The Relativity of Time — Question 9

Question 9

A periodically varying speed leaves a second-order clock deficit

An ideal clock has laboratory velocity $v(t)=v_0\sin(\omega t)$ with $v_0\ll c$. Observe one complete cycle $T=2\pi/\omega$. 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. Expand the proper-time integral through order $v_0^2/c^2$.

2. Find the fractional clock deficit and explain why zero average velocity does not make it vanish.

Question 9 - Solution

A periodically varying speed leaves a second-order clock deficit

(a) Method and calculation

$$\tau=\int_0^T\sqrt{1-\frac{v_0^2}{c^2}\sin^2\omega t}\,dt.$$$$\tau\simeq T-\frac{v_0^2}{2c^2}\int_0^T\sin^2\omega t\,dt=T\left(1-\frac{v_0^2}{4c^2}\right).$$

(b) Interpretation and further calculation

$$\boxed{\frac{T-\tau}{T}\simeq\frac{v_0^2}{4c^2}}.$$$$\langle v\rangle=0,\qquad\langle v^2\rangle=v_0^2/2>0.$$

Physical check

The leading correction is controlled by mean squared speed. The neglected fractional terms are of order $(v_0/c)^4$.