Question 705024

SCQHARD

A square plate of dimension $a \times a$ makes an angle $\theta = \pi/4$ with the $x$ axis in its rest frame $S$ . It is moving with a speed $v = \sqrt{2/3} c$ along the $x$ axis with respect to an observer $S'$ . The value of the interior angle $\phi$ indicated in the figure as measured in $S'$ is

(A)

$\pi/3$

(B)

$2\pi/3$

(C)

$3\pi/6$

(D)

$4\pi/3$

Detailed Solution

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 $\phi = \pi - 2\theta_s = \pi - 2\pi/3 = \pi/3$ .

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