University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.9
Electron-muon scattering: Mott limit and magnetic term
Advanced QM derived Mott's formula for an electron in the frozen Coulomb field of a nucleus. Here that field becomes a particle. Evaluate topic 9.08's electron-muon amplitude in the muon's rest frame and Mott returns as a limit; let the muon recoil and its spin shows, the template every electron-proton experiment is read against.
Build the model
Connect the measurement to the mechanism.
Mott's formula in UAQM-05.09 needed an external field: a nucleus pinned at the origin with its A⁰ prescribed. Quantum field theory has no prescribed fields, only particles, so a static Coulomb source has to emerge as the limit of something dynamical. Electron-muon scattering is the cleanest test, one photon exchanged between two point Dirac particles, and topic 9.08 already has the amplitude.
Evaluate it in the muon's rest frame with mₑ kept and let mμ grow past the momentum transfer: the muon's trace collapses to a spin-blind charge density 2mμ, its mass cancels against the flux, and Rutherford times 1 − β² sin²(θ/2) appears, the electron's own spin at work. Now let the muon recoil. Two things change: the electron leaves with less energy, E′ instead of E, and a new term (Q²/(2mμ²)) sin²(θ/2) joins cos²(θ/2), a term a spin-0 target does not have.
It is the muon's Dirac magnetic moment, and it is the only way a helicity-conserving electron can bounce straight back. The model is lowest order in α and assumes a point target with g = 2 exactly. That holds for the muon to a few parts in a thousand and fails badly for the proton, whose charge and moment are smeared into two form factors; and it omits the radiative corrections, of order (α/π) ln(Q²/mₑ²), that Unit 12 supplies.
- Simple definition
- Electron-muon scattering at lowest order is single-photon exchange between two point Dirac particles; in the muon's rest frame it becomes Mott scattering when the muon is heavy, and gains a recoil factor and a magnetic term when it is not.
- Example
- A 2.0 GeV electron scattered straight back off a muon at rest leaves with E′ = 51.5 MeV. Mott's cos²(θ/2) is zero there, yet dσ/dΩ = 0.615 nb sr⁻¹, all of it from the muon's magnetic term.
The last two terms are new to topic 9.08's bracket: the electron-muon mirror of −mμ²(p₁⋅p₃) and the product of the two mass terms. They fix the low-velocity limit.
e⁻(p₁) μ⁻(p₂) → e⁻(p₃) μ⁻(p₄), t = (p₁ − p₃)²; momenta and masses in GeV, ⟨|M|²⟩ a pure number. Derived.
Rutherford with relativistic p and v, times the electron's spin factor: UAQM-05.09's Mott formula with Z = 1. Multiply by 0.3894 mb GeV² to compare with data.
Muon rest frame, Q ≪ mμ; p and β the electron's momentum (GeV) and speed; dσ/dΩ in GeV⁻² sr⁻¹. Derived.
One E′/E is the outgoing electron's smaller momentum in its phase space; the other comes from dividing by mμE/(E′E₄), the Jacobian of the energy δ function, whose E₄ cancels the muon's phase-space 1/E₄. As mμ → ∞ both tend to 1, the fixed-centre phase space of the Mott limit.
Massless electron E → E′ at angle θ off a muon at rest; E′/E = 1/(1 + (2E/mμ) sin²(θ/2)); energies in GeV. Derived.
Rutherford at β = 1, a recoil factor, and a bracket whose second term is the muon's magnetic moment. mμ → ∞ returns the Mott limit at β = 1.
Q² = −t = 4EE′ sin²(θ/2) = 2mμ(E − E′) in GeV²; E and E′ in GeV; dσ/dΩ in GeV⁻² sr⁻¹. Derived.
The tan² term is the Dirac moment, g = 2. It grows as Q²/mμ² and takes over at backward angles, where cos²(θ/2) → 0.
Same mass, charge and kinematics; the spin-0 target couples through (p₂ + p₄)μ, quoted. Ratio derived, a pure number.
GE = GM = 1 returns the point Dirac result. The proton has GE(0) = 1 and GM(0) = μₚ = 2.793, both falling with Q².
τ = Q²/(4M²), M the target mass in GeV; GE, GM pure-number Sachs form factors; σMott at β = 1. Quoted.
Redo topic 9.08's trace with mₑ kept
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Crossing in topic 9.08 left M = (e²/t)[ū(p₃)γμu(p₁)][ū(p₄)γμu(p₂)], t = (p₁ − p₃)², squared with mₑ = 0. At a few MeV that is not allowed, because the electron's speed is exactly what Mott's factor measures. Keep the mass in Casimir's trick: the electron tensor becomes 4[p₃μp₁ν + p₃νp₁μ − gμν(p₁⋅p₃ − mₑ²)], the mirror of the muon's, and contracting the two gives ⟨|M|²⟩ = (8e⁴/t²)[(p₁⋅p₂)(p₃⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) − mμ²(p₁⋅p₃) − mₑ²(p₂⋅p₄) + 2mₑ²mμ²], derived. Two terms are new. −mₑ²(p₂⋅p₄) is the image of −mμ²(p₁⋅p₃) under electron ↔ muon, a symmetry the bracket must have because the photon couples to both lines alike. 2mₑ²mμ² comes only from the two gμν mass terms meeting. Dimensions: the bracket is mass⁴, so is t², and ⟨|M|²⟩ is a pure number, as a 2 → 2 amplitude must be.
Freeze the muon and a Coulomb field appears
Go to the muon's rest frame, p₂ = (mμ, 0), and let mμ ≫ Q. The muon then takes momentum but almost no energy, Q²/(2mμ), so p₄ ≈ p₂ and E′ ≈ E: for p = 2.00 MeV at 60° the loss is 0.019 MeV, under 1% of E. Each product of an electron momentum with p₂ or p₄ becomes Emμ, p₂⋅p₄ = mμ², and p₁⋅p₃ = mₑ² + 2p² sin²(θ/2), so the bracket is 2mμ²(E² − p² sin²(θ/2)). The new terms earn their place here: −mₑ²mμ² + 2mₑ²mμ² cancels the mₑ²mμ² inside mμ²(p₁⋅p₃), and without them the bracket at low speed is half its true value. With t = −4p² sin²(θ/2) and fixed-centre phase space, dσ/dΩ = ⟨|M|²⟩/(64π²mμ²), the muon's mass cancels between trace and flux, leaving dσ/dΩ = (α²E²/(4p⁴ sin⁴(θ/2)))(1 − β² sin²(θ/2)), derived. Since E²/p⁴ = 1/(p²β²), this is UAQM-05.09's Mott formula with Z = 1, with no external field assumed: only a heavy second particle.
Mott's factor belongs to the electron
Read the limit both ways. As β → 0, p²β² → mₑ²v⁴ and the result becomes α²/(4mₑ²v⁴ sin⁴(θ/2)), Rutherford's classical law, its 1/sin⁴ now traced to the photon propagator's 1/t². As β → 1 the factor becomes cos²(θ/2) and kills scattering at 180°: the vector coupling preserves a fast electron's helicity (UAQM-05.08), so an electron turned straight back must change its Jz along the beam by one unit, and a spin-blind Coulomb field cannot absorb it. Nothing in the factor refers to the muon. In this limit ūₛ′(p₂)γμuₛ(p₂) = 2mμ δₛ′s for μ = 0 and zero for the spatial components: no spin flip, no spin dependence, a pure charge density. For p = 2.00 MeV and θ = 60°: β² = 0.939, Rutherford 22.1 mb sr⁻¹, Mott factor 0.765, Mott 16.9 mb sr⁻¹. The same factor is the measured reference for electron scattering off spin-0 nuclei such as ⁴He and ¹²C, whose data are read as Mott times a form factor squared.
Let the muon recoil
Now take E ≫ mₑ, set mₑ = 0 and keep mμ finite. Energy and momentum conservation for a massless electron on a muon at rest give E′ = E/(1 + (2E/mμ) sin²(θ/2)) and Q² = −t = 4EE′ sin²(θ/2) = 2mμ(E − E′): the electron's energy loss is the muon's kinetic energy. Two-body phase space in the lab, with flux 4Emμ and the Jacobian mμE/(E′E₄) of the energy δ function, gives dσ/dΩ = (E′/E)²⟨|M|²⟩/(64π²mμ²), derived from Unit 5's phase space. In the bracket, p₄ = p₁ + p₂ − p₃ gives p₁⋅p₄ = Emμ − Q²/2 and p₃⋅p₄ = E′mμ + Q²/2; with E − E′ = Q²/(2mμ) it collapses to 2EE′mμ²[cos²(θ/2) + (Q²/(2mμ²)) sin²(θ/2)]. Assembled: dσ/dΩ = (α²/(4E² sin⁴(θ/2)))(E′/E)[cos²(θ/2) + (Q²/(2mμ²)) sin²(θ/2)], derived. Limits: mμ → ∞ sends E′/E → 1 and the second term to zero, the Mott result at β = 1; θ → 0 returns Rutherford for any mass. At E = 400 MeV and 90°, E′ = 83.6 MeV, Q² = 0.0669 GeV² and the bracket is 0.50 + 1.50.
The new term is the muon's magnetic moment
Replace the muon by a spin-0 particle of the same charge and mass, coupling through (p₂ + p₄)μ, scalar QED's vertex, quoted. With P = p₂ + p₄ one has p₁⋅P = p₃⋅P = mμ(E + E′) and P² = 4mμ² + Q², and the contraction leaves only 2EE′mμ² cos²(θ/2), derived. So a spinless target gives Mott times E′/E, and the Dirac muon multiplies that by 1 + (Q²/(2mμ²)) tan²(θ/2). The Gordon identity names the extra piece: ū′γμu = ū′[(p₂ + p₄)μ + iσμν qν]u/(2mμ), q = p₄ − p₂, and the σμν part couples a magnetic moment g(e/(2mμ))S with g = 2. It is down by Q/mμ in amplitude, so the static limit never saw it, and it flips the muon's spin, which is why it survives at 180° where cos²(θ/2) vanishes. At 400 MeV and 90° it is three quarters of the cross-section. Against measurement: gμ = 2.002 331 8, so the tree-level term is short by (gμ/2)² − 1 = 0.23%; that anomaly arrives at one loop, in Unit 12.
A proton is not a point: GE, GM and Rosenbluth
The muon result has no free parameter: charge e and moment e/(2mμ) are fixed by the Dirac equation, and there is no size. Put a proton in its place, M = 938.27 MeV, and experiment disagrees twice: its moment is μₚ = 2.793 nuclear magnetons, not 1, and its charge is spread over about 0.84 fm, both measured. Lorentz invariance and current conservation leave the vertex two functions of Q², and the cross-section takes Rosenbluth's form σMott(E′/E)[(GE² + τGM²)/(1 + τ) + 2τGM² tan²(θ/2)], τ = Q²/(4M²), quoted, with Sachs form factors GE(0) = 1 and GM(0) = μₚ. Check it against this lesson: GE = GM = 1 makes the first term (1 + τ)/(1 + τ) = 1 and the second (Q²/(2M²)) tan²(θ/2), the point Dirac result. At fixed Q² the bracket is linear in tan²(θ/2), so two angles separate GE from GM. Measured GE and GM/μₚ follow the dipole (1 + Q²/0.71 GeV²)⁻² to about ten per cent below 1 GeV², quoted; at 800 MeV and 50° that puts the proton at 0.38 of a point Dirac proton.
Change one variable at a time
Make the relationship visible.
Start at M = 1 mμ and E = 400 MeV: past 90° the solid curve stays near 0.4 while Mott's falls to zero, the muon's spin flipping to send the electron back. Then set M = 10 mμ and E = 20 MeV, and solid and faint close onto the dashed Mott curve.
RECOIL E′/E0.209
Q = √(−t)259 MeV
MAGNETIC SHARE75 %
dσ/dΩ AT θ54.1 nb sr⁻¹
Live interpretationRECOIL E′/E: 0.209. Q = √(−t): 259 MeV. MAGNETIC SHARE: 75 %. dσ/dΩ AT θ: 54.1 nb sr⁻¹
Catch the common trap
Explain before calculating.
A 1.00 GeV electron scatters through 90° off a muon at rest, mₑ neglected. At lowest order, by what factor does dσ/dΩ exceed that for a hypothetical spin-0 particle with the muon's mass and charge, at the same E and θ?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron with momentum p = 2.00 MeV scatters through θ = 60° off a muon at rest. Treat the muon as static: find the Mott factor and dσ/dΩ in mb sr⁻¹, compare with Rutherford's cross-section with relativistic p and v, and check that ignoring the muon's recoil is fair.
- The electron is relativistic but not massless: E = √(p² + mₑ²) = √(4.000 + 0.261) MeV = 2.064 MeV, so β = p/E = 0.9689 and β² = 0.9387.
- Rutherford with relativistic p and v: α²/(4p²β² sin⁴30°) = (1/137.036)²/(4 × 4.000 MeV² × 0.9387 × 0.0625) = 5.673 × 10⁻⁵ MeV⁻². Restore ħc: times (197.327 MeV fm)² this is 2.209 fm², and 1 fm² = 10 mb, so 22.09 mb sr⁻¹.
- Mott factor: 1 − β² sin²30° = 1 − 0.9387 × 0.25 = 0.7653. So dσ/dΩ = 0.7653 × 22.09 mb sr⁻¹ = 16.9 mb sr⁻¹, about a quarter below Rutherford, because the electron's spin already resists large-angle scattering.
- Recoil: Q² ≈ 4p² sin²30° = 4.00 MeV², so the muon takes Q²/(2mμ) = 4.00/211.3 MeV = 0.019 MeV, 0.9% of E. The magnetic term is smaller still: (Q²/(2mμ²)) tan²30° = 6 × 10⁻⁵. Keeping the recoil exactly lowers the answer by 0.9%.
- Checks: α² is a pure number and p² carries MeV², so the result is MeV⁻², an area once (ħc)² is restored. As β → 0, p²β² → mₑ²v⁴ and the factor tends to 1: the classical Rutherford law.
AnswerMott factor 0.765; dσ/dΩ ≈ 16.9 mb sr⁻¹ against Rutherford's 22.1 mb sr⁻¹. The muon's recoil, 0.019 MeV, changes this by about 1%.
MediumA 400 MeV electron scatters through 90° off a muon at rest; neglect mₑ. Find E′ and Q², split the bracket into its charge and magnetic terms, and give dσ/dΩ in nb sr⁻¹. Compare it with the static Mott value and with a spin-0 target of the muon's mass.
- Recoil: 2E/mμ = 800/105.66 = 7.572, so E′ = 400/(1 + 7.572 × 0.5) MeV = 400/4.786 MeV = 83.6 MeV. Q² = 4EE′ sin²45° = 2 × 400 × 83.58 MeV² = 6.686 × 10⁴ MeV² = 0.0669 GeV². Check: 2mμ(E − E′) = 211.3 × 316.4 MeV² gives the same.
- Bracket: the charge term is cos²45° = 0.500. The magnetic term is (Q²/(2mμ²)) sin²45° = (6.686 × 10⁴/2.233 × 10⁴) × 0.5 = 2.995 × 0.5 = 1.497. Total 1.997, three quarters of it magnetic.
- Prefactor α²/(4E² sin⁴45°) = 5.325 × 10⁻⁵/(4 × 0.160 GeV² × 0.25) = 3.328 × 10⁻⁴ GeV⁻². Times E′/E = 0.2090 and the bracket 1.997 gives 1.389 × 10⁻⁴ GeV⁻²; with 1 GeV⁻² = 0.3894 mb that is 5.41 × 10⁻⁵ mb, or 54.1 nb sr⁻¹.
- Static Mott at the same E and θ: 3.328 × 10⁻⁴ GeV⁻² × 0.500 = 1.664 × 10⁻⁴ GeV⁻² = 64.8 nb sr⁻¹. A spin-0 target recoils identically but lacks the magnetic term: 64.8 × 0.2090 = 13.5 nb sr⁻¹. Dirac over spin-0: 1 + 2.995 tan²45° = 3.99.
- Checks: keeping mₑ exactly changes nothing to four figures, since mₑ²/Q² = 4 × 10⁻⁶. And the answer sits only 17% below Mott because recoil (×0.21) and the magnetic term (×4.0) nearly cancel: Mott's formula is close here by accident.
AnswerE′ = 83.6 MeV, Q² = 0.0669 GeV², dσ/dΩ = 54.1 nb sr⁻¹: 0.83 of static Mott (64.8 nb sr⁻¹) and 4.0 times a spin-0 target (13.5 nb sr⁻¹); 75% of the bracket is magnetic.
HardAn 800 MeV electron scatters through 50° off a proton at rest, M = 938.27 MeV. Predict dσ/dΩ for a point Dirac proton from this lesson's formula, then from Rosenbluth's form (quoted) with the dipole fit GE = GD, GM = μₚGD, GD = (1 + Q²/0.71 GeV²)⁻², μₚ = 2.793. Which of the proton's two departures from a point Dirac particle wins?
- Kinematics: sin²25° = 0.1786 and 2E/M = 1.7053, so E′ = 800/(1 + 1.7053 × 0.1786) MeV = 800/1.3046 MeV = 613.2 MeV. Then Q² = 4EE′ sin²25° = 0.3505 GeV², τ = Q²/(4M²) = 0.0995 and tan²25° = 0.2174.
- Static Mott at β = 1: σMott = α² cos²25°/(4E² sin⁴25°) = (5.325 × 10⁻⁵ × 0.8214)/(4 × 0.640 × 0.03190) GeV⁻² = 5.356 × 10⁻⁴ GeV⁻², which is 208.6 nb sr⁻¹.
- Point Dirac proton: bracket 1 + (Q²/(2M²)) tan²25° = 1 + 2τ tan²25° = 1 + 2 × 0.0995 × 0.2174 = 1.0433, so dσ/dΩ = 208.6 × 0.7665 × 1.0433 nb sr⁻¹ = 166.8 nb sr⁻¹. At this forward angle the magnetic term adds only 4%.
- Rosenbluth with the dipole fit: GD = 1.4936⁻² = 0.4482, GE² = 0.2009, GM² = 7.800 × 0.2009 = 1.567. First term (0.2009 + 0.0995 × 1.567)/1.0995 = 0.3246; second 2 × 0.0995 × 1.567 × 0.2174 = 0.0678. Bracket 0.3924, so dσ/dΩ = 208.6 × 0.7665 × 0.3924 = 62.7 nb sr⁻¹.
- Separate the effects. The moment alone (GE = 1, GM = 2.793, no size) gives a bracket of 1.953, 1.87 times the point value; the size alone (GE = GM = GD) multiplies it by GD² = 0.201. Together 1.87 × 0.201 = 0.376, which is 62.7/166.8.
- Size wins: at Q² = 0.35 GeV² the photon resolves ħc/Q = 0.197 GeV fm/0.592 GeV = 0.33 fm, well inside the proton's 0.84 fm, so the form factor costs a factor of five while the anomalous moment gains less than two. Every bracket is a pure number and σMott an area per steradian, as it must be.
AnswerPoint Dirac proton 167 nb sr⁻¹; Rosenbluth with dipole form factors 62.7 nb sr⁻¹, 0.38 of it. The finite size (×0.20) outweighs the anomalous moment (×1.87).