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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.

Buzz
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Bedabrata Choudhury
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  • Does this https://physics.stackexchange.com/q/605567/ and this https://physics.stackexchange.com/q/529682/ help? – Cryo Jan 23 '21 at 00:39
  • 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 $B$. – Claudio Saspinski Jan 23 '21 at 00:53
  • What would collinear mean in Minkowski space ? I am confused about what's the equation of a straight line is in Minkowski space. – Bedabrata Choudhury Jan 23 '21 at 01:06
  • @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. – Bedabrata Choudhury Jan 23 '21 at 01:07
  • @justcurious92 In a diagram $t \times x$, each point is an event. Straight (timelike) lines are paths between events at constant velocity. – Claudio Saspinski Jan 23 '21 at 01:12
  • @Claudio Sapinski constant four-velocity ? I was trying to come at an intuitive solution from what I know so far. There is an action that should be maximized in some non-euclidean(in this case Minkowski) space to arrive at the solution of a straight line in said space. – Bedabrata Choudhury Jan 23 '21 at 01:21
  • @justcurious92, can you state more concisely what you are trying to prove? That straight line is the between two events is the one that maximizes the proper time along this line? How do you define straight line? By maximizing the proper time? Then you have a tautology. – Cryo Jan 23 '21 at 01:25

1 Answers1

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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.

Valter Moretti
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    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. – Andrew Steane Jan 23 '21 at 10:06
  • 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... – Valter Moretti Jan 23 '21 at 12:46