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Physics V course sequence · Interactive lesson

Cavity modes and 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.

UPV-02.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Cavity modes and the Planck lawSpectral energy density of cavity radiation against frequency at T = 6000 K. Solid: Planck's law, peaking at 353 THz. Dashed: Rayleigh–Jeans — the same ν² mode count paid a full kT per mode — running off the top of the frame: the ultraviolet catastrophe. At the probe frequency the mean occupancy is n̄ = 0.008 photons.spectral energy density u(ν, T)solid: Planck ν³/(eˣ − 1)dashed: Rayleigh–Jeans ν²kTpeak at 353 THz, height ∝ T³probe: x = 4.8, n̄ = 0.00807501500 THz
Start with a picture. Change the model in the lab below.
1

Build the model

The mode count was right; the kT per mode was wrong.

The key idea

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.

The complete physical model and assumptions

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.

1Name 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.

2Count 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.

3Equipartition 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.

4Restrict 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).

5Take 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.

6What 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Cavity eigenvalue problem

−∇²u = (ω/c)² u with u = 0 on the walls, on [0, L]³

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

Weyl mode count

g(ν) dν = (8πV/c³) ν² dν

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

Geometric sum over the ladder

Z = Σₙ e(−nhν/kT) = 1/(1 − e(−hν/kT)) → ⟨E⟩ = hν/(e(hν/kT) − 1)

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

Planck's law

u(ν, T) = (8πhν³/c³) · 1/(e(hν/kT) − 1)

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)

Stefan–Boltzmann from the integral

∫ u dν = aT⁴, a = 8π⁵k⁴/(15h³c³) = 7.566×10⁻¹⁶ J m⁻³ K⁻⁴

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

Wien displacement

νₘₐₓ = 2.821 kT/h = (58.8 GHz K⁻¹) Tλₘₐₓ T = 2.898 mm K

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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

6000 K
4000 to 8000 K; steps of 200.
600 THz
50 to 1400 THz; steps of 25.

Try this. 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.

Live results

PEAK 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

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelSpectral energy density of cavity radiation against frequency at T = 6000 K. Solid: Planck's law, peaking at 353 THz. Dashed: Rayleigh–Jeans — the same ν² mode count paid a full kT per mode — running off the top of the frame: the ultraviolet catastrophe. At the probe frequency the mean occupancy is n̄ = 0.008 photons.spectral energy density u(ν, T)solid: Planck ν³/(eˣ − 1)dashed: Rayleigh–Jeans ν²kTpeak at 353 THz, height ∝ T³probe: x = 4.8, n̄ = 0.00807501500 THz
Spectral energy density of cavity radiation against frequency at T = 6000 K. Solid: Planck's law, peaking at 353 THz. Dashed: Rayleigh–Jeans — the same ν² mode count paid a full kT per mode — running off the top of the frame: the ultraviolet catastrophe. At the probe frequency the mean occupancy is n̄ = 0.008 photons.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Temperature T: 6000 K; Probe frequency ν: 600 THz

Baseline 6000; choose 4000 to 8000, in steps of 200.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The ultraviolet catastrophe was not a wrong mode count, and Planck did not fix it by deleting modes. The density g(ν) = 8πVν²/c³ is a theorem about the Dirichlet Laplacian — Weyl's law — and it stands unchanged inside Planck's own formula; if the count were the culprit, the Rayleigh–Jeans limit that experiment confirms at low frequency would fail too. What failed was equipartition: the assumption that every oscillator, whatever its frequency, holds kT. A mode with hν = 10 kT is still present and still in the sum; its occupancy is 1/(e¹⁰ − 1) ≈ 4.5×10⁻⁵, frozen out by the Boltzmann weight, not removed by a cutoff. Quantisation changed the energy each mode may hold, never how many modes there are.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

HardExample 3

Show 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?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.