Timeline for In GR, how do particles know how to fall in instead of out of a gravitational well?
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7 events
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| May 16, 2011 at 17:06 | comment | added | Ted Bunn | I agree completely. What I meant was something like this: Imagine an observer at some location in spacetime. He watches a particle move through his neighborhood on a geodesic. He then imagines a particle moving along the time-reversed version of that geodesic -- that is, a geodesic that "looks like" the original, running backwards in time, keeping himself and the local geometry unchanged. Is the result a physically possible path for a particle? The answer is yes outside the horizons, but no inside. That's because, inside the horizons, the spacetime lacks a "local" time-reversal symmetry. | |
| May 16, 2011 at 14:09 | comment | added | Zo the Relativist | Well, I'd argue that the "real" (I hesitate to say this in such an unphysical spacetime) geodesic is one that extends from the white hole and goes into the black hole, and we only see a portion of it when we look at the space in Schwarzschild coordinates. Then, $t\rightarrow -t$ $\tau\rightarrow -\tau$ flips the white hole and the black hole and reverses the direction of time flow, and you have the exact same picture. It's only when you cut off the solution with the collapsing matter that this changes. | |
| May 16, 2011 at 12:42 | comment | added | Ted Bunn | The entire spacetime is time-symmetric, but it doesn't have a time-reversal symmetry at each point: there's not a symmetry that reverses future and past, preserves a given spacetime event, and preserves the geometry. It seems to me that that's what's relevant here: if you want to replace a geodesic with another one by substituting $\tau\to -\tau$, and think of the new geodesic as "just as good" as the previous one, then you need to move to a different part of your spacetime: instead of one falling into the BH horizon, you get one popping out of the WH horizon, which is at a different time. | |
| May 15, 2011 at 23:08 | comment | added | Zo the Relativist | Well, the extended Kruskal solution is certainly time-symmetric, too. | |
| May 15, 2011 at 20:54 | comment | added | Ted Bunn | That's true, and it's an important point to add. What's going on here is that the spacetime metric itself isn't time-symmetric, so the forward-in-time and backward-in-time solutions are physically different. (The Schwarzschild solution is often thought of as time-symmetric, because the metric doesn't change under $t\to -t$, but at the horizon, $t$ is no longer a timelike coordinate, so this symmetry is not a time-reversal symmetry.) | |
| May 15, 2011 at 18:44 | comment | added | Zo the Relativist | The only caveat here is that if you have a particle falling into a black hole, the time-reversed path would be the particle falling out of a white hole, which we expect to be unphysical | |
| May 15, 2011 at 15:59 | history | answered | Ted Bunn | CC BY-SA 3.0 |