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When will they meet?

  1. Jun 2, 2010 #1
    In the Schwarzschild solution consider two test point particles, p1 and p2, both at the same distance from the black hole (e.g. both particles have the same r value) but separated a distance d from each other (so they make an angle with respect to the black hole)

    So now the question: when will their paths cross? At the event horizon or at the singularity?

    If you say at the singularity then what is their distance in terms of d when they both cross the event horizon?
     
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  3. Jun 2, 2010 #2

    George Jones

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    You haven't been specific enough. Are the particles falling radially? Falling initially from rest (with respect to Schwarzschild coordinates)? Falling form rest at infinity?

    Distance is tricky and ambiguous in general relativity. What does "separated a distance d from each other" mean?
     
  4. Jun 2, 2010 #3
    Yes they are falling radially, you can take any condition you like: falling from rest at infinity (problem then would be how to define any notion of distance between them) or initially falling from rest. In other words the two particles are exactly equivalent (the Schwarzschild solution is symmetrical after all). With regards to d, you are free to choose. I understand that talking about distance is tricky. However one could perhaps envision the definition of a function (valid for the Schwarzschild solution) that translates a given d at the initial position to a d at the event horizon. Alternatively you can express the situation with an angle or if everything else fails just pick some coordinate distance that you prefer. Also consider the extreme case, two test particles opposite to each other with the black hole exactly in the middle (this one could in principle be expressed in terms of 'falling from rest at infinity' using symmetry).

    However if it is not possible to express any kind of distance then the question becomes: how can you prove they do not meet at the event horizon.

    :smile:
     
    Last edited: Jun 2, 2010
  5. Jun 2, 2010 #4

    Ich

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    If you mean the distance as measured along an angular direction (i.e. a long a line of constant r), then it's simply the constant angle between the particles times the Schwarzschild radius r.
    In case you're referring to Antoci: r=2m at the horizon, not 0.
     
  6. Jun 2, 2010 #5

    DrGreg

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    One way of looking at this is not to try to draw a "straight line" between the two points, but instead to consider the arc length of the circle that they both lie on (centred on the black hole). If the two points are close together (i.e. d is tiny compared with the radius), that's a good approximation anyway.

    Choose your coordinate system so that both points are on the "equator" [itex]\theta = \pi / 2[/itex]. The circumferential distance between the points is just [itex]r (\phi_2 - \phi_1)[/itex] and both particles move along worldlines of constant [itex]\phi[/itex] and [itex]\theta[/itex]. So the circumferential distance in the limit as they approach the event horizon is just [itex]r_s (\phi_2 - \phi_1)[/itex] where [itex]r_s[/itex] is the Schwarzschild radius.

    Oops, Ich beat me to it!
     
  7. Jun 2, 2010 #6

    George Jones

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    If any event P is on the worldlines of both particles, then they meet. If there is no event P common to both worldlines, they never meet. If event P is on the worldlines of both particles, and event P is on the event horizon, then they meet on the event horizon.

    Schwarzschild coordinates are really two disjoint coordinate patches, and neither patch covers the event horizon. Another coordinate patch that does cover the event horizon should be used to determine if particles meet on the event horizon.
     
  8. Jun 2, 2010 #7

    DrGreg

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    Just had a thought. The picture of Flamm's paraboloid seems to indicate that as you approach the event horizon (where the paraboloid becomes vertical), the circumference really is the geodesic distance (using the Schwarzschild coordinates to decompose spacetime into space+time).
     
  9. Jun 2, 2010 #8
    So you are saying they meet on the event horizon?

    No problem, you could use any chart for the Schwarzschild metric that is suitable (provided of course there are no "voodoo" mathematical translations).
     
  10. Jun 2, 2010 #9

    George Jones

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    If I understand the initial conditions correctly, I say that the particles do not meet on the event horizon.
    What translations to you consider to be'"voodoo" mathematical translations'?
     
  11. Jun 2, 2010 #10
    I did not have anything particular in mind when I wrote that.
     
  12. Jun 2, 2010 #11

    Dale

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    Then they do not meet until the singularity.
     
  13. Jun 2, 2010 #12
    Can that really be inferred, at any point past the event horizon? To my mind, anything that occurs beyond the EH is "who knows". Perhaps they never meet?
     
  14. Jun 2, 2010 #13

    Dale

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    The event horizon is a coordinate singularity only and can easily be removed through suitable choice of coordinates. So to solve this problem you would simply pick some such coordinate system that is well-behaved everywhere except the singularity.
     
  15. Jun 2, 2010 #14

    JesseM

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    There would certainly be a well-defined theoretical answer given by general relativity. Are you asking about the possibility that GR might give inaccurate predictions in this scenario?
     
  16. Jun 2, 2010 #15
    This is what I was getting at, but as DaleSpam pointed out that doesn't really apply here, so I will bow out. :)
     
  17. Jun 3, 2010 #16

    JesseM

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    Well, DaleSpam's comment was just about the fact that there's no problem deriving the answer theoretically in general relativity, he wasn't addressing the possibility that GR might be wrong one way or another.
     
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