once formed, so that the horizon never grew. Now consider a case still somewhat simpler than yours, where a single golf ball falls in, but it has non-negligible mass ##m##, so it does increase the size of the hole as it falls in. Suppose you are sitting somewhere well away from the hole, watching the golf ball fall past you. Then, if the hole's mass, as you measure it, is ##M## at some instant just before the golf ball of mass ##m## falls past you, then just after the golf ball passes you, you will measure the hole's mass to be ##M + m##. That is, you measure the hole's mass increase as soon as the golf ball's mass ##m## is closer to the hole's horizon than you are. (Note that this is a general rule not limited to black holes; it applies to any massive object that gains mass by things falling into it.)

    In other words, in the case of a hole into which mass is falling, the "mass of the hole" is not just a function of time; it's a function of time and radius. If you are at radius ##R## (note that the radial coordinate in the chart I'm using is the same as in the Schwarzschild chart; only the time coordinate is different), then the mass that you attribute to the hole is a function ##F(T, R)##; if a golf ball of mass ##m## passes you at time ##T##, then ##F(T - \epsilon, R) = M## and ##F(T + \epsilon, R) = M + m##. Someone at a different radius ##R'## would see that particular golf ball pass at some different time ##T'##, and would observe the mass increase from ##M## to ##M + m## at that time.

    Now consider an observer who is "hovering" just above the horizon an instant before the golf ball we've been discussing falls past him. The mass function ##F## at his radius ##R = 2M + \delta r## will switch values from ##M## to ##M + m## at some time ##T_h##; i.e., ##F(T_h - \epsilon, 2M + \delta r) = M## and ##F(T_h + \epsilon, 2M + \delta r) = M + m##. But suppose that ##\delta r## is just equal to the increase in horizon radius due to the mass ##m## falling in; then this observer, who was just outside the horizon at ##T < T_h##, will be exactly on the horizon at ##T = T_h##--and at that instant, which is the instant the golf ball falls past him, he will no longer be able to maintain altitude, but will start falling inward towards the singularity.

    Now, what about light from this observer, and from the golf ball that falls in? Take the latter first. If its mass were negligible, then the horizon radius would not increase when it falls in, so it would reach the horizon at some time ##T > T_h##; but since the horizon radius does increase, because the golf ball's mass is not negligible, the ball reaches the horizon at ##T = T_h##; what happens is that, at some time ##T_g##, a short time before ##T_h##, the horizon starts growing from radius ##2M## to radius ##2M + \delta r = 2(M + m)##, such that it reaches radius ##2(M + m)## just at ##T_h##, which is the instant that the golf ball falls to that same radius. To a distant observer, it will look as though the ball's light redshifts a little faster than it would have if the hole's mass had not increased. (We're assuming a classical BH that never evaporates, btw.) That is, at some time ##T##, the frequency of light being received by a distant observer from the ball will be a bit less than it would have been if the hole's mass had not increased (because the last light visible from the ball is emitted just before ##T_h##, instead of just before some time greater than ##T_h##).

    Light from the observer that was hovering at radius ##2M + \delta r = 2(M + m)## was redshifted too, as seen by a distant observer; but before the golf ball fell in, that redshift was constant. However, when the horizon starts growing, at time ##T_g##, the redshift of the light emitted by this observer starts increasing; and by time ##T_h##, when that observer is at the horizon, the light he emits can no longer escape. So a distant observer will, at some time ##T > T_g## (the exact time will depend on how much time delay the hole's mass imposes on outgoing light), see the hovering observer's redshift start to increase, and it will keep increasing forever, since light the hoverer emits as ##T \rightarrow T_h## will be received by an observer at infinity at ##T \rightarrow \infty##, just as any other light emitted by an object falling through the horizon.
     
  18. Sep 30, 2014 #17
    Then we've shifted the contradiction to the definition of "event horizon" because we are still able to get information about the golf balls arbitrarily close to R = 0, correct?
     
  19. Sep 30, 2014 #18

    PeterDonis

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    The description I gave in my post is, AFAIK, qualitatively correct. I don't think there are exact solutions for asymmetric collapse; the only exact ones I'm aware of are for spherically symmetric collapse, such as a spherical shell of matter falling into a Schwarzschild black hole. But AFAIK the qualitative features of those solutions, which are what I described, are still the same in numerical models of more realistic scenarios like an object falling into a hole from some particular direction.

    [Edit: See further post below for a qualifier to the above.]
     
  20. Sep 30, 2014 #19

    PeterDonis

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    I don't think that's what PAllen meant; I think he was describing light emitted closer and closer to the object crossing the horizon, not closer and closer to it hitting the singularity.
     
  21. Sep 30, 2014 #20

    PeterDonis

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    I just realized what you meant here--yes, if all the golf balls are present and close together, that changes things, because it means you don't have a spacetime with an isolated black hole and mass falling in, you have a spacetime with a long line of mass extending to infinity, which is a very different solution. What I posted previously doesn't apply to that case; it only applies to a single golf ball falling in, or to individual ones falling in far enough apart that they can each be considered isolated objects.
     
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