# Help in recovering the Rutherford formula for $ep\to ep$ scattering in the nonrelativistic limit

If the proton were a point charge like the muon, then $$ep\to ep$$ scattering, the differential scattering crossection is $$\frac{d\sigma}{d\Omega}\Bigg|_{\rm lab}=\Bigg(\frac{\alpha^2}{4E^2\sin^4(\theta/2)}\Bigg)\Bigg\{1+\frac{2E}{m_p}\sin^2\theta\Bigg\}^{-1}\Bigg[\cos^2(\theta/2)-\frac{q^2}{2m^2_p}\sin^2(\theta/2)\Bigg]$$ 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 $$m_p\to \infty$$, or more accurately, $$E,q^2\ll m_p^2,$$ the term inside the second bracket becomes unity and second term inside the third bracket vanishes. In this limit, we get $$\frac{d\sigma}{d\Omega}\Bigg|_{\rm lab}\rightarrow\Bigg(\frac{\alpha^2}{4E^2\sin^4(\theta/2)}\Bigg)\cos^2(\theta/2)$$ 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

$$\frac{2E}{m_p}\sin^2\theta$$

proton recoil is neglected. In this case, helicity is conserved, and the $$\cos^2\theta/2$$ 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.

• So, "the overlap of the spin wave functions of the initial and final electron. It is still electric scattering." ? Strange. Neutral particles must have the same "electric scattering" too? I feel something is missing here, for example an inclusive approach - to sum and average spin states. Oct 3, 2020 at 16:45
• @JEB But in going from the first formula to the second formula, I used $M\to \infty$ limits or $q^2\ll M^2$ limit. Why is it not the nonrelativistic limit? Oct 3, 2020 at 20:07
• @mithusengupta123 because $E \gg m_e$.
– JEB
Oct 4, 2020 at 14:13
• So is there no way I can recover the Rutherford limit from this formula? Oct 4, 2020 at 15:09
• In the center of mass reference frame you must recover the Rutherford formula, but in the laboratory RF there may be factors conncting different definition of scattering angles, I guess. Oct 5, 2020 at 8:55