Lagrangian field theory and Noether's theorem

Units are SI, with metric diag(1, −1, −1, −1), x⁰ = ct and ∂_μ = (∂/∂(ct), ∇). Take beads of mass M a distance a apart, displacements qⱼ(t), neighbours joined by springs k and each bead tied to its rest position by an on-site spring κ: L = Σⱼ[½Mq̇ⱼ² − ½k(qⱼ₊₁ − qⱼ)² − ½κqⱼ²]. Pull out one factor of a: L = Σⱼ a[½(M/a)q̇ⱼ² − ½(ka)((qⱼ₊₁ − qⱼ)/a)² − ½(κ/a)qⱼ²]. Let a → 0 holding ρ = M/a, Y = ka and σ = κ/a fixed. The sum becomes ∫dx, the difference quotient becomes ∂ₓq, and L = ∫dx[½ρ(∂ₜq)² − ½Y(∂ₓq)² − ½σq²], derived. The label j has become the coordinate x, and q(x, t) is a degree of freedom at every point: a field. Rescale φ = √Y q and the density is ½[(1/v²)(∂ₜφ)² − (∂ₓφ)² − (ω₀/v)²φ²], with v = √(Y/ρ) and ω₀ = √(σ/ρ) = √(κ/M): the one-dimensional Klein-Gordon density, v in place of c and ω₀/v in place of mc/ħ.

Newton's law for bead j, Mq̈ⱼ = k(qⱼ₊₁ − 2qⱼ + qⱼ₋₁) − κqⱼ, is solved by qⱼ = cos(Kja − ωt) when ω² = ω₀² + (4v²/a²)sin²(Ka/2), derived. For Ka ≪ 1 the sine is Ka/2 and ω² = ω₀² + v²K²; multiply by ħ², read ħω as E, ħK as p, v as c and ħω₀ as mc², and this is E² = p²c² + m²c⁴. Without on-site springs ω = vK, the massless wave equation; with them a gap opens and no mode oscillates below ω₀, as no Klein-Gordon quantum has energy below mc². The lattice adds what the field lacks: no modes beyond K = π/a, where the group velocity v² sin(Ka)/(aω) falls to zero, and frequencies lower than the field's by a fraction of about v²K²(Ka)²/(24ω²). The chain of the definition has v = 2.24 m s⁻¹ and ω₀ = √(g/ℓ) = 4.95 s⁻¹, with g measured; a 1.0 m wave has Ka = 0.31, and the field gives 14.90 s⁻¹ against the chain's 14.84 s⁻¹, 0.37% high.

The action of a field is S = ∫dt∫d³x ℒ(φ, ∂_μφ), in J s when ℒ is in J m⁻³. Vary φ by δφ, zero on the boundary: δS = ∫dt∫d³x[(∂ℒ/∂φ)δφ + (∂ℒ/∂(∂_μφ))∂_μδφ]. Integrate the second term by parts; the total divergence becomes a surface term and vanishes, leaving the integral of [∂ℒ/∂φ − ∂_μ(∂ℒ/∂(∂_μφ))]δφ. It is zero for every δφ only if ∂_μ[∂ℒ/∂(∂_μφ)] − ∂ℒ/∂φ = 0, derived. Apply it to ℒ = ½[∂_μφ∂^μφ − (mc/ħ)²φ²], where ∂_μφ∂^μφ = (1/c²)(∂ₜφ)² − (∇φ)². Then ∂ℒ/∂(∂_μφ) = ∂^μφ, the raised index carrying the minus sign of the space terms, and ∂ℒ/∂φ = −(mc/ħ)²φ, so ∂_μ∂^μφ + (mc/ħ)²φ = 0: Unit 3's Klein-Gordon equation, derived. The density itself is chosen, not derived: a Lorentz scalar quadratic in φ with first derivatives only can be no more than C₁∂_μφ∂^μφ + C₂φ², an energy bounded below fixes the signs, C₁ > 0 and C₂ ≤ 0, and the size of C₁ is a normalisation. Dimensions: ℒ in J m⁻³ makes φ J¹⁄² m^(−1/2), and every term is φ per m². Static limit: away from a source e^(−r/λ)/r solves it, λ = ħ/mc; the measured π± mass, 139.57 MeV/c², gives 1.41 fm, the range of the nuclear force's tail (quoted).

Suppose φ → φ + εΔφ, with ε a constant, leaves ℒ unchanged. Then 0 = δℒ = ε[(∂ℒ/∂φ)Δφ + (∂ℒ/∂(∂_μφ))∂_μΔφ]. On a solution the field equation replaces ∂ℒ/∂φ by ∂_μ(∂ℒ/∂(∂_μφ)), and the two terms merge into one derivative: 0 = ε∂_μ[(∂ℒ/∂(∂_μφ))Δφ]. So j^μ = (∂ℒ/∂(∂_μφ))Δφ obeys ∂_μj^μ = 0, derived; if ℒ changes by a total divergence ε∂_μK^μ instead, subtract K^μ. Write j^μ = (cρ_N, j). Then ∂ₜρ_N + ∇⋅j = 0, and integrating over all space with fields that vanish far away gives dQ/dt = 0 for Q = ∫ρ_N d³x. Three limits are built in. The symmetry must be continuous, so φ → −φ or parity gives no current. The theorem fixes j^μ only up to a constant factor, since any multiple of a conserved current is conserved. And conservation holds only on solutions: off them, ∂_μj^μ equals the field equation times Δφ, a quick test of any candidate current.

A complex field φ = (φ₁ + iφ₂)/√2 holds two real fields, and ℒ = ∂_μφ*∂^μφ − (mc/ħ)²φ*φ, varied with φ and φ* independent, gives the Klein-Gordon equation for each. ℒ is unchanged by φ → e^(−iα)φ, so Δφ = −iφ and Δφ* = iφ*. Noether's current divided by ħ is J^μ = (i/ħ)(φ*∂^μφ − φ∂^μφ*), with J⁰ = cn, n = (i/ħc²)(φ*∂ₜφ − φ∂ₜφ*) and flux j = (1/iħ)(φ*∇φ − φ∇φ*), derived. Put φ = ħψ/√(2m) and this is Unit 3's ρ = (iħ/2mc²)(ψ*∂ₜψ − ψ∂ₜψ*) and its current, now explained by a symmetry. For φ = Ae^{(−i(ωt − K⋅x))}, n = 2ω|A|²/(ħc²) and j/n = c²K/ω, the velocity pc²/E of a particle with p = ħK and E = ħω. The wave Ae^{(+i(ωt − K⋅x))} gives n = −2ω|A|²/(ħc²). Unit 3 stated Pauli and Weisskopf's reading of that sign; here the symmetry accounts for it: the conserved quantity is the Noether charge of a phase rotation, with charge density qn, and nothing requires it to be positive; reading it as particles minus antiparticles waits for topic 9.03. A real field with m > 0 has only the discrete internal symmetry φ → −φ, so it carries no such charge.

ℒ has no explicit x-dependence, so shifting the argument of φ by a constant ε^ν, δφ = ε^ν∂_νφ, changes ℒ by the total divergence ε^ν∂_νℒ. Noether's theorem with that K^μ gives four conserved currents, T^μ_ν = (∂ℒ/∂(∂_μφ))∂_νφ − δ^μ_νℒ, derived. For the real Klein-Gordon field T^μν = ∂^μφ∂^νφ − g^μνℒ, so T⁰⁰ = ½[(1/c²)(∂ₜφ)² + (∇φ)² + (mc/ħ)²φ²]. That equals π∂ₜφ − ℒ with π = ∂ℒ/∂(∂ₜφ) = (1/c²)∂ₜφ: the Hamiltonian density ℋ, derived, a sum of squares. Momentum density is T⁰ⁱ/c, so g = −(1/c²)∂ₜφ∇φ and the energy flux is S = c²g. Check directly: ∂ₜT⁰⁰ + ∇⋅S = ∂ₜφ[(1/c²)∂ₜ²φ − ∇²φ + (mc/ħ)²φ], zero on solutions. For φ = A cos(K⋅x − ωt) the averages give ⟨S⟩/⟨T⁰⁰⟩ = c²K/ω = pc²/E, the speed of the particle, below c whenever m > 0. For either complex wave of the last section, energy and momentum per unit of |n| are ħω and ħK, so the Noether quantities obey (ħω)² = (ħKc)² + (mc²)².