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Black hole surface gravity question

  1. Jul 7, 2011 #1

    PeterDonis

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    The "surface gravity" of a black hole is defined as the force required "at infinity" to hold an object at rest at the horizon. The Wikipedia page gives a brief discussion:

    http://en.wikipedia.org/wiki/Surface_gravity

    This has also been discussed on PF before, for example in this thread:

    https://www.physicsforums.com/showthread.php?t=405941

    I've also seen this subject treated in textbooks; I believe it's covered in Wald's GR textbook (I actually encountered it first in his book "Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics"). I understand the basic derivation and why the "surface gravity" is finite even though the acceleration required "locally" to hover at rest diverges at the horizon.

    My question is whether the idea of "holding an object at rest at the horizon", even if we measure the force required "at infinity" so it's finite, makes sense. The horizon is a null surface, so any object "at rest" there would have to travel on a null worldline. The idea of holding "at rest" an object like a light beam doesn't really make sense to me. Is this just an indication that the "surface gravity" of the black hole can't be directly measured physically? Or does it indicate that the concept itself doesn't really make sense? I'd be interested in PF members' opinions on this.
     
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  3. Jul 7, 2011 #2

    bcrowell

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    An object falling radially into a Schwarzschild black hole asymptotically approaches being "at rest" as it approaches the event horizon, as determined by a distant observer. So in that sense, it's not only possible for an object to be at rest at the horizon, but the force required at infinity is zero, because the object will be "at rest" even without any force being applied.

    But I think what the WP article is saying is that if you measure the "surface gravity" g by this definition at distance r, then the function g(r) has a certain definite limit as r approaches the Schwarzschild radius rs -- even though the operational definition of the function g(r) doesn't actually make sense at r=rs.
     
  4. Jul 7, 2011 #3

    PeterDonis

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    This is technically true, but it wasn't really what I was concerned about when I was contemplating the idea of an object being "held at rest" at the horizon.

    Okay, this makes sense; the "surface gravity at the horizon" is a limit of the "surface gravity measured at infinity" at a given finite radius r, as r goes to 2M. The "measured at infinity" part is needed so that the limit is finite (because of the redshift factor between a finite radius r and infinity).
     
  5. Jul 7, 2011 #4

    PAllen

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    I've understood this generally as follows:

    The proper acceleration is what would be needed to maintain hovering by material body; e.g. a rocket would really need asymptotically infinite thrust to hover ever closer to the horizon. However, an observer at infinity sees this compensated by ever slower time (as they observe the horizon hoverer). Thus they see a finite limit on thrust consumption. However, the infinite value is the one relevant for the rocket, indicating the impossibility of avoiding infall.
     
  6. Jul 8, 2011 #5

    tom.stoer

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    The WP article mentions that the surface gravity cannot be defined as usual. As discussed this is due to the fact that the horizon is a null surface which results in infinite thrust for a rocket hovering at the horizon.

    So I think we agree on the wikipedia statement "In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which can only be truly treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. "

    Now wikipedia says "Because of this, a renormalized value is used that corresponds to the Newtonian value in the non-relativistic limit. The value used is generally the local proper acceleration (which diverges at the event horizon) multiplied by the gravitational redshift factor (which goes to zero at the event horizon). For the Schwarzchild case, this value is mathematically well behaved for all non-zero values of r and M."

    The problem is that the Newtonian limit does not exist at a null surface.

    So my question regarding this definition is: what is the physical meaning of this 1/4M for a Schwarzschild black hole? is there any experiment via which this can be measured?
     
  7. Jul 8, 2011 #6

    pervect

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    The physically interesting part of it to me is the subsequent relation developed by Wald showing that k is the ratio of the rate of change of area of the event horizon and the mass of the black hole, holding the angular momentum constant (or doing some more work to accomodate some other terms if you insist on varying J).
     
  8. Jul 8, 2011 #7

    bcrowell

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    I believe a thought experiment that would measure the thing they're defining is essentially the same as the one described here http://en.wikipedia.org/wiki/Komar_mass as an interpretation of the Komar mass (in the initial section, where the metric is assumed to be not only stationary but a Schwarzschild metric). You tie a test particle of mass m to the end of a string and measure the force F needed to support it from far away. The "surface gravity" is the limit of F/m as you let the test particle approach the event horizon. Even though the string is assumed to have negligible mass, I believe that its tension slacks off with distance by a factor that is equal to the redshift factor.
     
  9. Jul 8, 2011 #8

    PAllen

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    Does this say the if you tether an object to distant platform, you can bring it arbitrarily close to the event horizon and back?
     
  10. Jul 8, 2011 #9

    bcrowell

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    I think so, in theory. Of course any real-world material for the rope would be far too weak, and it would be much easier to do a fly-by.
     
  11. Jul 8, 2011 #10

    PAllen

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    Ah, but that's the interesting point. The flyby would take infinite thrust to graze the event horizon and escape. The world line of the rocket would experience the proper acceleration of its path, which goes to infinity. Yet (in theory) a distant platform could pull a probe away with finite force. The difference in time makes it mathematically consistent (the probe will have spent zero proper time at the horizon when lowered and pulled away by the distant platform).

    Still, I am having trouble believing this. I keep thinking there is some missing piece to the analysis which would show why the tethered probe cannot really be lowered arbitrarily close to the horizon and back with 'modest' force.
     
  12. Jul 8, 2011 #11

    PeterDonis

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    This was the kind of thinking that prompted my OP in this thread, but as I noted in post #3, I agree with bcrowell that the "surface gravity at the horizon" is supposed to be a limit only and is not supposed to mean that a real object could actually be held "at rest" *at* the horizon, by a tether from a distant platform or any other means.

    I think the missing piece is that, while the force exerted on the tether at the distant platform might be "modest", the tension in the tether would increase as you went from the platform (at a large radius) to the suspended object (at a radius arbitrarily close to the horizon), so by the time you were at the suspended object, the force exerted by the tether on the object would have to be equal to the object's mass times the local proper acceleration (which increases without bound as the horizon is approached) in order to hold the object at rest.

    This means, of course, that for the tether to not break, its tensile strength would have to approach the relativistic limit (speed of sound in the material = speed of light) as the suspended object approached the horizon. As bcrowell noted, any real material has a tensile strength orders of magnitude smaller than this, so a real tether would break long before the horizon was approached. (I'm not sure that a flyby would be any easier, however, since any real rocket engine is also incapable of producing an arbitrarily large acceleration; the ones we have now are only good for a few g's.)
     
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