Question

A rod flies with constant velocity past a mark which is stationary in the reference frame $K$. In the frame $K$ it takes $\Delta t=20ns$ for the rod to fly past the mark. In the reference frame fixed to the rod the mark moves past the rod for $\Delta t^{\prime }=25ns$. Find the proper length of the rod.

Answer 1

In the frame $K$ the length $l$ of the rod is related to the time of flight $\Delta t$ by
$l=v\Delta t$
In the reference frame fixed to the rod (frame $K^{\prime }$) the proper length $l_0$ of the rod is given by
$l_0=v\Delta t^{\prime }$
But $l_0=\frac{l}{\sqrt{1-{\beta }^2}}=\frac{v\Delta t}{\sqrt{1-{\beta }^2}}$, $\beta =\frac{v}{c}$
Thus, $v\Delta t^{\prime }=\frac{v\Delta t}{\sqrt{1-{\beta }^2}}$
So
$1-{\beta }^2={\left(\frac{\Delta t}{\Delta t^{\prime }}\right)}^2$ or $v=c\sqrt{1-{\left(\frac{\Delta t}{\Delta t^{\prime }}\right)}^2}$
and $l_0=c\sqrt{{\left(\Delta t^{\prime }\right)}^2-{\left(\Delta t\right)}^2}=c\Delta t^{\prime }\sqrt{1-{\left(\frac{\Delta t}{\Delta t^{\prime }}\right)}^2}$