Lorentz generators: rotations J and boosts K

Work in SI units with x^μ = (ct, x) and the metric g = diag(1, −1, −1, −1), kept for the whole unit. A change of inertial frame is a real 4 × 4 matrix, x′ = Λx, and relativity asks only that it keep the interval xᵀgx = c²t² − r² for every x, which is Λᵀ g Λ = g. Products pass, since (Λ₁Λ₂)ᵀgΛ₁Λ₂ = Λ₂ᵀgΛ₂ = g, and so do inverses, Λ⁻¹ = gΛᵀg: the solutions form a group. The determinant gives (det Λ)² = 1, and the 00 entry gives (Λ⁰₀)² = 1 + Σᵢ(Λⁱ₀)² ≥ 1. Neither det Λ = ±1 nor the sign of Λ⁰₀ can change along a continuous path, so the group falls into four pieces. The identity, every rotation and every boost sit in the proper orthochronous piece, det Λ = +1 and Λ⁰₀ ≥ 1. Parity P = diag(1, −1, −1, −1) has det Λ = −1, time reversal T = diag(−1, 1, 1, 1) has Λ⁰₀ = −1, and PT = −1 has det Λ = +1 yet Λ⁰₀ = −1. No chain of small steps reaches them; they must be added by hand, as topic 5.06 does for parity.

Near the identity write Λ = 1 + ω. To first order Λᵀ g Λ = g becomes ωᵀg + gω = 0, so gω is antisymmetric: six independent numbers. In ω itself the spatial block is antisymmetric, three rotations, while the time row and column are equal, three boosts, because lowering a spatial index flips a sign and lowering the time index does not. Take each generator as i times the derivative at the identity. A turn about z gives J_z, with −i in the (x, y) slot and +i in (y, x), Hermitian as in Quantum Mechanics II. An active boost along x gives Kₓ, with i in both the (t, x) and (x, t) slots, so Kₓ† = −Kₓ. Finite transformations are R = e^{(−iθ n̂⋅J)} and B = e^{(−iη n̂⋅K)}. On the (t, x) block (−iηKₓ)² = η² times the unit matrix, so the series for B splits into cosh η on the diagonal and sinh η off it: a boost is a hyperbolic rotation, with tanh η = v/c and cosh η = γ. Check: all six generators are dimensionless, and on quantum states ħJ is the angular momentum.

Multiply the generators as Quantum Mechanics II multiplied rotation matrices. The J's are the spin-1 matrices bordered by a row and column of zeros, so [Jᵢ, Jⱼ] = iεᵢⱼₖJₖ. Mixed products give [Jₓ, K_y] = iK_z, and in general [Jᵢ, Kⱼ] = iεᵢⱼₖKₖ: K is a vector operator, so a boost direction turns like any arrow. The new relation is between boosts. The product KₓK_y runs through the time slot, so its only entry is i × i = −1 at (x, y), and K_yKₓ has −1 at (y, x). Their difference is −iJ_z, so [Kₓ, K_y] = −iJ_z and cyclically [Kᵢ, Kⱼ] = −iεᵢⱼₖJₖ, derived. The minus sign is the metric's: in a Euclidean four-space the same product would give +iJ_z. Passive boosts reverse K and leave all three relations unchanged, so the algebra does not hang on that convention. One corollary prepares topic 5.08: the combinations (J ± iK)/2 are Hermitian, each obeys the angular-momentum algebra, and the two commute, so the Lorentz algebra is two angular-momentum algebras working side by side.

Two boosts along one axis are generated by one matrix, and exponents of a single matrix add: Bₓ(η₁)Bₓ(η₂) = e^{(−i(η₁ + η₂)Kₓ)} = Bₓ(η₁ + η₂). Collinear boosts form a one-parameter subgroup in which rapidity is the additive quantity, as angle is for turns about one axis. Through tanh η = v/c and the identity for tanh of a sum, this is Einstein's v = (v₁ + v₂)/(1 + v₁v₂/c²), derived. For 0.60c and 0.80c the rapidities 0.693 and 1.099 add to 1.792, so v = 0.946c, where Galileo would give 1.40c. Limits: for v ≪ c, η ≈ v/c and the Galilean sum returns, and if v₂ = c then η₂ is infinite and the result is c for every v₁. The measured test is old. Light moving at c/n through water flowing at w along the beam gains rapidity tanh⁻¹(w/c) ≈ w/c, so its lab speed rises by c sech²η × w/c = (1 − 1/n²)w, since tanh η = 1/n. For n = 1.333 the coefficient is 0.437, and Michelson and Morley measured 0.434 ± 0.02 in 1886, where a Galilean sum predicts 1.

Now boost along x by η₁, then along y by η₂, both along fixed lab axes: Λ = B_y(η₂)Bₓ(η₁). Expanding both sides to second order proves e^A e^B = e^{(A + B + ½[A, B] + …)}, and with A = −iη₂K_y and B = −iη₁Kₓ the extra term is −(η₁η₂/2)[K_y, Kₓ] = −i(η₁η₂/2)J_z. So Λ is a boost towards the final velocity times a rotation through θ ≈ η₁η₂/2 ≈ v₁v₂/2c² about +z = x̂ × ŷ, derived. The exact matrix shows it plainly: Λ^yₓ = γ₁β₁γ₂β₂ while Λ^x_y = 0, and every pure boost is a symmetric matrix. Multiplying out the matrices and splitting Λ = L(u)R, with L(u) the pure boost to the final velocity u, derives the exact angle, cos θ = (γ₁ + γ₂)/(1 + γ₁γ₂). Checks: θ is a pure number, it vanishes when either speed does, and at 0.10c and 0.10c the exact 0.2879° and the estimate 0.2865° agree to 0.5%. Inverting Λ and conjugating with parity, which reverses every boost and no rotation, gives the opposite order, Λ′ = L(R⁻¹u)R⁻¹: the same speed turned back by θ, so the leftover rotation shows even in the velocities.

Carry a gyroscope through the two boosts and compare it with one given a single boost to the same velocity: their axes differ by R. Take small boosts round a closed loop and the commutator returns a rotation equal, at low speed, to the area the loop encloses in the velocity plane divided by c², as the Hard example shows for a square. An electron whose velocity swings round a circle of radius v once per orbit therefore collects about πv²/c² per orbit. Unit 6 turns this into the Thomas precession, whose factor ½ in the spin–orbit energy makes hydrogen's 2p splitting the measured 10.97 GHz rather than about 21.9 GHz (quoted here). Three limits stay in view. P and T sit in other pieces and must be imposed separately; on states T must be antiunitary (quoted). Because K is anti-Hermitian this representation is not unitary, and topic 5.03 quotes that no non-trivial finite-dimensional one can be. And the algebra fixes the generators, not every global sign: a 2π turn is 1 on four-vectors but −1 on the spinors of topic 5.04.