# I Why do particles move along longest proper time trajectories

1. Jun 22, 2016

### pervect

Staff Emeritus
A slightly more precise statement is one that says the curve of a body in "natural motion", i.e. force-free motion, extremizes proper time. Note that this is what Hartle said, formally. (The excrutiatingly correct statement, according to the referece I just read, is that the curve of natural motion is a curve of stationary action).

http://www.eftaylor.com/leastaction.html has some helpful articles on the action principle, specifically "When action is not least" addresses the issue of why we say the action is stationary rather than minimal.

As discussed in "When action is not least", the idea that action is "least" rather similar to the idea that a spatial geodesic minimizes length. This is always true on a plane, it's true only in a sufficiently small local region on a curved spatial geometry such as the surface of the sphere. I'll leave it at that rather than try to explain further, and refer the interested reader to the article.

As far as your particular question goes, I'd say that the short and simple version is that proper time is maximized only if a) one restricts oneself to a set of time-like curves, and b) consider only a "sufficiently small" region of space-time. The problem in your case is a). Things do break down when you try to apply the principle of least action to light-like curves. The solution to this is not to do it - instead, apply the correct principle of stationary action.

2. Jun 22, 2016

### MeJennifer

Time taken for which observer?

For instance consider the following thought experiment.

In a Schwarzschild solution stationary observers Ar1 and Br2 (r2 > r1 > r-event horizon) send each other light signals and clock the time it takes light to go from A to B and back to A and from B to A and back to B.

Would you think light takes exactly the same round trip time for both observers?

3. Nov 5, 2017

### Gerry99

Only need to prove the proper time between p0 and p2 is larger than p0 to p1 + p1 to p2.
We can make p0 to p1: {t1, x1} = τ1{cosh a1, sinh a1}, p1 to p2: {t2, x2}=τ2{cosh a2, sinh a2}, so p0 to p2 is {t3, x3}={τ1cosh a1 + τ2 cosh a2, τ1 sinh a1 + τ2 sinh a2}
only need to prove |S1| + |S2| < |S3|