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Black Holes - help needed to settle a bet

  1. Aug 18, 2004 #1
    Please verify that this explanation of black holes is consistent with the mainstream theory. (I have a $10 bet riding on this.)

    The Schwarzschild radius is 2GM/C-squared. If the mass M is within R spatial units of the center, then there will be a black hole, with an event horizon having the time-like Schwarzschild radius, in pure time units with no spatial component because the event horizon's radius is within the event horizon, where space-time has only a time component. This means that the spatial radius of a black-hole event horizon is zero. The event horizon is positioned in space in the same mathematical point as the singularity. In other words, the spatial position of the event horizon is at the singularity, not R spacial units out from the singularity. Every point in space, other than the singularity (which is out of space) is outside the event horizon.

    This means that an object can approach arbitrarily close (spatially) to the singularity and still be outside the event horizon. Thus the special time-like calculations that apply within the event horizon are not applicable at any point in space (other than the singularity point). So escape velocities at points arbitrarily close spatially to the singularity can still be analyzed like ordinary escape velocities.

    At every point within the spatial Schwarzschild radius, the escape velocity is greater than the speed of light. This means that a beam of light originating just inside the spatial Schwarzschild radius and directed radially outward will go a very long, but finite, distance before it becomes infinitely red-shifted. A beam originating just outside the spatial Schwarzschild radius will be more and more red-shifted as it travels, but it will never become infinitely red-shifted. A beam originating very close to the singularity/event horizon will become infinitely red-shifted after a very short, but non-zero, distance.

    A rocket ship (with unlimited but finite thrust to overcome gravity, unlimited strength to withstand tidal forces, and piloted by Superman) can approach arbitrarily close, spatially, to a singularity and at any time (or, from the pilot's reference frame, at any time within the finite time interval before his spatial intersection with the event horizon, which is sooner than his temporal intersection with the singularity) the pilot can fire his rockets and accelerate away from the singularity. If he continues to fire his rockets (just enough to slightly overcome the gravity at each point), he will eventually reach the region outside the spatial Schwarzschild radius. In that outer region, the escape velocity is less than the speed of light. If the pilot continues to fire his rockets just enough to overcome gravity, he will eventually reach a point where the escape velocity is less than his current speed. Then the pilot can cut off his rockets and coast forever away from the black hole, always decelerated by the black hole's gravity, but never losing all of his outward velocity. This is how Superman escaped from the black hole of Krypton.

    When an object is very close to the singularity/event horizon, it is in a very intense gravitational field, with a very steep gravitational gradient. Thus it experiences strong tidal forces, increasing without limit as the object approaches the singularity/event horizon. (For a very large black hole, the tidal forces at the spatial Schwarzschild radius might be moderate. But remember that the event horizon is not a the spatial Schwarzschild radius, it is at the singularity point, separated from the singularity only by a time interval which is equal to the Schwarzschild radius in time units.) Also, relative to distant observers, the object undergoes a slowing down of time, approaching a stand-still at the singularity/event horizon. I'm not sure whether an observer looking outward from the vincinity of the singularity/event horizon would see red-shifting and time-slowing or blue-shifting and time-speeding (can someone out there tell me?), but I am sure that after returning from near the singularity/event horizon he would find that his distant stationary colleague had experienced a much longer time interval than what he had experienced.

    A distant observer (with super infra-red vision) sees all the mass of the black hole between the spatial Schwarzschild radius and the temporal Schwarzschild radius (the event horizon), with most of it "frozen" just outside the event horizon/singularity. The mass never crosses the event horizon, but it doesn't have to cross the event horizon in order to form a black hole. It is a black hole because the mass is inside the spatial Schwarzschild radius, even though the mass never (from a fixed frame of reference) crosses to within the time-like Schwarzschild radius of the event horizon.
    Last edited: Aug 18, 2004
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  3. Aug 18, 2004 #2


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    This is false. The event horizon has non-zero spatial radius. I don't know what "time-like Schwarzschild radius" means, but it's not part of normal relativity parlance.
    This would be correct if you replaced 'singularity' with 'event horizon.'
    This would be correct if you replaced 'singularity' with 'event horizon.'
    This would be correct if you replaced 'singularity' with 'event horizon.'
    This would be correct if you replaced 'singularity' with 'event horizon.'
    I have no idea what this means, but it's the same misconception as that in your first paragraph.
    An observer very close to the event horizon of a black hole will see the rest of the universe blueshifted and running in fast-forward.
    There is no such distinction. The event horizon is a locus of points in space, independent of time, for a black hole that is not changing in mass.
    The mass never crosses the event horizon as viewed from the outside world, that much is true. Once again, there is no such separation of two different event horizons, however.

    - Warren
    Last edited: Aug 18, 2004
  4. Aug 18, 2004 #3


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    This got seriously mangled and isn't correct as stated. What you can say is that inside the event horizion, the Scwarzschild coordinate 'r' is timelike, and the rest of the Scwarzschild coordinates, including time, are spacelike. Outside the event horizion, though, r is a spatial coordinate.

    The event horizon is definitely NOT located at the singularity, as you seem to think, however. The circumference of the event horizion is 2*Pi*Rs, where Rs is the scwarzschild radius you gave earlier, Rs=2GM/c^2.

    The "embedding diagram" of a black hole might help you visualize the geometry of space there. There's an applet at


    The embedding is just to help you understand the geometry - it's not even necessarily unique, it's just a "visual aid".
  5. Aug 18, 2004 #4
    Not to mention that "spatial radius" is not a very meaningful term. The value r which appears in the Schwarzchild metric is not a distance from r to 0. It is a coordinate which represents the spatial location of a point. It is sometimes refered to as the reduced circumference. The physical meaning of r is that you measure the circumference of a great circle which passes through the location with the singularity at the center and divide the circumference by 2*pi then you get r. The Schwarzchild radius, Rs has a similar meaning.

    If you were to measure the spatial distance from the location to the singularity then it would be an infinite distance, wouldn't it?

  6. Aug 18, 2004 #5
    Seems like everyday that I hear/see something about GR that I never have before. That's what I love about physics. :smile:

    What, exactly, does "timelike coordinate" mean?



    ps - I believe that I will know everything about GR 1 year after I've passed away, since then I can talk to Einstein who will then have been corrected by God. :biggrin:
  7. Aug 18, 2004 #6
    I've learned a long time ago never to place a bet with a person on anything, especially after a few beers. :rofl: But that's another, long, story.
    Let me rephrase this for you. If an amout of mass, M, is confined within a distance, Rs = Schwarzschild "radius" (really "reduced circumference") which represents a sphere, then the object is a black hole. Burt sorry, I don't know what "time-like Schwarzschild radius" means or what the rest of that statement means.
    No. It does not mean that.
    No. It is not. Note that there are two singularities which pertain to a black hole. One is a coordinate singularity which, as I recall, used to be refered to as the Schwarzchild singularity. This is no longer true. When people use the term Schwarzchild singularity now they are refering to the physical singularity which is at the center of a black hole and is distinct from the coordinate singularity which is the event horizon.
    No. The physical singularity is at a different locationn than the event horizon. Every point in space, other than the singularity (which is out of space) is outside the event horizon.
    I don't believe that this is possible. A beam of light which originates inside the event horizon cannot be directed radially outward.

    Only if he is outside the event horizon. If he is inside the event horizon then he's doomed. However I don't believe that anything can actually reach the singulartity, i.e. nothing can reachg r = 0.
    The last term is not meaningful since you're equating the physical singularity with the coordinate singularity and these are two different things.
    True if you deleted the term "singularity".
    I wouldn't phrase it like that since this is the "frozen star" paradigm and that is not a very useful way to look at this. Since you can't measure the location of that matter if you're outside the event horizon then its useless to say that its there. You don't know if the black hole is the result of a collapsing star or if it was "born" that way when the universe was created.

    I think you lost $10 dude.

    Note: A black hole doesn't necessarily have a physical singularity at r = 0. Recall the definition of black hole - A finite region of space into which signals can enter, but from which no signal can ever emerge - Hans C. Ohanian. This does not mean that all the mass is located at r = 0. Ohanian gives a pretty cool example of this. Suppose that the nulceus of the Galaxy consists of 1011 stars, each of which is one solar mass, which is uniformly distributed over a spherical volume contained within the Schwarzschild radius. The density of matter would then be 10-6 g/cm3. This would eventually collapse to a singularity though.

    Last edited: Aug 18, 2004
  8. Aug 18, 2004 #7


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    If we use a -1, -1, -1, 1 metric signature, the spacelike coordinates have a negative g_ii, and the timelike coordinates have a positve g_ii.

    I *think* the terminology is standard. There's a long section in MTW about "dynamic wormholes" where this issue is at least talked about.

    The easy way to describe this is to just look at the Scwarzschild metric and note the change in sign of g_tt and g_rr as one passes through the Scwarzschild radius.
  9. Aug 18, 2004 #8
    I don't understand that statment. What do you mean by "the spacelike coordinate have a negative g_ii.."
    Absolutely, it is standard terminology. Ohanian defines these terms too. That doesn't mean that I understand it. :confused: :smile:
    What does the change in sign of a component of the metric tensor have to do with the spacelike/timeline nature of a coordinate? That's the part that I don't see.

    Thanks for your patience.

  10. Aug 18, 2004 #9


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    I'm not sure how to answer more fully. If we have a diagonal metric, we can write

    ds^2 = g_{00} dx_{0}^2 + g_{11} dx_{1}^2 + g_{22} dx_{2}^2 + g_{33} dx_{3}^2

    One of these coordinates will have a different sign for the metric coefficient than the others. That coordinate is a time coordinate in that basis.
  11. Aug 19, 2004 #10
    Never mind. I asked a buddy of mine and he explained to me that the terms "spacelike" and "timelike" don't have the same meaning as they do when used to describe spacetime intervals. That was the part that was confusing me

    Note regarding other comments - There are some comments above which seem to indicate that tidal forces at the event horizon are finite. What that means is that an object in free-fall will experience finite tidal forces. It does not mean that all the components of the Riemann tensor (aka tidal force tensor, aka curvature tensor) are finite at the event horizon in all coordinate systems. Take R0101 as an example. Assuming Ohanian & Ruffini didn't make a mistake in their text Gravitation and Spacetime, in Schwarzschild coordinates R0101 is

    [tex]R^0_{101} = \frac {r_s}{r^3}\frac{1}{1-r_s/r}[/tex]

    This is obviously infintite when r = rs. However in a free-fall frame

    [tex]R'^0_{101} = \frac {r_s}{r'^3}[/tex]

    This obviously doesn't vanish when r' = rs.

    Last edited: Aug 19, 2004
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