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Speed of a particle inside an event horizon

  1. Oct 27, 2008 #1
    Consider a particle that has fallen inside the event horizon of a black hole. You can show that
    it must have a minimum radial velocity that scales as [tex] \frac{1}{\sqrt{r}} [/tex] for small r. Where, by radial velocity I mean [tex] \frac{dr}{d \tau} [/tex] and tau is the proper time. Doesn't this mean that as the particle approaches the singularity its speed surpasses c?
     
  2. jcsd
  3. Oct 27, 2008 #2
    Only because you're using a weird, non-local system of coordinates. To any local observer, particles will have velocities less than c.

    In other words, you're using proper time, but co-ordinate radius. Inside the black hole, one meter of coordinate radius has basically no relationship with an actual meter as measured by a meter stick by a local observer.
     
  4. Oct 27, 2008 #3
    I see, thanks.
     
  5. Oct 28, 2008 #4

    DrGreg

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    The variable r represents distorted distance outside the event horizon, but inside the event horizon it represents distorted time. Similarly, the variable t represents distorted time outside the event horizon, but inside the event horizon it represents distorted distance.

    I find it hard to mentally picture what that really means, but the maths is clear on this point.
     
  6. Oct 28, 2008 #5
    Is there a way to reasonably define "proper velocity" inside the event horizon? What would be the relation between this coordinate system and the standard Schwarzschild coordinates?
     
  7. Dec 22, 2008 #6
    This is an old thread but I recently found an equation specifically referred to as the 'radial proper velocity of a zero angular momentum observer dropped from infinity' in relation to rotating black holes-

    [tex]
    v=-\frac{\sqrt{2Mr(r^2+a^2)}}{r^2+a^2cos^2\theta}\ c[/tex]

    where

    a=J/mc

    M=Gm/c^2

    and θ is the angle between the zenith and the radial line.

    The entire equation reduces to [itex]v=c\sqrt{r_s/r}[/itex] for a static black [itex](r_s=2M)[/itex]. Interestingly, while proper velocity equals c at the event horizon for a static black hole (relative to infinity), for a rotating black hole, radial proper velocity seems to exceed c not just outside the outer event horizon but outside the ergosphere also.

    source-
    http://arxiv.org/PS_cache/arxiv/pdf/0805/0805.0206v1.pdf page 2
     
    Last edited: Dec 22, 2008
  8. Dec 22, 2008 #7
    Boy am I glad to see someone else say that!! I've read that interpretation several places and it still boggles my mind!!

    But it convinces me even more that space and time have underlying origins and fundamental relationships we do not understand.
     
  9. Dec 23, 2008 #8
    A couple of equations regarding velocity from 'Exploring Black Holes' by Taylor & Wheeler (chapter 3: Plunging, project B: Inside the Black Hole) based on static black holes-

    Rain frame (object dropped from rest at infinity) time between given radii-

    [tex]\tau_{2\ rain} - \tau_{1\ rain}\ =\ \frac{1}{3}\sqrt{\frac{2}{M}}\left(r_1^{3/2} - r_2^{3/2}\right)[/tex]

    where

    [itex]M=Gm/c^2[/itex]

    Divide the answer by c to get time, multiply by c (over 1m increments) to get velocity. Unity (i.e. 1) at EH. When you put in the Schwarzschild radius for [itex]\tau_{2\ rain}[/itex] and zero for [itex]\tau_{1\ rain}[/itex], you basically get the horizon to crunch distance for an object falling radially from rest at infinity-

    [tex]\tau_{rain}(2M \rightarrow 0)=\frac{4}{3}M[/tex]

    divide by c to get the horizon to crunch time for an object falling radially from rest at infinity.


    velocity of a free-falling object as clocked by the shell observer (r>2M)-

    [tex]v_{shell}=\frac{dr_{shell}}{dt_{shell}}=-\sqrt{\frac{2M}{r}}[/tex]

    (minus square root because the expression describes a decreasing radius as the object falls toward the black hole outside the horizon)


    proper velocity (can apply to r<2M)-

    [tex]\frac{dr}{d\tau_{rain}}=-\sqrt{\frac{2M}{r}}[/tex]


    velocity as viewed from infinity (v=0 at event horizon)-

    [tex]\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}[/tex]


    Horizon to crunch distance for an object falling radially from rest at the event horizon (drip frame)-

    [tex]\tau_{max}(2M \rightarrow 0)=\pi M[/tex]

    divide by c to get the horizon to crunch time for an object falling radially from rest at the event horizon.

    It appears the whole concept of distance inside a black hole is relative to how fast you cross the event horizon.


    Draft copy of second edition of EBH-
    http://www.eftaylor.com/comments/
     
    Last edited: Dec 24, 2008
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