The transverse field as a set of oscillators

Units are SI throughout; where Unit 9's covariant forms are recalled, the metric is diag(1, −1, −1, −1). Topic 11.01 is quoted, not repeated: with no charges present, Coulomb gauge ∇⋅A = 0 sets the scalar potential to zero, so E_⊥ = −∂A/∂t, B = ∇ × A and ∇²A = (1/c²)∂²A/∂t². Put the field in a cube of side L, volume V = L³, repeating from face to face. Periodic faces are bookkeeping, not mirrors: they keep the translation symmetry of free space, so the plane waves e^(ik⋅r) with k = (2π/L)(n₁, n₂, n₃) form a complete set, orthogonal in the sense ∫_V e^{(ik⋅r − ik′⋅r)} d³r = Vδₖₖ′. Transversality then acts term by term: each vector coefficient must be perpendicular to its own k. Choose two real orthonormal vectors ε₁(k) and ε₂(k) perpendicular to k, and take ε_λ(−k) = ε_λ(k), allowed because k and −k share one line. There is no third. The lattice is fine: for L = 1.00 m the spacing 2π/L is 6.28 m⁻¹, while light of wavelength 1.00 μm has |k| = 6.28 × 10⁶ m⁻¹, a million cells out. The k = 0 term, a uniform field with no restoring force, carries no radiation and is dropped.

Write A = (1/√(ε₀V)) Σₖ Σ_λ Q_λ(k, t) ε_λ(k) e^(ik⋅r). In the wave equation ∇² brings down −k², and orthogonality isolates each term: d²Q_λ(k)/dt² = −c²k² Q_λ(k), derived. Every coefficient oscillates at ωₖ = c|k|, the free-space dispersion relation, now the frequency of a mode. The Q's are not independent, though: A is real only if Q_λ(−k) = Q_λ*(k), so half of them are fixed by the other half. Split each into its two frequencies, Q_λ(k) = α_λ(k) + α_λ*(−k), with α_λ(k, t) = α_λ(k, 0) e^{(−iωₖ t)}. Relabelling −k as k in the second piece gives the expansion on the first formula card, and now every α_λ(k) is free: one complex number per wavevector and polarisation fixes the free field completely, and α_λ(k) e^{(ik⋅r − iωₖ t)} is a plane wave running along +k at speed c. The prefactor 1/√(ε₀V) is chosen so that the energy comes out clean. Since A is in T m, it gives α the unit J¹⁄² s, which the next step checks.

Differentiate the expansion: E_⊥ = (1/√(ε₀V)) Σ iωₖ ε_λ(k)[α_λ(k) e^(ik⋅r) − α_λ*(k) e^(−ik⋅r)], and B has i(k × ε_λ(k)) in place of iωₖ ε_λ(k). Square, integrate over the box, and orthogonality leaves two kinds of product. A mode with its own conjugate gives ωₖ²|α_λ(k)|² from the electric term and c²k²|α_λ(k)|² from the magnetic one, since |k × ε|² = k². A mode with its mirror image, α_λ(k)α_λ(−k), oscillates as e^{(−2iωₖ t)}. It enters with weight −ωₖ² from E², where i² stands alone, and +c²k² from c²B², where i² meets (k × ε)⋅(−k × ε) = −k². Because ωₖ = ck these cancel exactly, leaving (ε₀/2)∫_V (E_⊥² + c²B²) d³r = Σₖ Σ_λ 2ωₖ²|α_λ(k)|², derived. Dimensions: s⁻² times J s² is J. Single-wave limit: one travelling wave has E = cB at every point, so its energy is ε₀∫E² d³r = ε₀E₀²V/2, and with E₀ = 2ω|α|/√(ε₀V) that is 2ω²|α|². Every term is constant in time, as an isolated oscillator's energy must be.

The α's are free but complex, and the raw Q's are worse: written in them the Lagrangian is Σ½[Q̇_λ(k)Q̇_λ(−k) − ωₖ² Q_λ(k)Q_λ(−k)], which couples k to −k. Pair them into real standing waves. Keep one k from each ±k pair and write A = Σₘ qₘ uₘ, where m runs over k, λ and two mode functions, √(2/(ε₀V)) ε_λ(k) cos(k⋅r) and √(2/(ε₀V)) ε_λ(k) sin(k⋅r), normalised so that ε₀∫uₘ⋅uₙ d³r = δₘₙ. Then E_⊥ = −Σ q̇ₘ uₘ and ε₀∫(∇ × uₘ)⋅(∇ × uₙ) d³r = kₘ² δₘₙ. The Maxwell Lagrangian density, quoted from electromagnetism, is ℒ = (ε₀/2)(E_⊥² − c²B²) for this free field, and it integrates to L = Σₘ ½(q̇ₘ² − ωₘ² qₘ²), derived: oscillators of unit mass. Unit 9's recipe gives π = ∂ℒ/∂(∂A/∂t) = −ε₀E_⊥, so pₘ = q̇ₘ, and H = Σₘ ½(pₘ² + ωₘ² qₘ²) is the energy of the previous step. The electric energy is the oscillators' kinetic energy and the magnetic energy their potential energy. A travelling wave is a cos and a sin oscillator a quarter-cycle apart: α_λ(k) = [(q_c + pₛ/ω) + i(p_c/ω − qₛ)]/(2√2).

Each allowed wavevector owns a cell of volume (2π/L)³ = (2π)³/V in k-space. When L is long compared with the wavelengths of interest, the number of wavevectors with |k| ≤ ω/c is the volume of that sphere over the cell, (4π/3)(ω/c)³V/(2π)³, and two polarisations double it: N(ω) = Vω³/(3π²c³). Then dN/dω = Vω²/(π²c³), so the density of modes per unit volume per unit angular frequency is ρ(ω) = ω²/(π²c³), derived, or 8πν²/c³ per unit ordinary frequency. At 1.00 μm that is 1.33 × 10⁴ s m⁻³. V has gone, and it always goes from a free-space quantity: an intensive sum becomes (1/V)Σₖ → ∫d³k/(2π)³, and the box's size and periodic faces leave no trace. Two cautions. The replacement needs a box many wavelengths across, L ≫ 2πc/ω: a 10.0 cm box has 340 oscillators at or below 10.0 GHz where the integral says 311. And a closed cavity with conducting walls has a different lattice of standing waves, whose density tends to the same ω²/(π²c³) at high frequency (Weyl's theorem, quoted). The gap between the exact sum and the integral is what topic 11.06 turns into the Casimir force.

The model already makes a thermal prediction. In the canonical ensemble each oscillator holds ½k_B T in pₘ²/2 and ½k_B T in ωₘ² qₘ²/2, so the radiation energy per unit volume per unit angular frequency is u(ω) = ω²k_B T/(π²c³), the Rayleigh-Jeans law, derived. Its integral over ω diverges: a classical field in equilibrium holds infinite energy. The mode count is not at fault. Keep ρ(ω), replace k_B T by Planck's mean energy ħω/(e^x − 1) with x = ħω/k_B T, quoted from statistical mechanics, and use the integral of x³/(e^x − 1) from 0 to ∞, π⁴/15, quoted from analysis. The energy density becomes π²k_B⁴T⁴/(15ħ³c³), and the Stefan-Boltzmann constant σ = π²k_B⁴/(60ħ³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴, as measured. That number tests the factor of 2: a third polarisation would make it 8.51 × 10⁻⁸ W m⁻² K⁻⁴. Limiting case: for ħω ≪ k_B T Planck's energy tends to k_B T and the classical result returns; for the 2.7255 K cosmic background at 1.42 GHz, x = 0.025 and the two differ by 1.2%. What fails is the oscillator's classical energy, which is why the next topic quantises each pair qₘ, pₘ.