University Physics IV · Limits of Classical Physics · 1.3
Mode counting and the ultraviolet catastrophe
Two ingredients build the classical cavity spectrum: a mode count that is pure geometry, and a mean energy per mode that is pure assumption. Multiplying them gives a law exact in the far infrared and infinitely wrong in the ultraviolet — and the arithmetic tells you which factor to blame.
Build the model
Connect the measurement to the mechanism.
Take a cavity in thermal equilibrium and ask how its energy is spread over frequency. Two ingredients answer it, and they come from different places. The first is a count: conducting walls force standing waves, the allowed wavevectors form a cubic lattice of spacing π/L, and counting lattice points in a shell — positive octant only, doubled for the two transverse polarisations — gives 8πν²/c³ modes per unit volume per unit frequency, independent of the cavity's shape and of which boundary condition you impose.
That is geometry, and nothing in it can fail. The second is a physical assumption: each mode is a harmonic oscillator with two quadratic energy terms, so classical equipartition awards it kT whatever its frequency. Multiply and you have u(ν, T) = 8πν²kT/c³, the Rayleigh–Jeans law — parameter-free, and exact at low frequency, where measurement confirms it to fractions of a percent.
It also carries nothing that could stop it. The energy density integrated to a cut-off grows as ν³, so the total diverges, and the cost of the derivation is that one of its two ingredients must be false. Since the count is geometry, equipartition is the ingredient that has to go.
- Simple definition
- The Rayleigh–Jeans law is the classical cavity spectrum: the number of electromagnetic modes per unit volume per unit frequency, 8πν²/c³, multiplied by the equipartition energy kT that classical statistics assigns to every one of them.
- Example
- At 2000 K and 500 nm it gives u(λ, T) = 8πkT/λ⁴ = 1.11 × 10⁷ J m⁻⁴, while a real cavity holds about 90 J m⁻⁴ there — an overshoot of 1.2 × 10⁵.
Turns a wave equation into lattice-point counting in n-space.
L in m, ν in Hz; conducting cube, tangential E zero at every wall; one index may also be zero, at a single polarisation
A 1.0 litre cavity holds 4.0 × 10¹⁴ modes in a 1 nm band at 500 nm.
m⁻³ Hz⁻¹; the two transverse polarisations are already inside the 8π
Frequency-independent — the one factor a successor theory can replace.
k = 1.381 × 10⁻²³ J K⁻¹; two quadratic terms, amplitude and its rate
Geometry × assumption, with no adjustable constant left to fit.
J m⁻³ Hz⁻¹; exact as hν/kT → 0, and within 1% for hν/kT < 0.02
Skip it and you are wrong by 21 orders of magnitude at 500 nm.
J m⁻⁴; the Jacobian c/λ² is what turns ν² into λ⁻⁴, not λ⁻²
At 2000 K it has already overtaken the whole measured budget at 2.67 µm.
J m⁻³; the total goes as the cube of the cut-off, and per unit volume
The count is geometry, and geometry does not fail
Put the field in a cubical cavity of side L with conducting walls. Tangential E must vanish on every wall, which forces Eₓ ∝ cos(kₓ x) sin(ky y) sin(kz z) and its two cyclic partners — a sine in each of the two directions along which that component runs tangential — and the allowed wavevectors are k = (π/L)(nₓ, ny, nz) with the nᵢ integers. Every mode is one point of a cubic lattice in n-space, one point per unit cell, with frequency ν = (c/2L)√(nₓ² + ny² + nz²). Two features of the count are worth defending, because both get challenged. Only the positive octant counts, since −n labels the same standing wave as +n. And each lattice point with all three indices non-zero carries two modes rather than three: ∇⋅E = 0 kills the longitudinal polarisation and leaves two transverse states. Set one index to zero and two of the three components collapse with their sines, leaving one surviving polarisation instead of two — those are the lowest resonances a real cavity has, and they are why the nᵢ ≥ 1 bookkeeping used below is an asymptotic count rather than an exact one. Swap the conducting walls for periodic boundary conditions and the lattice spacing doubles to 2π/L while all eight octants open up — the same answer. Weyl's theorem goes further: for any cavity large compared with the wavelength, the leading count depends on the volume alone and not on the shape.
From lattice points to a smooth density
Counting every mode below ν means counting lattice points in the positive octant of a sphere of radius R = 2Lν/c. Replace the point count by the octant's volume and double it for polarisation: N(ν) = 2 × (1/8)(4π/3)R³ = (8π/3)(ν³/c³)V, and differentiating gives the density everything else is built on, dN/(V dν) = 8πν²/c³. That smoothing step needs R ≫ 1, and it is worth watching it fail. Take a 1.0 cm cube below 30 GHz, where R = 2.00: the smooth formula claims 8.4 modes, while an exact tally finds five — three at (c/2L)√2 = 21.2 GHz, each with one index zero and so one polarisation, and two at (c/2L)√3 = 26.0 GHz. Go up to R = 8.0 and the tally gives 515 against the formula's 537, 4% high; the surface corrections cancel for a conducting cavity, so what survives is only a term of order R, and its share dies away as R grows. At optical frequencies there is nothing left to worry about, since R = 2L/λ reaches 4 × 10⁵ for a 10 cm cube at 500 nm. And the density doubles for every 41% rise in frequency, without ever turning over.
Equipartition hands every mode the same kT
A mode of the field is dynamically a harmonic oscillator. Expand the field in the cavity as a sum over modes and the energy separates into terms ½q̇² + ½ω²q² in the mode amplitudes — equivalently, half electric and half magnetic. Two quadratic terms, so classical equipartition gives ⟨E⟩ = 2 × ½kT = kT. It is worth seeing where that comes from, because that is where the trouble lives: the mean energy is ∫E e(−E/kT)dE ÷ ∫e(−E/kT)dE taken over a continuous E from 0 to ∞, and the integral returns kT no matter what ω is. Nothing classical says a fast mode should be harder to excite than a slow one — the Boltzmann factor cares only about energy, and a classical oscillator may carry any energy at any frequency. So the 10¹⁵ Hz modes are stocked exactly as generously as the 10¹⁰ Hz ones: at 2000 K, 2.76 × 10⁻²⁰ J apiece.
The law, and the range over which it is exact
Multiply the two factors: u(ν, T) = (8πν²/c³)kT, in joules per cubic metre per hertz. There is no adjustable constant — geometry and equipartition fix it completely, which makes it a prediction rather than a fit. And in the far infrared and radio it is right. Against the measured spectrum the classical value is too large by the factor (ex − 1)/x with x = hν/kT, which expands as 1 + x/2 + …, so Rayleigh–Jeans is within 1% for x < 0.02. At 2000 K that means wavelengths beyond about 360 µm. Radio astronomy lives entirely inside that window and still quotes source strengths as brightness temperatures defined from this law. Convert to wavelength carefully: u(λ, T) = u(ν, T)|dν/dλ| = u(ν, T)c/λ², and that Jacobian is what turns ν² into λ⁻⁴. Substituting ν = c/λ without it returns λ⁻² — a different power law, not merely a different constant.
Integrating it destroys it
Now add the modes up. The energy density below a frequency ν is 8πkTν³/3c³, growing without limit because neither classical electromagnetism nor classical statistics supplies a highest frequency, and growing per unit volume, so it is not an artefact of a large box. Numbers show the scale of the failure. At 2000 K the measured total is aT⁴ = 1.21 × 10⁻² J m⁻³. The classical running total passes that value at 1.12 × 10¹⁴ Hz — a wavelength of 2.67 µm, still in the near infrared — and by the time the integral reaches the violet edge at 400 nm it stands at 3.61 J m⁻³, about three hundred times the entire measured budget. Carry it to X-rays at 3 × 10¹⁸ Hz and it is 2 × 10¹¹ J m⁻³. Ehrenfest named this the ultraviolet catastrophe in 1911, and the name is precise: the trouble is not at any one frequency but in the tail, where a ν² density of modes times a fixed kT can never converge.
Which ingredient has to go
A derivation with two ingredients and a fatal conclusion has exactly two suspects, and they can be tried separately. The mode count is checkable on its own: resolve the resonances of a microwave cavity and count them, one line at a time. They are there, in the predicted number. Equipartition gets no such direct test inside the cavity, and its second failure was already on the table elsewhere. Molar heat capacities of solids sit at 3R near room temperature, exactly as equipartition predicts, then collapse towards zero on cooling; diatomic gases shed their vibrational and then their rotational contributions the same way. Both failures share one signature: degrees of freedom stop accepting energy once kT is small compared with something. Naming that something, and the constant that sets its scale, is the next step.
Change one variable at a time
Make the relationship visible.
Push the cut-off past the ring and the classical total exceeds every joule the cavity actually holds. Then raise T: the ring slides right in proportion to T, so the frequency where classical bookkeeping runs out is not a property of the box, and no fixed cut-off can rescue it.
CUT-OFF WAVELENGTH2.50 µm
CLASSICAL TOTAL19.3 mJ m⁻³
MEASURED TOTAL aT⁴34.6 mJ m⁻³
CLASSICAL ÷ MEASURED0.56 ×
Live interpretationCUT-OFF WAVELENGTH: 2.50 µm. CLASSICAL TOTAL: 19.3 mJ m⁻³. MEASURED TOTAL aT⁴: 34.6 mJ m⁻³. CLASSICAL ÷ MEASURED: 0.56 ×
Catch the common trap
Explain before calculating.
Planck's radiation law can be written u(ν, T) = (8πν²/c³) · hν/(e(hν/kT) − 1), and its first factor is exactly the Rayleigh–Jeans prefactor. What does that shared factor tell you about where the classical derivation went wrong?
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 = 12.0 cm. Find its lowest resonant frequency, estimate the number of electromagnetic modes below 10.0 GHz, and find the average frequency spacing between neighbouring modes there.
- Conducting walls force k = (π/L)(nₓ, ny, nz), so ν = (c/2L)√(nₓ² + ny² + nz²) with c/2L = (2.998 × 10⁸)/(0.240) = 1.249 GHz. The lowest resonance is not (1,1,1) but the triple with one index zero and the other two equal to 1: ν = (1.249 × 10⁹)(1.414) = 1.77 GHz, carrying one polarisation. The lowest triple with all three indices non-zero, (1,1,1), sits at (1.249 × 10⁹)(1.732) = 2.16 GHz and carries two.
- Modes below ν are lattice points in the positive octant of a sphere of radius R = 2Lν/c = 2(0.120 m)(1.00 × 10¹⁰ Hz)/(2.998 × 10⁸ m s⁻¹) = 8.005.
- Count them by volume and double for the two polarisations: N = 2 × (1/8)(4π/3)R³ = (π/3)R³ = (1.047)(513) = 537 modes. An exact tally of this cavity gives 515, so at R = 8 the smooth count runs 4% high — trustworthy here, useless at R = 2.
- The density at 10.0 GHz is 8πν²/c³ = 9.33 × 10⁻⁵ m⁻³ Hz⁻¹, so this cavity gains (9.33 × 10⁻⁵)(1.728 × 10⁻³ m³) = 1.61 × 10⁻⁷ modes per hertz — one new resonance every 6.2 MHz. That sparseness is why the count can be checked one resonance at a time in the microwave, and why nobody checks it that way at 500 nm.
AnswerLowest resonance 1.77 GHz; about 537 modes below 10.0 GHz from the smooth count, 515 from an exact tally; roughly one new resonance every 6.2 MHz there. Only the volume and the frequency enter the smooth count — 2 for polarisation, 1/8 for the octant, one lattice point per cell.
MediumA cavity is held at 2000 K. Write the Rayleigh–Jeans law per unit wavelength, evaluate it at λ = 500 nm, and compare it with the measured value of about 90 J m⁻⁴. Then show what happens if the conversion from ν to λ is done by substitution alone.
- Start from u(ν, T) = 8πν²kT/c³. A spectral density is energy per interval, so the change of variable needs the Jacobian: u(λ, T) = u(ν, T)|dν/dλ| = u(ν, T)⋅c/λ².
- Substitute ν = c/λ in the first factor and multiply: u(λ, T) = 8π(c²/λ²)kT⋅c/(c³λ²) = 8πkT/λ⁴. The ν² has become λ⁻⁴, not λ⁻².
- Numbers: kT = (1.381 × 10⁻²³ J K⁻¹)(2000 K) = 2.762 × 10⁻²⁰ J and λ⁴ = (5.00 × 10⁻⁷ m)⁴ = 6.25 × 10⁻²⁶ m⁴, so u(λ, T) = (25.13)(2.762 × 10⁻²⁰)/(6.25 × 10⁻²⁶) = 1.11 × 10⁷ J m⁻⁴.
- Measurement gives about 90 J m⁻⁴, so the classical value is too large by 1.2 × 10⁵. Here hν/kT = hc/λkT = 14.4, far outside the window where the classical law is trustworthy.
- Now the wrong route. Substituting ν = c/λ and stopping there gives 8πkT/(cλ²) = (6.941 × 10⁻¹⁹)/[(2.998 × 10⁸)(2.50 × 10⁻¹³)] = 9.3 × 10⁻¹⁵ J m⁻⁴ — wrong by 10²¹, and in the direction that would hide the catastrophe rather than reveal it.
Answeru(λ, T) = 8πkT/λ⁴ = 1.11 × 10⁷ J m⁻⁴ at 500 nm and 2000 K, about 1.2 × 10⁵ times the measured 90 J m⁻⁴. The Jacobian c/λ² is not cosmetic: omitting it returns 9.3 × 10⁻¹⁵ J m⁻⁴.
HardFor a cavity at 2000 K, integrate the Rayleigh–Jeans law from zero up to a cut-off νₘₐₓ. Evaluate the total for a cut-off at 400 nm and compare it with the measured aT⁴, taking a = 7.566 × 10⁻¹⁶ J m⁻³ K⁻⁴. Then find the cut-off that would make the classical total come out right, and say why no such cut-off can be the explanation.
- ∫₀(νₘₐₓ) (8πν²/c³)kT dν = 8πkT νₘₐₓ³/3c³. The total grows as the cube of the cut-off and has no limit as νₘₐₓ → ∞ — the divergence in one line, and it is per unit volume, so a bigger box is not the cause.
- At 400 nm, νₘₐₓ = c/λ = (2.998 × 10⁸)/(4.00 × 10⁻⁷) = 7.495 × 10¹⁴ Hz. With kT = 2.762 × 10⁻²⁰ J: u = (25.13)(2.762 × 10⁻²⁰)(4.210 × 10⁴⁴)/(3 × 2.694 × 10²⁵) = (2.922 × 10²⁶)/(8.083 × 10²⁵) = 3.61 J m⁻³.
- The measured total is aT⁴ = (7.566 × 10⁻¹⁶)(2000 K)⁴ = (7.566 × 10⁻¹⁶)(1.60 × 10¹³) = 1.21 × 10⁻² J m⁻³. The classical account has already overspent by a factor of about 300 before the integral even reaches the ultraviolet.
- Set the two equal and solve: ν³ = 3c³aT⁴/(8πkT) = 3c³aT³/(8πk) = (8.083 × 10²⁵)(7.566 × 10⁻¹⁶)(8.00 × 10⁹)/(3.470 × 10⁻²²) = 1.41 × 10⁴² Hz³, so ν = 1.12 × 10¹⁴ Hz — a wavelength of 2.67 µm, in the near infrared.
- Two things kill the cut-off story. It scales as T, since ν³ ∝ T³, and no cavity wall has a cut-off frequency that rises when you heat it. And 1.12 × 10¹⁴ Hz sits at hν/kT = 2.69 while the real spectrum peaks at hν/kT = 2.82, so the cut-off would have to be planted essentially on the brightest part of the radiation, where plainly nothing is missing.
Answeru = 8πkTνₘₐₓ³/3c³; at a 400 nm cut-off it is 3.61 J m⁻³ against a measured 1.21 × 10⁻² J m⁻³, about 300× too much. Matching the true total needs a cut-off at 1.12 × 10¹⁴ Hz (2.67 µm) — one that would scale with T, so it is a fit, not a mechanism.