If you hear an astrophysicist screaming, it might be my fault

I just dropped this one on Ask an Astrophysicist since they cover both black holes and relativity.

Take a mass M, compressed to very near its Schwarzschild radius (translation for non-physicists: squeezed to almost but not quite the density at which it must become a black hole). Move it at a relativistic velocity (translation: fast enough that Einstein’s Special Relativity comes into play on a measurable scale).

Observer A, outside M’s frame of reference, will see it compress along the axis upon which it’s moving (the Lorenz contraction: x‘=(xvt)/(1-v2/c2)1/2) to a radius smaller than the Schwarzschild radius, and/or will observe a relativistic mass increase (m=m0/(1-v2/c2)1/2), either or both of which mean that M must form an event horizon and become a black hole.

Decelerate M to non-relativistic velocities. Does it remain a black hole? Or does it somehow un-collapse, lose its event horizon, and return to ‘normal’ space?

And what does Observer B, moving along with M, see? Especially if M does undergo a collapse and un-collapse?

So far I have managed to convince myself that it will collapse to a black hole and stay that way because it’s a black hole, not collapse at all, collapse and un-collapse because the initial collapse was virtual, and finally that I must be seriously messed in the head to even think of this.

You shoulda seen the look on my face when I thought of the question in the first place. One doesn’t often get to “WTF?” one’s self…


17 comments so far

  1. chilayse on

    . . . o.O

    …..too much free time.

    • The Rev Dr Sherwood Forrester on

      Not enough free time to learn the math and solve it myself. Just enough to think of the problem in the first place. :)

      • chilayse on

        Wonder if they’ll try to answer it?

        I can’t imagine it’s their usual request…

        • The Rev Dr Sherwood Forrester on

          My main worry is that they’ll think it’s some sort of assignment or paper. That would be funny… I’m just curious!

  2. jayteeone on

    I think once the event horizon opens up and the black hole forms there is no going back. Essentially you cannot unform or un-collapse a black hole.

    • The Rev Dr Sherwood Forrester on

      Yeah, but the question is: does it really become a black hole? From Observer A’s standpoint, it must. From Observer B’s, who is moving with M and therefore sees A suffer relativistic chance and not M, it must not. By golly, it’s Schrödinger’s Hole!

    • johnpalmer on

      I may have heard that it is possible for something to be a black hole only from a particular frame of reference. No promises, because I don’t understand it all myself. I believe it had to do with accelerating in such a way that light would never reach you. (at constant velocity, light will always reach you, but not necessarily if you’re accelerating.) But I have no idea if I heard right, understood right, or if it matters.

      • The Rev Dr Sherwood Forrester on

        So someone behind a relativistically accelerating mass would see a black hole form as it departed, and someone alongside or ahead wouldn’t… actually, that’s sounds similar to the case of travelling near c, there will be a virtual black hole behind you, which would vanish as you return to a non-relativistic velocity.

  3. thattallguy201 on


    Ow. Ow ow. Ow ow ow ow ow. Uh… ow.

    OK, try this on for size (ow):

    Since the shrinkage along the direction of travel is infinite, this question applies not just to large masses but to *every* object: just move faster (closer to c) and compression will (apparently) shrink the object enough to get within the SR.

    So I have to say that no, it won’t collapse — because if the answer is yes, then any single object moving infinitely near c would cause every mass in the universe to collapse because every mass in the universe would appear to be shrunk as a result.

    (“There are some people who say this has already happened.”)

    Another relevant q: It’s clear what happens to a spherical object whose radius shrinks to its SR. What about an object part of whose mass stays outside the radius? Do we just count the part of the mass within the radius as the “collapser” and the rest of the mass happens to be attached? (Side note: also consider the counteracting outward-directed gravitational force!) Because if you accelerate a spherical mass you’re going to end up with a disc-shaped object with the center apple-core-like cylinder a black hole and the rest unaffected (if collapse does occur.)

    Another relevant q: if collapse does occur, is a potential un-collapse affected by how long near-c speed was maintained, without additional acceleration? Collapse is a continuing process. How can a constant X amount of deceleration undo C1 amount of collapse in one case and C2 amount of collapse in another?

    • The Rev Dr Sherwood Forrester on

      Re: Ow.

      Well, we know that moving closer and closer to c (theoretically) causes a virtual black hole to appear behind the traveller, but that’s not due to the observed gravitational collapse of a massive object, and it goes away as one decelerates to non-relativistic speeds. It’s not a physical object being observed.

      Oh, I love stuff like this. :) I don’t have a good answer for any of these questions, that’s why I posed it. I suspect that they will revert to normal from A’s perspective, just like the virtual black hole behind a relativistic traveller also vanishes as they return to non-relativistic speeds.

      The problem is that that not only inherently implies that a black hole need not be a terminal state (ourside of considerations of Hawking radiation), but that information is not destroyed–otherwise it couldn’t return to its pre-collapse state. And that, I fear, makes big, gaping holes in physics.

      Hee. I destroyed the fabric of spacetime, and all I got was this lousy LJ thread. XD

    • The Rev Dr Sherwood Forrester on

      Re: Ow.

      Ohyeah. And it’s not infinite shrinkage along the X axis. An object with mass cannot ever actually reach c, so it must therefore always have some measurable length. I suppose one could argue that it’s limited to the Planck length, if only because it’s probably impossible to measure anything on a scale smaller than that.

      That leads to a whole ‘nother area–whether objects have a natural maximum velocity v such that in the equation x‘=(xvt)/(1-v2/c2)1/2, x‘ equals the Planck length… ;)

      • thattallguy201 on

        Re: Ow.

        So this leads to:

        1) how massy is an object whose Swartzchild radius is the Planck length? Does this make a “quantum” black hole?

        and 2) why do you assume the length must remain measurable (by current theoretical limits) and does not actually disppear from view (become immeasurable)?

        • The Rev Dr Sherwood Forrester on

          Re: Ow.

          An object with a Schwarzschild radius of the Planck length has π Planck mass — roughly 0.6836 µg. If I read the definitions right–in any case, the Planck mass is something rather more comprehensible than Planck time and Planck length and all–and unlike the others, it doesn’t theoretically form a natural boundary, it just falls out of the natural units.

          The Planck length determines the limits of observability because (it gets a little messy here, but roughly paraphrasing) a photon of energy of sufficient wavelength to resolve finer detail than that would collide with the object to be measured with sufficient energy to cause the object to collapse into a mini-black hole… which would then eat the photon. (Wikipedia: Planck Length explains better than I, and this is where I’m paraphrasing from).

          So it’s not that I assume it must remain measurable–it’s that we can’t measure anything on a scale smaller than the Planck length. So it may be that reality is inherently grainy.

  4. bass_o_matic on

    Hmm. I have some ideas on this, but given that I am supposed to be working, I’ll come back later with something.. :-)

    • The Rev Dr Sherwood Forrester on

      You know, it just occurred to me that I could give my weight by giving my Schwarzschild radius… I mean, it’s a simple enough conversion, yes? XD

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