Question

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' axis.
Required:
a. Show that the length of the rod as measured by a stationary observer is L=L0[1-v^2/c^2 cos^2(θ)0].
b. Show that the angle the rod makes iwth the x-axis is given by the expression tan(theta)=tan(θ)0/(1-v^2/c^2)^.5

Answer 1

Answer:

From the question we are told that

  The length of the rod is  $L_o$

    The  speed is  v  

     The angle made by the rod is  $\theta$

     

Generally the x-component of the rod's length is  

     $L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       $L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=>     $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length  is mathematically represented as

      $L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e $L_y$

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     $L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=>  $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=>  $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=>   $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

       $L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

   Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=>  $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation: