PC1144 Homework 1

Qi Ji (A0167793L)

3rd Feb 2021


Let event 1 be lamp \(A\) turning on and event 2 for lamp \(B\). Let frame \(F'\) be the moving along the rod, at velocity \(V\) away from frame \(F\) that is stationary with respect to \(O\).


Start from the frame \(F\) where rod 1 is at rest and the other is moving at speed \(V\). We want to find a frame \(F'\) moving at velocity \(u\) away from rod 1 such that \[ V' = \frac{V - u}{1 - \frac{u}{c^2}V} \] and \(V' = u\), this will make both rods have the same length. Solving for \(u\), we have \[ u = \frac{c^2}{v}\left(1 - 2\sqrt{1 - \frac{v^2}{c^2}}\right) \] reject the other quadratic solution since that will make \(u\geq c\).


Total energy twice as rest energy, this means \[ \gamma mc^2 = 2 mc^2 \implies \gamma = 2. \]

Solving \[ \frac12 = \sqrt{1 - \beta^2} \] we have \[ u = \frac{\sqrt{3}}2 c. \]