It seems if I reverse velocities then things begin orbiting backwards, at least in classical mechanics.

From here:

Every orbit and trajectory outside atmospheres is in principle reversible, i.e., in the space-time function the time is reversed. The velocities are reversed and the accelerations are the same, including those due to rocket bursts. Thus if a rocket burst is in the direction of the velocity, in the reversed case it is opposite to the velocity. Of course in the case of rocket bursts there is no full reversal of events, both ways the same delta-v is used and the same mass ratio applies.

What's up when I put relativistic effects into the mix?

So for example I watch a super light test particle orbiting a black hole in a highly precessing flower shaped orbit. Then I put a bouncy wall into it's path that's at rest from my viewpoint when the particle hits it, so the particle bounces back reversing it's velocity from my viewpoint.

Would it begin running its orbits backwards?

The actual reason I'm asking this, because I want to know whether I can use backwards ray-tracing to render a black hole.

  • $\begingroup$ Without knowing in any detail (I'm not a relativist) I would imagine that you have to reverse the gravitational radiation as well as the motion of the massive bodies for the answer to be "yes", though in the weak field approximation you can reasonably ignore this for most purposes. $\endgroup$ – dmckee --- ex-moderator kitten Jul 26 '13 at 17:06
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    $\begingroup$ I think your question is answered here. physics.stackexchange.com/q/9881 Take a look at the comments as well. $\endgroup$ – joshphysics Jul 26 '13 at 17:44
  • $\begingroup$ @joshphysics I think Ben Crowell's (new) answer there leaves sufficient doubt as to how this scenario plays out that this warrants a separate question. Time reversibility is shown there to be on shaky ground, but then what does happen if you reflect a massive particle around a BH? Is there anything we could say about its orbit? What if it's a Kerr BH? $\endgroup$ – user10851 Jul 27 '13 at 0:21
  • $\begingroup$ @ChrisWhite Sounds reasonable to me. $\endgroup$ – joshphysics Jul 27 '13 at 0:39
  • $\begingroup$ Addressing just the last sentence - there is a whole subfield of astronomy concerned with the paths light takes near strongly gravitating bodies. This is because we observe many binary systems with at least one member being a white dwarf, neutron star, or black hole, and at least one member a glowing object. We would like to understand the amount of light we receive as a function of time. Often this analysis is done with some form of ray tracing. $\endgroup$ – user10851 Jul 27 '13 at 1:02

Yes, the Schwarzchild space-time is reversible. Closed orbits and the like will stay closed in the time-reversed system.

There is, of course one obvious flaw: what about the horizon? Things go in, but they do not go out. Well, the answer to that is that the full spacetime doesn't JUST include a black hole, it includes a white hole/black hole pair. If you time reverse the spacetime, you tranform a particle falling into the black hole into a particle falling out of the white hole (and eventually, back into the black hole). This doesn't come up, because we typically don't consider spacetimes containing a white hole as real physical solutions. But in the idealized, mathematical case, you have to include it.


For the simple case of a black hole in a locally flat spacetime(so that it does not accelerate) you don't have to worry about losing precision due to relativistic effects, because Schwarzschild and Kerr black holes emit exactly zero gravitational radiation(because they are both cylindrically symmetric), that is: the metric does not change over time so you can reverse the photons and get arrive at the same path.


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