Mott Scattering

The leading Mott cross section is the relativistic Coulomb Born result for a spin-half projectile. Relative to the spinless Born baseline, it contains the factor $1-\beta^2\sin^2(\theta/2)$, where $\beta=v/c$. Here the source is infinitely heavy, pointlike, and spinless: it can absorb momentum without recoil energy. The derivation retains the projectile’s exact free relativistic kinematics but uses only the first order potential amplitude.

Required background. Relativistic Normalization supplies unit box waves and flux; Free Dirac Spinors supplies spin sums; First Born Approximation explains the perturbative step. Helpful background. Validity of the Born Approximation sets error-control expectations, and Dirac Dynamics in Electromagnetic Fields identifies the electrostatic coupling.

Coulomb matrix element for a fixed source

First use $\hbar=c=1$. Let the projectile and source charges be $q$ and $Q$, and define the signed coupling $g=qQ/(4\pi)$ in rationalized electromagnetic units. Then

$$\begin{aligned} V(r)&=\frac gr,\\ \widetilde V(\mathbf k) &=\int d^3x\,e^{-i\mathbf k\cdot\mathbf x}V(r) =\frac{4\pi g}{\mathbf k^2}. \end{aligned}$$

Here $\mathbf k=\mathbf p'-\mathbf p$ is a momentum transfer, not a wave number with a hidden factor of $\hbar$. To justify the Fourier transform, one may first use $g e^{-\mu r}/r$, whose transform is $4\pi g/(\mathbf k^2+\mu^2)$, and then take $\mu\downarrow0$ at fixed nonzero scattering angle.

The source is static, so $E'=E=\sqrt{p^2+m^2}$ and $p'=p$. For scattering angle $\theta$,

$$|\mathbf k|=2p\sin(\theta/2).$$

Use $u^\dagger u=2E$, $\bar uu=2m$ and unit box waves $u e^{-ip\cdot x}/\sqrt{2E\mathcal V}$. The interaction $V I_4$ in the Hamiltonian gives

$$V_{fi}= \frac{\widetilde V(\mathbf k)} {2\sqrt{E'E}\,\mathcal V}\, \bar u_{s'}(p')\gamma^0u_s(p).$$

For covariantly normalized continuum states the corresponding first-order $S-I$ matrix element is

$$-i\,2\pi\delta(E'-E)\, \widetilde V(\mathbf k)\, \bar u_{s'}(p')\gamma^0u_s(p).$$

There is one energy delta. The external source breaks spatial translation invariance, so this is not a two-body invariant amplitude multiplied by a four-dimensional conservation delta.

From transition rate to cross section

For fixed initial and final spin, the golden-rule rate into final momenta is

$$d\Gamma_{s's} =2\pi|V_{fi}|^2\delta(E'-E)\, \frac{\mathcal V\,p'^2dp'\,d\Omega}{(2\pi)^3}.$$

The radial delta integral is $\int p'^2dp'\,\delta(E'-E)=pE$, since $dE'/dp'=p'/E'$. Therefore

$$\frac{d\Gamma_{s's}}{d\Omega} =\frac{p}{16\pi^2E\mathcal V} |\widetilde V|^2 |\bar u_{s'}\gamma^0u_s|^2.$$

Divide by the incident number flux $p/(E\mathcal V)$. The arbitrary box volume and group velocity cancel:

$$\frac{d\sigma_{s's}}{d\Omega} =\frac{|\widetilde V|^2}{16\pi^2} |\bar u_{s'}(p')\gamma^0u_s(p)|^2.$$

This derivation explains why inserting an extra final-state $1/(2E')$ after already using unit box waves would give a wrong answer.

The unpolarized spin trace

For an unpolarized incident beam with final spin unobserved, average over the two initial spins and sum over the two final spins. The spin sums give

$$\mathcal S= \frac12\operatorname{tr} \left[(p'\!\!\!/+m)\gamma^0 (p\!\!\!/+m)\gamma^0\right].$$

Odd gamma traces vanish. Using the four-gamma trace,

$$\mathcal S=2\left[2E^2-p'\cdot p+m^2\right].$$

Since $p'\cdot p=E^2-p^2\cos\theta$ and $E^2=p^2+m^2$,

$$\mathcal S=4E^2 \left[1-\beta^2\sin^2(\theta/2)\right], \qquad \beta=\frac pE.$$

Substitution gives the leading Mott formula:

$$\frac{d\sigma_{\rm Mott}^{(1)}}{d\Omega} = \left[\frac{g}{2p\beta\sin^2(\theta/2)}\right]^2 \left[1-\beta^2\sin^2(\theta/2)\right].$$

The superscript records a first-order amplitude, whose squared contribution is second order in $g$. DeGrand’s section 8.7 gives the same cross section with a different Fourier-transform convention. The spin effect cannot be recovered by changing only the nonrelativistic energy–momentum relation.

In SI units define $\kappa_C=qQ/(4\pi\epsilon_0)$ and $\zeta=\kappa_C/(\hbar c)$. With physical momentum $p$ and $v=pc^2/E$, the equivalent prefactor is

$$\left[\frac{\kappa_C}{2pv\sin^2(\theta/2)}\right]^2 = \left[\frac{\zeta\hbar}{2p\beta\sin^2(\theta/2)}\right]^2.$$

For an electron and charge $Ze$, $\zeta=-Z\alpha$. Its sign disappears in this leading cross section, but need not disappear in higher-order Coulomb corrections.

Limiting checks and limits of the approximation

For $v\ll c$, the spin factor tends to one and $p\simeq mv$. The SI result becomes

$$\frac{d\sigma}{d\Omega} \longrightarrow \left[\frac{\kappa_C}{2mv^2\sin^2(\theta/2)}\right]^2,$$

the Rutherford formula. In the massless limit at fixed $p>0$, the spin factor becomes $\cos^2(\theta/2)$. Leading Born backscattering vanishes at $\theta=\pi$. For a massive projectile the backscattering factor is instead $1-\beta^2=m^2c^4/E^2$ in SI units.

At small angle the differential cross section behaves as

$$\frac{d\sigma}{d\Omega} \sim\frac{4\zeta^2\hbar^2}{p^2\beta^2\theta^4}.$$

The unscreened total cross section diverges. A finite detector acceptance, screening model, or other specified infrared treatment is needed for an integrated count.

At fixed angle away from the forward singularity, $|\zeta|/\beta\ll1$ is a useful conservative perturbative regime. It is not a uniform bound over the unscreened forward limit. Exact Dirac–Coulomb scattering, finite target recoil, nuclear form factors, and radiative corrections are distinct refinements. The equality between Born and exact nonrelativistic Rutherford magnitudes does not make the Coulomb Born amplitude exact; see Coulomb Scattering.

Exercises

1. For $\beta=4/5$, find the ratio of the Mott result to the spinless Born baseline at $90^\circ$ and $180^\circ$.

Solution

At $90^\circ$ the ratio is $1-(16/25)/2=17/25$. At $180^\circ$ it is $1-16/25=9/25$. The comparison holds at the same momentum, speed, and charge coupling.

2. Retain the screened potential $g e^{-\mu r}/r$. Which part of the leading calculation changes?

Solution

Only the potential Fourier factor changes: $|\widetilde V|^2=(4\pi g)^2/(4p^2\sin^2(\theta/2)+\mu^2)^2$. The Dirac spin trace and box-to-flux conversion stay the same. For $\mu>0$ the forward differential value is finite, although the small-$\mu$ integrated limit is not finite.

3. Does this leading real Coulomb potential polarize an initially unpolarized beam when only one scattering direction is selected? Use fixed-axis two-spinors.

Solution

The spin matrix in $u^\dagger(p')u(p)$ is

$$W=(E+m)\left[ I+\frac{\mathbf p'\cdot\mathbf p +i\boldsymbol\sigma\cdot(\mathbf p'\times\mathbf p)} {(E+m)^2}\right].$$

At a fixed direction it is $aI+i b\boldsymbol\sigma\cdot\widehat{\mathbf n}$ with real $a,b$. Consequently $WW^\dagger=(a^2+b^2)I$ and the outgoing spin density from $I/2$ is still unpolarized. The Dirac spin structure changes the angular result relative to the spinless baseline, but the spin-summed cross section has no polarization analyzing power at this order. A nonzero analyzing effect requires additional relative phases beyond this leading calculation.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill, 1964. External-field scattering and spin sums.
• DeGrand, Thomas. A One-Semester Course on Quantum Field Theory. University of Colorado lecture notes, 30 December 2025, section 8.7, especially equation 8.143. Lecture text.
• Greiner, Walter. Relativistic Quantum Mechanics: Wave Equations. 3rd edition, Springer, 2000. doi:10.1007/978-3-662-04275-5. Relativistic Coulomb scattering and its approximation regimes.