The Relativity of Length — Question 5

Question 5

A longitudinal light clock reconciles unequal laboratory legs

A rod of proper length $L_0=300\,\mathrm m$ moves at $v=0.60c$. A pulse leaves its rear, reflects immediately at its front, and returns to its rear. Both endpoints move with the rod. Use $\gamma=(1-v^2/c^2)^{-1/2}$ and $c=3.00\times10^8\,\mathrm{m/s}$.

1. Find the laboratory outward and return travel times.

2. Find the elapsed proper time on the rear clock and compare with $2L_0/c$.

Question 5 - Solution

A longitudinal light clock reconciles unequal laboratory legs

(a) Method and calculation

$$L=L_0/\gamma=240\,\mathrm m,\quad t_+=L/(c-v)=\boxed{2.00\,\mu\mathrm s}.$$$$t_-=L/(c+v)=\boxed{0.500\,\mu\mathrm s}.$$

(b) Interpretation and further calculation

$$T=t_++t_-=2.50\,\mu\mathrm s,\quad\tau=T/\gamma.$$$$\boxed{\tau=2.00\,\mu\mathrm s=2L_0/c}.$$

Physical check

The moving rear meets the returning pulse, so its return distance is shorter in the laboratory. A fixed-source mirror problem has different endpoints.