Black-body radiation and Planck's quantum hypothesis

The first dead end from the previous section: classical theory computes black-body radiation and returns infinity.

This section is about how Planck strangled it — and about the fact that he did not reason his way to the formula, but the other way round.

First, what the measured curve looks like

Through the 1890s the Imperial Physical-Technical Institute in Berlin produced the most accurate black-body measurements then available. The curve has a distinctive shape:

What Planck did

In October 1900 Planck used interpolation to stitch Wien’s formula and Rayleigh–Jeans together — a purely mathematical move, aimed at producing an expression correct at both ends:

$$u(\nu,T)=\frac{8\pi h\nu^3}{c^3}\cdot\frac{1}{\ee^{h\nu/k_BT}-1}\tag{1.2.3}$$

That same evening someone tested it against the data. The agreement was perfect.

But why should the formula hold? For the next two months Planck was, in his own words, working “desperately” to find a reason. What he found was this:

Why “one unit at a time” defeats the catastrophe

This is the step in the section most worth understanding, and it takes no calculation at all.

The answer: that mode cannot even afford its first unit, so it stays at $n=0$ and its average energy is essentially zero.

There are still just as many high-frequency modes, but they are all empty. The divergence is cut off at its source.

Derivation: from E = nhν to Planck's lawadvanced~7 minexpand

Restrict the oscillator energies to $E_n=nh\nu$ and compute the average with the Boltzmann distribution. Write $x\equiv h\nu/k_BT$:

$$\langle E\rangle=\frac{\sum_{n=0}^{\infty}nh\nu\,\ee^{-nx}}{\sum_{n=0}^{\infty}\ee^{-nx}}\tag{1.2.5}$$

The denominator is a geometric series:

$$\sum_{n=0}^{\infty}\ee^{-nx}=\frac{1}{1-\ee^{-x}}\tag{1.2.6}$$

For the numerator use a standard trick — differentiate the denominator:

$$\sum_{n=0}^{\infty}n\,\ee^{-nx}=-\frac{\dd}{\dd x}\frac{1}{1-\ee^{-x}}=\frac{\ee^{-x}}{(1-\ee^{-x})^2}\tag{1.2.7}$$

Divide:

$$\langle E\rangle=h\nu\cdot\frac{\ee^{-x}}{1-\ee^{-x}}=\frac{h\nu}{\ee^{h\nu/k_BT}-1}\tag{1.2.8}$$

That is the key result. Now substitute it for $k_BT$ in the classical derivation; the mode density $8\pi\nu^2/c^3$ needs no change at all:

$$u(\nu,T)=\frac{8\pi\nu^2}{c^3}\cdot\frac{h\nu}{\ee^{h\nu/k_BT}-1}\tag{1.2.9}$$

Check the two limits.

Low frequency ($h\nu\ll k_BT$, i.e. $x\ll1$): using $\ee^x\approx1+x$,

$$\langle E\rangle\approx\frac{h\nu}{x}=k_BT\tag{1.2.10}$$

Equipartition returns intact, so Rayleigh–Jeans is the low-frequency limit of Planck’s law.

High frequency ($h\nu\gg k_BT$): the $-1$ in the denominator is negligible, so

$$\langle E\rangle\approx h\nu\,\ee^{-h\nu/k_BT}\tag{1.2.11}$$

Exponential suppression. This is Wien’s formula, and this is where the ultraviolet catastrophe is cured: exponential decay beats the $\nu^2$ growth easily, and the integral converges.

The second dead end falls out for free

The low-temperature specific heat problem is solved here too, using the very same sentence.

Atomic vibrations in a solid are oscillators as well. To excite a mode of frequency $\nu$ costs at least $h\nu$. When the temperature is low enough that $k_BT\ll h\nu$, the mode is frozen and contributes nothing to the heat capacity.

Raise the temperature and more modes thaw out, so the heat capacity climbs towards the classical $3R$. Einstein built the first quantum model of specific heat on this in 1907, and Debye refined it in 1912 to the point of quantitative agreement.

An exercise in orders of magnitude

Estimate: what a body, a filament and the Sun emitbasic~4 minexpand

Using Wien’s displacement law $\lambda_\text{max}T=2.898\times10^{-3}\ \text{m}\cdot\text{K}$:

• A human body (310 K): $\lambda_\text{max}=9.3\ \mu\text{m}$, far infrared. That is the band thermal cameras work in, and why you cannot see a person in the dark.
• An incandescent filament (2800 K): $\lambda_\text{max}=1.0\ \mu\text{m}$, near infrared. The peak is not in the visible at all — a light bulb turns most of its energy into heat, which is exactly why it is so inefficient.
• The Sun’s surface (5800 K): $\lambda_\text{max}=0.50\ \mu\text{m}$, green light, right in the middle of the visible spectrum.

The last one is no coincidence: the sensitivity band of the human eye was selected by the Sun’s radiation peak.

∎Key formulas

• Planck's quantum hypothesis

  $$E_n=nh\nu,\qquad n=0,1,2,\dots$$

  Energy is exchanged in whole units whose size grows with frequency

• Mean oscillator energy

  $$\langle E\rangle=\frac{h\nu}{\ee^{h\nu/k_BT}-1}$$

  Low frequency → k_BT (equipartition); high frequency → exponentially suppressed

• Planck's radiation law

  $$u(\nu,T)=\frac{8\pi h\nu^3}{c^3}\cdot\frac{1}{\ee^{h\nu/k_BT}-1}$$

  Reduces to Rayleigh–Jeans and to Wien at the two ends

• Wien's displacement law

  $$\lambda_{\text{max}}T=2.898\times10^{-3}\ \text{m}\cdot\text{K}$$

  Hotter means the peak moves to shorter wavelengths

?Self-check4 questions

1. 1.

   How does the quantum hypothesis E = nhν remove the ultraviolet catastrophe?

   • For high-frequency modes one unit hν far exceeds k_BT, so they cannot even be excited once and their mean energy is exponentially suppressed
   • It reduces the number of high-frequency modes
   • It imposes a maximum radiated frequency
   • It restores conservation of energy

   Submit
2. 2.

   Which statements about Planck's law and the two older formulas are correct? (Select all that apply.)

   Select all that apply

   • For hν ≪ k_BT it reduces to the Rayleigh–Jeans law
   • For hν ≫ k_BT it reduces to Wien's formula
   • It matches experiment across the whole spectrum
   • It overturns the mode count 8πν²/c³

   Submit
3. 3.

   Why is diamond already well below 3R at room temperature while lead is not?

   • Stiff bonds and light atoms give diamond high vibrational frequencies, so room temperature is effectively "cold" for it and most modes remain frozen
   • Diamond is denser
   • Diamond is an insulator with no free electrons to conduct heat
   • Lead has a higher atomic number

   Submit
4. 4.

   A furnace at 1500 K radiates with a peak wavelength of how many micrometres? (Wien constant 2.898×10⁻³ m·K)

   μm10% relative tolerance

   Submit

What comes next

Planck held that “one unit at a time” was a property of the oscillators in the cavity wall; light itself remained a continuous wave.

Five years later a patent clerk pushed that sentence somewhere Planck did not want it to go: what if light itself comes in units?