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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Oct 3, 2020 at 16:45
  • $\begingroup$ @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? $\endgroup$ Commented Oct 3, 2020 at 20:07
  • $\begingroup$ @mithusengupta123 because $E \gg m_e$. $\endgroup$
    – JEB
    Commented Oct 4, 2020 at 14:13
  • $\begingroup$ So is there no way I can recover the Rutherford limit from this formula? $\endgroup$ Commented Oct 4, 2020 at 15:09
  • $\begingroup$ 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. $\endgroup$ Commented Oct 5, 2020 at 8:55

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