3
$\begingroup$

I've started studying general relativity course and now I have a question about proper time. Consider functional $$S[x]=-\int_A^B ds,$$ where $A$, $B$ are fixed points of the space-time and $ds^2=dt^2-dx^2$ (let our space-time be two-dimensional without loss of generality). Finding its minimum is equivalent to maximizing proper time $s=\int_A^B ds$ and it a well-known fact that maximizers of proper time are straight lines, as it can be easily checked by writing Lagrange equation for $\mathcal{L}(t,x,\dot{x})=-\sqrt{1-\dot{x}^2}$. So my question is: what are minimizers of proper time?

I've heard that there are, hm, lots of them, but I can't write down any. If one wants to minimize $s$, he takes lagrangian $\mathcal{L}=\sqrt{1-\dot{x}^2}$ and he obtains that $$\ddot{x}=0,$$ so minimizers necessarily are straight lines. But we have already proved that they are maximizers, so I come to contradiction. Where am I wrong?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

  1. Given two (timelike separated) points in Minkowski space, there is a unique timelike curve that maximizes the proper time, namely the straight line, as OP already mentions. However there is no timelike curve that minimizes the proper time. Nevertheless proper time does have an infimum, namely zero. This is essentially because a massive point particle can always fly a bit closer to the speed of light without reaching it.

  2. For a general action functional, the Euler-Lagrange (EL) equations yield stationary configurations. Extremal configurations might not exist.

    Example. For a differentiable function $f:I\to \mathbb{R}$ on an (open or closed) interval $I$, recall that a stationary point is neither a necessary nor a sufficient condition for an extremum for $f$. Similar statements are true in calculus of variations.

Improve this answer
$\endgroup$
6
  • $\begingroup$ well, again: if a function $f\colon I\to\mathbb{R}$ is differentiable and $x\in I$ is its extremum point, then it is stationary point. It is known as Fermat theorem. For Euler-Lagrange equation, if $x$ is local extremum of an action $\int_{t_A}^{t_B}\mathcal{L}(t,x,\dot{x})dt$, THEN it is a solution of EL-equation, that's the point. I use exactly this statement in my reasoning. $\endgroup$
    – igor.tsutsurupa
    Commented Sep 10, 2019 at 13:53
  • $\begingroup$ Not necessarily if $x$ is an endpoint of $I$. $\endgroup$
    – Qmechanic ♦
    Commented Sep 10, 2019 at 13:55
  • $\begingroup$ ok, let's talk about open $I$. Anyway, the question concerns to the statement about EL-equation $\endgroup$
    – igor.tsutsurupa
    Commented Sep 10, 2019 at 14:02
  • $\begingroup$ Then the main issue is that an extremum might not exist. $\endgroup$
    – Qmechanic ♦
    Commented Sep 10, 2019 at 14:37
  • $\begingroup$ That's the point! In my case, I realize that if minimizer of proper time exists, it has to be straight line, but it's absurd. However, I'm confused because my lecturer told me that minimum exists and it is attained on infinite set of curves. $\endgroup$
    – igor.tsutsurupa
    Commented Sep 10, 2019 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.