Question

Prove that a stick of proper length l has a length l’ in a frame in which it moves withspeed v along a line that makes an angle q withits length, where

Answer 1

Length along the line of motion contracts but along the perpendiculat. It remains the same.

So Initial length $\overrightarrow{L} = {L}_{x}\widehat{x}+ {L}_{y}\widehat{y}$

Lx = Lcosq

Ly = Lsinq

When moving with a velocity v in x direction. contraction occurs only in x direction by factor $\gamma$

So Lx becomes Lcosq * ($\sqrt(1-{{(\frac{v}{c})}}^2)$

So total length L' becomes L$\gamma$cosq i +​ L sinq j

So L'= L sqrt{(cos^2 q - (v/c)^2 * cos^2 q + sin^2 q)}

L'= L sqrt{1 - (vcosq/c)^2}

From given above Let (v^2/c^2) is k . So (1-k^2)/ (1-k^2sin^2q) Multiply above and below by (1- K^2 cos^2 q) We get (1- k^2sin^2q) (1- K^2 cos^2 q) = 1 - k^2 ( cos^2 q +sin^2 q) + k^4 sin^2q cos^2 q This is approximately equal to 1 - k^2 ; Because k is largely less than 1 and sin^2 q * cos^2 q is less than 1 So both (1- K^2) in numerator and denominator is cancelled.