The Relativity of Simultaneity — Question 1

Question 1

Simultaneous laboratory flashes acquire a signed time separation

Events A and B have $\Delta x=x_B-x_A=900\,\mathrm m$ and $\Delta t=0$. Let $v=0.60c$. 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. Calculate $t_B\prime-t_A\prime$.

2. State which event occurs first in the moving frame and find the result for a frame moving at $-0.60c$.

Question 1 - Solution

Simultaneous laboratory flashes acquire a signed time separation

(a) Method and calculation

$$\gamma=(1-0.60^2)^{-1/2}=1.25.$$$$\Delta t\prime=-1.25(0.60)(900/c)=\boxed{-2.25\,\mu\mathrm s}.$$

(b) Interpretation and further calculation

$$\Delta t\prime<0\quad\Rightarrow\quad\boxed{B\text{ occurs first in }S\prime}.$$$$v\to-v:\quad\boxed{\Delta t\prime=+2.25\,\mu\mathrm s}.$$

Physical check

The sign follows from the chosen event difference and boost direction; it is not an optical travel-time delay.