University Physics V · Foundations of Quantum Mechanics · 2.2
Cavity Modes & the Planck Law
The first quantum calculation is a linear-algebra exercise: diagonalise a box, count the eigenvalues, then ask how much energy each mode may hold. Its structure — a density of states multiplied by a mean occupancy — is the template every statistical calculation in this course reuses.
Build the model
Connect the measurement to the mechanism.
Radiation in a reflecting box is not a mysterious substance; it is the Dirichlet eigenvalue problem of the Laplacian, and each eigenmode is a harmonic oscillator of frequency ω = c|k|. Two theorems and one assumption then fix the spectrum. Weyl's law counts the modes — 8πVν²/c³ per unit frequency, whatever the cavity's shape — and classical equipartition pays each one kT, so the integral over an ever-denser ultraviolet count must diverge.
The count is a fact about the operator, so the only adjustable assumption is the energy a mode may hold. Planck's restriction to the ladder Eₙ = nhν turns the partition function into a geometric series and the mean energy into hν/(e(hν/kT) − 1): a full kT for modes far below kT/h, exponential defunding above it. The cost is a new constant of nature, h = 6.626×10⁻³⁴ J s, fitted to the measured spectrum, and the loss of the classical energy continuum.
The purchase is a finite integral giving the T⁴ law, a peak at 2.82 kT/h that thermometers stars, and — read forward — the Bose–Einstein occupancy n̄ = 1/(eˣ − 1) that quantum field theory later derives from the ladder operators Planck never knew he was using.
- Simple definition
- Planck's law u(ν, T) = (8πhν³/c³)/(e(hν/kT) − 1) is the energy per volume per frequency of thermal radiation: the cavity's mode count multiplied by the mean energy the Boltzmann-weighted ladder Eₙ = nhν allows each mode.
- Example
- At the Sun's T = 5800 K the spectrum peaks near ν = 340 THz, where x = hν/kT = 2.82: that mode holds n̄ = 1/(e2.82 − 1) ≈ 0.063 photons — a mean energy of 0.18 kT, where equipartition promised the full kT to this and every higher mode.
Turns radiation in a box into a discrete, countable spectrum: one mode per lattice point k = (π/L)(nₓ, ny, nz), doubled for polarisation.
Dirichlet Laplacian for one field component; eigenfunctions sin(nₓπx/L) sin(nyπy/L) sin(nzπz/L), nᵢ = 1, 2, … — the vector problem with n̂ × E = 0 lives on the same k lattice
The ν² that every spectrum shares. Rayleigh–Jeans and Planck multiply this same count by different mean energies per mode.
V in m³, ν in Hz, c = 3.00×10⁸ m s⁻¹; the leading Weyl term — cavity shape enters only via surface corrections
Replaces equipartition's kT. Low modes still get ≈ kT; a mode at hν = 5 kT gets 0.034 kT — priced out by its Boltzmann weight, not deleted from the count.
Eₙ = nhν; h = 6.626×10⁻³⁴ J s, k = 1.381×10⁻²³ J K⁻¹; ⟨E⟩ = −∂ ln Z/∂β with β = 1/kT
eˣ − 1 ≈ x below kT/h returns Rayleigh–Jeans; dropping the −1 above it returns Wien — both old laws survive as limits of one.
spectral energy density in J m⁻³ Hz⁻¹ = count × quantum hν × occupancy n̄ = 1/(eˣ − 1)
The classically divergent integral now converges, and the T⁴ constant is built from h, k and c — measure σ and you constrain h.
uses ∫₀^∞ x³/(eˣ − 1) dx = π⁴/15; emitted flux is σT⁴ with σ = ac/4 = 5.670×10⁻⁸ W m⁻² K⁻⁴
One peak reads a temperature: 500 nm gives the Sun's 5800 K; 160 GHz gives the CMB's 2.725 K.
2.821 solves 3(1 − e⁻ˣ) = x; the λ form solves 5(1 − e⁻ˣ) = x at 4.965, so λₘₐₓ ≠ c/νₘₐₓ
Name the operator: the cavity is an eigenvalue problem
Inside a perfectly conducting box each Cartesian field component obeys the wave equation, and the walls impose E∥ = 0. Separating out time leaves the Helmholtz equation −∇²u = (ω/c)²u on [0, L]³, and modelling one component with Dirichlet conditions — the same operator, basis and boundary-condition language as every Schrödinger problem to come — gives eigenfunctions sin(nₓπx/L) sin(nyπy/L) sin(nzπz/L) with nᵢ = 1, 2, 3, …, so the allowed wavevectors form the lattice k = (π/L)(nₓ, ny, nz) in the positive octant, each carrying two transverse polarisations. The full vector problem with n̂ × E = 0 swaps sines for cosines component by component and admits modes with a single zero index, but it lives on the same lattice and changes only the sub-leading terms of the count. Substitute one mode back into Maxwell's equations and its amplitude obeys q̈ = −ω²q: a harmonic oscillator of frequency ω = c|k|. Black-body radiation is therefore a countable family of independent oscillators indexed by three integers and a polarisation, and this whole topic is statistical mechanics applied to that basis.
Count the modes: Weyl's law gives the ν² density
How many lattice points lie within |k| ≤ k? Each owns a cell of volume (π/L)³ and only the positive octant counts, so N(k) = (1/8)(4πk³/3)/(π/L)³ = Vk³/6π², doubled to Vk³/3π² for polarisation. With k = 2πν/c and one derivative, g(ν) = 8πVν²/c³ modes per unit frequency. Weyl's theorem is what licenses the sphere approximation: the leading term of the Laplacian's eigenvalue count depends only on the volume, with corrections of order the surface area times k² — the box could be a kettle and the ν² would survive. The count is enormous where it matters: a 1 cm³ cavity holds 8π×10⁻⁶ × (6×10¹⁴)³/(3c³) ≈ 7×10¹³ modes below the frequency of green light, and the density keeps growing as ν². Whatever mean energy each mode is assigned multiplies this same relentless count.
Equipartition pays every mode kT — and the ledger diverges
Each mode's energy is quadratic in its amplitude and its conjugate momentum, so the classical canonical ensemble assigns it ⟨E⟩ = 2 × kT/2 = kT, independent of frequency. Multiplying by the count gives the Rayleigh–Jeans law, u(ν, T) = 8πν²kT/c³. It is genuinely correct at low frequency — radio engineers still quote antenna noise as a temperature because the long-wavelength tail of every thermal spectrum obeys it — but the total energy density ∫ 8πν²kT/c³ dν diverges as ν³. Ehrenfest later named this the ultraviolet catastrophe. Note where the failure is allowed to be: the ν² count is a theorem about the Dirichlet Laplacian, and the multiplication is bookkeeping. The only physical assumption in the chain is that a mode may hold any energy from a continuum — so that is what has to give.
Restrict the ladder, then sum the geometric series
Planck's move: let a mode of frequency ν hold only Eₙ = nhν, n = 0, 1, 2, …. With β = 1/kT the partition function collapses to a geometric series, Z = Σ e(−nβhν) = 1/(1 − e(−βhν)), and ⟨E⟩ = −∂ ln Z/∂β = hν/(e(hν/kT) − 1). Write x = hν/kT and read it as ⟨E⟩ = n̄hν with n̄ = 1/(eˣ − 1) — the Bose–Einstein occupancy, derived from nothing but Boltzmann weights on a ladder. The numbers show the crossover: at x = 0.1 the mode gets 0.95 kT, essentially the classical answer; at x = 3 it gets 0.16 kT; at x = 10 it gets 4.5×10⁻⁴ kT. Nothing removed the high-frequency modes — their first rung simply costs more than the thermal budget, so the Boltzmann factor prices them out. Multiplying by the unchanged count gives Planck's law, u(ν, T) = (8πhν³/c³)/(eˣ − 1).
Take the limits, then do the integral
A new law must contain the old ones. For hν ≪ kT, eˣ − 1 ≈ x and ⟨E⟩ → kT: Rayleigh–Jeans returns as the low-frequency limit, so everything it explained stays explained. For hν ≫ kT the −1 is negligible and u → (8πhν³/c³)e(−hν/kT): Wien's empirical exponential falls out. The full integral now converges: substituting x = hν/kT gives u = (8πk⁴T⁴/h³c³)∫x³/(eˣ − 1)dx, the integral is π⁴/15, and u = aT⁴ with a = 7.566×10⁻¹⁶ J m⁻³ K⁻⁴ — Stefan–Boltzmann with its constant now built from h, k and c (radiated flux σT⁴, σ = ac/4 = 5.670×10⁻⁸ W m⁻² K⁻⁴). Maximising ν³/(eˣ − 1) gives 3(1 − e⁻ˣ) = x, so xₚₑₐₖ = 2.821 and νₘₐₓ = 58.8 GHz K⁻¹ × T. The cosmic microwave background at 2.725 K peaks at 160 GHz and follows the whole curve to better than one part in 10⁴ — the best-measured Planck spectrum in nature.
What Planck quantised — and what the modes became
Read the 1900 derivation closely and the quantum sits in the walls, not the field: Planck restricted the energy exchanged by the material resonators in the cavity walls, treated hν as a bookkeeping device he hoped to remove, and left the field itself classical. Einstein's 1905 light quantum moved the restriction into the radiation; Bose's 1924 counting derived the same law from photon statistics; and quantum electrodynamics closes the loop by promoting each cavity mode to a quantum harmonic oscillator, Ĥ = ℏω(â†â + ½), whose number states are exactly the ladder Eₙ = nhν Planck guessed. The zero-point ½ℏω adds a temperature-independent constant per mode, so it cancels from every thermal average here — but it is not fiction: summed over the same Dirichlet modes it returns as the Casimir force between the cavity's own walls.
Change one variable at a time
Make the relationship visible.
Slide T from 4000 to 8000 K: the peak slides right in proportion to T and its height grows as T³, while the Rayleigh–Jeans curve always escapes the frame. Then park the probe at 1200 THz and watch n̄ stay near zero at every temperature — the Boltzmann factor prices those modes out, but they never leave the count.
PEAK ν = 2.82 kT/h353 THz
PROBE x = hν/kT4.80
PROBE n̄ = 1/(eˣ − 1)0.008 photons
ENERGY DENSITY aT⁴981 mJ m⁻³
Live interpretationPEAK ν = 2.82 kT/h: 353 THz. PROBE x = hν/kT: 4.80. PROBE n̄ = 1/(eˣ − 1): 0.008 photons. ENERGY DENSITY aT⁴: 981 mJ m⁻³
Catch the common trap
Explain before calculating.
A single cavity mode has frequency ν with hν = 4 kT. Equipartition assigns it a mean energy kT. What does the Planck distribution assign, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA cubical microwave cavity has side L = 2.0 cm. Use the Weyl count N(ν) = 8πVν³/(3c³) to estimate how many electromagnetic modes, both polarisations included, lie below ν = 30 GHz — and say how far the estimate can be trusted.
- V = (0.020 m)³ = 8.0×10⁻⁶ m³. Below 30 GHz: ν³ = (3.0×10¹⁰ Hz)³ = 2.7×10³¹ Hz³, and c³ = (3.0×10⁸ m s⁻¹)³ = 2.7×10²⁵ m³ s⁻³.
- N = 8π × (8.0×10⁻⁶ × 2.7×10³¹)/(3 × 2.7×10²⁵) = 8π × 8.0×10⁻⁶ × 10⁶/3 = 8π × 8/3 ≈ 67.
- Sanity-check the low end. In the scalar Dirichlet model the lowest mode is (1,1,1) at ν = √3 c/(2L) = 1.732 × 3.0×10⁸/0.040 ≈ 13.0 GHz; the vector field also admits one zero index, so a conducting cube first resonates in the TE₁₀₁ family at c/(√2 L) = 3.0×10⁸/(1.414 × 0.020) ≈ 10.6 GHz. Either way, modes below 30 GHz exist and the count only begins a third to a half of the way up the range.
- Trust: Weyl's Vk³ term is only the leading one. An explicit count is quick here — integer triples (nₓ, ny, nz) with nₓ² + ny² + nz² ≤ (2Lν/c)² = 16, two polarisations for the 17 triples with no zero index and one for the 24 with a single zero — and gives 58, 13% below 67. Sub-leading Weyl terms and the integer jumpiness of the exact count matter at that level; the formula becomes precise where N is large — exactly the black-body regime.
AnswerN ≈ 67 modes below 30 GHz by Weyl's leading term, against 58 by explicit lattice count — good to about 13% here, superb at the 10¹³-mode densities of a hot cavity.
MediumA furnace holds thermal radiation at T = 6000 K. For (a) an infrared mode at ν = 30 THz and (b) an ultraviolet mode at ν = 1500 THz, find x = hν/kT, the mean photon occupancy n̄, and the mean energy as a fraction of the kT that equipartition would assign.
- kT = 1.381×10⁻²³ × 6000 = 8.29×10⁻²⁰ J (0.517 eV). This is the thermal budget every mode competes for.
- (a) hν = 6.626×10⁻³⁴ × 3.0×10¹³ = 1.99×10⁻²⁰ J, so x = 1.99/8.29 = 0.240. Then n̄ = 1/(e0.240 − 1) = 1/0.271 = 3.69 photons, and ⟨E⟩/kT = x n̄ = 0.885.
- (b) hν = 6.626×10⁻³⁴ × 1.5×10¹⁵ = 9.94×10⁻¹⁹ J, so x = 12.0. Then e¹² − 1 = 1.63×10⁵, so n̄ = 6.1×10⁻⁶ photons and ⟨E⟩/kT = 12 × 6.1×10⁻⁶ = 7.4×10⁻⁵.
- Read the pair together: at x = 0.24 the classical kT is only 13% high, which is why Rayleigh–Jeans works in the infrared and radio; at x = 12 it is wrong by four orders of magnitude, and the ν² count would multiply that phantom kT into a divergent total.
Answer(a) x = 0.240, n̄ = 3.7, ⟨E⟩ = 0.89 kT. (b) x = 12.0, n̄ = 6.1×10⁻⁶, ⟨E⟩ = 7.4×10⁻⁵ kT. Equipartition is nearly right below kT/h and catastrophically wrong above it.
HardShow that the frequency-form Planck density u(ν, T) ∝ ν³/(e(hν/kT) − 1) peaks where 3(1 − e⁻ˣ) = x, solve for x by iteration, and predict the peak frequency of the cosmic microwave background at T = 2.725 K. Why is the answer not c/λₘₐₓ from the wavelength form?
- With x = hν/kT, maximise f(x) = x³/(eˣ − 1): f′ = 0 gives 3x²(eˣ − 1) = x³eˣ, i.e. 3(eˣ − 1) = xeˣ, i.e. 3(1 − e⁻ˣ) = x.
- Iterate x ← 3(1 − e⁻ˣ) from x = 3: → 2.851 → 2.827 → 2.822 → 2.8216 → 2.8215 → 2.8214. Converged: xₚₑₐₖ = 2.8214.
- νₘₐₓ = xₚₑₐₖ kT/h = 2.8214 × (1.381×10⁻²³/6.626×10⁻³⁴) × T = 2.8214 × 2.084×10¹⁰ Hz K⁻¹ × T = (5.88×10¹⁰ Hz K⁻¹) T.
- At T = 2.725 K: νₘₐₓ = 5.88×10¹⁰ × 2.725 = 1.602×10¹¹ Hz = 160.2 GHz — the peak COBE's FIRAS instrument measured.
- The wavelength form u(λ, T) carries an extra factor from |dν/dλ| = c/λ², so it peaks where 5(1 − e⁻ˣ) = x, at x = 4.965: λₘₐₓ = 2.898 mm K ÷ 2.725 K = 1.06 mm, while c/νₘₐₓ = 1.87 mm. Both are correct — a peak belongs to a density, and per-hertz and per-metre are different measures.
Answerxₚₑₐₖ = 2.8214, so νₘₐₓ = (58.8 GHz K⁻¹) T = 160.2 GHz at 2.725 K. The λ-form peak (1.06 mm) differs from c/νₘₐₓ (1.87 mm) because the two densities weight the spectrum differently.