Question

Find the proper length of a rod if in the laboratory frame of reference its velocity is $v=\frac{c}{2}$, the length $l=1.00m$, and the angle between the rod and its direction of motion is $\theta ={45}^{\circ }$.

Answer 1

In the rest frame, the coordinates of the ends of the rod in terms of proper length $l_0$
$A:(0,0,0)B:(l_0\cos {\theta }_0,l_0\sin {\theta }_0,0)$
at time $t$. In the laboratory frame the coordinates at time $t^{\prime }$ are
$A:(v{t}^{\prime },0,0),B:(l_0\cos {\theta }_0\sqrt{1-{\beta }^2}+v{t}^{\prime },l_0\sin {\theta }_0,0)$; where $\beta =\frac{v}{c}$
Therefore we can write,
$l\cos {\theta }_0=l_0\cos {\theta }_0\sqrt{1-{\beta }^2}$ and $l\sin {\theta }_0=l_0\sin {\theta }_0$
Hence $l_0^2=\left(l^2\right)\left(\frac{{\cos }^2\theta +(1-{\beta }^2){\sin }^2\theta }{1-{\beta }^2}\right)$
or, $=\sqrt{\frac{1-{\beta }^2{\sin }^2\theta }{1-{\beta }^2}}$

$\because \theta ={45}^0$ and $v=\frac{c}{2}$

$l_0=1.08m$