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Is it impossible for a particle (with zero angular momentum) to free-fall from rest at infinity into the ergosphere of a Kerr black hole? It seems like it is very easy to show this is the case, but most of what I've read seems to skirt over this.

Qmechanic
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John Davis
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    Free fall from infinity? That might take a while. :-) – userLTK May 09 '15 at 19:49
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    Yeah from infinity, that cant really be answered because you're an infinite distance away...technically at that point the gravitational force on you would be zero – Triatticus May 09 '15 at 20:04
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    @Dan the concept of free-fall from infinity at rest is not a problem as it refers to the asymptotic properties of the particle. – John Davis May 09 '15 at 21:05
  • Oh yes im very aware of that, most time i hear this problem it is phrased as an observer at infinity observes an additional observer approach the black hole from a much closer distance...like say the schwarzchild radius – Triatticus May 09 '15 at 21:07
  • However what i know about the kerr metric is that it depends on the angle of approach to a kerr black hole. If you come in along the axis of rotation you hit both the ergosphere and schwarzchild radius at the same time, elsewhere you hit the ergosphere first. There isnt anything stopping one from entering the ergosphere but the only requirement is that the particle will begin to rotate with the black hole as spacetime is dragged along with the blackhole – Triatticus May 09 '15 at 21:20
  • @Dan I'm not finding the comments very helpful, my question relates to particles having specific properties, I'm not asking whether it is generally possible for a particle to enter the ergoregion. I think the answer to my question is that from the point of view of an observer at infinity it is not possible for such a particle to enter the ergoregion as the Killing vector field of asymptotic time translational symmetry becomes spacelike, but I have a nagging feeling I may have made an error. – John Davis May 09 '15 at 22:08
  • What properties are you interested for these particles – Triatticus May 09 '15 at 22:16
  • That they are at free-falling, have zero angular momentum and are at rest at infinity – John Davis May 09 '15 at 22:18
  • What do you think would happen to an object dropped into a black hole from infinity if it didn't enter the ergosphere? – Peter Shor May 09 '15 at 23:20
  • John , why don't you simply show us the eqs you have made and we start talking from there? This is not a philosophy Q&A place. – magma May 09 '15 at 23:31
  • It sounds as you're asking whether the coordinate time for a particle on an initially radial trajectory with zero total energy to reach the ergosphere is infinite. Is this a fair rephrasing of your question? – John Rennie May 10 '15 at 09:37

1 Answers1

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From this paper$^1$ we have the equations for a particle of zero total energy on an infalling trajectory in the equatorial plane:

$$\begin{align} \Sigma\frac{d\theta}{d\tau} &= 0 \\ \Sigma\frac{dr}{d\tau} &= -\sqrt{2Mr(r^2 + a^2)} \\ \Sigma\frac{dt}{d\tau} &= -a^2\sin^2\theta + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} \\ \Sigma\frac{d\phi}{d\tau} &= -a \left( 1 - \frac{r^2 + a^2}{r^2-2Mr+a^2} \right) \\ \Sigma &= r^2 + a^2\cos^2\theta \\ a &= \frac{J}{M} \end{align}$$

If you're happy to assume that the proper time remains finite, then to find out if the coordinate time becomes infinite anywhere we simply ask if $dt/d\tau$ becomes infinite anywhere. If we use the simplification that the trajectory is equatorial ($\theta = \pi/2$) then $dt/d\tau$ simplifies to:

$$ r^2\frac{dt}{d\tau} = -a^2 + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} $$

and this becomes infinite when:

$$ r^2-2Mr+a^2 = 0 $$

which has the two solutions:

$$ r = M \pm \sqrt{M^2 - a^2} $$

or in a more familiar form:

$$ r = \frac{r_s \pm \sqrt{r_s^2 - 4a^2}}{2} $$

But the ergosphere is at $r = r_s$ in the equatorial plane, so $dt/d\tau$ is not infinite at the ergosphere and this implies the infalling particle will reach the ergosphere in finite coordinate time.

$^1$ there's nothing special about this paper, it was just the first paper I found when doing a Google search.
John Rennie
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