The Relativity of Simultaneity — Question 6

Question 6

Differential simultaneity can determine the direction of motion

A laboratory-simultaneous pair is separated by $\Delta x=800\,\mathrm m$. A moving frame measures $\Delta t\prime=-2.00\,\mu\mathrm s$. 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. Solve for the signed quantity $\gamma\beta$, where $\beta=v/c$.

2. Infer v and check it reproduces the measured time difference.

Question 6 - Solution

Differential simultaneity can determine the direction of motion

(a) Method and calculation

$$\Delta t\prime=-\gamma\beta\Delta x/c\quad\Rightarrow\quad\boxed{\gamma\beta=0.750}.$$$$(\gamma\beta)^2=\frac{\beta^2}{1-\beta^2}\quad\Rightarrow\quad\beta=\frac{0.750}{\sqrt{1+0.750^2}}.$$

(b) Interpretation and further calculation

$$\boxed{v=+0.600c},\qquad\gamma=1.250.$$$$-1.250(0.600)(800/c)=\boxed{-2.00\,\mu\mathrm s}.$$

Physical check

The negative transformed time difference fixes the positive boost direction when the spatial difference is positive.