1

I only have a very basic understanding of general relativity, so this might sound stupid. But as I see that gravitational objects orbiting each other, even if one of them is massive, as long as their paths don't lead to collision, they curve back, like a sling shot.

So since singularity is supposed to have zero radius, it seems nothing should actually collide with it. And even if it's non-zero, only near zero, there could still be a possibility that a particle falling (depending on its initial orbit) could eventually just get a sling shot around it, never colliding with it or falling into it.

Does that make sense? Or do all paths really lead directly at the exact the center of the singularity?

Qmechanic
  • 201,751
laggingreflex
  • 862
  • 1
    You imagine the inside of the black hole in terms of Euclidean geometry, something like the inside of a sphere with singularity at the centre. That is very misleading. – MBN Jun 09 '15 at 09:20

1 Answers1

1

For a non-rotating uncharged black hole (the Schwarzschild metric) once you have crossed the event horizon there is no timelike trajectory that will increase the distance between you and the singularity. By this I mean that to increase your distance from the singularity would require you to be moving faster than light, which is of course impossible. So all paths lead ultimately to the singularity - note I say ultimately rather than directly because the paths are generally not just a straight line to the singularity.

Understanding why this happens requires a grasp of general relativity beyond the non-nerd. However there are various papers that can give you some idea why this happens, though they will still need some effort on your part. I quite liked No Way Back by Lewis and Kwan. The river model of black holes is also an interesting read.

For charged (Reissner-Nordstrom) or rotating black holes (Kerr/Kerr-Newman) there are in principle time-like paths that lead in through the event horizon then back out again. See for example Entering a black hole, jumping into another universe---with questions. However I should note that the physical reality of such trajectories is controversial. The geometries where they exist describe eternal black holes i.e. black holes that have existed forever and will continue to exist forever. In the real universe black holes can't be older than 13.7 billion years (and counting) and we believe they will eventually evaporate due to Hawking radiation.

Community
  • 1
John Rennie
  • 355,118
  • That is true for Schwarzschild black holes, but Kerr black holes do allow you to avoid crashing in the singularity (at the expense of acceleration). – Slereah Jun 09 '15 at 11:15
  • @Slereah: yes, good point. I've modified my answer. – John Rennie Jun 09 '15 at 11:25
  • 1
    I think this is the correct explanaition for the way the question is asked, but strictly speaking you cannot talk about the distence between you and the singularity. – MBN Jun 09 '15 at 12:03
  • @MBN: to avoid ambiguity I mean all trajectories inside the event horizon have $dr/dt < 0$, where $r$ and $t$ are the Gullstrand-Painlevé coordinates. So by distance I mean the G-P $r$ coordinate (which is actually the same as the Schwarzschild $r$ coordinate anyway). – John Rennie Jun 09 '15 at 14:34
  • 1
    @JohnRennie: I understand that, I think it is a good explanation, thus +1. But pedantically $r$ is not even space like in that region, so it is not a distance. – MBN Jun 09 '15 at 18:21