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For an azimuthally symmetric classical scattering map from impact parameter $b$ to scattering angle $\theta$,When several impact parameters produce the same angle, their nonnegative contributions are summed.

Scattering that resolves the constituents of a hadron at large spacelike momentum transfer, with an inclusive sum over final hadron states.

In electron scattering from a proton, write the reaction as $e^{-}p\to e^{-}X$: the outgoing electron is measured, while $X$ denotes all allowed final hadron states. Large momentum transfer supplies a short-distance probe; a large final invariant mass separates the deep continuum from elastic scattering and individual particle resonances. The useful surprise is that an extended proton can have a response characteristic of much smaller constituents. Deep inelastic scattering turns measurements outside the proton into constraints on its internal parton distribution functions.

The SLAC-MIT deep inelastic electron scattering experiment connects this definition to an actual measurement. Its evidence came from distributions of outgoing electron energies and scattering angles, rather than a photograph of a quark. Bjorken scaling concerns deep-inelastic structure functions after known scattering factors have been accounted for; it does not say that the raw differential scattering cross-section is constant.

In electron-phonon scattering, an electron changes its energy by $\pm \hslash {\omega }_{\mathbf{q}}$ while its crystal momentum changes by a phonon wavevector modulo a reciprocal lattice vector. Charge-transport relaxation weights a scattering angle $\theta$ by $1-\cos \theta$. Small-angle events can therefore relax heat efficiently while relaxing electrical current weakly. At temperatures above the phonon excitation scale, scattering becomes approximately elastic in the electron thermal shell; residual elastic disorder often dominates at very low temperature.

Work in natural units with $c=1$, neglect the electron mass, and put the initial proton at rest with mass $M$. Let $k$ and $k^{\prime }$ be the initial and final electron four-momenta, $q=k-k^{\prime }$, and $Q^2=-q^2>0$. Define the laboratory energy transfer ${\nu }_{\mathrm{lab}}=E-E^{\prime }=P\cdot q/M$. This ${\nu }_{\mathrm{lab}}$ has units of energy; the invariant $P\cdot q$ has units of energy squared. The Bjorken scaling variable and the final hadron invariant mass obeyThese follow by squaring $k-k^{\prime }$ and $P+q$, using $k^2=k^{\prime 2}=0$ and $P=(M,0)$. Solving the Bjorken x definition gives the outgoing energy and then the scattering angle:Alternatively, substituting $Q^2=4E{E}^{\prime }{\sin }^2(\theta /2)$ before solving gives $E^{\prime }=E/[1+2E{\sin }^2(\theta /2)/(Mx)]$. A chosen $x,E,Q^2$ must satisfy $E^{\prime }>0$ and $0\le {\sin }^2(\theta /2)\le 1$. In addition, $W$ must exceed the lightest allowed inelastic hadron threshold; positive $W^2$ alone is insufficient.

For an original numerical example, take $M=0.938\mathrm{GeV}$, $x=0.3$, $E=15\mathrm{GeV}$ and $Q^2=3{\mathrm{GeV}}^2$. Then ${\nu }_{\mathrm{lab}}\simeq 5.330\mathrm{GeV}$, $E^{\prime }\simeq 9.670\mathrm{GeV}$, $\theta \simeq 8.25^{\circ }$, and $W\simeq 2.807\mathrm{GeV}$. Varying $2\le Q^2/{\mathrm{GeV}}^2\le 5$ and $10\le E/\mathrm{GeV}\le 20$ at this same Bjorken x keeps every setting physical and gives $W^2>5.54{\mathrm{GeV}}^2$, above the proton-pion threshold and the stricter chosen continuum selection $W>2\mathrm{GeV}$. These are calculated settings, not historical measurements.

When a particle of mass $M_1$ and initial speed $u_1$ strikes a stationary mass $M_2$, the centre-of-momentum frame moves at $\mathbf{V}=M_1{u}_1\hat{\mathbf{x}}/(M_1+M_2)$. If the relative speed retention in an oblique collision is $q$, its possible outgoing velocity vectors lie on the displayed circle in a scattering plane. Intersect a ray of fixed laboratory direction with that circle to obtain the quadratic formula for outgoing speed. A negative root is inadmissible, and the origin has no defined scattering angle. Tangency gives the maximum laboratory angle when the circle excludes the origin.

Let a projectile have mass $m$, incident speed $v$, and incident kinetic energy $E=m{v}^2/2$. The stationary target has mass $M$. In a head-on elastic collision, the signed outgoing projectile velocity $v^{\prime }$ and target velocity $u$ obey momentum conservation and conservation of energy:Factor the second equation and eliminate $Mu=m(v-v^{\prime })$. The nontrivial collision branch has $u=v+v^{\prime }$, henceThis is the maximum over scattering angles, not the energy of every recoil. To see the bound without assuming a head-on event, set the recoil momentum magnitude to $q$ and its laboratory angle to the incident direction to $\theta$. Conservation of energy with outgoing projectile momentum $\mathbf{p}-\mathbf{q}$ givesThus $T=T_{\max }{\cos }^2\theta$ for a nonzero recoil, and the endpoint occurs at $\theta =0$.

For an original numerical example, stipulate $E=4\mathrm{MeV}$ and $m=m_p$, with $m_p$ the proton mass. Approximate hydrogen, helium and nitrogen target masses by $M=m_p$, $4m_p$ and $14m_p$ respectively. Their maximum fractions and recoil energies areThese stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_p{c}^2\simeq 938\mathrm{MeV}$, the incident ratio $E/(m_p{c}^2)\simeq 0.0043$ gives $v/c\simeq 0.092$, consistent with the nonrelativistic limit used here. A million electronvolts is an energy unit, whereas a mass expressed in $\mathrm{MeV}/c^2$ requires the factor $c^2$ when inserted into an energy equation.

The hydrogen endpoint permits essentially complete transfer because the projectile and target masses match. For a very heavy target, $M\gg m$, the fraction approaches $4m/M$, which is small: the target can take substantial momentum with little kinetic energy. The exact inequality follows from $(m-M{)}^2\ge 0$, which implies $(m+M{)}^2\ge 4mM$.

Consider a nonrelativistic particle of mass $m$ undergoing an elastic collision with a stationary free target of mass $M=Am$, where $A\ge 1$. Assume the outgoing direction is uniformly distributed over solid angle in the center of mass frame; this assumption concerns the scattering angle distribution and need not hold for every target or energy. Let $\mu$ be the cosine of that frame's scattering angle. Transforming the outgoing velocity back to the laboratory givesThe uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha ,1]$. For $A>1$, integration yieldsFor equal masses, take the limit $\alpha \to 0$: $\xi =1$. The expected value of the fractional energy after a collision is instead $\mathbb{E}[x]=(1+\alpha )/2$. These are different averages: averaging a logarithm does not give the logarithm of an average.

For independent collisions with the same distribution, the logarithms of successive energy ratios add. ThusHere $E_{\mathrm{geom}}$ is the geometric-mean energy, and $n_{\log }$ is a characteristic collision count for reaching an energy scale $E_{\ast }$. It is not an exact first-passage expected value. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.

In an original chosen example, set $E_0=10^6\mathrm{eV}$ and $E_{\ast }=0.025\mathrm{eV}$. Approximate hydrogen by $A=1$ and carbon by $A=12$. ThenThese are illustrative model values, not measured collision counts from 1934. The energy scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero energy.

For an ultrarelativistic electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.

For nonrelativistic elastic scattering of a WIMP of mass $m_{\chi }$ from a stationary nucleus of mass $m_N$, the reduced mass is ${\mu }_{\chi N}=m_{\chi }{m}_N/(m_{\chi }+m_N)$. In the centre-of-momentum description, a scattering angle $\vartheta$ gives momentum transfer squared $q^2=2{\mu }_{\chi N}^2{v}^2(1-\cos \vartheta )$. The target kinetic energy is $E_R=q^2/(2m_N)$, giving the displayed maximum at backscattering. In SI units, using masses in kilograms and velocity in metres per second gives joules. Low recoil energy and a falling recoil spectrum make threshold and response calibration important for dark-matter direct detection.

An electron losing energy to an acoustic phonon has, in a normal process, ${\mathbf{k}}^{\prime }={\mathbf{k}}_F-{\mathbf{k}}_{\mathrm{ph}}$, with possible addition of a reciprocal-lattice vector for an Umklapp process. The energy transfer is at most of order $\hslash {\omega }_D=k_B{\theta }_D$, much smaller than $E_F=k_B{T}_F$. Hence the magnitudes change very little even though the momentum direction can change substantially. At $T\sim {\theta }_D\ll T_F$, thermal vacancies near the Fermi surface allow emission into some lower-energy states; exactly at zero temperature emission into an occupied Fermi sea would be blocked by the Pauli exclusion principle.

Using $k^{\prime }=(1-\delta )k_F$ and the free-electron dispersion, energy conservation givesSince $\delta \ll 1$, the bracket is $2\delta +O({\delta }^2)$. Rearranging yieldsEquivalently $\delta \simeq \hslash c{k}_{\mathrm{ph}}/(2E_F)$. For a Debye-cutoff phonon in the preceding model, $\delta \simeq {\theta }_D/(2T_F)=1/480=0.00208$, confirming the approximation. Near-elastic scattering constrains the radial distance in wave-vector space, not the scattering angle alone.

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

The parton model treats a fast proton as an ensemble of approximately collinear, nearly free quarks and antiquarks. A hard deep inelastic scattering collision resolves one parton on a time scale short compared with its binding motion. The probabilities are parton distribution functions; the simplest approximation neglects intrinsic transverse momentum, quark masses and the proton mass in the hard collision. Quantum chromodynamics introduces logarithmic violations of Bjorken scaling, so the parton distribution functions depend on a resolution scale as well as a parton momentum fraction.

Write the incident electron and proton four-momenta as $k,P$, the scattered electron four-momentum as $k^{\prime }$, and set $q=k-k^{\prime }$. With the $(+---)$ Minkowski metric, the Bjorken scaling variable and the inelasticity areAn incident parton with four-momentum $\xi P$ emerges on shell, giving $(\xi P+q{)}^2\simeq 2\xi P\cdot q-Q^2=0$. Thus $\xi =x_B$: the observable Bjorken scaling variable measures the struck parton momentum fraction in this approximation.

For clarity, measure the scattering angle $\vartheta$ from the incoming electron direction, and denote the outgoing electron energy by $E_e^{\prime }$. Direct four-momentum products giveThe expression uses the known beam energies and two measured quantities. An electromagnetic calorimeter measures $E_e^{\prime }$; a tracking detector, the collision interaction vertex and the position of the electromagnetic shower determine $\vartheta$. Coverage close to the incoming electron direction matters at small $Q^2$. A hadronic calorimeter supplies an independent consistency check and rejects backgrounds. If the angle is instead measured from the incoming proton, its sine and cosine half-angle factors interchange.

With the four-momenta in a, $q^2=-2E{E}^{\prime }(1-\cos \theta )$, $p_2\cdot q=m_p(E-E^{\prime })$, and $p_2\cdot p_1=m_{p}E$. Thus the measured beam energy $E$, scattered electron energy $E^{\prime }$ and scattering angle $\theta$ determineHere $Q^2$ measures the spacelike momentum transfer, $y$ is the fractional energy transfer, and the Bjorken scaling variable $x$ is the struck parton's longitudinal momentum fraction in the leading collinear description. The hadronic invariant mass satisfies $W^2=(p_2+q{)}^2=m_p^2+Q^2(1/x-1)$; large $Q^2$ and $W^2$ select deep inelastic scattering.

A beam spectrometer measures $E$ before a thin stationary hydrogen target. Tracking layers on both sides of a magnetic spectrometer measure the outgoing electron direction, charge and curvature; an electromagnetic calorimeter measures its energy. Matching the track to its particle shower, with downstream muon rejection where useful, separates the scattered electron from hadronic tracks. A downstream hadronic detector measures the hadronic activity and supports event-quality checks. The sketch in a shows these separate measurements.

The scattered electron provides a clean, single-particle reconstruction. Reconstructing the entire hadronic system requires summing many charged and neutral fragments, correcting unequal particle calorimeter response and losses through beam holes, and sometimes accounting for escaping neutrinos. Thus the leptonic definition is experimentally convenient. It is not a fundamental requirement of Lorentz invariance: exact reconstruction of $p_X$ gives the same $q=p_X-p_2$, and hadronic or combined reconstructions can be preferable in some kinematic regions.

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.

Let the projectile's initial laboratory velocity be $u\hat{\mathbf{x}}$. The centre-of-momentum frame moves atIn that frame, the projectile's initial speed is $w=u-V=mu/(M+m)$. An elastic collision preserves that speed, but can change its direction. Transforming the outgoing projectile velocity back by Galilean velocity addition givesThis is the velocity-circle construction for elastic scattering: the outgoing tip lies on a circle of radius $w$, centred a distance $V$ from the origin in the velocity plane. Since $M>m$, $V>w$ and the origin lies outside this circle. The largest scattering angle occurs when the ray from the origin is tangent to it. The resulting right triangle givesEquality is kinematically possible; in the centre-of-momentum frame it corresponds to $\cos \chi =-w/V=-m/M$, where $\chi$ is the outgoing projectile angle to its initial direction. In particular, a heavier projectile cannot reverse its laboratory direction in an isolated elastic collision with an initially stationary lighter target.

In the laboratory frame, the initial target is at rest and the outgoing directions are directly related to detector positions. The centre-of-momentum frame has zero total momentum, so the two outgoing momenta are equal and opposite; it cleanly separates bulk motion from relative kinetic energy. Transforming back is needed to predict laboratory angles and speeds. Neither reference frame removes the need for an appropriate collision law.

Take the incident direction as $\hat{\mathbf{x}}$ and put $S=M_1+M_2$. Momentum conservation fixes the centre-of-momentum frame velocity, and a Galilean transformation gives the initial velocities in that frame:After the collision, $M_1{\mathbf{w}}_1^{\prime }+M_2{\mathbf{w}}_2^{\prime }=0$. Hence both particles have the same final-to-initial speed ratio,The relative kinetic energy is multiplied by $q^2$. This common ratio exists independently of how the outgoing direction is selected.

The meaning of restitution matters here. The usual coefficient of restitution $e_n$ is a ratio of normal separation and approach speeds, not total speeds. In a smooth oblique collision, the tangential relative velocity is unchanged. If the incoming relative velocity makes angle $\chi$ to the contact normal, thenFor example, $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}\ne 1/2$. Therefore the requested identification is not generally true with the standard normal definition. The printed equation is valid if its $e$ denotes the relative speed retention in an oblique collision, namely $e=q$, or if an additional collision model supplies that equality. The elastic cases have $q=e_n=1$ and are unaffected by this qualification.

To derive the scattering formula under that convention, write ${\mathbf{v}}_1^{\prime }=\mathbf{V}+{\mathbf{w}}_1^{\prime }$. Its components give the laboratory scattering circleSolving the resulting quadratic for the nonnegative speed yieldsWith $q=e$, this is the requested formula. A real square root and $v_1\ge 0$ are both necessary: a negative algebraic root is not a second physical speed, and $v_1=0$ does not define a scattering angle. The subsequent limits are taken using this laboratory scattering circle.

Normalization and angular pattern. A scattering phase function describes the distribution of single-scattered power over outgoing solid angle, for a specified incident direction, wavelength and polarization convention. Use the common normalizationThe scattering asymmetry parameter is the mean cosine of the scattering angle: forward scattering has positive $g$, backward scattering negative $g$, and a forward–backward symmetric pattern has $g=0$.

Let $C=E_0(2\pi /\lambda {)}^4[(n^2-1)/(n^2+2){]}^2{r}^6$. The given small-sphere angular intensity is $I=C(1+{\mu }^2)/2$, where $\mu =\cos \Theta$. Its total scattered power isNormalizing givesDimensionally the supplied $I$ is radiant intensity, power per solid angle. If interpreted as detector irradiance at a distance, the appropriate inverse-square distance factor is also needed; it cancels out of the normalized angular pattern.

The polar plot has equal forward and backward lobes, each with radius $3/2$, and lateral values $3/4$. A small nonabsorbing particle behaves as an induced electric dipole. Radiation transverse to the dipole and the average over two orthogonal incident polarizations produce the $1+{\cos }^2\Theta$ form. Its odd angular moment vanishes:As the size parameter $2\pi r/\lambda$ grows, Rayleigh scattering in the electric-dipole approximation no longer suffices. Higher multipoles and interference produce structure in the angular pattern, and large particles generally develop a strong narrow forward diffraction peak. Their $g$ is normally positive, although detailed values and angular features depend on refractive index, size distribution and shape; they are not fixed by radius alone. Large spheres may show rainbows and backscattering structure, while nonspherical ice introduces further angular features.

Optically thick transmission. In a nonabsorbing two-stream or transport approximation, it is the transport optical depth $(1-g)\tau$ that controls the loss of forward directional memory. Using the same idealized albedo closure as in option 1 gives total forward-hemisphere transmittanceFor the same large but finite extinction optical depth,The respective transport-depth factors are $0.15$, $0.30$ and $1$. The water cloud scatters most strongly forward and returns the least light backward. All transmittances tend to zero in the infinite-depth limit. The strictly unscattered direct-beam transmittance is instead $e^{-\tau }$, identical in the three cases: the ranking concerns total transmitted radiation, including forward-scattered light, not that direct component alone. Full radiative transfer can also depend on higher angular moments beyond $g$.

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.

Integrate the squared partial-wave expansion of a scattering amplitude over the solid angle. Legendre polynomial orthogonality eliminates terms with unequal angular-momentum indices, givingAt zero scattering angle, $P_{\ell }(1)=1$ and $\mathrm{Im}(e^{i{\delta }_{\ell }}\sin {\delta }_{\ell })={\sin }^2{\delta }_{\ell }$. ConsequentlyThis is the optical theorem. In the present elastic problem the total and elastic cross sections coincide. For an absorptive interaction the inclusive total still obeys the optical theorem, but need not equal the integral of the elastic differential cross section.

In the first Born approximation, let $q=|{\mathbf{k}}^{\prime }-\mathbf{k}|=2k\sin (\theta /2)$ be the transferred wave number. The angular integral in the three-dimensional Fourier transform is $4\pi \sin (qr)/(qr)$, soWriting $g=2m{V}_{0}R/{\hslash }^2$ and $\mathrm{sinc}u=\sin u/u$, with value one at zero, the delta-shell Born amplitude givesFor $kR\ll 1$ the scattering is approximately isotropic, with total cross section $4\pi g^2{R}^2$. For $kR\gg 1$ it has a forward lobe of angular width of order $(kR{)}^{-1}$ and oscillatory zeros where $2kR\sin (\theta /2)=n\pi$. At a fixed nonzero scattering angle, its envelope decreases as $k^{-2}$. The right-hand panel of the figure shows this angular dependence.

For real $V_0$ the first-order scattering amplitude is real, so inserting it literally in the optical theorem would predict zero total cross section, contrary to its nonzero squared modulus. The first Born amplitude alone does not satisfy the optical theorem at order $V_0^2$. This is an order-of-approximation issue: the squared first-order amplitude is already second order, and the second Born term supplies the required second-order imaginary forward amplitude. At first order both sides vanish. A weak-shell regime is needed for the Born truncation itself to be accurate.