In flat, free, Euclidean space, the shortest path and the zero acceleration path are the same path, which is a straight line. However, in general relativity, is the zero acceleration path also the shortest path between two points? I am assuming that free fall is zero acceleration.

  • $\begingroup$ Related: physics.stackexchange.com/q/24359/2451 $\endgroup$ – Qmechanic Sep 12 '14 at 20:54
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    $\begingroup$ The answer to this question is a bit subtle, and depends on what you mean by "two points". General relativity is a theory of spacetime, and the points in spacetime are places in space <b>at</b> an exact instant. The geometry of spacetime says, then, that two events fall into three categories, being spacelike seperated if the distance between them is greater than zero, timelike seperated if the distance between them is less than zero, and null seperated if it is zero. I'm going to come back and write a proper answer to this later, but you have to consider each of these cases, $\endgroup$ – Jerry Schirmer Sep 12 '14 at 20:56
  • $\begingroup$ as well as the commonsense idea of "distance between two points", which becomes "the distance between two timelike curves through spacetime" $\endgroup$ – Jerry Schirmer Sep 12 '14 at 20:56

In general relativity, you're dealing with a 4D spacetime, so the "points" in spacetime are events, and the measures that you can make coordinate-independent statements about are intervals instead of distances.

The rule that applies is that the world line with the longest possible proper time between two events is a world line that involves zero proper acceleration. Such a world line is called a "time-like geodesic".

There's a similar concept for space-like curves. A "space-like geodesic" is a curve with a stationary proper length between two events with a space-like separation. A space-like geodesic is locally straight.

For more information, see the Wikipedia article section "Geodesics as curves of stationary interval"

And yes, free fall means zero proper acceleration.


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