Physics Asked on May 27, 2021
If the proton were a point charge like the muon, then $epto ep$ scattering, the differential scattering crossection is $$frac{dsigma}{dOmega}Bigg|_{rm lab}=Bigg(frac{alpha^2}{4E^2sin^4(theta/2)}Bigg)Bigg{1+frac{2E}{m_p}sin^2thetaBigg}^{-1}Bigg[cos^2(theta/2)-frac{q^2}{2m^2_p}sin^2(theta/2)Bigg]$$ where $theta$ is angle scattering, $E,E’$ are the initial and final energies of the scattered electron, $q$ is the momentum transfer in the scattering and $m_p$ is the mass of the proton. In the limit of $m_pto infty$, or more accurately, $E,q^2ll m_p^2,$ the term inside the second bracket becomes unity and second term inside the third bracket vanishes. In this limit, we get $$frac{dsigma}{dOmega}Bigg|_{rm lab}rightarrowBigg(frac{alpha^2}{4E^2sin^4(theta/2)}Bigg)cos^2(theta/2)$$ where the quantity inside the first bracket is the Rutherford scattering formula.
Why does the factor $cos^2(theta/2)$ survive in this limit? Why we do not recover the exact Rutherford formula in this limit nonrelativistic limit?
The Rutherford cross section is a non-relativistic limit, and is equivalent to a particle scattering from a static electric potential $V(r)$ without any consideration of the interaction of intrinsic magnetic moments. The angular dependence, $1/sin^4{theta/2}$, arises entirely from the $1/q^2$ propagator. In this low energy limit, all four helicity amplitudes contribute and the cross section looks like the spin-0 alpha-scattering.
The formula you show is for a relativistic electron (Mott scattering), though the
$$frac{2E}{m_p}sin^2theta $$
proton recoil is neglected. In this case, helicity is conserved, and the $cos^2theta/2$ term comes from the overlap of the spin wave functions of the initial and final electron. It is still electric scattering.
The $-frac{q^2}{2m_p}sin^2{theta/2} $ term is the magnetic (spin-spin) interaction.
Correct answer by JEB on May 27, 2021
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