Mr Rossi goes to the theatre

Problem

Mr Rossi leaves home at $t=0$ and has an appointment at the theatre, which is $10$~sl from home, at $t=20$~s. (1) At what speed must he travel by uniform straight-line motion to arrive on time? (2) And to arrive $5$~s early? (3) In both cases, how much proper time passes on his watch?

Solution

Natural units $c=1$. The distance is $L=10$~sl.

(1) Speed to arrive on time. He must cover $L=10$~sl in laboratory time $t=20$~s: $v_1=\frac{L}{t}=\frac{10}{20}=0{,}5 \;\Rightarrow\; \ev{v_1=0{,}5\,c}$

(2) Speed to arrive 5 s early. Now the available time is $t'=20-5=15$~s: $v_2=\frac{L}{t'}=\frac{10}{15}=\frac{2}{3} \;\Rightarrow\; \ev{v_2=\tfrac{2}{3}\,c\approx 0{,}67\,c}$

(3) Proper time. On Rossi’s watch the proper time elapsed is $\Delta\tau=\sqrt{\Delta t^2-\Delta s^2}$ (the invariant between departure and arrival, with $\Delta s=L=10$~sl): $\Delta\tau_1=\sqrt{20^2-10^2}=\sqrt{400-100}=\sqrt{300}=10\sqrt{3}\approx 17{,}3\ \text{s}$ $\Delta\tau_2=\sqrt{15^2-10^2}=\sqrt{225-100}=\sqrt{125}=5\sqrt{5}\approx 11{,}2\ \text{s}$ $\ev{\Delta\tau_1\approx 17{,}3\ \text{s}\qquad \Delta\tau_2\approx 11{,}2\ \text{s}}$ Travelling faster (to arrive early) shortens the proper time: this is time dilation as seen from the traveller’s point of view.

Links

Topics: Special relativity Concepts: Time dilation · Spacetime invariant Skills: Symbolic setup Methods: Natural units