Question

In a triangle the proper length of each side equals $a$. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity $V$ along one of its sides.

Answer 1

Length of the rod moving parallel to the direction of velocity $\left(v\right)$ gets reduced by a factor of $\sqrt{1-{\beta }^2}$ of its proper length whereas the length perpendicular to the direction of velocity remains the same.

Given : $|AB|=|AC|=|BC|=a$ (proper length)

In the moving frame :

$\therefore$ $|A^{\prime }{B}^{\prime }|=a\sqrt{1-{\beta }^2}$

$\overrightarrow{AC}=acos\left(60\right)\hat{i}+asin\left(60\right)\hat{j}$

$\therefore$ $\overrightarrow{A^{\prime }{C}^{\prime }}=\left[\right(acos60)\sqrt{1-{\beta }^2}\hat{i}+asin60\hat{j}]=\frac{a\sqrt{(1-{\beta }^2)}}{2}\hat{i}+\frac{a\sqrt{3}}{2}\hat{j}$ where $\beta =v/c$

$⟹$ $|A^{\prime }{C}^{\prime }|=\frac{a\sqrt{4-{\beta }^2}}{2}$

Similarly $|B^{\prime }{C}^{\prime }|=$$\frac{a\sqrt{4-{\beta }^2}}{2}$

$\therefore$ Perimeter of the triangle $P^{\prime }=|A^{\prime }{B}^{\prime }|+|A^{\prime }{C}^{\prime }|+|B^{\prime }{C}^{\prime }|=a\sqrt{1-{\beta }^2}+$$\frac{a\sqrt{4-{\beta }^2}}{2}+$ $\frac{a\sqrt{4-{\beta }^2}}{2}$

$⟹$ $P^{\prime }=a(\sqrt{1-{\beta }^2}+\sqrt{4-{\beta }^2})$