Neutrino scattering

A neutrino is the most reluctant particle we know: it carries no charge, no colour, and feels only the weak force, so it can drill through a light-year of lead with an even chance of passing clean through. Yet fire enough of them at a target and this very reluctance becomes a scalpel. Because the neutrino talks only to the weak charged current — and that current has a rigid V−A structure that couples exclusively to left-handed particles and right-handed antiparticles — every neutrino interaction comes stamped with a fixed helicity. There is no averaging over spin states to blur the picture; the beam selects one handedness at the source and the target quark's spin is the only thing left to sum over. That single fact drives everything in this article. The parton-level reactions $\nu_\mu d\to\mu^- u$ and $\bar\nu_\mu u\to\mu^+ d$ each admit exactly one helicity combination, and the total spin of the colliding pair along the beam — $S_z=0$ for a neutrino on a quark, $S_z=1$ for an antineutrino on a quark — fixes the angular distribution directly, with no dynamics beyond the contact vertex. The $S_z=0$ case is isotropic; the $S_z=1$ case must vanish backward to conserve angular momentum, giving a $(1-y)^2$ falloff, so antineutrino cross sections on quarks come out three times smaller than neutrino ones. Phenomenologically the neutrino is just a differently-coloured probe of the proton — quasi-elastic at low $Q^2$ , resonant at a few GeV, and deeply inelastic at high energy, the same machinery as electron–proton scattering with the photon swapped for a $W$ . Folding the parton cross sections with the quark distributions turns the helicity rule into the celebrated linear rise of the neutrino cross section with energy — the cleanest signature of a point-like contact interaction in all of particle physics.

Key takeaways

Neutrinos probe nucleons through the weak charged current; because the current is V−A, each parton reaction $\nu q\to\ell q'$ has exactly one allowed helicity combination — there is no spin averaging over the neutrino.
The colliding-pair spin fixes the angular distribution: $\nu q$ is an $S_z=0$ state and scatters isotropically, $\sigma_{\nu q}=G_F^2\hat{s}/\pi$ ; $\bar\nu q$ is an $S_z=1$ state with a $\tfrac14(1+\cos\theta^*)^2=(1-y)^2$ shape and $\sigma_{\bar\nu q}=G_F^2\hat{s}/3\pi$ — three times smaller.
The inelasticity $y=\tfrac12(1-\cos\theta^*)$ is the master observable; $\mathrm{d}\sigma/\mathrm{d}y$ is flat for $\nu q$ and $\propto(1-y)^2$ for $\bar\nu q$ , and its shape counts quarks versus antiquarks.
Since $s\approx2m_N E_\nu$ and $Q^2=sxy\le2m_N E_\nu\ll M_W^2$ even at 400 GeV, the $W$ propagator collapses to Fermi's contact term and the cross section rises linearly with beam energy.
Measured slopes: $\sigma^{\nu N}/E_\nu=0.677\times10^{-38}\,\mathrm{cm^2/GeV}$ and $\sigma^{\bar\nu N}/E_\nu=0.334\times10^{-38}\,\mathrm{cm^2/GeV}$ ; the ratio $\approx0.5$ (not $1/3$ ) is direct evidence of the antiquark sea.
Key idea: the V−A weak current dictates a single allowed helicity combination for each neutrino–quark process, so the spin state of the colliding pair (S_z=0 versus S_z=1) directly fixes the angular distribution and hence the factor-of-three difference between neutrino and antineutrino cross sections.

Where this lives today

The helicity rule and the linear rise are now precision tools. The NuTeV, CCFR, CDHS and CHARM experiments turned the neutrino/antineutrino cross-section difference into one of the sharpest measurements of the weak mixing angle $\sin^2\theta_W$ via the Paschos–Wolfenstein ratio, and NuTeV's $\sim3\sigma$ anomaly there remains a cautionary tale about nuclear and isospin corrections. The beam technology Schwartz invented now runs at the intensity frontier: Fermilab's NuMI and LBNF and J-PARC's beam fire megawatt-class proton pulses at graphite targets, focus the pions with magnetic horns exactly as in the overview, and send $\nu_\mu$ across hundreds of kilometres of rock to NOvA, T2K, and the coming DUNE — where the goal is no longer the cross section but flavour oscillation and leptonic CP violation. At the opposite scale, IceCube has pushed the linear-rise story into the PeV regime, where the neutrino energy is finally high enough that the Earth turns opaque and the $W$ -propagator damping predicted by the contact approximation's breakdown appears as an attenuation of upward-going neutrinos. And the same charged-current vertex, run in reverse, is how every detector sees a neutrino at all: a muon born from $\nu_\mu N\to\mu^- X$ , the reaction this article computes.

The physical picture

Strip away the beamline and the phenomenology and neutrino scattering is one contact vertex acting on two spins. The neutrino arrives already polarized — left-handed, no choice — and the $W$ it exchanges refuses any other handedness. So the only freedom is the target quark's spin, and the total spin of the incoming pair projected on the beam axis is the whole story. The first figure sets the two cases side by side: a left-handed neutrino meeting a left-handed quark makes a spin-zero state that scatters every which way, while a right-handed antineutrino meeting a left-handed quark makes a spin-one state that cannot fire straight backward.

Figure

Helicity bookkeeping for the two parton reactions. Left: ν_μ (left-handed) on a quark (left-handed) forms an S_z = 0 state — scattering is isotropic, σ = G_F² ŝ/π. Right: ν̄_μ (right-handed) on a quark (left-handed) forms an S_z = 1 state — backward scattering (θ* = 180°) is forbidden by angular momentum, giving a (1−y)² shape and σ = G_F² ŝ/3π, three times smaller

The rest is bookkeeping — but bookkeeping with a spectacular payoff. Because the neutrino must borrow a virtual $W$ some eighty-five proton-masses heavy, and the momentum it can transfer even at 400 GeV is tiny beside $M_W$ , the $W$ propagator freezes to a constant and the interaction is genuinely point-like. A point interaction has no length scale to set a falling form factor, so the cross section simply grows with the available energy: $\sigma\propto\hat{s}\propto E_\nu$ . Fold the parton reactions over the quark and antiquark content of the nucleon and that linear rise survives all the way to the nucleon level. The second figure is the quantity a neutrino experiment actually measures — the cross section per unit energy, flat across two decades of beam energy, with the neutrino line sitting almost exactly twice as high as the antineutrino line.

Figure

The measured total charged-current cross section per nucleon divided by beam energy, σ/E_ν, plotted against E_ν for an isoscalar target. Both stay flat — the hallmark σ ∝ E_ν of a point interaction — at σ^νN/E_ν = 0.677×10⁻³⁸ cm²/GeV and σ^ν̄N/E_ν = 0.334×10⁻³⁸ cm²/GeV. Their ratio 0.49 exceeds the pure-valence value 1/3 because the antiquark sea lifts the antineutrino rate

The core equations

Neutrino scattering inherits its kinematics from deep inelastic scattering and its dynamics from the weak charged current; both are recalled here and specialized. The first pair of relations fixes the energy and momentum scales. For a neutrino of energy $E_\nu$ striking a nucleon of mass $m_N$ at rest, the centre-of-mass energy squared and the momentum transfer are set entirely by the beam energy and the two dimensionless scaling variables.

s=(p_\nu+p_N)^2=m_N^2+2m_N E_\nu\approx 2m_N E_\nu,\qquad Q^2=2m_N E_\nu\,x\,y\le 2m_N E_\nu \qquad (1)

where: $s$ — squared CM energy; $m_N\approx0.94$ GeV $/c^2$ — nucleon mass; $x$ — Bjorken scaling variable (the struck quark's momentum fraction); $y$ — inelasticity, the fractional energy lost by the lepton in the target frame. Read it as: even at $E_\nu=400$ GeV, $Q^2\le750$ GeV $^2\ll M_W^2=6460$ GeV $^2$ , so the $W$ is never close to on-shell. Because $Q^2\ll M_W^2$ always, the massive- $W$ propagator collapses exactly as it does for muon decay, and the interaction is Fermi's four-fermion contact with strength $G_F$ .

\frac{-i\,g_{\mu\nu}}{q^2-M_W^2}\;\xrightarrow{\,q^2\ll M_W^2\,}\;\frac{i\,g_{\mu\nu}}{M_W^2},\qquad \frac{G_F}{\sqrt2}=\frac{g_w^2}{8M_W^2} \qquad (2)

where: $q$ — four-momentum through the $W$ ; $g_w\approx0.65$ — weak coupling; $G_F/(\hbar c)^3=1.166\times10^{-5}$ GeV $^{-2}$ — Fermi constant. The propagator carries no $q^2$ dependence in this limit, which is precisely why the cross section will grow with energy instead of falling. The heart of the matter is the parton cross section. For a neutrino on a quark the pair is in an $S_z=0$ state, the amplitude is isotropic, and the differential cross section in $y$ is flat.

\frac{\mathrm{d}\sigma_{\nu q}}{\mathrm{d}y}=\frac{G_F^2\,\hat{s}}{\pi},\qquad \sigma_{\nu q}=\frac{G_F^2\,\hat{s}}{\pi},\qquad \hat{s}=x\,s \qquad (3)

where: $\hat{s}$ — the quark-level CM energy squared, a fraction $x$ of the nucleon-level $s$ ; the flatness in $y$ is the direct fingerprint of the spin-zero (isotropic) initial state. Integrating the flat distribution over $y\in[0,1]$ just returns the same constant, $G_F^2\hat{s}/\pi$ . For an antineutrino on a quark the pair is in an $S_z=1$ state; angular momentum forbids backward scattering, imposing a $(1-y)^2$ weight and shrinking the total by exactly three.

\frac{\mathrm{d}\sigma_{\bar\nu q}}{\mathrm{d}y}=\frac{G_F^2\,\hat{s}}{\pi}(1-y)^2,\qquad \sigma_{\bar\nu q}=\frac{G_F^2\,\hat{s}}{3\pi} \qquad (4)

where: $(1-y)^2=\tfrac14(1+\cos\theta^*)^2$ is the square of the $S_z=1$ rotation ( $d$ -)matrix element; $\int_0^1(1-y)^2\mathrm{d}y=\tfrac13$ produces the factor-of-three suppression. The same $(1-y)^2$ appears for a neutrino on an antiquark, because there the pair spin is again $S_z=1$ . Convolving the parton cross sections with the quark and antiquark distributions gives the nucleon-level double-differential cross section — the master formula of neutrino deep inelastic scattering.

\frac{\mathrm{d}^2\sigma^{\nu N}}{\mathrm{d}x\,\mathrm{d}y}=\frac{G_F^2\,s}{\pi}\,x\Big[q(x)+\bar q(x)\,(1-y)^2\Big] \qquad (5)

where: $q(x)$ — momentum-weighted distribution of quarks the neutrino can strike (down-type quarks $d,s$ ); $\bar q(x)$ — the corresponding antiquarks ( $\bar u,\bar c$ ). For antineutrinos the roles of $q$ and $\bar q$ swap. The quark term is flat in $y$ ; the antiquark term carries the $(1-y)^2$ , so the shape of the $y$ distribution literally counts the sea. Integrating over both variables and using $s=2m_N E_\nu$ yields the total cross section, which is strictly linear in the beam energy.

\frac{\sigma^{\nu N}}{E_\nu}=\frac{2G_F^2 m_N}{\pi}\Big[Q+\tfrac13\bar Q\Big],\qquad Q=\!\int_0^1\! x\,q(x)\,\mathrm{d}x,\;\; \bar Q=\!\int_0^1\! x\,\bar q(x)\,\mathrm{d}x \qquad (6)

where: $Q,\bar Q$ — the total momentum fractions carried by the struck quarks and antiquarks; the prefactor evaluates to $3.16\times10^{-38}$ cm $^2$ /GeV. With $Q\approx0.20$ , $\bar Q\approx0.04$ this gives $\sigma^{\nu N}/E_\nu\approx0.67\times10^{-38}$ cm $^2$ /GeV — the measured slope.

Derivation AWhy neutrino–quark scattering is isotropic and antineutrino–quark is not

The goal is to derive, from angular-momentum conservation alone and without evaluating a single trace, the two angular distributions that control everything downstream: a flat (isotropic) distribution for $\nu q$ scattering and a $(1-y)^2$ distribution for $\bar\nu q$ , and hence the factor-of-three ratio of their total cross sections. The entire argument rests on one input — the V−A current couples only to left-handed particles and right-handed antiparticles — so once the incoming helicities are fixed, the total spin projected on the beam axis is determined, and the scattering amplitude is just the rotation matrix that carries that spin state from the incoming to the outgoing direction. This is the cleanest place in the Standard Model where a cross-section ratio is fixed by kinematics rather than dynamics.

Step 1. Fix the incoming helicities. In $\nu_\mu d\to\mu^- u$ the V−A vertex admits only a left-handed neutrino and a left-handed quark. Take the neutrino moving in $+z$ and the quark in $-z$ in the CM frame. A left-handed neutrino has spin along $-z$ ; a left-handed quark moving in $-z$ has spin along $+z$ . The two spins are antiparallel, so the total spin projection is $S_z=-\tfrac12+\tfrac12=0$ . Why: with both particles ultrarelativistic, helicity equals chirality, so V−A does not merely prefer these helicities — it forbids all others, leaving a single, pure spin state to scatter.

Step 2. Scatter the $S_z=0$ state. A spin-0 combination is rotationally invariant: the amplitude to emerge at CM angle $\theta^*$ is the Wigner element $d^0_{00}(\theta^*)=1$ , independent of angle. The differential cross section is therefore flat, $\mathrm{d}\sigma_{\nu q}/\mathrm{d}\Omega^*=\text{const}$ , and integrating over $4\pi$ gives $\sigma_{\nu q}=G_F^2\hat{s}/\pi$ . Why: an isotropic outgoing distribution is not an approximation here — it is forced by the spherical symmetry of a spin-zero initial state, which carries no preferred axis.

Step 3. Repeat for the antineutrino. In $\bar\nu_\mu u\to\mu^+ d$ the vertex now demands a right-handed antineutrino and a left-handed quark. The right-handed antineutrino moving in $+z$ has spin along $+z$ ; the left-handed quark moving in $-z$ has spin along $+z$ . The spins are parallel: $S_z=+\tfrac12+\tfrac12=+1$ . The amplitude is the spin-1 element $d^1_{11}(\theta^*)=\tfrac12(1+\cos\theta^*)$ , which vanishes at $\theta^*=180^\circ$ . Why: a backward-scattered final state would require $S_z\to-1$ , flipping one unit of angular momentum that nothing in the contact interaction can supply, so the backward amplitude is strictly zero.

Step 4. Square and integrate. The angular weight is $|d^1_{11}|^2=\tfrac14(1+\cos\theta^*)^2$ ; with $y=\tfrac12(1-\cos\theta^*)$ this is exactly $(1-y)^2$ . Integrating, $\int_0^1(1-y)^2\mathrm{d}y=\tfrac13$ , so $\sigma_{\bar\nu q}=G_F^2\hat{s}/3\pi=\tfrac13\sigma_{\nu q}$ . Why: the factor of three is nothing but the angular integral of the $S_z=1$ distribution against the flat $S_z=0$ one — a purely geometric statement about spin, measured to confirm V−A. *In summary: $\int_0^1 1\,\mathrm{d}y=1$ (flat) versus $\int_0^1(1-y)^2\mathrm{d}y=1/3$ ; the $S_z=0$ state is isotropic and the $S_z=1$ state is backward-forbidden; the ratio $\sigma_{\nu q}/\sigma_{\bar\nu q}=3$ follows directly.*

Go deeper · Derivation B · 2 minThe inelasticity y and the parton cross section

Derivation BThe inelasticity y and the parton cross section

The aim is to connect the abstract CM scattering angle to the laboratory observable — the inelasticity $y$ , the fraction of the neutrino's energy transferred to the hadronic system — and to write the parton cross sections as compact functions of $y$ that an experiment can histogram directly. Establishing $y=\tfrac12(1-\cos\theta^*)$ turns the angular distributions of Derivation A into $\mathrm{d}\sigma/\mathrm{d}y$ shapes, and shows that a single measured distribution, the $y$ spectrum, separates the quark from the antiquark contribution because the two enter with different, orthogonal-looking $y$ dependences. This is the observable that made neutrino beams a quantitative tool for proton structure.

Step 1. Define $y$ invariantly. In terms of four-momenta, $y=(p_N\cdot q)/(p_N\cdot p_\nu)$ , where $q$ is the four-momentum transfer to the lepton. In the target rest frame this reduces to $y=(E_\nu-E_\mu)/E_\nu=\nu/E_\nu$ , the fractional lab energy given to the hadrons. Why: writing $y$ as a ratio of invariants guarantees the same number in every frame, so the CM computation and the lab measurement refer to one quantity.

Step 2. Relate $y$ to the CM angle. Working in the massless-parton CM frame, elementary kinematics give $Q^2=\hat{s}\,\tfrac12(1-\cos\theta^*)$ , and since $Q^2=\hat{s}\,y$ at fixed $x$ , one reads off $y=\tfrac12(1-\cos\theta^*)$ . Thus $y=0$ is forward scattering ( $\theta^*=0$ ) and $y=1$ is backward ( $\theta^*=180^\circ$ ). Why: this is the bridge between Derivation A's angular language and the experimenter's $y$ — the backward direction that the $S_z=1$ state forbids is exactly $y=1$ , where $(1-y)^2$ vanishes.

Step 3. Convert the distributions. The isotropic $\nu q$ amplitude is flat in $\cos\theta^*$ , hence flat in $y$ : $\mathrm{d}\sigma_{\nu q}/\mathrm{d}y=G_F^2\hat{s}/\pi$ . The $\bar\nu q$ amplitude carries $\tfrac14(1+\cos\theta^*)^2=(1-y)^2$ : $\mathrm{d}\sigma_{\bar\nu q}/\mathrm{d}y=(G_F^2\hat{s}/\pi)(1-y)^2$ . Why: because $\mathrm{d}y\propto\mathrm{d}\cos\theta^*$ with constant Jacobian, the $y$ spectrum is a faithful, undistorted image of the CM angular distribution — no reweighting is hidden in the change of variable.

Step 4. Read off the mean inelasticities. For the flat quark distribution $\langle y\rangle_\nu=\int_0^1 y\,\mathrm{d}y/\int_0^1\mathrm{d}y=\tfrac12$ ; for the antiquark $(1-y)^2$ distribution $\langle y\rangle=\int_0^1 y(1-y)^2\mathrm{d}y/\int_0^1(1-y)^2\mathrm{d}y=(1/12)/(1/3)=\tfrac14$ . Why: the measured mean- $y$ is a direct quark/antiquark meter — a target of pure quarks would give $\langle y\rangle=1/2$ for neutrinos, and any downward shift toward $1/4$ betrays the antiquark sea. *In summary: $y=\tfrac12(1-\cos\theta^*)$ ; the flat distribution gives $\langle y\rangle=1/2$ ; the $(1-y)^2$ distribution gives $\langle y\rangle=1/4$ via $\int_0^1 y(1-y)^2\mathrm{d}y=1/12$ and $\int_0^1(1-y)^2\mathrm{d}y=1/3$ .*

Go deeper · Derivation C · 2 minFolding partons into the nucleon cross section and the linear rise

Derivation CFolding partons into the nucleon cross section and the linear rise

The goal is to assemble the nucleon-level cross section from the parton pieces and to prove that it rises linearly with beam energy — the single most striking feature of neutrino scattering and the clearest experimental proof that the weak interaction is point-like at these energies. We weight each parton process by the probability of finding a quark of momentum fraction $x$ , integrate over $x$ and $y$ , and track the energy dependence, arriving at a slope $\sigma/E_\nu$ that is a pure number times the quark momentum fraction. The linearity is not a small-angle approximation; it holds exactly as long as $Q^2\ll M_W^2$ , and its eventual failure at very high energy is itself a prediction.

Step 1. Weight by the parton density. The probability of striking a quark carrying fraction $x$ is $q(x)\,\mathrm{d}x$ , and that quark carries $\hat{s}=xs$ . Multiplying the parton cross sections (3) and (4) by the densities and summing quark and antiquark channels gives $\mathrm{d}^2\sigma^{\nu N}/\mathrm{d}x\,\mathrm{d}y=(G_F^2 s/\pi)\,x[q(x)+\bar q(x)(1-y)^2]$ . Why: this is the parton-model factorization — a hard, calculable parton cross section times a soft, measured probability distribution — and it works because at high $Q^2$ the neutrino resolves individual quarks that scatter incoherently.

Step 2. Integrate over $y$ . Using $\int_0^1\mathrm{d}y=1$ and $\int_0^1(1-y)^2\mathrm{d}y=\tfrac13$ , the single-differential cross section is $\mathrm{d}\sigma^{\nu N}/\mathrm{d}x=(G_F^2 s/\pi)\,x[q(x)+\tfrac13\bar q(x)]$ . Why: the $y$ integral is where the factor-of-three between the quark and antiquark contributions enters the nucleon cross section, so the total already encodes the helicity rule of Derivation A.

Step 3. Integrate over $x$ and expose the energy dependence. Define $Q=\int x\,q\,\mathrm{d}x$ and $\bar Q=\int x\,\bar q\,\mathrm{d}x$ ; then $\sigma^{\nu N}=(G_F^2 s/\pi)[Q+\tfrac13\bar Q]$ . Substituting $s=2m_N E_\nu$ gives $\sigma^{\nu N}=(2G_F^2 m_N/\pi)[Q+\tfrac13\bar Q]\,E_\nu$ , linear in $E_\nu$ . Why: every factor of energy in the cross section came from $\hat{s}=xs\propto E_\nu$ in the point coupling; with no form-factor falloff to oppose it, the growth is pristinely linear — the signature the data confirm across two decades.

Step 4. Put in numbers. The prefactor is $2G_F^2 m_N/\pi=3.16\times10^{-38}$ cm $^2$ /GeV. With the neutrino striking down-type quarks and up-type antiquarks of an isoscalar target, $Q\approx0.20$ and $\bar Q\approx0.04$ , so $\sigma^{\nu N}/E_\nu\approx3.16\times10^{-38}\times0.213=0.67\times10^{-38}$ cm $^2$ /GeV. For antineutrinos the same fractions give $[\tfrac13 Q+\bar Q]\approx0.107$ and $0.34\times10^{-38}$ cm $^2$ /GeV. Why: these reproduce the measured slopes $0.677$ and $0.334$ and, crucially, the ratio $0.49$ rather than the pure-valence $1/3$ — the excess is a direct, quantitative sighting of the antiquark sea. *In summary: the prefactor $2G_F^2 m_N/\pi=3.16\times10^{-38}$ cm $^2$ /GeV; the resulting slopes are $\sigma^{\nu N}/E_\nu=0.67\times10^{-38}$ cm $^2$ /GeV and $\sigma^{\bar\nu N}/E_\nu=0.34\times10^{-38}$ cm $^2$ /GeV, with ratio $0.49$ — the pure-valence prediction would give $1/3$ .*

Limiting cases

Case	Result	Meaning
Neutrino on quark ( $S_z=0$ )	flat $\mathrm{d}\sigma/\mathrm{d}y$ , $\sigma_{\nu q}=G_F^2\hat{s}/\pi$	isotropic; $\langle y\rangle=1/2$
Antineutrino on quark ( $S_z=1$ )	$\mathrm{d}\sigma/\mathrm{d}y\propto(1-y)^2$ , $\sigma_{\bar\nu q}=G_F^2\hat{s}/3\pi$	backward forbidden; $\langle y\rangle=1/4$ ; three times smaller
Pure valence target (no sea)	$\sigma^{\bar\nu N}/\sigma^{\nu N}=1/3$	the naive quark-only prediction
Realistic target (with sea)	ratio $\approx0.5$	antiquarks lift the antineutrino rate above $1/3$
$Q^2\ll M_W^2$ (all fixed-target energies)	contact interaction, $\sigma\propto E_\nu$	point coupling, no form-factor falloff
$Q^2\to M_W^2$ ( $E_\nu\gtrsim$ few TeV)	propagator damping sets in, $\sigma$ flattens	the $W$ resolves; linear rise finally tames

Common trapAveraging over four spin states as if the neutrino were unpolarized

The tempting mistake is to compute $\overline{|\mathcal{M}|^2}$ with the usual QED spin-average factor of $1/4$ — one quarter for the four combinations of two initial spins. That is wrong for a neutrino. The neutrino is produced left-handed at the source and the V−A vertex admits no other handedness, so there is nothing to average over on the neutrino leg: its spin is fixed, not summed. You average only over the two spin states of the target quark, a factor of $1/2$ , not $1/4$ . Using $1/4$ halves every neutrino cross section and quietly destroys the factor-of-three ratio that is the whole point — because that ratio comes precisely from the single, definite helicity of the incoming neutrino meeting a definite-helicity quark. The rule of thumb: count only the spins you are genuinely ignorant of. A polarized beam removes one factor of two; a purely left-handed beam removes it completely.

Worked examples

1 · The neutrino cross section and a neutrino's mean free path (fully worked)

Using the measured slope $\sigma^{\nu N}/E_\nu=0.677\times10^{-38}$ cm $^2$ /GeV, find the total charged-current cross section on a nucleon at $E_\nu=100$ GeV, and estimate how far a 100 GeV neutrino travels in solid iron before interacting.

Step 1. Cross section at 100 GeV. Linearity makes this trivial: $\sigma^{\nu N}=0.677\times10^{-38}\times100=\mathbf{6.77\times10^{-37}\ cm^2}$ per nucleon. For comparison a typical strong-interaction cross section is $\sim10^{-26}$ cm $^2$ , so the neutrino is about ten orders of magnitude more penetrating.

Step 2. Nucleon number density of iron. With density $\rho=7.87$ g/cm $^3$ and very nearly 1 g per mole of nucleons, the nucleon density is $n=\rho N_A=7.87\times6.022\times10^{23}=\mathbf{4.74\times10^{24}\ cm^{-3}}$ .

Step 3. Mean free path. $\lambda=1/(n\sigma)=1/(4.74\times10^{24}\times6.77\times10^{-37})=1/(3.21\times10^{-12}\ \mathrm{cm^{-1}})=3.1\times10^{11}$ cm. Converting, $\lambda\approx\mathbf{3.1\times10^{6}\ km}$ — about three million kilometres of solid iron, roughly eight times the Earth–Moon distance, for a single 100 GeV neutrino to have an even chance of interacting. This is why neutrino detectors weigh thousands of tonnes and why the linear growth of $\sigma$ with energy matters so much: at these energies the neutrino is finally interacting often enough to study. σ = 6.8×10⁻³⁷ cm², λ ≈ 3×10⁶ km of iron.

2 · Reading the antiquark sea from the cross-section ratio

A neutrino experiment measures $\sigma^{\bar\nu N}/\sigma^{\nu N}=0.49$ on an isoscalar target. Show why this exceeds the naive $1/3$ , and estimate the fraction of the struck momentum carried by antiquarks.

Step 1. Write the ratio from equation (6). With $R\equiv\bar Q/Q$ the fraction of antiquark to quark momentum, $\dfrac{\sigma^{\bar\nu N}}{\sigma^{\nu N}}=\dfrac{\tfrac13 Q+\bar Q}{Q+\tfrac13\bar Q}=\dfrac{\tfrac13+R}{1+\tfrac13 R}$ . A pure-valence target ( $R=0$ ) gives exactly $\tfrac13$ ; any antiquarks push the ratio up because they enter the antineutrino cross section with the flat (large) weight and the neutrino cross section with the suppressed $(1-y)^2$ weight.

Step 2. Solve for $R$ . Setting $\dfrac{\tfrac13+R}{1+\tfrac13 R}=0.49$ gives $\tfrac13+R=0.49+0.163R$ , so $0.837R=0.157$ , $R=\mathbf{0.19}$ . About one part in five of the neutrino-struck momentum sits in antiquarks — a clean, model-independent count of the sea.

Step 3. Cross-check with $\langle y\rangle$ . The measured $y$ distribution independently fixes the sea: a pure-quark neutrino sample has $\langle y\rangle=1/2$ , and the antiquark admixture (weighted $(1-y)^2$ , $\langle y\rangle=1/4$ ) pulls the neutrino mean below $1/2$ while pushing the antineutrino mean above $1/4$ . The two measurements — total-rate ratio and mean inelasticity — agree, which is what turned neutrino DIS into a precision probe of parton content. R ≈ 0.19: antiquarks carry ~19% of the struck momentum, lifting the ratio from 1/3 to ~0.49.

3 · Checking the contact approximation at the highest fixed-target energy

Confirm that the point-interaction picture holds for the highest-energy fixed-target neutrino beam, $E_\nu=400$ GeV.

Step 1. Maximum momentum transfer. From equation (1), $Q^2_{\max}=2m_N E_\nu=2\times0.938\times400=\mathbf{750\ GeV^2}$ , reached at $x=y=1$ .

Step 2. Compare to the $W$ mass. $M_W^2=(80.4)^2=6460$ GeV $^2$ , so $Q^2_{\max}/M_W^2=750/6460=0.116$ . Even in the most extreme kinematics the propagator denominator changes by about 12%, and for typical events ( $x\sim0.2$ , $y\sim0.5$ ) the shift is under 2%. Q²_max/M_W² ≈ 0.12: the contact approximation is good to a few percent, and σ ∝ E_ν holds until multi-TeV energies where the propagator finally bites.

Physics of the portrait

Neutrino scattering became a science the day someone worked out how to aim neutrinos. That someone realized the two-body decay of a fast pion would spray neutrinos forward in a tight cone — and that a wall of steel would strip away everything but the neutrinos, leaving a beam.

Origins · 1 minMelvin Schwartz and how this was discovered

Melvin Schwartz

1932–2006 · the accelerator neutrino beam and the two-neutrino experiment

Melvin Schwartz was a young physicist at Columbia in 1959 when Bruno Pontecorvo's and his own colleagues' talk of studying the weak interaction directly ran into an apparent wall: neutrinos interact so feebly that no source seemed strong enough. The story Schwartz told is that the idea came to him overnight — use the neutrinos that stream from the decay of energetic pions produced at an accelerator. Because a pion decays $\pi^+\to\mu^+\nu_\mu$ in flight, and because $m_\pi\ll E_\pi$ , the neutrinos are boosted into a forward cone; pile up enough shielding to absorb the muons and hadrons and what emerges is the first controlled neutrino beam. With Leon Lederman and Jack Steinberger, Schwartz built the experiment at the Brookhaven AGS, firing protons into a beryllium target, letting the pions decay, and stopping everything in thirteen metres of steel armour plate salvaged from a scrapped warship. In 1962 their spark chambers caught neutrinos producing muons but never electrons — proof that the neutrino accompanying the muon, $\nu_\mu$ , is a distinct particle from the electron's $\nu_e$ . This was the discovery of the second neutrino flavour and the first demonstration of lepton-family structure, and it won the three of them the 1988 Nobel Prize. But the deeper legacy was the method: Schwartz had turned the weak interaction from a nuisance into an instrument. Every subsequent neutrino programme — Gargamelle's discovery of neutral currents, the CDHS and CHARM measurements of nucleon structure and $\sin^2\theta_W$ , the long-baseline oscillation experiments running today — is built on the accelerator neutrino beam he conceived. The reluctant particle became a probe because Schwartz asked how to make a lot of them and point them the same way.

History

1956

— Reines and Cowan detect the (anti)neutrino directly at a reactor

1960

— Schwartz proposes the accelerator neutrino beam from pion decay in flight

1962

— the Brookhaven two-neutrino experiment shows ν_μ is distinct from ν_e

1973

— Gargamelle at CERN discovers weak neutral currents in neutrino scattering

1979

— CDHS and CHARM neutrino DIS confirm V−A, count the sea, and measure sin²θ_W

Going further

Texts. Thomson, Modern Particle Physics, §12.2 Neutrino scattering, §12.2.1 Neutrino–quark scattering cross section, §12.2.2 Antineutrino–quark scattering, §12.2.3 Neutrino deep inelastic scattering, Eqs. (12.6)–(12.19) (the source this article grounds) · Halzen & Martin, Quarks and Leptons, §12.5–12.7 (neutrino–nucleon scattering) · Griffiths, Introduction to Elementary Particles, 2nd ed., §9.2–9.4 · Peskin & Schroeder, An Introduction to QFT, §17.5 (deep inelastic scattering) · Schwartz, QFT and the Standard Model, Ch. 32. The key numerical checks for this article are: (1) $S_z=0$ isotropic $\Rightarrow\sigma_{\nu q}=G_F^2\hat{s}/\pi$ and $S_z=1\Rightarrow(1-y)^2$ , $\sigma_{\bar\nu q}=G_F^2\hat{s}/3\pi$ ; (2) $\langle y\rangle=1/2$ (flat) and $1/4$ ( $(1-y)^2$ ); (3) prefactor $2G_F^2 m_N/\pi=3.16\times10^{-38}$ cm $^2$ /GeV, slopes $0.677$ and $0.334\times10^{-38}$ cm $^2$ /GeV, ratio $0.49$ ; (4) mean free path in iron at 100 GeV $\approx3\times10^6$ km; (5) $Q^2_{\max}=2m_N E_\nu=750$ GeV $^2$ at 400 GeV $\ll M_W^2$ .

Prerequisites. The Weak Charged Current · The V−A Interaction · Deep Inelastic Scattering · The Fermi Constant · Spin and Helicity States. Leads to. Weak Interactions of Quarks · Parton Distribution Functions · The Weak Neutral Current · Long-Baseline Neutrino Experiments.

Ask the tutor: Ask me to derive the $(1-y)^2$ antineutrino shape from angular momentum, or to reproduce the measured $0.677$ and $0.334$ cross-section slopes from the quark momentum fractions for any target you name.

Why does a neutrino interact so rarely if the weak coupling is not actually small?

The feebleness is the $W$ mass, not the charge. The intrinsic weak coupling $\alpha_w=g_w^2/4\pi\approx1/30$ is larger than the electromagnetic $1/137$ , but every low-energy amplitude carries the $W$ propagator, which at $Q^2\ll M_W^2$ is frozen at $1/M_W^2\sim1/(80\,\mathrm{GeV})^2$ . Squaring for a rate brings $1/M_W^4$ , an enormous suppression, so a nucleon presents a cross section of only $\sim10^{-38}$ cm $^2$ per GeV of beam energy. That is why a 100 GeV neutrino travels millions of kilometres of iron before interacting. Raise the energy toward the TeV scale, though, and $Q^2$ approaches $M_W^2$ : the propagator stops being constant, the linear rise flattens, and the neutrino at last becomes a "normal" strength probe — the regime IceCube now explores.

What exactly is the inelasticity y, and why is it so useful?

The inelasticity $y=(E_\nu-E_\mu)/E_\nu$ is the fraction of the neutrino's energy handed to the struck hadron system in the target frame; equivalently $y=\tfrac12(1-\cos\theta^*)$ in the parton CM frame. Its usefulness is that the two things a neutrino can scatter from — quarks and antiquarks — imprint different $y$ shapes. Scattering off a quark gives a flat $y$ distribution for neutrinos (mean $y=1/2$ ); scattering off an antiquark gives a $(1-y)^2$ falloff (mean $y=1/4$ ). So the shape of the measured $y$ spectrum, quite apart from the total rate, separates the valence quarks from the sea antiquarks. It was by histogramming $y$ that neutrino experiments first counted the antiquark content of the proton model-independently.

Where does the factor of three between neutrino and antineutrino cross sections come from?

Purely from angular momentum. A neutrino and a quark, both left-handed, combine to total spin projection $S_z=0$ along the beam — a spherically symmetric state that scatters isotropically, so its $y$ distribution is flat and integrates to $G_F^2\hat{s}/\pi$ . An antineutrino (right-handed) and a quark (left-handed) combine to $S_z=1$ , a state that cannot scatter backward without flipping a unit of angular momentum the contact interaction cannot supply; its distribution is $(1-y)^2$ , which integrates to one-third of the flat result, $G_F^2\hat{s}/3\pi$ . The factor of three is therefore not dynamical at all — it is the ratio $\int_0^1 1\,\mathrm{d}y : \int_0^1(1-y)^2\mathrm{d}y = 1 : \tfrac13$ , a statement about spin geometry that experiment then confirms.

If quark-only scattering predicts a ratio of 1/3, why is the measured ratio about 0.5?

Because the nucleon contains antiquarks. A neutrino scatters off quarks with the flat distribution and off antiquarks with the $(1-y)^2$ distribution; for an antineutrino the assignment reverses. If the target were pure valence quarks (no antiquarks) the neutrino cross section would be three times the antineutrino one, ratio $1/3$ . But real nucleons carry a sea of quark–antiquark pairs, and the antiquarks contribute to the antineutrino cross section with the large flat weight and to the neutrino cross section with the small $(1-y)^2$ weight — lifting the ratio above $1/3$ . The measured $\approx0.5$ implies antiquarks carry roughly a fifth of the struck momentum, a direct sighting of the sea.

How is a nearly pure neutrino beam actually made?

Fire an intense proton beam into a light target to produce a spray of hadrons, mostly charged pions. Magnetic "horns" — pulsed toroidal lenses — focus pions of one sign and defocus the other, selecting $\pi^+$ (for a $\nu_\mu$ beam) or $\pi^-$ (for $\bar\nu_\mu$ ). The selected pions fly down a long evacuated tunnel and decay, $\pi^+\to\mu^+\nu_\mu$ , with the neutrinos boosted into a forward cone because $m_\pi\ll E_\pi$ . At the end of the tunnel a thick absorber — Schwartz used steel from a scrapped warship — stops the muons and any surviving hadrons, so that only neutrinos emerge. The result is a collimated, nearly pure $\nu_\mu$ (or $\bar\nu_\mu$ ) beam whose flavour and sign are chosen by the horn polarity.

Why can neutrino scattering measure things electron scattering cannot?

Because the weak charged current is chiral and flavour-selective in ways the photon is not. The photon couples to charge and sees all quark flavours weighted by charge squared, symmetrically for quarks and antiquarks. The $W$ , by contrast, couples only to left-handed particles, changes quark flavour ( $d\to u$ ), and treats quarks and antiquarks with the different flat versus $(1-y)^2$ weights. That asymmetry is exactly what lets a neutrino beam separate quark from antiquark distributions and, by comparing neutrino and antineutrino rates, extract combinations of structure functions — including the parity-violating one, $xF_3$ , which measures the valence quark distribution directly and is simply invisible to electromagnetic probes.