Question 705024

In the rest frame, $l_x = a \cos(\pi/4) = a/\sqrt{2}$ and $l_y = a \sin(\pi/4) = a/\sqrt{2}$.

Due to Lorentz contraction, the length in the $x$-direction in $S'$ becomes:

$l'_x = l_x \sqrt{1 - v^2/c^2} = (a/\sqrt{2}) \sqrt{1 - 2/3} = (a/\sqrt{2})(1/\sqrt{3}) = a/\sqrt{6}$. $l'_y = l_y = a/\sqrt{2}$.

Then $\tan \theta_s = l'_y / l'_x = (a/\sqrt{2}) / (a/\sqrt{6}) = \sqrt{3}$. Thus $\theta_s = \pi/3$. The interior angle $\phi = \pi - 2\theta_s = \pi - 2\pi/3 = \pi/3$.