The Relativity of Time — Question 6

Question 6

Piecewise motion requires adding proper times rather than averaging speeds

An ideal clock spends 5.0 s at speed $0.60c$ and then 5.0 s at speed $0.80c$, all durations measured in one inertial laboratory. Transition time is negligible. Use $d\tau=dt\sqrt{1-v^2/c^2}$ for an ideal clock, with $c=3.00\times10^8\,\mathrm{m/s}$ when needed.

1. Find the total clock advance.

2. Find the constant speed that would give the same clock advance over 10.0 laboratory seconds.

Question 6 - Solution

Piecewise motion requires adding proper times rather than averaging speeds

(a) Method and calculation

$$\tau=5\sqrt{1-0.60^2}+5\sqrt{1-0.80^2}.$$$$\boxed{\tau=4+3=7.00\,\mathrm s}.$$

(b) Interpretation and further calculation

$$7=10\sqrt{1-\beta_{\mathrm{eff}}^2}\quad\Rightarrow\quad\boxed{\beta_{\mathrm{eff}}=\sqrt{0.51}\simeq0.714143}.$$$$\tfrac12(0.60+0.80)=0.70\ne\beta_{\mathrm{eff}}.$$

Physical check

Proper time is nonlinear in speed; an arithmetic mean speed generally does not preserve the clock advance.