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Black holes and exotic matter

  1. Sep 25, 2013 #1
    A thought experiment. C is considered to be a constant speed, that can not be exceeded. Once light passes the event horizon of a black hole however, the gravitational pull is strong enough to exceed, C. This is indicated by the supposition, that not even light can escape the event horizon of a black hole.
    So what about Exotic Matter? If a particle with negative mass were to somehow pass the event horizon of a black hole, would it not then be repelled from the black hole at a velocity greater then C?
     
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  3. Sep 25, 2013 #2

    Simon Bridge

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    No. All observers measure the same speed for light in a vacuum. We say that c is "invariant", not "constant". The difference is quite subtle though.

    That sentence does not mean anything.
    The event horizon of a black hole is where the escape velocity is greater than the speed of light. Not gravity.

    Which depends on what you mean by "escape the event horizon".

    Consider: classically, on the Earth, if you fire a projectile slower than the escape velocity at the surface of the Earth then it may still escape the surface. i.e. it flies through the air. Of course this escape is temporary - unless it gets fired into an orbit. Fired at the escape velocity, and the projectile does not come back down ...

    You are considering the possibility of antigravity.
    The short answer is "no" - exotic matter would not exceed the speed of light either, it would just get a lot of kinetic energy. But, since gravity is repulsive for the object you are considering, you would have to do work on it to bring it close to the BH. The same reasoning that says that light does not escape also means the exotic-matter does not get in.

    The relationships are not so simple though - I'm sure one of the others will be keen to talk about where I am cringingly oversimplifying things :)
     
  4. Sep 27, 2013 #3
    I'm just a novice. Thank you.
     
  5. Sep 28, 2013 #4

    Simon Bridge

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    No worries: we've all been there.
    Being new does not exclude you from the discussion and gives you the advantage of being able to ask "naive" questions without losing face :)

    Anyway - newcomers often come up with questions that more experienced people would never think of.
     
  6. Sep 28, 2013 #5
    I think in SR "invariant" and "constant" are the same thing, aren't they? In GR, it would mean a local c has the same "invariant" numerical value of c, though it is not an Universal constant, as a non-local value of c could be different. Is that the subtle difference?

    But of course, this is not a full analogy in GR terms? A projectile fired at the 'escape velocity' at the event horizon never spatially leaves the event horizon I imagine.

    In fact, this raises an interesting question. If something were fired radially outward at the event horizon at 'escape velocity', it would always continue to hover at the event horizon and never get drawn into the black hole. I assume that is correct?
     
  7. Sep 28, 2013 #6

    PAllen

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    Even at an event horizon, in the local frame of every inertial observer, light moves at c.

    What could escape the event horizon, if they exist, are tachyons. These have imaginary mass not negative mass. If they exist (all searches are negative), they travel on spacelike trajectories, and these readily escape
    the event horizon.
     
  8. Sep 28, 2013 #7

    Simon Bridge

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    It's more to do with how you think of it.

    Consder: a constant speed is something one sees other things travel at - if something travelled at a constant velocity of 30m/s and you travelled at a constant velocity, same direction, of 20m/s, then the "something" would be travelling at a constant velocity of 10m/s right?

    But if the 30m/s were invarient, then, no matter how fast you went, the "something" would always be travelling at 30m/s.

    I think your other questions have already been answered by PAllen.
    If you were hovering at the horizon - goodness knows how, and you fired a photon radially away from the hole, would it just hover there next to you?

    Basically your questions are getting confused because you keep leaving out the observer.
    Where the observer is and what they are doing there makes a difference.
     
  9. Sep 28, 2013 #8

    PAllen

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    Hovering at the horizon means you are following a lightlike world line, that is 'you are a photon'. Thus the statement becomes, from an inertial frame (crossing the horizon) what happens if a photon splits or decays? If this were an allowed process, the inertial observer sees one photon become two, moving along the same lightlike world line. Put in this correct local framework, the question becomes a general question about light in SR.

    Strictly classically, a model of a massless particle with energy E splitting into 2 comoving massless particles each with energy E/2 (for example) is allowed within SR: 4 momentum is conserved.

    Within QFT, this turns out to be a prohibited process. There was a long thread on this:

    https://www.physicsforums.com/showthread.php?t=512811
     
  10. Sep 28, 2013 #9

    Simon Bridge

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    Yeah - it's really annoying to talk about and only reinforces the point about naive thought experiments... the another approach is to have a free-falling observer fire the photon radially outwards.

    There's probably a document someplace that handles these sorts of things ... orbits, ballistic trajectories etc in the situations people keep asking about - close to or at the event horizon.
     
  11. Sep 28, 2013 #10

    PAllen

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    Free falling body emitting radial outward photon at horizon is easy to analyze. To an observer hovering outside the BH, it never arrives. In the sense that in the limit of ever closer hovering observers, this photon is never detected, but also never reaches singularity, it is valid to describe it globally as trapped on the horizon.

    For a free faller, such a photon appears to recede at c, nothing special.

    Consider also two successive free fallers emitting photons the moment each crosses the horizon, each photon trapped in place. This corresponds to the following everyday phenomenon in an inertial frame: A emits a photon toward B, B emits a photon simultaneous with the arrival of photon from A (could be stimulated emission!). The two photons proceed passed B, in tandem.

    Everything that happens at the horizon has a perfectly ordinary local SR description. It is only when you look at the global picture that you discover remarkable features.
     
  12. Sep 28, 2013 #11

    Simon Bridge

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    Which is why I'm trying to encourage OP to include where the observer is and what it's doing when considering these sorts of questions... check to see if the question makes sense.
     
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