 Solution

For $m>0$, $E={\gamma }_{u}m{c}^2$ and $\mathbf{p}={\gamma }_{u}m\mathbf{u}$, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$. ThusA photon obeys $E=pc$; $E=hf$ and $p=h/\lambda$ agree because $f\lambda =c$ in vacuum.

Take the incident photon along $+x$, and write $\beta =v/c$ for the boost. In the boosted frame the initial momenta areTheir sum vanishes when $1-2\beta =0$, giving $v_{\mathrm{ZM}}=c/2$ and ${\gamma }_v=2/\sqrt{3}$. The initial photon energy there is $m{c}^2/\sqrt{3}$, and total energy is $\sqrt{3}m{c}^2$.

For right-angle Compton scattering in the centre-of-momentum frame, choose the outgoing ZM photon along $+y$. The electron has the opposite momentum $-p^{\prime }\hat{\mathbf{y}}$. Conservation of energy determines its magnitude:Thus ${\mathbf{u}}_e^{\prime }=-(c/2)\hat{\mathbf{y}}$. Inverting the velocity transformation givesConsequentlyThe electron recoils below the incident direction for our choice of photon scattering above it; the reflected configuration reverses both transverse signs. Finally,As a check, $E_{e,\mathrm{lab}}=4m{c}^2/3$, so the final energies sum to the initial $2m{c}^2$. The outgoing laboratory photon makes $60^{\circ }$ with $+x$.