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Moving faster than light in Schwarzshield radius

  1. Jul 15, 2013 #1
    When I watch the spacetime diagram of a particle inside the Scharzshield radius, it seems that the particle can move 'horizontal' (they move dr while X0 stays constant).

    Does this mean the particle can move faster than light inside the Schwarzschield sphere? Or does the meaning of space/time components change (I read somewhere that timecomponent becomes spacelike)?
     
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  3. Jul 15, 2013 #2

    PeterDonis

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    Do you have a reference? What kind of spacetime diagram? There are a number of possible ones, and not all of them have this apparent property; which particular spacetime diagram you use depends on the coordinates you choose.

    No. The particle always moves inside the local light cones. What happens as you get closer and closer to the singularity is that the light cones get tilted further and further inward. (The horizon is the point at which the outgoing side of the light cone is tilted straight up, so to speak--so moving outward at the speed of light keeps you at the same radius.) However, exactly how this is visualized also depends on which coordinates you choose

    This depends on the coordinates as well. In the standard Schwarzschild coordinates, yes, the ##r## coordinate becomes timelike inside the horizon and the ##t## coordinate becomes spacelike. But there are other coordinates (for example, Painleve coordinates) in which ##r## stays spacelike inside the horizon, while the "time" coordinate also becomes spacelike. And there are coordinates (for example, Kruskal coordinates) in which the coordinates don't change character at all inside the horizon.
     
  4. Jul 15, 2013 #3
    It was one in my course, based on schwarzschildcoordinates.

    Thanks for the explanation, very clear :)
     
  5. Jul 15, 2013 #4

    WannabeNewton

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    Schwarzschild coordinates are not valid within the event horizon. They only cover the portion of space-time outside the event horizon. If you want to analyze the behavior within the event horizon, you need a chart valid in that space-time region, such as what Peter mentioned (the Kruskal maximal extension).
     
  6. Jul 15, 2013 #5
    What about Eddington-Finkelstein coordinates? I see that they 'make' the worldline from a particle continues when going over the Schwarzradius. Are they also valid inside the event horizon? How do I see if they are valid?


    Thanks in advance! :)
     
  7. Jul 15, 2013 #6

    WannabeNewton

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    The advanced Eddington coordinates (which follow ingoing null geodesics) can describe the interior black hole region, yes. To see it, just look at the definition of the advanced Eddington coordinates. See here: https://www.physicsforums.com/showthread.php?t=481831
     
  8. Jul 15, 2013 #7

    PeterDonis

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    Yes. They are similar to Painleve coordinates, in that the ##r## coordinate remains spacelike inside the horizon, while the "time" coordinate also becomes spacelike. (Actually, what I just said applies to one form of E-F coordinates; there is another form which uses two null coordinates instead of a "time" and "space" coordinate. That form is also valid at and inside the horizon.)
     
  9. Jul 16, 2013 #8
    Ok, thanks a lot guys! :)


    One 'final' question.

    Let's say an astronaut is moving towards the Schwarzradius. He has a lamp with him and he sends out pulses. I read that the time between this pulses become longer while the astronaut reaches the SR.
    Does this means that we never actually see the astronaut pass the border, as that would take an infinite time for us? Or do we simply stop seeing the pulses and the astronaut turns 'black' as the light doesn't reach us any more?
     
  10. Jul 16, 2013 #9

    WannabeNewton

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    Well the light rays will suffer infinite redshift relative to infinity as the lamp approaches the horizon and the time it takes the light rays to reach our eyes will also approach infinity as the lamp approaches the horizon.
     
  11. Jul 16, 2013 #10

    PeterDonis

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    No, this is not correct. The correct statement is that Schwarzschild coordinates inside the horizon are a separate, disconnected coordinate patch from Schwarzschild coordinates outside the horizon, which means you can't use Schwarzschild coordinates to continuously describe processes that cross the horizon (like objects falling in)*. But the separate Schwarzschild coordinate patch inside the horizon is perfectly valid as a coordinate patch.

    (* - Actually you *can* do this in some cases, if you're willing to go through a lot of extra mathematical gymnastics, taking limits and so on. It's easier, IMO, to just pick a better chart. :wink:)
     
  12. Jul 16, 2013 #11

    WannabeNewton

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    You're right, I should have added the qualifier 'exterior'. My apologies.
     
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