The Relativity of Length — Question 1

Question 1

The simultaneity condition selects the contracted endpoint separation

A rod at rest in $S\prime$ has length $L_0=120\,\mathrm m$ along x. It moves at $v=0.60c$ in S. Use $\Delta x\prime=\gamma(\Delta x-v\Delta t)$ and $\Delta t\prime=\gamma(\Delta t-v\Delta x/c^2)$. Use $\gamma=(1-v^2/c^2)^{-1/2}$ and $c=3.00\times10^8\,\mathrm{m/s}$.

1. Derive the laboratory length using simultaneous laboratory endpoint events.

2. Find the time separation of those same endpoint events in the rod rest frame.

Question 1 - Solution

The simultaneity condition selects the contracted endpoint separation

(a) Method and calculation

$$\Delta t=0,\quad\Delta x=L,\quad\Delta x\prime=L_0\quad\Rightarrow\quad L_0=\gamma L.$$$$\boxed{L=L_0/\gamma=96.0\,\mathrm m}.$$

(b) Interpretation and further calculation

$$\Delta t\prime=-\gamma vL/c^2=-vL_0/c^2.$$$$\boxed{\Delta t\prime=-0.240\,\mu\mathrm s}.$$

Physical check

The rod rest frame permits different endpoint times because its endpoints are stationary there. The laboratory length explicitly uses equal laboratory times.