= Solution
Use the <Minkowski metric> convention $+---$. A particle's <four-momentum> is $P=(E/c,\mathbf p)$, with $P\cdot P=m^2c^2$ for <rest mass> $m$. For a <photon>, its energy and <wavelength> obey $E=hc/\lambda$, and its <four-momentum> is null. Here $h$ is the <Planck constant>.
The initially stationary electron has $P=(mc,0,0,0)$. Choose the scattering plane as the $xy$ plane. The incoming and outgoing <photon> <four-momenta> are
$$
K_1=\frac h{\lambda_1}(1,1,0,0),\qquad K_2=\frac h{\lambda_2}(1,\cos\theta,\sin\theta,0).
$$
By <four-momentum conservation>, the final electron has $P'=P+K_1-K_2$. Its <rest mass> is unchanged, so the <relativistic energy-momentum relation> gives $P'^2=P^2=m^2c^2$. Since $K_1^2=K_2^2=0$, expansion yields
$$
P\cdot(K_1-K_2)=K_1\cdot K_2,
$$
or
$$
mc h\left(\frac1{\lambda_1}-\frac1{\lambda_2}\right)=\frac{h^2}{\lambda_1\lambda_2}(1-\cos\theta).
$$
Multiplying through gives the <Compton scattering> shift:
$$
\boxed{\lambda_2-\lambda_1=\frac h{mc}(1-\cos\theta)=\frac{2h}{mc}\sin^2(\theta/2).}
$$
It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the <photon>.
Back to article page
