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Physics IV course sequence · Interactive lesson

Cavity radiation and 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.

UPIV-01.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Cavity radiation and the pre-Planck lawsLog–log cavity spectrum against wavelength, each curve divided by the measured peak so that changing T only slides it sideways. Solid is the measured spectrum, dashed Wien's fitted exponential. At T = 1200 K the peak sits at 2.41 µm; at the cursor x = hc/λkT = 1.20, and Wien returns 0.698 of the measurement.cavity spectrum, log–log, scaled to the measured peakT = 1200 Ksolid: measured · dashed: Wien 1896λmax = 2898/T = 2.41 µm1 µm10 µm100 µm110⁻⁴across: wavelength λ in µm · up: uλ ÷ its peak, four decades
Start with a picture. Change the model in the lab below.
1

Build the model

Thermodynamics fixes the shape of the law, never the function inside it.

The key idea

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.

The complete physical model and assumptions

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.

1Kirchhoff'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.

2Radiation 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.

3Turning 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.

4Squeeze 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.

5Where 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.

6The 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Kirchhoff's law

eν(T) / aν(T) = Iν(T), one function for every body

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.

Radiation pressure of an isotropic field

U = u(T)V, P = u/3

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.

Stefan–Boltzmann from the two laws

T du/dT = 4u ⇒ u = aT⁴, M = (c/4)u = σT⁴

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.

Wien's scaling law

uν = ν³ f(ν/T) ⇔ uλ = λ⁻⁵ g(λT)

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.

The peak, and how it moves

λₘₐₓ T = b = 2.898 × 10⁻³ m K

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.

Wien's fitted distribution

uν = (8πhν³/c³) e(−hν/kT)

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.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1200 K
300 to 2400 K; steps of 25.
10.0 µm
0.5 to 200 µm; steps of 0.5.

Try this. 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.

Live results

PEAK λmax = 2898/T
2.41 µm
x = hc/λkT AT CURSOR
1.20
WIEN ÷ MEASURED
0.698
EXITANCE σT⁴
117.6 kW m⁻²

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

Interactive Physics IV modelLog–log cavity spectrum against wavelength, each curve divided by the measured peak so that changing T only slides it sideways. Solid is the measured spectrum, dashed Wien's fitted exponential. At T = 1200 K the peak sits at 2.41 µm; at the cursor x = hc/λkT = 1.20, and Wien returns 0.698 of the measurement.cavity spectrum, log–log, scaled to the measured peakT = 1200 Ksolid: measured · dashed: Wien 1896λmax = 2898/T = 2.41 µm1 µm10 µm100 µm110⁻⁴across: wavelength λ in µm · up: uλ ÷ its peak, four decades
Log–log cavity spectrum against wavelength, each curve divided by the measured peak so that changing T only slides it sideways. Solid is the measured spectrum, dashed Wien's fitted exponential. At T = 1200 K the peak sits at 2.41 µm; at the cursor x = hc/λkT = 1.20, and Wien returns 0.698 of the measurement.

Predict → change one variable → check

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

Saved baseline. Cavity temperature T: 1200 K; Cursor wavelength λ: 10.0 µm

Baseline 1200; choose 300 to 2400, in steps of 25.
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

Two different results are both called Wien's law, and only one of them is a theorem. The displacement law λₘₐₓ T = b — really the scaling law uλ = λ⁻⁵g(λT) — follows from the second law applied to a slowly compressed mirrored cavity, and holds whatever g turns out to be. Wien's distribution law uν ∝ ν³e(−βν/T) is a guess at g, made by analogy with the Maxwell speed distribution and fitted to short-wave photometry. So "Wien's law failed in 1900" is backwards about which one failed: the displacement law is still exact and still in the data books, while the exponential survives only as the hν/kT ≳ 4.6 limit of Planck's. A second slip rides along. λₘₐₓ and νₘₐₓ are not the same point of the spectrum, because converting uν to uλ carries a Jacobian c/λ². At 5772 K the wavelength peak is 502 nm; the frequency peak is 3.39 × 10¹⁴ Hz, which is 884 nm.

5

Practice & worked examples

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

EasyExample 1

A 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.

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.

MediumExample 2

A 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.

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.

HardExample 3

Test 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.

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

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • 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.