Four events and the reversed inequality

Problem

In the SR of a laboratory, consider four events $(t,s)$ in seconds and light-seconds: $E_1=(3,2)\quad E_2=(7,3)\quad E_3=(11,5)\quad E_4=(15,8).$ A traveller can go from $E_1$ to $E_4$ in a straight line, or by breaking the journey at points $E_2,E_3$. Calculate the two proper times and verify that the uniform straight-line trajectory maximises the proper time.

Solution

Natural units $c=1$. On each straight segment the proper time is the invariant $\Delta\tau=\sqrt{\Delta t^2-\Delta s^2}$.

Straight-line trajectory $E_1\to E_4$. $\Delta t = 15-3 = 12$~s, $\Delta s = 8-2 = 6$~ls: $\Delta\tau_\text{straight}=\sqrt{12^2-6^2}=\sqrt{144-36}=\sqrt{108}=6\sqrt{3}\approx 10{,}39\ \text{s}$

Broken trajectory $E_1\to E_2\to E_3\to E_4$. Segment by segment: $\Delta\tau_{12}=\sqrt{(7-3)^2-(3-2)^2}=\sqrt{16-1}=\sqrt{15}\approx 3{,}87\ \text{s}$ $\Delta\tau_{23}=\sqrt{(11-7)^2-(5-3)^2}=\sqrt{16-4}=\sqrt{12}=2\sqrt{3}\approx 3{,}46\ \text{s}$ $\Delta\tau_{34}=\sqrt{(15-11)^2-(8-5)^2}=\sqrt{16-9}=\sqrt{7}\approx 2{,}65\ \text{s}$ Summing: $\Delta\tau_\text{broken}=\sqrt{15}+2\sqrt{3}+\sqrt{7}\approx 3{,}87+3{,}46+2{,}65 = 9{,}98\ \text{s}$

Comparison. $\ev{\Delta\tau_\text{straight}\approx 10{,}39\ \text{s} \;>\; \Delta\tau_\text{broken}\approx 9{,}98\ \text{s}}$ The uniform straight-line trajectory gives the larger proper time: it is the “reversed triangle inequality” of relativity. Every departure from uniform straight-line motion (every change of SR) shortens the proper time — the root of the twin paradox.

Links

Topics: Special relativity Concepts: Spacetime invariant · Time dilation Skills: Constructing Minkowski diagrams Methods: Minkowski diagram · Natural units