General Relativity: An Introduction for Physicists · Problem 5.13

Special relativity revisited — Hobson, Efstathiou & Lasenby

Physics · compton-scattering

Problem

A photon with energy EE collides with a stationary electron whose rest mass is m0m_0. As a result of the collision the direction of the photon's motion is deflected through an angle θ\theta and its energy is reduced to E′E'. Show that m0c2(1E′−1E)=1−cos⁡θ.m_0 c^2\left(\frac{1}{E'} - \frac{1}{E}\right) = 1 - \cos\theta. Deduce that the wavelength of the photon is increased by Δλ=2hm0csin⁡2 ⁣(θ2),\Delta\lambda = \frac{2h}{m_0 c}\sin^2\!\left(\frac{\theta}{2}\right), where hh is Planck's constant. At what angle to the initial photon direction does the electron move? Show that, if the photon is deflected through a right angle, and the photon energy satisfies E≪m0c2E \ll m_0 c^2, then after the interaction the angle of the electron's motion to the direction of the photon's initial motion is α=−π/4\alpha = -\pi/4.

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