Thomas precession derived from non-commuting boosts

Units are SI throughout, with Unit 5's metric signature (+, −, −, −). A particle on a curved path has lab velocity v(t) and acceleration a = dv/dt. Call its rest frame at time t, R(t), the frame reached from the lab by the pure boost L(v(t)): its axes are as nearly parallel to the lab's as relativity allows, and the rest-frame spin s lives there. Now ask what a spin with no torque on it does. Between t and t + dt the particle only ever experiences a small boost, by the velocity change it sees in R(t), with nothing to turn it, so s keeps its components on a second family of axes built step by step from pure boosts out of the particle's own frame (Fermi-Walker transport). The families coincide only if L(v + dv)L(v)⁻¹ is itself a pure boost. For dv parallel to v it is, because collinear rapidities add. For any part of dv perpendicular to v, Unit 5's [Kₓ, K_y] = −iJ_z says it cannot be.

At low speed the rapidity is v/c and L(v) = exp(−i v⋅K/c). Compose with the Baker-Campbell-Hausdorff series, e^X e^Y = exp(X + Y + ½[X, Y] + …), taking X = −i(v + a dt)⋅K/c and Y = +i v⋅K/c. The first-order part, X + Y = −i(a dt)⋅K/c, is a boost by the velocity change. The second-order part is ½[X, Y] = (dt/(2c²)) aᵢvⱼ[Kᵢ, Kⱼ] = (dt/(2c²)) aᵢvⱼ(−iεᵢⱼₖJₖ) = −i(dt/(2c²))(a × v)⋅J, a rotation generator. So stepping from R(t) to R(t + dt) is a boost plus a turn through dθ = (dt/(2c²))(a × v), which carries the spin's components to s + dθ × s. Relative to the lab-aligned rest frames a torque-free spin therefore precesses at ω_T ≈ (1/(2c²)) a × v, derived. Dimensions: (m s⁻²)(m s⁻¹)/(m² s⁻²) = s⁻¹. What this drops, [X, [X, Y]] and beyond, is smaller by a further v²/c².

The commutator fixes the low-speed coefficient; the exact one comes from multiplying the finite matrices. L(v) has time-space entries −γvᵢ/c and spatial block δᵢⱼ + ((γ − 1)/v²)vᵢvⱼ, and L(v)⁻¹ = L(−v). In the spatial block of L(v + dv)L(−v), every term built from one velocity alone is symmetric in i and j, so only the cross terms, one factor from each velocity, can rotate anything. Antisymmetrised and kept to first order in dv they give −((γ − 1)/v²)(dvᵢvⱼ − vᵢdvⱼ), which sends s to s + ((γ − 1)/v²)(dv × v) × s. Since γ² − 1 = γ²v²/c², (γ − 1)/v² = γ²/((γ + 1)c²), and so ω_T = (γ²/(γ + 1))(a × v)/c², derived. Its transverse time-space entries, −γ dv⊥/c, give the boost: the transverse velocity γ dv⊥ that R(t) sees. Checks: γ²/(γ + 1) → ½ as v → 0, recovering the commutator; a ∥ v gives zero; at γ = 5/3 the coefficient is 25/24, already about twice its low-speed value.

On a circle a ⊥ v and |a × v| = v²ω, and a × v points opposite to the orbital angular velocity ω. So ω_T = −(γ²β²/(γ + 1))ω = −(γ − 1)ω: the spin turns backwards through 2π(γ − 1) per lap, derived. At β = 0.6, γ = 1.25 and each lap costs 90°; at β = 0.866, γ = 2 and the spin turns once backwards per lap while the velocity turns once forwards. At low speed 2π(γ − 1) → πβ², and that shortcut is already 13% low for a 50 keV electron with γ = 1.098. In hydrogen's Bohr ground orbit v = αc and ħω = α²mc² = 27.2 eV, so the spin falls behind by only πα² = 1.67 × 10⁻⁴ rad per orbit. Yet ħω_T = α⁴mc²/2 = 0.72 meV, the size of the fine structure and half the magnetic precession energy α⁴mc² it competes with, which the next step turns into an energy shift.

Any extra precession ω × S of a spin is generated by an energy ω⋅S, since (i/ħ)[ω⋅S, S] = ω × S. The lab-aligned rest axes turn at −ω_T against the torque-free ones, so the energy written in them gains U_T = S⋅ω_T. For a central force ma = −(1/r)(dV/dr) r, so a × v = −(1/(m²r))(dV/dr) L and, at low speed, U_T = −(1/(2m²c²))(1/r)(dV/dr) S⋅L, derived. The moment μ = g(q/2m)S in the rest-frame field B′ = −v × E/c², quoted from Quantum Mechanics II, adds U_B = −μ⋅B′ = (g/(2m²c²))(1/r)(dV/dr) S⋅L, using qE = −(1/r)(dV/dr) r. The sum has coefficient g − 1, topic 6.07's result reached without the Dirac equation. Hydrogen's 2P splitting is then (g − 1)α⁴mc²/32: the magnetic term alone gives 21.9 GHz, the Thomas term brings it to 10.95 GHz, and the anomalous moment raises g − 1 to 1 + 2aₑ = 1.0023, giving 10.97 GHz against the measured 10.969 GHz.

ω_T knows only a and v, so the −1 is the same for every spinning body, while g belongs to the particle. A gyroscope pushed round a track keeps only the Thomas term, and so does the kinematic part of a nucleon's spin-orbit energy, whatever else the strong force adds. Topic 6.07's Lorentz-scalar potential makes no rest-frame magnetic field and gives exactly the Thomas term, hence its opposite sign. A neutron crossing a nucleus's electric field has a moment but no acceleration, so it keeps the whole magnetic term and no Thomas term. In a storage ring with B ⊥ v and no E, the time-dilated rest-frame Larmor rate (g/2)qB/m and the backward (γ − 1)ω_c leave the spin outrunning the momentum by a qB/m at any γ, derived in the Hard example. The covariant law for all speeds in slowly varying fields is the Bargmann-Michel-Telegdi equation of 1959, named and quoted: an electric field adds a term in a − 1/(γ² − 1), which vanishes at the magic γ = √(1 + 1/a) = 29.3, the 3.094 GeV/c of the muon g − 2 rings.