research33 min studyLesson 624 in Physics

Infer proton structure from inclusive lepton scattering

Define DIS invariants, separate measured structure functions from fitted parton densities, and derive the leading electromagnetic flavor weights.

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What you’ll understand

  • Compute Bjorken x, inelasticity and the hadronic invariant mass.
  • Connect photon-exchange structure functions to charge-weighted quark densities.
  • State the mass, higher-twist and perturbative limits of the parton relation.

Deep inelastic scattering measures an outgoing lepton while summing over the hadronic final state. This inclusive choice makes short-distance proton structure accessible without reconstructing every fragment. The exchanged four-momentum determines the resolution, while the hadronic invariant mass distinguishes a deep inelastic event from elastic scattering or a low-lying resonance. Structure functions describe the measured response. Interpreting them as parton densities requires a factorization framework and an accuracy statement, rather than identifying every structure function directly with a constituent count.

Use invariants to identify the probed regime

Let P be the target momentum and q the incoming minus outgoing lepton momentum. For spacelike exchange, Q squared is minus q squared and is positive. Bjorken x is Q squared divided by twice P dot q; inelasticity y compares P dot q with P dot k for the incoming lepton. In the target rest frame, the energy transfer nu is the lepton energy loss and y is nu divided by its initial energy.

The inclusive hadronic system has momentum P plus q. Expanding its square gives a useful consistency check relating its mass to x and Q squared. Elastic scattering has that mass equal to the target mass and x equal to one. A large Q alone does not ensure an inclusive continuum if x is too near one. Deep inelastic reasoning needs both adequate virtuality and adequate final-state mass, with target-mass and resonance effects checked for the selected kinematics.

Q2=−q2>0,x=Q22P⋅q,y=P⋅qP⋅k,ν=P⋅qM,W2=(P+q)2=M2+Q2(1x−1)Q^2=-q^2>0,\quad x=\frac{Q^2}{2P\cdot q},\quad y=\frac{P\cdot q}{P\cdot k},\quad \nu=\frac{P\cdot q}{M},\qquad W^2=(P+q)^2=M^2+Q^2\left(\frac1x-1\right)
P²=M², with M the target mass, and q=k−k′. In natural units Q²,W² have energy-squared units, ν has energy units, and x,y are dimensionless. Lepton masses are neglected in the simple rest-frame interpretation.↔ Scroll wide equations horizontally.

A virtual photon weights electric charges squared

At leading order with massless spin-half partons and electromagnetic exchange, the transverse response counts quarks and antiquarks weighted by their electric charges squared. The density q of x is defined per unit momentum fraction, not as the probability that the whole proton consists of only that quark. Up-type flavors have weight four ninths and down-type flavors one ninth. A photon cannot separately determine every flavor combination from a single proton structure function, because several flavors share the same charge.

In this simplest limit the Callan–Gross relation sets F two equal to twice x F one, corresponding to zero longitudinal structure function after neglecting target-mass effects. QCD radiation produces a nonzero longitudinal response and logarithmic scaling violations. Weak charged- and neutral-current measurements supply other flavor and parity combinations. A neutron comparison can help isolate up and down content, but extracting a neutron from nuclear data introduces nuclear corrections that are not part of the free-proton leading-order identity.

F2γp(x,Q2)=x∑fef2 [qf(x,Q2)+qˉf(x,Q2)],F2=2xF1,FL=F2−2xF1=0at massless LOF_2^{\gamma p}(x,Q^2)=x\sum_f e_f^2\,[q_f(x,Q^2)+\bar q_f(x,Q^2)],\qquad F_2=2xF_1,\qquad F_L=F_2-2xF_1=0\quad\text{at massless LO}
F1,F2,FL and PDFs per unit dimensionless x are dimensionless. ef is the quark charge in units of the positron charge. These leading-order relations neglect target mass, heavy-quark masses, electroweak Z exchange and higher-twist terms.↔ Scroll wide equations horizontally.

Separate a measurable response from a factorization convention

The photon-exchange double differential cross section contains a known leptonic weight multiplying F two and another multiplying F longitudinal. Changing the beam energy at fixed x and Q squared changes y and can separate these responses. At higher order the structure functions are convolutions of coefficient functions with PDFs. The physical response is independent of the factorization prescription when all terms are included consistently, whereas the individual fitted densities and coefficients depend on scheme and scale.

Approximate Bjorken scaling follows when the hard probe resolves effectively pointlike partons, but exact scale independence is not expected in QCD. Evolution accounts for leading-twist logarithms; corrections suppressed by powers of the hard scale describe additional long-distance correlations. Heavy-flavor thresholds and small-x logarithms may require specialized treatment. A toy charge-weighted calculation tests the algebra, not a global PDF extraction. Likewise the projection lab can illustrate weights and components, but contains no hadronic tensor, lepton detector or proton model.

d2σγdx dQ2=2παem2xQ4[Y+F2−y2FL],Y+=1+(1−y)2\frac{d^2\sigma^{\gamma}}{dx\,dQ^2}=\frac{2\pi\alpha_{\mathrm{em}}^2}{xQ^4}\left[Y_+F_2-y^2F_L\right],\qquad Y_+=1+(1-y)^2
This unpolarized photon-exchange formula neglects lepton and target-mass corrections and electroweak Z exchange. In natural units d²σ/(dx dQ²) has energy dimension−4; F2 and FL are dimensionless. At higher order FL need not vanish.↔ Scroll wide equations horizontally.

Check a candidate deep inelastic event

Use an illustrative target mass M=1GeV, incident lepton energy100GeV, measured energy loss25GeV, and Q²=20GeV². Find x, y and W², and compare the hadronic mass with the target mass.

  1. In the target rest frame P dot q=Mν=25GeV². Hence x=20/(2×1×25)=0.40, within the physical inclusive range for this target.

  2. Inelasticity is y=ν/E=25/100=0.25. The outgoing lepton energy is75GeV, consistent with a positive transferred energy.

  3. The hadronic invariant is W²=1+50−20=31GeV², so W≈5.57GeV. It is well above the target mass; Q² and W² both exceed M², unlike elastic x=1 kinematics.

The result

The event has x=0.40, y=0.25 and W²=31GeV², with W≈5.57GeV in the stated illustrative kinematics.

Explore the model

Interactive lab

Interactive lab · vectors

Keep component and total weight distinct

For fresh comparisons Reset and set BOTH angle0°, magnitude1; then Reset and set angle60°, magnitude1. Predict x projections1 and0.5 while the magnitude stays1. This is a geometric weighting analogy, not a computation of PDFs, flavor charges, DIS invariants or a structure function.

Keep component and total weight distinctPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA weighted measured component is not automatically the full underlying distribution.

Check your understanding

At massless electromagnetic leading order, which combination gives proton F2?

Try it yourself

At x=0.20 in a massless leading-order toy proton, take u=3, d=1.5 and all antiquark and other-flavor densities zero. Compute F2, F1 and FL. Assuming exact isospin exchange u↔d for a free neutron, find neutron F2 and its ratio to the proton value.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use weights4/9 and1/9, then the Callan–Gross relation; the supplied numbers are densities at one x.

Compare with the solution

Proton F2=0.20[(4/9)3+(1/9)1.5]=0.30. Therefore F1=0.30/(2×0.20)=0.75 and FL=0 at the stated order. Exchanging up and down gives neutron F2=0.20[(4/9)1.5+(1/9)3]=0.20, so Fn2/Fp2=2/3. These local toy densities are not a fitted global proton model and do not establish their normalization over all x.

Take it with you

  • Both virtuality and hadronic invariant mass delimit the DIS regime.
  • Leading photon exchange weights flavors by electric charge squared.
  • Structure functions are observable responses; PDFs carry factorization conventions.

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