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Assuming Minkowski spacetime, we know that the longest proper time curve joining two points is the rect joinining both events, While the shortest time-like curve is not a compact set (because there are sequences with limits that do not belong to the set) the proper acceleration is unbounded in such curves.

My question is this: is there an easy way to prove that, time-like curves between $p_0$ and $p_1$, with

$$ \bigl |\bigl| \frac{d^2 z}{d \tau^2} \bigr|\bigr| \le K $$

Have a shortest proper time curve joining two given points belonging to the set, and here it comes the hard part: That the minimal proper time curve is not of constant acceleration.

Is this assertion even true? How can i see it intuitively.

The curves must be at rest at $p_0$ and $p_1$ in some reference frame where both planets are at rest. So, the curve needs to accelerate in order to reach destination

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the integral is important, otherwise the statement is trivially wrong; since the curve with constant proper acceleration K will be already an extremal curve – lurscher Jun 10 '13 at 3:50
I know what you mean, but this is a different problem, thanks for the clarification though – diffeomorphism Jun 10 '13 at 3:53
I can't answer your question, but to my intuition it seems like the minimal proper time curve will "spiral around" an inertial trajectory, with the acceleration always perpendicular to the direction of motion, and always of constant magnitude. Without the bound on the acceleration, you can probably always make a light-like trajectory that way, i.e. zero proper time. – Nathaniel Jun 10 '13 at 3:56
Nathaniel, that is the point in this exercise, finding the minimal proper time that is subject to realistic constraints, i.e: finite proper acceleration. – diffeomorphism Jun 10 '13 at 3:59
Sure. I just meant that I expect the solution with finite acceleration will have this form as well. It's just intuition, though, I can't prove it. – Nathaniel Jun 10 '13 at 4:00

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