Question

check my work

Answer 1

Consider a rod kept on the ground, it is at rest with respect to a frame of reference S
of an observer standing at a distance of x1 from the rod(since both r moving along with the earth, they are at rest wrt the earth). He thus measure it's length (L) to be x2-x1, it is the rod's length in a frame of reference respect to which is at rest, so it is the rod's proper length

\[L=x_{2}-x_{1}\]
Now consider the observer to be standing in the same place, but the rod is now kept in a vehicle that is moving with a velocity v in the positive x direction w.r.t. the observer
The rod is now at rest with respect to the frame of reference of the bus, and it's length now measured by an observer in the bus is given as
\[L'=x_{2}'-x_{1}'\]

But for the observer on ground, the coordinates are given by using inverse lorentz transformation equation(we use inverse as we want to find coordinates in S frame)
\[x_{1}^{new}=\frac{x_{1}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}} \space \space , \space \space x_{2}^{new}=\frac{x_{2}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}\]
\[L^{new}=x_{2}^{new}-x_{1}^{new}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x_{2}'+vt'-x_{1}'-vt')=\gamma(x_{2}'-x_{1}')=\gamma L'\]
But the new length observed by the observer on the ground should be same as before, as he has not moved from his original position
so we have
\[L=L^{new} \space \space \space ; \space \space \space L=\gamma L' \implies L'=\sqrt{1-\frac{v^2}{c^2}}L\]

@IrishBoy123