University Physics IV · Limits of Classical Physics · 1.2
Cavity Radiation & the Pre-Planck Laws
Before any quantum, thermodynamics alone forces the spectrum of a hot cavity into a very tight box: universal, integrating to T⁴, and a function of one variable rather than two. Learn exactly what that argument buys, then watch Wien fill the last gap with a guess that fitted the data of the day and broke in the infrared.
Build the model
Connect the measurement to the mechanism.
Seal a cavity at temperature T, drill a hole too small to disturb the equilibrium inside, and what leaks out is a universal function of frequency and temperature: nothing about the walls survives. Kirchhoff proved that in 1859 from the second law alone — if emissive power and absorptivity failed to stand in the same ratio at some frequency, two bodies at one temperature in an isothermal enclosure would exchange net energy, and heat would flow with no temperature difference driving it. Treat the trapped radiation as a working substance, with U = uV and P = u/3 because its carriers obey E = pc, and the Maxwell relation (∂U/∂V)T = T(∂P/∂T)V − P collapses to T du/dT = 4u: hence u = aT⁴, and the hole radiates (c/4)u = σT⁴.
Squeeze the cavity slowly between perfect mirrors and every mode frequency falls as 1/L while T falls as 1/L too, so ν/T cannot change and the spectrum is boxed into uν = ν³f(ν/T) — one unknown function of one variable in place of a surface over two. That is everything thermodynamics can pay for. It cannot supply f, because fixing f takes a constant with the dimensions of action and neither law contains one.
Wien's exponential ν³e(−βν/T) was a fit to short-wave photometry, not a theorem: right to 0.7% on the displacement constant, 7.6% low on σ, and low by a factor of six at 51 µm.
- Simple definition
- Cavity (blackbody) radiation is the electromagnetic field in thermal equilibrium with the walls of a sealed enclosure, and its spectrum depends only on temperature — never on the material, the size, or the shape of the cavity.
- Example
- A kiln cavity at 2000 K holds u = aT⁴ = 1.21 × 10⁻² J m⁻³ of radiation, presses on its walls with u/3 = 4.04 × 10⁻³ Pa, and pours σT⁴ = 907 kW m⁻² out through a small hole — identical numbers for a graphite lining or an alumina one.
A good absorber is a good emitter at that frequency. A cavity hole has aν = 1, so it emits the universal Iν itself.
eν spectral emissive power in W m⁻² Hz⁻¹; aν absorptivity, dimensionless; both at the same T.
The one mechanical input the argument needs. It follows from E = pc, which is why it is 1/3 and not the ideal gas's 2/3.
u is energy density in J m⁻³ and P is in Pa; the factor 1/3 needs isotropic radiation.
Thermodynamics fixes the exponent 4 exactly and leaves a to experiment. Only h, c and k together predict its value.
a = 4σ/c = 7.566 × 10⁻¹⁶ J m⁻³ K⁻⁴; σ = 5.670 × 10⁻⁸ W m⁻² K⁻⁴.
It cuts a function of two variables down to one, and integrating it returns u ∝ T⁴ with nothing extra assumed.
f and g are dimensionless functions of a single variable that thermodynamics leaves unknown.
5772 K → 502 nm; 300 K → 9.66 µm. Since the peak sits at fixed λT, the λ⁻⁵ prefactor alone sets its height, which therefore climbs as T⁵.
b in m K. The peak in λ and the peak in ν are different points: νₘₐₓ = 2.821 kT/h.
Fitted to Paschen's short-wave data, not derived. It puts b 0.70% low and σ 7.6% low, yet returns 17% of the truth at 51.2 µm and 1500 K.
Its ratio to the measured law is exactly 1 − e⁻ˣ with x = hν/kT; within 1% only for x ≳ 4.6.
Kirchhoff's equality, and why the walls drop out
Put two bodies of any materials inside a sealed enclosure at temperature T and wait. Each must absorb exactly what it emits, or its temperature would drift and heat would move between objects already at the same temperature. Slide a narrow-band filter between them and the balance must hold frequency by frequency, which forces eν = aν Iν(T) with the same Iν for every body: a good absorber is a good emitter, and the ratio is universal. A body with aν = 1 at every frequency is a blackbody, and no real surface is one — lamp soot reaches about 0.95, polished tungsten far less. A small hole in a large cavity is one: light entering is scattered many times before it can find the hole again. For walls of emissivity 0.5 and a hole one thousandth of the internal area, the standard isothermal-cavity estimate ε/[ε + (1 − ε)Aₕ/A] gives an effective emissivity of 0.999. That is why the radiation standard is a hole and not a paint.
Radiation as a working substance: why P = u/3
To do thermodynamics on the field you need one mechanical fact. Radiation carries momentum p = E/c, and a beam striking a wall at angle θ delivers normal momentum with one factor of cos θ from the projection and another from the flux; averaging cos²θ over an isotropic hemisphere gives 1/3, so P = u/3. Compare a monatomic ideal gas, where E = p²/2m rather than pc and the same average returns P = 2u/3. The factor is small; its consequences are not. At 300 K, u = aT⁴ = 6.13 × 10⁻⁶ J m⁻³ and P = 2.0 × 10⁻⁶ Pa, some 5 × 10¹⁰ times below atmospheric — which is why nobody stumbled on radiation pressure by accident. At 1.5 × 10⁷ K, the temperature of the Sun's core, the same formula gives P = 1.3 × 10¹³ Pa. The two pressures differ by a factor of 6 × 10¹⁸, almost nineteen orders of magnitude, because T⁴ is a violent function.
Turning the crank: why the exponent is exactly four
Write U = u(T)V and use dU = T dS − P dV. The Maxwell relation that follows is (∂U/∂V)T = T(∂P/∂T)V − P. The left side is simply u, since doubling the volume at fixed T doubles the energy. The right side, with P = u/3, is (T/3)(du/dT) − u/3. Rearranged, 4u/3 = (T/3)(du/dT), so du/u = 4 dT/T and u = aT⁴; the same integration gives S = (4/3)aT³V. Only the outward hemisphere escapes a small hole and the cos θ projection costs another half, so the flux is M = (c/4)u = σT⁴ with σ = ac/4. Notice what was never used: no model of the walls, no counting of modes, no h. Notice too where the 4 came from. For P = γu the exponent is (1 + γ)/γ, so radiation's 1/3 gives 4 where a monatomic gas's 2/3 would have given 5/2.
Squeeze the cavity: why ν/T cannot change
Now compress the cavity quasi-statically between perfect mirrors. Entropy is constant and S = (4/3)aT³V, so T³V is constant and T ∝ V(−1/3) ∝ 1/L. Meanwhile each standing mode satisfies ν = nc/2L with n fixed — a slow squeeze creates and destroys no modes — so ν ∝ 1/L as well; equivalently, each bounce off the retreating mirror Doppler-shifts the light by exactly that amount. So ν/T survives untouched, and so does the classical adiabatic invariant E/ν of each mode. Combine those with a mode count going as ν² and the spectrum is forced into uν = ν³f(ν/T), or uλ = λ⁻⁵g(λT) after the Jacobian c/λ². Two consequences come free. The peak of λ⁻⁵g(λT) sits wherever λT takes one fixed value, whatever g is, so λₘₐₓ T = b exists as a theorem even though its value 2.898 mm K does not. And ∫ν³f(ν/T) dν = T⁴∫x³f(x) dx returns Stefan–Boltzmann, a check on the whole chain.
Where thermodynamics stops, and what Wien put in the gap
Thermodynamics cannot supply f, and the reason is dimensional: pinning the shape of a spectrum takes a constant carrying the dimensions of action, and neither Maxwell's equations nor the two laws contain one. Wien filled the gap in 1896 by analogy. He supposed each frequency was emitted by molecules of one speed and weighted them by the Maxwell distribution's e(−mv²/2kT); to make that exponent depend on ν/T rather than on ν²/T he had to take the emitted frequency proportional to the molecule's kinetic energy, ν ∝ v², which turns the factor into e(−βν/T). The result uν = αν³e(−βν/T) already has the right scaling form, so it passes automatically every test thermodynamics can set — and it fitted Paschen's 1897 photometry from roughly 1 to 8 µm at 700–1500 K to within the errors of the day. Its displacement constant comes out at hc/5k = 2.878 mm K against the true 2.898, low by 0.70%; its Stefan constant is low by the factor 6/(π⁴/15) = 0.924, or 7.6%. Neither number is a fair test, because neither reaches the region where the guess is wrong: the peak sits at hν/kT ≈ 5, where Wien's curve runs only 0.7% below the truth, and the Stefan integral draws a mere 18% of its weight from hν/kT < 2.
The break: the far infrared, and what comes next
Wien's law falls below the truth by exactly the factor 1 − e⁻ˣ, with x = hν/kT: 0.7% at x = 5, 5% at x = 3, 37% at x = 1. The defect only becomes obvious below x ≈ 1, which means long wavelengths or high temperatures, and Rubens and Kurlbaum reached them — reststrahlen mirrors isolated a band near 51.2 µm while the source was swung from 85 K to about 1770 K. At 51.2 µm and 1500 K, x = 0.187 and Wien's formula returns 17% of the energy actually measured. The shape of the failure matters more than its size. Hold ν fixed and raise T: e(−hν/kT) → 1, so Wien's law saturates at a ceiling αν³. The data instead ran dead straight, uν ∝ T, with no ceiling in sight. That linear law is what classical equipartition predicts, and it is the next topic — Wien's guess and Rayleigh–Jeans fail at opposite ends of the same axis, which is the clue Planck read.
Change one variable at a time
Make the relationship visible.
Slide T: the curve moves bodily along the log axis without changing shape, the peak tracking λmax = 2898/T µm. Then set T = 300 K and the cursor to 120 µm — the dashed Wien curve has dropped to a third of the measured one, which is the gap Rubens and Kurlbaum found.
PEAK λmax = 2898/T2.41 µm
x = hc/λkT AT CURSOR1.20
WIEN ÷ MEASURED0.698
EXITANCE σT⁴117.6 kW m⁻²
Live interpretationPEAK λmax = 2898/T: 2.41 µm. x = hc/λkT AT CURSOR: 1.20. WIEN ÷ MEASURED: 0.698. EXITANCE σT⁴: 117.6 kW m⁻²
Catch the common trap
Explain before calculating.
Kirchhoff's law, the mechanical relation P = u/3, and quasi-static adiabatic compression of a mirrored cavity are the only inputs allowed. Which of these results does that thermodynamics NOT deliver?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA furnace cavity is held at 1450 K. Taking σ = 5.6704 × 10⁻⁸ W m⁻² K⁻⁴ and c = 2.9979 × 10⁸ m s⁻¹, find the radiation energy density inside, the pressure it exerts on the walls, and the power leaving through a viewing hole 4.0 mm across.
- The radiation constant follows from σ: a = 4σ/c = 4 × 5.6704 × 10⁻⁸ ÷ 2.9979 × 10⁸ = 7.566 × 10⁻¹⁶ J m⁻³ K⁻⁴.
- T⁴ = 1450⁴ = 4.4205 × 10¹² K⁴, so u = aT⁴ = 7.566 × 10⁻¹⁶ × 4.4205 × 10¹² = 3.344 × 10⁻³ J m⁻³.
- The radiation is isotropic, so P = u/3 = 1.115 × 10⁻³ Pa — about 9 × 10⁷ times smaller than the 1.013 × 10⁵ Pa outside.
- The hole emits the universal flux: M = σT⁴ = 5.6704 × 10⁻⁸ × 4.4205 × 10¹² = 2.507 × 10⁵ W m⁻².
- Its area is π(2.0 × 10⁻³ m)² = 1.257 × 10⁻⁵ m², so the power out is 2.507 × 10⁵ × 1.257 × 10⁻⁵ = 3.15 W.
Answeru = 3.34 × 10⁻³ J m⁻³, P = 1.11 × 10⁻³ Pa, and 3.1 W leaves the hole. By Kirchhoff's law none of it depends on what the walls are made of.
MediumA tungsten strip lamp runs at 2450 K and is treated as a cavity source. (a) Where does its spectrum peak in wavelength? (b) Wien's fitted law returns a fraction 1 − e⁻ˣ of the true spectral radiance, with x = hc/λkT. Evaluate that fraction at 900 nm and at 10 µm, taking hc/k = 1.4388 × 10⁻² m K and b = 2.898 × 10⁻³ m K.
- (a) λₘₐₓ = b/T = 2.898 × 10⁻³ ÷ 2450 = 1.183 × 10⁻⁶ m, that is 1.18 µm — in the near infrared, which is why the lamp looks orange rather than white.
- (b) At 900 nm: x = 1.4388 × 10⁻² ÷ (900 × 10⁻⁹ × 2450) = 1.4388 × 10⁻² ÷ 2.205 × 10⁻³ = 6.525.
- So the fraction is 1 − e⁻⁶·⁵²⁵ = 1 − 0.00147 = 0.9985. Wien's law is 0.15% low here, far inside 1890s photometric error.
- At 10 µm: x = 1.4388 × 10⁻² ÷ (10 × 10⁻⁶ × 2450) = 1.4388 × 10⁻² ÷ 2.450 × 10⁻² = 0.5873.
- The fraction is 1 − e⁻⁰·⁵⁸⁷³ = 1 − 0.5558 = 0.4441, so Wien's law is 56% low at the same temperature, on the same curve.
Answerλₘₐₓ = 1.18 µm. Wien's law returns 0.9985 of the truth at 900 nm and 0.444 of it at 10 µm — only a detector reaching the far infrared could expose the defect.
HardTest Wien's fitted law uν = (8πh/c³)ν³e(−hν/kT) against the two integral results thermodynamics had already fixed. (a) Locate its peak in wavelength and compare with b = 2.8978 × 10⁻³ m K. (b) Integrate it over all frequencies and compare its Stefan constant with the true one, given ∫x³e⁻ˣ dx = 6 while the true integrand gives ∫x³/(eˣ − 1) dx = π⁴/15.
- (a) Convert with uλ = uν c/λ²: uλ ∝ λ⁻⁵ e(−A/λ), with A = hc/kT = 1.43878 × 10⁻² ÷ T metres.
- Differentiate: d/dλ[λ⁻⁵e(−A/λ)] = λ⁻⁶e(−A/λ)(A/λ − 5), which vanishes at λₘₐₓ = A/5.
- So λₘₐₓ T = hc/5k = 1.43878 × 10⁻² ÷ 5 = 2.8776 × 10⁻³ m K, against the measured 2.8978 × 10⁻³ m K — low by 0.70%.
- (b) Substitute x = hν/kT: ∫uν dν = (8πh/c³)(kT/h)⁴ ∫x³e⁻ˣ dx = 48πk⁴T⁴/(c³h³). The true law replaces 6 by π⁴/15 = 6.4939, giving 8π⁵k⁴T⁴/(15c³h³).
- The ratio is 6 ÷ 6.4939 = 0.9239, so Wien's law understates σ by 7.6% — inside the calorimetric uncertainty of the 1890s.
- Neither test looks where the guess fails. The peak test lives at x ≈ 5, where 1 − e⁻ˣ = 0.993: Wien's maximum sits at exactly x = 5 against 4.965 for the measured curve, and that gap is the 0.70%. The Stefan integral takes only 18% of its weight from x < 2, yet that sliver alone supplies two thirds of the 7.6% shortfall.
AnswerWien's guess gives λₘₐₓ T = 2.8776 × 10⁻³ m K, 0.70% low, and a σ 7.6% low. Both sit within 1890s error, which is why only fixed-wavelength far-infrared work could refute it.