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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.

Calmarius
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2 Answers2

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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.

Zo the Relativist
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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.

Jani Kovacs
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