The Relativity of Simultaneity — Question 10

Question 10

Programmed emission delays can create simultaneity in a chosen frame

Two fixed laboratory emitters at $x_A=0$ and $x_B=L=1500\,\mathrm m$ are programmed to flash simultaneously in a frame moving at $v=0.60c$. Set $t_A=0$. Use $\Delta t\prime=\gamma(\Delta t-v\Delta x/c^2)$, $\gamma=(1-v^2/c^2)^{-1/2}$; $S\prime$ moves at +v along x. Take $c=3.00\times10^8\,\mathrm{m/s}$.

1. Find the required laboratory delay at B.

2. Find when both flashes reach a detector fixed at $x=L/2$, and compare their arrival-time difference.

Question 10 - Solution

Programmed emission delays can create simultaneity in a chosen frame

(a) Method and calculation

$$0=\gamma(t_B-vL/c^2)\quad\Rightarrow\quad\boxed{t_B=3.00\,\mu\mathrm s}.$$$$\boxed{t_B\prime-t_A\prime=0}.$$

(b) Interpretation and further calculation

$$t_{R,A}=L/(2c)=2.50\,\mu\mathrm s,\quad t_{R,B}=t_B+L/(2c)=5.50\,\mu\mathrm s.$$$$\boxed{t_{R,B}-t_{R,A}=3.00\,\mu\mathrm s}.$$

Physical check

The midpoint detector is stationary in the laboratory; equality of emission times in a different frame does not imply equal receptions there.