The rabbit race

Problem

Matilda and Roger set off simultaneously (for the spectator) from positions $-1000$~ls and $+1000$~ls and run toward the opposite finish line. In the spectator’s frame (SRS), both reach their respective finish line at instant $t=2500$~s. How much proper time passes for Matilda between departure and arrival? At what speed did she run?

Solution

Natural units $c=1$. We work in the spectator’s frame (SRS).

Matilda’s data in SRS. She starts at $s=-1000$~ls and arrives at $s=+1000$~ls, so $\Delta s = 2000$~ls, in time $\Delta t = 2500$~s.

Speed. In the same frame (SRS) the speed is the ratio of distance and time both measured in SRS: $v_M=\frac{\Delta s}{\Delta t}=\frac{2000}{2500}=0{,}8 \;\Rightarrow\; \ev{v_M=0{,}8\,c}$

Proper time. The time elapsed on Matilda’s clock is the invariant between departure and arrival: $\Delta\tau_M=\sqrt{\Delta t^2-\Delta s^2}=\sqrt{2500^2-2000^2}=\sqrt{6\,250\,000-4\,000\,000}=\sqrt{2\,250\,000}$ $\ev{\Delta\tau_M=1500\ \text{s}}$ (Roger is perfectly symmetric: same numbers, $v_R=0{,}8\,c$ and $\Delta\tau_R=1500$~s.)

"I travelled faster than light!"? $\Delta s/\Delta\tau_M = 2000/1500 \approx 1{,}33\,c$ and conclude that Matilda exceeded the speed of light. Wrong! You cannot combine a distance measured in one frame (the spectator's $2000$~ls) with a time measured in another frame (Matilda's proper $1500$~s): the two numbers are expressed in "units" of different frames. The correct speed is always $\Delta s^{(A)}/\Delta t^{(A)}$, distance and time in the same frame: here $2000/2500=0{,}8\,c<c$.

One might be tempted to compute

Links

Topics: Relatività ristretta Concepts: Dilatazione dei tempi · Invariante spazio-temporale Skills: Costruzione di diagrammi di Minkowski · Cambio di sistema di riferimento Methods: Diagramma di Minkowski · Unità naturali