Failure of Classical Theory
Key idea: Explain where the Rayleigh–Jeans blackbody model succeeds, why it diverges at short wavelength, and how experiment motivates energy quantisation.
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The core idea
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Learning objectives
- Connect blackbody evidence, Planck quantisation, photon momentum, and Compton scattering.
The Rayleigh–Jeans model correctly describes the long-wavelength part of a blackbody spectrum, but it predicts unbounded energy at short wavelength. The failure is not merely a poor fitted constant: it exposes a wrong classical assumption about how thermal energy is shared among electromagnetic modes.
1. Classical prediction
For spectral energy density per unit wavelength, the Rayleigh–Jeans law is
u_λ(λ,T) = (8π kT)/λ⁴
where u_λ dλ is energy per unit volume in the wavelength interval [λ,λ + dλ].
At fixed T,
u_λ ∝ λ⁻⁴
so the prediction diverges as λ → 0.
Blackbody spectra and the Rayleigh–Jeans failure
Hotter and cooler blackbody spectra each rise to a peak and fall at short wavelength. The hotter peak is at shorter wavelength and has greater area. A dashed Rayleigh–Jeans curve diverges toward zero wavelength.
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View figure data
| Series | Wavelength (relative units) | Wavelength uncertainty | Spectral intensity (relative units) | Spectral intensity uncertainty |
|---|---|---|---|---|
| Hotter body | 0.5 | 0.01 | ||
| Hotter body | 1 | 0.12 | ||
| Hotter body | 1.5 | 0.45 | ||
| Hotter body | 2 | 0.78 | ||
| Hotter body | 2.5 | 0.96 | ||
| Hotter body | 3 | 1 | ||
| Hotter body | 3.5 | 0.94 | ||
| Hotter body | 4 | 0.83 | ||
| Hotter body | 5 | 0.62 | ||
| Hotter body | 6 | 0.46 | ||
| Hotter body | 7 | 0.34 | ||
| Hotter body | 8 | 0.25 | ||
| Hotter body | 9 | 0.18 | ||
| Hotter body | 10 | 0.13 | ||
| Cooler body | 0.5 | 0 | ||
| Cooler body | 1 | 0.01 | ||
| Cooler body | 1.5 | 0.04 | ||
| Cooler body | 2 | 0.1 | ||
| Cooler body | 2.5 | 0.2 | ||
| Cooler body | 3 | 0.32 | ||
| Cooler body | 3.5 | 0.44 | ||
| Cooler body | 4 | 0.53 | ||
| Cooler body | 5 | 0.6 | ||
| Cooler body | 6 | 0.57 | ||
| Cooler body | 7 | 0.49 | ||
| Cooler body | 8 | 0.4 | ||
| Cooler body | 9 | 0.32 | ||
| Cooler body | 10 | 0.25 | ||
| Rayleigh–Jeans prediction | 0.5 | 1.45 | ||
| Rayleigh–Jeans prediction | 1 | 0.9 | ||
| Rayleigh–Jeans prediction | 1.5 | 0.58 | ||
| Rayleigh–Jeans prediction | 2 | 0.4 | ||
| Rayleigh–Jeans prediction | 2.5 | 0.29 | ||
| Rayleigh–Jeans prediction | 3 | 0.22 | ||
| Rayleigh–Jeans prediction | 4 | 0.14 | ||
| Rayleigh–Jeans prediction | 5 | 0.1 | ||
| Rayleigh–Jeans prediction | 6 | 0.07 | ||
| Rayleigh–Jeans prediction | 8 | 0.04 | ||
| Rayleigh–Jeans prediction | 10 | 0.025 |
2. Why the divergence occurs
Classical equipartition assigns an average energy of order kT to every allowed electromagnetic standing-wave mode in the cavity. The number of modes per unit frequency interval grows as ν². With no upper limit on frequency, infinitely many high-frequency modes each receive finite average energy.
The total predicted energy therefore diverges:
U ∝ ∫₀^∞ ν² kT dν → ∞
This unphysical prediction is the ultraviolet catastrophe.
3. Comparison with experiment
Measured spectra have a finite peak and fall toward zero at high frequency or short wavelength. The Rayleigh–Jeans result remains useful when hν≪ kT, which is the low-frequency, long-wavelength limit. A strong answer therefore says both where the model works and where it fails.
Planck’s resolution changes the energy-exchange assumption. An oscillator of frequency f can exchange energy only in quanta hf. When hf≫ kT, thermal excitation of that mode becomes exponentially unlikely, suppressing the high-frequency contribution.
4. Common mistakes
- Saying classical physics predicts infinite intensity at every wavelength. The divergence occurs in the short-wavelength limit.
- Treating the catastrophe as an experimental divergence. Experiment shows the opposite: intensity falls.
- Claiming quantisation removes high-frequency modes. The modes remain; their average thermal occupation is suppressed.
- Mixing wavelength and frequency densities without changing variables correctly.
6. Worked Examples
Modelled example 1
Shortening wavelength in the classical model
Problem
Study the worked solution
Apply the wavelength dependence
Method
The ratio is 16.Reason
u_λ ∝ λ⁻⁴ in the classical model.Working
(u_λ(λ/2))/u_λ(λ) = (λ/(λ/2))⁴ = 16Interpret repeated halving
Method
The predicted density grows without bound.Reason
Every halving multiplies the result by another factor of 16.Working
λ → 0 ⇒ u_λ → ∞
Common misconception 2
Comparing quantum and thermal energy scales
Learner claim
Try this before viewing the solution
View solution step by step
Calculate both scales
Method
hf = 6.63 × 10⁻¹⁹ J and kT = 4.14 × 10⁻²⁰ J.Reason
These are the quantum spacing and available thermal scale.Working
hf = (6.63 × 10⁻³⁴)(10¹⁵), kT = (1.38 × 10⁻²³)(3000)Compare
Method
hf/(kT) = 16.0.Reason
The quantum is far larger than the thermal energy scale.Working
hf/(kT) = 16.0Repair the claim
Method
Classical equipartition is inappropriate; occupation is strongly suppressed.Reason
Planck’s model prevents the ultraviolet divergence when hf≫ kT.Working
hf≫ kT
7. Limiting-case check
Mind stretcher 1: Required limiting casesExtension
Any replacement law must recover Rayleigh–Jeans when hf/(kT) → 0 and must suppress the spectrum when hf/(kT)≫1. Planck’s distribution satisfies both limits.
Next: Planck’s Hypothesis.
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Course and syllabus information
- Course
- Advanced Physics
- Edition
- Advanced Physics