Classical Electrodynamics · Problem 11.26

Radiating Systems, Multipole Fields and Radiation — John David Jackson

Physics — E&M · elastic-scattering-energy-transfer

Problem

In an elastic scattering process the incident particle imparts energy to the stationary target. The energy ΔE\Delta E lost by the incident particle appears as recoil kinetic energy of the target. In the notation of Problem 11.23, m3=m1m_3 = m_1 and m4=m2m_4 = m_2, while ΔE=T4=E4−m4\Delta E = T_4 = E_4 - m_4. (a) Show that ΔE\Delta E can be expressed in the following different ways, ΔE=m2W2pLAB2(1−cos⁡θ′)ΔE=2m2pLAB2cos⁡2θ4W2+pLAB2sin⁡2θ4ΔE=Q22m2\begin{aligned} \Delta E &= \frac{m_2}{W^2} p^2_{\rm LAB}(1 - \cos\theta')\\ \Delta E &= \frac{2m_2 p^2_{\rm LAB} \cos^2\theta_4}{W^2 + p^2_{\rm LAB}\sin^2\theta_4}\\ \Delta E &= \frac{Q^2}{2m_2} \end{aligned} where Q2=−(p1−p3)2=(p1−p3)2−(E1−E3)2Q^2 = -(p_1 - p_3)^2 = (\mathbf{p}_1 - \mathbf{p}_3)^2 - (E_1 - E_3)^2 is the Lorentz invariant momentum transfer (squared). (b) Show that for charged particles other than electrons incident on stationary electrons (m1≫m2m_1 \gg m_2) the maximum energy loss is approximately ΔEmax≈2γ2β2me\Delta E_{\rm max} \approx 2\gamma^2\beta^2 m_e where γ\gamma and β\beta are characteristic of the incident particle and γ≪(m1/me)\gamma \ll (m_1/m_e). Give this result a simple interpretation by considering the relevant collision in the rest frame of the incident particle and then transforming back to the laboratory. (c) For electron-electron collisions, show that the maximum energy transfer is ΔEmax(e)=(γ−1)me\Delta E^{(e)}_{\rm max} = (\gamma - 1)m_e

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