# In GR, how do particles know how to fall in instead of out of a gravitational well?

The geodesic equation (let's suppose that we're talking about massive particles, so I'll parameterize the path by proper time $\tau$)

$\frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu_{\rho \sigma}\frac{d x^\rho}{d \tau} \frac{d x^\sigma}{d \tau}=0$

is invariant under $\tau \rightarrow -\tau$.

However, falling particles clearly have a direction, they always fall in instead of out. Formally, how does a falling particle 'know' which way to move in $\tau$?

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I don't think I know what you mean by saying that particles "always fall in instead of out." Tell that to the Voyager spacecraft, still merrily moving "out" after all these decades! – Ted Bunn May 15 '11 at 15:53

If you take any solution to the geodesic equation, the time-reversal of that will also be a solution. If one describes a rock falling down in the Earth's gravitational field, the other will describe a rock that was tossed up at some point in the past. The situation is just like Newtonian gravity in that respect.

Mathematically, $\tau$ is just a parameter used to label points on the path. Physically, we have a clear understanding that one direction of $\tau$ is "toward the future" and one is "toward the past," but the fundamental equations of (non-statistical) physics don't distinguish between future and past. That's true for general relativity, Newtonian mechanics, electromagnetism, etc.

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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 – Jerry Schirmer May 15 '11 at 18:44
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.) – Ted Bunn May 15 '11 at 20:54
Well, the extended Kruskal solution is certainly time-symmetric, too. – Jerry Schirmer May 15 '11 at 23:08
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. – Ted Bunn May 16 '11 at 12:42
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. – Ted Bunn May 16 '11 at 17:06
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A geodesic parallel-transports its own tangent vector, so if you "start" the particle pointing into the local (time-like) future it will follow the geodesic in that direction by definition.

Actually, how does a particle "know" how to move forwards in time in plain Minkowski space for that matter? Microscopically it doesn't, this is also put in by definition as for example you can consider antiparticles to be particles travelling backwards through time. Then you have to decide which sort to call particles and which to call antiparticles..

Tau is just enumerating the points on a geodesic path in GR. You still have both forward and backward-tilted lightcones along the path, so somewhere you have to make a choice (which will be conserved by the geodesic as I wrote above).

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What you have noticed implies that both solutions (in and out) are acceptable physical solutions (that is, observing them would not be in violation of known principles). But which solution the particle happens to follow depends on the initial conditions of the motion. I would add that this is not restricted to GR.

Sensing the direction of the time is an "emerging" feature (we have found out that entropy increases, so we can "order" two snapshots taken at different times: the egg was whole, then it was broken and not viceversa).

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