In curved spacetime particles follow timelike geodesics, which should have maximal proper time (at least locally). I thought this path usually corresponds to a global maximum, and there are only strange exceptions. Propably I was wrong, or I really don't understand something, because it doesn't seem to be true for the case of orbitals in Schwarzschild geometry.

The Schwarzschild metric is (with +--- signature)

$$ g_{\mu\nu} = \textrm{diag}\left(1-\frac{2GM}{r},-\left(1-\frac{2GM}{r}\right)^{-1},-r^2,-r^2\sin^2\theta\right) $$

An observer with fixed $r,\theta,\phi$ coordinates will age according to $d\tau_1 = \sqrt{g_{tt}}dt$. An observer, which orbits the center at the same constant radius with $\theta=\pi/2$ will age according to $d\tau_2 = \sqrt{g_{tt}-\omega^2r^2\sin^2\theta}dt$, so he ages slower during the same amount of coordinate time $dt$. If the two observers start from the same point, they will meet again after a full rotation of the orbiting observer, and the orbiting observer will age less. What I don't understand is that the orbit is the geodesic (if the angular velocity is set properly), so why didn't he age more? He should have been on the path with maximal proper time.


1 Answer 1


I think the answer lies in the fact that the variational principle says is if there are two fixed points in space-time then there exists an open subset of space-time containing these two points such that amongst all curves contained in this open subset, a geodesic will be the curve of longest length between these two points.

So, the variational principle says geodesics maximizes length amongst all curves in space-time nearby the geodesic. The world line of the static observer in your example is certainly not nearby the world line of the observer in circular orbit between two fixed events for which you measure the elapsed proper time.

  • $\begingroup$ I also suspect this, but I find it hard to believe. You suggest that the orbital geodesic is only a local maximum. Is there a global maximum then as well? Does that solve the equations of motions as well? There can't be any other solution, otherwise the motion wouldn't be determined uniquely. On the other hand the proper time on any path between the two points should have an upper bound. I am confused. $\endgroup$
    – G_Almasi
    Commented Nov 20, 2014 at 15:39
  • $\begingroup$ Very good question; may I refer you to the following posted question. I think you may take something from the answer. physics.stackexchange.com/questions/18781/… $\endgroup$
    – Autolatry
    Commented Nov 20, 2014 at 16:23

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