The Relativity of Length — Question 9

Question 9

A measured rotation of a rod determines the boost magnitude

A rod makes angle $\theta_0=30^\circ$ to x in its rest frame and angle $\theta=45^\circ$ in simultaneous laboratory measurements. Motion is along x; transverse dimensions do not change. Use $\gamma=(1-v^2/c^2)^{-1/2}$ and $c=3.00\times10^8\,\mathrm{m/s}$.

1. Infer $\gamma$ and $|v|/c$ from the angle change.

2. Find the laboratory-to-proper length ratio.

Question 9 - Solution

A measured rotation of a rod determines the boost magnitude

(a) Method and calculation

$$\tan\theta=\gamma\tan\theta_0\quad\Rightarrow\quad\boxed{\gamma=\sqrt3}.$$$$\boxed{|v|/c=\sqrt{1-1/\gamma^2}=\sqrt{2/3}}.$$

(b) Interpretation and further calculation

$$\left(\frac L{L_0}\right)^2=\frac{\cos^2\theta_0}{\gamma^2}+\sin^2\theta_0=\frac14+\frac14.$$$$\boxed{L/L_0=1/\sqrt2}.$$

Physical check

The angular data determine speed magnitude, not the direction of translation.