The Relativity of Time — Question 10

Question 10

Received tick spacing contains a travel-time contribution

A receding clock emits two pulses toward a stationary laboratory receiver, separated by proper interval $\Delta\tau=1.0\,\mu\mathrm s$. Its speed is $v=0.60c$. Both emissions occur on the same side of the receiver. 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 laboratory interval between the emission events.

2. Find the interval between receptions and separate the clock-rate and increasing-distance contributions.

Question 10 - Solution

Received tick spacing contains a travel-time contribution

(a) Method and calculation

$$\Delta t_{\mathrm{emit}}=\gamma\Delta\tau=\boxed{1.25\,\mu\mathrm s}.$$$$\Delta x=v\Delta t_{\mathrm{emit}}.$$

(b) Interpretation and further calculation

$$\Delta t_R=\Delta t_{\mathrm{emit}}+\Delta x/c=\gamma(1+\beta)\Delta\tau.$$$$\boxed{\Delta t_R=2.00\,\mu\mathrm s=1.25\,\mu\mathrm s+0.75\,\mu\mathrm s}.$$

Physical check

Time dilation concerns assigned emission times; the directly observed arrival spacing also contains changing propagation delay.