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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

5 Answers 5

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

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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Reparametrizing the geodesic doesn't change the geodesic at all. The geodesic is simply a set of points, and that set of points gives a complete description of the motion. Reparametrizing doesn't change the set of points.

So reparametrizing like $\tau\rightarrow-\tau$ doesn't reverse the motion. You can see this in Minkowski space, where motion in opposite directions is represented by two different lines (two different point-sets) in the $x-t$ plane. Similarly, $\tau\rightarrow2\tau$ doesn't affect the validity of the solution to the geodesic equation, and it doesn't represent motion that's slowed down by a factor of 2 (which obviously wouldn't be physical in most cases).

GR doesn't even have a general notion of global time-reversal. For a given spacetime and a given geodesic, you're not guaranteed to have any well-defined notion of what the time-reversed version of the geodesic would be. For example, here's the Penrose diagram for a black hole:

enter image description here

The red line is a (null) geodesic. It is not even possible to define a time-reversed version of this geodesic. Locally, time-reversal would mean flipping it so it had the opposite slope. But globally, there is no specific line with the opposite slope that deserves to be called the time-reversal of the red line.

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The answer given doesn't refer to the OP question at all; notions like "direction" or "in instead of out" (of the OP question) don't appear recognizable in the answer. –  user12262 Jul 27 '13 at 8:52
@user12262: The OP's statement of the question clearly shows that the OP understands that asking about in/out is equivalent to asking about time reversal. –  Ben Crowell Jul 27 '13 at 14:17
"[...] the OP understands that asking about in/out is equivalent to asking about time reversal." -- Doubtful: OP rather suggests the idea of "in/out reversal" (or likewise "out/in reversal") being associated with a sign reversal "$\tau/-\tau$". (Perhaps OP has meanwhile learned that's wrong; so much the better.) However: OP's question wasn't about "in/out reversal" (or likewise about "out/in reversal") at all, but instead about how to determine whether "in" or "out". To that question your answer appears merely: "For a given spacetime and a given geodesic". –  user12262 Jul 28 '13 at 16:31

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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What you have noticed implies that both solutions (in and out) are acceptable physical solutions. No, as explained in my answer, the reparametrization doesn't change an in-going geodesic to an out-going one. –  Ben Crowell Jul 26 '13 at 21:37

A (stationary) "well" ($W$) has a certain geometric asymmetry (among a system $S_W$ of participants who remain rigid wrt. each other, "around" well $W$)
from which derives the "direction of (initial) free-fall upon being released from the rigid system" or short: the "direction straight into the well, ${\widehat {\mathbf a_W}}$"; vs. other directions.

If some (other) "particle" (${\mathbf P}$) met and passed some members of the rigid system $S_W$ in some particular order, and if they were asymmetric to each other in terms of direction ${\widehat {\mathbf a_W}}$ then particle ${\mathbf P}$ is said to have "fallen in" (or "risen out of") well $W$ (regardless of whether or not it had been otherwise "free").

If the order of those passages was entirely or predominantly towards direction ${\widehat {\mathbf a_W}}$, formally

$\langle {\widehat {\mathbf a_W}} \cdot {\mathbf P} \rangle \gt 0$,

where the ordered set ${\mathbf P}$ is understood as directed "from past to future",

then particle ${\mathbf P}$ is said (or "'knows'") to have fallen in the well, instead of risen out.

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This answer uses a lot of nonstandard, undefined, and irrelevant terminology. I don't see the relevance of "participants" to a discussion of geodesics. If "rigid" refers to the notion of Bohr rigidity (which AFAIK is the only widely discussed notion of rigidity in relativity), I don't see how that's relevant either. –  Ben Crowell Jul 27 '13 at 5:02
@Ben Crowell: "I don't see the relevance of "participants" to a discussion of geodesics" -- The OP question is stated in terms of "direction" and "falling particles"; not "geodesics". In addressing the question I don't see any need for discussing or even mentioning "geodesics" either. "If "rigid" refers to [...]" -- "rigid" refers to constant ping durations, cmp. physics.stackexchange.com/questions/38377/… This notion is useful for characterizing "wells" and associated "directions". –  user12262 Jul 27 '13 at 8:47
Oops, "Bohr" should have been "Born" in the comment above. –  Ben Crowell Jul 27 '13 at 14:34
There is a meta discussion going on about you : Please participate.. –  Dimensio1n0 Aug 17 '13 at 8:19

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