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Assume unpolarized beams, negligible electron and muon masses, a single scattering s-channel spin-one mediator with equal left and right vector couplings, and leading-order QED. Z boson exchange and radiative corrections are omitted. The angular factors are the Wigner D-matrix elements $d_{m{m}^{\prime }}^1({\theta }^{\star })$, with $m,m^{\prime }=\pm 1$. Up to common phases and normalization they are $(1+\cos {\theta }^{\star })$ for equal pair-spin projections and $(1-\cos {\theta }^{\star })$ for opposite projections. The four orthogonal helicity channels add incoherently, givingWith $s$ the squared center-of-mass energy, the normalized differential scattering cross-sections areThe second expression includes the solid angle Jacobian: a distribution per angle is different from one per $\cos {\theta }^{\star }$. Integration gives $\sigma =4\pi {\alpha }^2/(3s)$.

Parity reverses all momenta but leaves the spin axial vector unchanged, so it interchanges $L\leftrightarrow R$. The $LR\to LR$ and $RL\to RL$ amplitudes have equal magnitudes, as do $LR\to RL$ and $RL\to LR$. Thus every configuration has the same rate as its parity transform, proving parity conservation. The summed distribution is also forward-backward symmetric. Strictly, parity reverses both the beam and outgoing momenta, leaving their relative angle unchanged; forward-backward reflection by itself is not the definition of parity.

Count a reference process whose fiducial cross section is accurately calculable. At an electron-positron collider, small-angle Bhabha scattering, dominated by $t$-channel photon exchange, is abundant and provides a standard choice. Require coincident high-energy electron and Positron deposits on opposite sides of the interaction point, with consistent back-to-back directions, and impose precisely specified angular and energy cuts. ThenThe theoretical prediction must include radiation and other relevant electroweak effects under the same event definition. Finite angular resolution, alignment, beam position, live time, trigger efficiency and background contamination enter the uncertainty budget. For small scattering angle $\theta$, $t\simeq -s{\theta }^2/4$ and the leading differential scattering cross-section behaves approximately as $d\sigma /d\theta \propto {\theta }^{-3}$; the inner fiducial angle must therefore be particularly accurately known. Binning these counts in time can also measure the instantaneous collider luminosity.

At large $r$, $|\mathbf{r}-{\mathbf{r}}_j|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}_j+O(r^{-1})$. Retain the displacement in the phase, but replace the denominator by $r$. ThereforeThe differential scattering cross-section is $|f{|}^2$ for elastic scattering with this unit incident plane-wave normalization. interference depends on outgoing direction and centre separation as well as the incoming phases already contained in $c_j$.

For a short-range potential, an incident plane wave of unit amplitude and wave vector $k{\mathbf{e}}_z$ has the large-distance scattering wavefunctionThe coefficient $f$ is the scattering amplitude. For a central potential it depends only on the scattering angle $\theta$. The incident probability current is $\hslash k/m$, while the scattered radial current is $(\hslash k/m)|f{|}^2/r^2$ to leading order. Their ratio givesThese are the differential scattering cross-section and total cross section for elastic scattering with the same incident and outgoing wave number.

The expression is asymptotic: it specifies the outgoing wave's leading $1/r$ term, not its finite-distance radial corrections or its form inside the interaction region. In particular, for angle-dependent $f$, the spherical term by itself has an angular-Laplacian contribution of order $r^{-3}$ to the free wave equation, which is cancelled by subleading terms in the exact solution.

Assume a real, central, short-range potential and elastic quantum scattering. At large radius, normalize the stationary wavefunction aswhere $k=\sqrt{2mE}/\hslash$ and $\theta$ is the scattering angle. Equal incoming and outgoing speeds give the differential scattering cross-section $d\sigma /d\Omega =|f(\theta ){|}^2$.

The partial-wave scattering expansion separates the spherical harmonics. The incident plane wave has expansion ${\sum }_{\ell \ge 0}{i}^{\ell }(2\ell +1)j_{\ell }(kr)P_{\ell }(\cos \theta )$, with Spherical Bessel functions $j_{\ell }$ and Legendre polynomials $P_{\ell }$. The regular radial solution outside the potential has asymptotic form $u_{\ell }\propto \sin (kr-\ell \pi /2+{\delta }_{\ell })$. Its real scattering phase shift ${\delta }_{\ell }$ changes the ratio of outgoing to incoming waves to the Partial-wave S-matrix $S_{\ell }=e^{2i{\delta }_{\ell }}$. Matching outgoing coefficients yieldsHere $\ell$ labels the orbital angular momentum, with squared magnitude ${\hslash }^2\ell (\ell +1)$. The assumption of purely elastic scattering is what makes every $S_{\ell }$ have unit modulus.

Label the initial particle and antiparticle $p_1,p_2$ and the corresponding final momenta $p_3,p_4$, with $s=(p_1+p_2{)}^2$ and $t=(p_1-p_3{)}^2$. Charge flow permits an $s$-channel heavy-scalar annihilation graph and a $t$-channel heavy-scalar exchange graph. It excludes a third $u$-channel graph for these fixed charge assignments. The effective theory has one contact graph.

The two cubic Feynman vertices and the heavy propagator giveFor $|s|,|t|\ll M^2$,This independently checks the sign and the factor $1/2$ in solution. The next difference is the derivative correction of order $M^{-4}$ in the amplitude.

Let $E=\sqrt{s}$ denote the total center-of-mass energy, with $m\ll E\ll M$. The final particle and antiparticle are distinct, so no identical-particle factor is needed. The leading isotropic differential scattering cross-section and total cross section areCorrections are relatively of order $E^2/M^2$ and, in a high-energy expansion, $m^2/E^2$, in addition to loop corrections when they matter. If instead $E$ denotes each beam particle's energy, substitute $s=4E^2$ before comparing numerical prefactors.

The monostatic radar cross section is $4\pi$ times the backward differential scattering cross-section. Equivalently, ${\sigma }_{\mathrm{back}}={\lim }_{r\to \infty }4\pi r^2{S}_{\mathrm{back}}/S_{\mathrm{inc}}$. It is a directional scattering measure with dimensions of area, not generally the geometric area or the total scattering cross-section. The definition may additionally specify the transmitted and received polarizations.

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.

For an incident beam uniform across its relevant impact parameters, an incident annulus has area $d\sigma =2\pi b|db|$. Rotational symmetry maps it to outgoing solid angle $d\Omega =2\pi \sin \theta d\theta$. The central-potential classical differential cross-section and the Rutherford impact-parameter relation therefore giveThis is a differential scattering cross-section, with units of area per steradian, conventionally ${\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}$, not a detector count and not an area per degree. It applies for $0<\theta <\pi$, with a finite limiting value at $\pi$. Doubling incident kinetic energy divides it by four at fixed angle; doubling either electric charge number multiplies it by four.

A useful independent check integrates over all azimuths between two positive angles. For $0<{\theta }_1<{\theta }_2\le \pi$,This equals the incident annulus area $\pi (b_1^2-b_2^2)$ directly. A detector covering only a small patch at one angle does not collect this full azimuthal band.

Rutherford scattering is elastic scattering by a Coulomb potential. In the repulsive classical case, a positively charged projectile bends around a positively charged atomic nucleus. A close approach can turn it backwards without a contact collision. The calculation connects an invisible impact parameter with a measurable outgoing angle, then with a differential scattering cross-section and a detector count.

Assume a projectile of mass $m$ and electric charge $ze$, a stationary point atomic nucleus of electric charge $Ze$, and positive $z,Z$. Here $e$ is the elementary charge. Neglect nuclear recoil, electron screening effect, energy loss, and the nuclear force. With vacuum permittivity ${\epsilon }_0$, defineHere $E$ and $v$ are the incident kinetic energy and speed far from the atomic nucleus, $b\ge 0$ is the impact parameter, and $L$ is the magnitude of angular momentum. The nonrelativistic limit requires $v\ll c$, with $c$ the speed of light. The fixed-target approximation requires the target mass to be much greater than $m$. In SI, $\kappa$ has units $\mathrm{J}\mathrm{m}$, $E$ has units $\mathrm{J}$, and $b$ has units $\mathrm{m}$.

The plane spanned by incoming and outgoing propagation directions when those directions are not parallel or antiparallel. Polarization perpendicular or parallel to it is useful when expressing a differential scattering cross-section. The plane is not unique at exact forward or backward scattering.

A movable magnetic spectrometer selected scattered electrons by momentum and scattering angle. Magnetic fields dispersed momenta at the focal plane. With relativistic electrons, the inferred momentum determines outgoing energy $E^{\prime }$. Commissioning included steering the beam through the spectrometer at zero angle to check its optics and accepted solid angle. Counters sorted events into momentum and angle bins. The acceptance entered the conversion from counts to differential scattering cross-sections.^[ref]

Radiative corrections matter because photon emission can move events from an elastic scattering peak into lower outgoing-energy bins. The analyses subtracted the elastic radiative tail before unfolding inelastic radiation. Independent MIT and SLAC procedures provided a check. An uncorrected low-$E^{\prime }$ tail alone therefore cannot establish deep inelastic scattering.^[ref]

An ideal Bjorken scaling model has $F_2(x,Q^2)=f(x)$: varying $Q^2$ at fixed Bjorken x leaves this dimensionless deep-inelastic structure function unchanged. The comparison requires extracting a deep-inelastic structure function, rather than plotting raw event counts. Quantum chromodynamics predicts departures from exact Bjorken scaling; finite hadron masses also limit the idealization.

For an original shape comparison, choose $f(x)=x(1-x{)}^3$ and define $S(Q^2)=F_2(0.3,Q^2)/F_2(0.3,Q_0^2)$ with $Q_0^2=2{\mathrm{GeV}}^2$. The ideal model has $S=1$. To display a contrasting size-dependent suppression, define an illustrative elastic-form-factor shape $G(Q^2)=(1+Q^2/{\Lambda }^2{)}^{-2}$ with the freely chosen ${\Lambda }^2=1{\mathrm{GeV}}^2$. Its normalized squared magnitude isThus $A(5{\mathrm{GeV}}^2)=1/16$, whereas $S(5{\mathrm{GeV}}^2)=1$. The elastic-form-factor shape is a comparison of suppression mechanisms, not an elastic differential scattering cross-section assigned to $x=0.3$: actual elastic scattering from a proton has $W=M$ and $x=1$. Neither curve is measured data or a fit to the SLAC-MIT experiment.

The experimental interpretation required more than a flat curve. The angular dependence and small longitudinal response supported spin one-half partons, expressed at leading order by the Callan-Gross relation $F_2=2x{F}_1$. Identification with fractionally charged quarks also drew on later electron, muon and neutrino comparisons. A scaling plot alone does not determine the individual electric charges of constituents.^[ref]

The leading parton model formula already shows the ambiguity: $F_2=x{\sum }_f{Q}_f^2(q_f+{\bar{q}}_f)$ combines squared electric charges with unknown parton distribution functions. One measured function cannot independently fix all these quantities. Confinement also explains why an inclusive final hadron state need not contain an isolated outgoing quark.

Let $N_{\mathrm{inc}}$ be the number of projectiles actually incident on the foil during an exposure, $n_s$ its foil areal number density, and $\eta$ the probability of detecting an accepted particle. For a sufficiently small detector patch of area $A$ at distance $R$ from the scattering point, with its normal at angle $\psi$ to the line of sight,The product $n_s(d\sigma /d\Omega )\Delta \Omega$ is a dimensionless scattering probability in this approximation; $\mu$ is an expected value of counts, so it need not be an integer. The approximation requires a nearly constant differential scattering cross-section and efficiency across the patch and incident beam. For finite acceptance use $\mu =N_{\mathrm{inc}}{n}_s{\int }_{\mathrm{detector}}\eta (\Omega )(d\sigma /d\Omega )d\Omega$, including the incident beam's spread if appreciable. A screen facing the foil has $\psi =0$.

For an original numerical model, stipulate monoenergetic alpha particles with $z=2$, $E=5.00\mathrm{MeV}$, and a gold target with $Z=79$. Use the rounded model constant $e^2/(4\pi {\epsilon }_0)=1.440\mathrm{MeV}\mathrm{fm}$, where $1\mathrm{MeV}=10^6$ electronvolts and $1\mathrm{fm}=10^{-15}\mathrm{m}$. Thus $\kappa =227.52\mathrm{MeV}\mathrm{fm}$ and $a=22.752\mathrm{fm}$. Choose $N_{\mathrm{inc}}=10^{11}$, $n_s=10^{21}{\mathrm{m}}^{-2}$, $\eta =1$, and $\Delta \Omega =10^{-3}\mathrm{sr}$. For example, $A=1\mathrm{m}{\mathrm{m}}^2$ facing the foil at $R\simeq 31.6\mathrm{mm}$ gives this small-patch solid angle.

At $\theta =90^{\circ }$, $b=a$ and $d\sigma /d\Omega =a^2=517.654\mathrm{f}{\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}=5.17654\mathrm{b}\mathrm{s}{\mathrm{r}}^{-1}$, using $1$ barn $=100\mathrm{f}{\mathrm{m}}^2=10^{-28}{\mathrm{m}}^2$. The expected signal is $\mu \simeq 51.8$ counts. At $60^{\circ }$ it is four times larger, about $207$ counts. Doubling foil areal number density or detector solid angle doubles these expected signals as long as the approximations continue to hold.

An observation also contains backgrounds. For a stable source and matched geometry, an estimator is $C_{\mathrm{net}}=C_{\mathrm{foil}}-(T_{\mathrm{foil}}/T_{\mathrm{blank}})C_{\mathrm{blank}}$, where $T$ denotes exposure duration. If source activity changes, normalize by incident exposure as well. Background subtraction removes an estimated background; it does not make finite counting fluctuations disappear.