The Relativity of Length — Question 2

Question 2

A tilted rod contracts only in its longitudinal projection

A rod has rest length $L_0=10.0\,\mathrm m$ and rest orientation $\theta_0=45^\circ$ to x. Its frame moves at $v=0.80c$ along x. Simultaneous laboratory projections satisfy $L_x=L_0\cos\theta_0/\gamma$, $L_y=L_0\sin\theta_0$. Use $\gamma=(1-v^2/c^2)^{-1/2}$ and $c=3.00\times10^8\,\mathrm{m/s}$.

1. Find the laboratory length.

2. Find the laboratory angle and determine whether the rod appears more parallel or more transverse to the motion.

Question 2 - Solution

A tilted rod contracts only in its longitudinal projection

(a) Method and calculation

$$L_x=3\sqrt2\,\mathrm m,\quad L_y=5\sqrt2\,\mathrm m.$$$$\boxed{L=\sqrt{L_x^2+L_y^2}=2\sqrt{17}\,\mathrm m\simeq8.24621\,\mathrm m}.$$

(b) Interpretation and further calculation

$$\tan\theta=L_y/L_x=5/3.$$$$\boxed{\theta=59.0362^\circ>\theta_0}.$$

Physical check

This is an orientation from simultaneous endpoint coordinates, not a photograph with unequal light-travel times.