The Relativity of Simultaneity — Question 3

Question 3

Time-order reversal occupies an interval of allowed boost speeds

Events have $\Delta x=1500\,\mathrm m$ and $\Delta t=3.0\,\mu\mathrm s$. 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. Find all physical boost speeds for which B occurs before A.

2. Evaluate the transformed time difference at $v=0.80c$.

Question 3 - Solution

Time-order reversal occupies an interval of allowed boost speeds

(a) Method and calculation

$$\Delta t\prime<0\quad\Longleftrightarrow\quad v>c^2\Delta t/\Delta x=0.60c.$$$$\boxed{0.60c<v<c}.$$

(b) Interpretation and further calculation

$$\gamma(0.80c)=5/3,\quad v\Delta x/c^2=4.0\,\mu\mathrm s.$$$$\boxed{\Delta t\prime=-5/3\,\mu\mathrm s}.$$

Physical check

The equality at 0.60c is simultaneity. No physical inertial boost reaches c.