So it is accepted that the path that maximizes the proper time between two timelike separated events in Minkowski space is a straight line (in Minkowski space). I am having trouble deriving this from the expression of the proper time. The idea I have is that one should try to find the extremum of the proper time by solving an Euler-Lagrange type equation in Minkowski space (but I have no idea about how to do this) and arrive at an equation of motion for a straight line in Minkowski space. I also don't know what the equation of a straight line might look like in Minkowski space. (I know what a differential line element looks like, but not what a straight line looks like.)

So I tried solving the Euler-Lagrange equations for this action in one-dimensional Euclidean space: $$\int_{t_{0}}^{t_{1}}\sqrt{1-\dot{x}^{2}}\,{\rm d}t$$ and arrived at an equation of a straight line in Euclidean space, which I am sure isn't a straight line in Minkowski space.

Help on how to do this would be appreciated.

  • $\begingroup$ Does this physics.stackexchange.com/q/605567 and this physics.stackexchange.com/q/529682 help? $\endgroup$
    – Cryo
    Commented Jan 23, 2021 at 0:39
  • $\begingroup$ Have you tried to proof that the proper time of a straight line between $A$ and $B$ is always greater than $AC + CB$? Where $C$ is a non colinear point between $A$ and $B$. $\endgroup$ Commented Jan 23, 2021 at 0:53
  • $\begingroup$ What would collinear mean in Minkowski space ? I am confused about what's the equation of a straight line is in Minkowski space. $\endgroup$ Commented Jan 23, 2021 at 1:06
  • $\begingroup$ @Cryo thanks for the references. This is exactly the statement of the twin-paradox, but I am not sure the posts constitute of a proof. $\endgroup$ Commented Jan 23, 2021 at 1:07
  • $\begingroup$ @justcurious92 In a diagram $t \times x$, each point is an event. Straight (timelike) lines are paths between events at constant velocity. $\endgroup$ Commented Jan 23, 2021 at 1:12

1 Answer 1


Here is the proof. Consider two events $p$ and $q$ connected by a timelike segment, i.e., a timelike geodesic of Minkowski spacetime: $$\gamma(\tau) = p + \tau{\bf n}\:, \quad {\bf n}:= \frac{\vec{pq}}{\sqrt{- g(\vec{pq}, \vec{pq})}}\:.$$

Since the constant tangent unit vector ${\bf n}$ to the segment is future directed and timelike, we can fix a Minkowski reference frame $t,x$, where $x\in \mathbb{R}^3$, such that the segment is a portion of the $t$ axis from $t_p$ to $t_q$. In particular, its proper-time length turns out to be $$\sqrt{- g(\vec{pq}, \vec{pq})} = \sqrt{(t_q-t_p)^2 -0^2-0^2-0^2} = t_q-t_p\:.$$ With this choice, exploiting the said coordinates, the length of every timelike curve joining the two events takes the form $$\int_{t_{p}}^{t_{q}}\sqrt{1-\dot{x}^{2}}\,{\rm d}t\leq \int_{t_{p}}^{t_{q}} 1\,{\rm d}t= t_q-t_p\tag{1}$$ where I used the fact that the path is timelike so that its velocity satisfies $|\dot{x}|\leq 1$ and thus $$\sqrt{1-\dot{x}^{2}}\leq 1.$$ Inequality (1) implies that the proper time interval (the length of a curve joining two timelike connected events) attains its maximum along timelike geodesics, i.e., timelike segments in Minkowski spacetime.

  • 1
    $\begingroup$ This comment is to commend this proof to anyone reading. It adopts an approach which gets to the point without invoking unnecessarily complex or advanced methods. The heart of it is first the logic whereby $t_1 - t_0$ is the proper time along the straight path and then equation (1) which does the whole of the rest of the proof in one elegant step. $\endgroup$ Commented Jan 23, 2021 at 10:06
  • $\begingroup$ Thank you Andrew, I know your interest and competence in teaching issues, so I very appreciate your comment. This is the proof I present in first part of my course on mathematical foundations of Relativity. It is interesting also because its generalisation leads to the core of the proof of the so-called Gauss’ lemma which extends the established property to geodesics of general relativity... $\endgroup$ Commented Jan 23, 2021 at 12:46

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