Part I — The crisis

Why hot objects forced physics to change

Put a steel poker in a fire. It glows a dull red first, then orange, then a yellowish white. It never glows green on the way. Why that order, and why, however hot you make it, does it never glow violet?

The question looks like one for a blacksmith. Around 1900 it was one of the sharpest problems in physics, because the best theories of the day, applied with care, gave an answer that was not merely inaccurate but impossible. The repair introduced a new constant of nature and the first hint that energy comes in steps. That repair is the subject of this essay, and by the end we will have put numbers to the poker and to the Sun.

What a spectrum is, and a clean object to measure

Pass the glow of a hot object through a prism and measure how much energy arrives in each narrow band of wavelengths. The result, energy against wavelength, is the object's spectrum. Asking why a poker is red at one temperature and white at another is asking how this curve changes with temperature. The figure shows what such curves look like.

What we are trying to explain. The energy in the radiation, per cubic metre and per micrometre of wavelength, at three temperatures (the curves follow Planck's law, derived below, which matches measurement). Hotter means more energy at every wavelength and a peak at shorter wavelength. The dashed line is the classical Rayleigh–Jeans prediction for 5800 K: it closes in on the true curve at long wavelengths (they agree to within a few per cent beyond about 30 μm) and leaves the top of the plot near 1100 nm, still climbing. At 3500 K the visible band catches only the rising edge, strongest at the red end; at 5800 K the peak sits inside the band.

Each curve rises to a single peak and falls away. As the temperature goes up, every wavelength gets brighter and the peak slides towards shorter wavelengths. At 3500 K the visible band holds only the rising red edge; at 5800 K the peak is inside it. That slide is the poker changing colour, and it is what we have to explain. The dashed line is what classical physics predicts instead.

Real surfaces complicate matters: polished steel and soot at the same temperature do not glow identically. Physicists therefore chose an object with no surface properties at all: a hollow box held at a fixed temperature $T$, with a tiny hole in one wall. Light that enters the hole bounces around inside and is absorbed long before it finds its way out, so the hole is a perfect absorber, a blackbody. The light that leaks out of the hole is a sample of the radiation filling the box, which has been exchanging energy with the walls until the two are in equilibrium. Experiment shows, and thermodynamics demands, that this spectrum depends on the temperature alone, not on what the box is made of. Something that universal ought to be calculable. A poker, a lamp filament and the surface of the Sun all follow it closely.

The quantity to calculate is the spectral energy density $u_\nu(\nu,T)$: the energy of the radiation per cubic metre of the box, per unit of frequency. Multiplying it by a small frequency range $d\nu$ gives the energy per cubic metre carried by light with frequencies between $\nu$ and $\nu + d\nu$.

The classical calculation, and why it cannot be right

The classical calculation has two steps. First, list the ways the radiation in the box can oscillate. Like a guitar string, the box supports standing waves, each with its own frequency; each such pattern is called a mode. Short waves can be fitted into a box in many more ways than long ones, and counting them carefully (the count is in the optional section below) gives the number of modes per cubic metre with frequencies between $\nu$ and $\nu + d\nu$:

$$\frac{8\pi\nu^2}{c^3}\,d\nu .$$

The count grows as the square of the frequency, and it has no upper limit: there are always more modes at still shorter wavelengths.

Second, decide how much energy each mode holds. Thermal physics turns a temperature into an energy: multiplying $T$ (in kelvin) by Boltzmann's constant, $k_B = 1.381\times10^{-23}$ J/K, gives $k_BT$, the typical energy that random thermal motion hands to any one part of a system. Classical physics makes this precise with the equipartition rule: in equilibrium every oscillator holds, on average, exactly $k_BT$, whatever its frequency. A mode of the radiation is an oscillator, so it gets $k_BT$. Multiplying the two steps together gives the classical prediction, known as the Rayleigh–Jeans law:

$$u_\nu(\nu,T) \;=\; \frac{8\pi\nu^2}{c^3}\,k_BT .$$

At low frequencies, in the far infrared, this agrees well with measurement. At high frequencies it fails completely. It rises forever, so it predicts that a warm box is filled mostly with ultraviolet light, X-rays and beyond, and that the total energy, summed over all frequencies, is infinite. Every warm object would be a source of X-rays. Paul Ehrenfest later named this the ultraviolet catastrophe. The measured curves instead rise to a single peak and fall away again, as in the figure above. Since the mode count is plain geometry, the suspect is the second step: the claim that every mode, however high its frequency, gets its full share $k_BT$.

Planck's step: energy in units of $h\nu$

In 1900 Max Planck found a formula that fitted the measurements at every wavelength, and within weeks a derivation of it. The derivation needed one assumption with no basis in classical physics: an oscillator of frequency $\nu$ cannot hold any energy it likes, only whole multiples of a basic unit,

$$E = 0,\; h\nu,\; 2h\nu,\; 3h\nu,\;\dots$$

Here $h$ is a new constant, now called Planck's constant, $h = 6.626\times10^{-34}$ joule-seconds. The unit $h\nu$ is called a quantum of energy. Notice that its size depends on frequency: a high-frequency oscillator has tall steps, a low-frequency one has shallow steps.

This one assumption changes the average. Thermal physics says how likely each energy is: in equilibrium, the chance of finding something in a state of energy $E$ is proportional to the Boltzmann factor $e^{-E/k_BT}$, so states costing much more than $k_BT$ are rare, and exponentially so. The chance of holding $n$ quanta is therefore proportional to $e^{-nh\nu/k_BT}$. Averaging the energy $nh\nu$ over these chances (the sum is worked out just below) gives

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

in place of the classical $k_BT$. Everything in this essay turns on how this expression behaves, and it is governed by a single ratio, $h\nu/k_BT$: the height of one step compared with the typical thermal energy.

Shallow steps, $h\nu \ll k_BT$. The steps are too fine to notice. Using $e^x \approx 1 + x$ for small $x$, the average becomes $h\nu/(h\nu/k_BT) = k_BT$. The classical answer is recovered.
Tall steps, $h\nu \gg k_BT$. Even the first step costs far more than thermal motion typically supplies, so the oscillator almost always sits at zero. The average is about $h\nu\,e^{-h\nu/k_BT}$, exponentially small. The mode is still there; it is frozen out.

Optional: the average, step by step

Write $x = h\nu/k_BT$. The probability of holding $n$ quanta is $p_n = e^{-nx}/Z$, where $Z = \sum_{n=0}^{\infty} e^{-nx}$ makes the probabilities add up to one. This is a geometric series with ratio $e^{-x}$, so $Z = 1/(1-e^{-x})$.

The average energy is $\langle E\rangle = h\nu\sum_n n\,p_n$. The sum $\sum_n n\,e^{-nx}$ is what you get by differentiating $Z$ with respect to $x$ and changing the sign: $-dZ/dx = e^{-x}/(1-e^{-x})^2$. Dividing by $Z$,

$$\langle E\rangle = h\nu\,\frac{e^{-x}}{1-e^{-x}} = \frac{h\nu}{e^{x}-1}.$$

If the energy were continuous, the sums would become integrals and the same steps would give exactly $k_BT$. The difference between the two answers comes entirely from replacing an integral by a sum over steps.

Multiplying the new average by the unchanged mode count gives Planck's law:

$$u_\nu(\nu,T) \;=\; \frac{8\pi\nu^2}{c^3}\;\frac{h\nu}{e^{h\nu/k_BT}-1}.$$

Read it as a contest between its two factors. The first, the number of modes, rises as $\nu^2$. The second, the energy each mode actually holds, stays near $k_BT$ while the steps are shallow and then collapses exponentially once $h\nu$ exceeds a few $k_BT$. The product rises, peaks and falls. At low frequencies it coincides with the Rayleigh–Jeans law; at high frequencies it falls off as $\nu^3 e^{-h\nu/k_BT}$, a form Wilhelm Wien had already found to fit the short-wavelength data. The total energy is finite. And because the turnover happens where $h\nu$ is a few times $k_BT$, raising the temperature moves the peak to higher frequency in direct proportion. That is the poker changing colour.

The same law, plotted against wavelength

Spectra are usually measured and drawn against wavelength, so we need the energy per unit of wavelength, $u_\lambda$, instead of per unit of frequency. Substituting $\nu = c/\lambda$ is not quite enough, because equal steps in wavelength are not equal steps in frequency: from $\nu = c/\lambda$, a small range $d\lambda$ covers a frequency range $d\nu = (c/\lambda^2)\,d\lambda$. Including that factor,

$$u_\lambda(\lambda,T) \;=\; \frac{8\pi hc}{\lambda^5}\;\frac{1}{e^{hc/\lambda k_BT}-1}.$$

This form peaks at a wavelength that is inversely proportional to the temperature. The relation is Wien's displacement law:

$$\begin{aligned}\lambda_{\max}\,T &= \frac{hc}{4.965\,k_B}\\[4pt] &= 2.898\times10^{-3}\ \text{m K}.\end{aligned}$$

The number 4.965 comes from finding where the slope of $u_\lambda$ is zero; the optional section below shows how. What matters for the argument is what it says: the peak sits where one quantum, $hc/\lambda$, costs about five times the thermal energy $k_BT$. Double the temperature and the peak wavelength halves.

One caution: because of the factor $c/\lambda^2$ above, the peak of $u_\lambda$ and the peak of $u_\nu$ fall at different places. "The peak of the spectrum" only means something once you say which variable the spectrum is plotted against. Here it is always wavelength.

Optional: where 4.965 comes from

Write $x = hc/\lambda k_BT$, so $u_\lambda \propto x^5/(e^x - 1)$ at fixed temperature. Setting the derivative with respect to $x$ to zero gives $5x^4(e^x-1) = x^5e^x$, which simplifies to

$$x = 5\,(1 - e^{-x}).$$

This has no solution in closed form. Trying values works quickly: $x = 5$ gives a right-hand side of $4.966$; $x = 4.966$ gives $4.965$, and it settles there. Then $\lambda_{\max} = hc/(4.965\,k_BT)$.

Worked example: the poker and the Sun

The size of a quantum. We need $c = 2.998\times10^{8}$ m/s and a convenient unit, the electronvolt: $1\ \text{eV} = 1.602\times10^{-19}$ J, the energy an electron gains crossing one volt. For red light at $\lambda = 700$ nm,

$$\begin{aligned} \nu &= \frac{c}{\lambda} = \frac{2.998\times10^{8}\ \text{m/s}}{700\times10^{-9}\ \text{m}}\\[4pt] &= 4.28\times10^{14}\ \text{Hz},\\[4pt] h\nu &= 2.84\times10^{-19}\ \text{J} = 1.77\ \text{eV}. \end{aligned}$$

For violet light at 400 nm the same steps give $\nu = 7.49\times10^{14}$ Hz and $h\nu = 3.10$ eV. A violet quantum costs nearly twice what a red one does.

The poker at 1000 K. The thermal energy is $k_BT = 1.381\times10^{-20}$ J $= 0.0862$ eV. Both visible quanta are enormous by comparison: $h\nu/k_BT = 20.6$ for red and $36.0$ for violet. All visible modes are deep in the frozen-out regime, which is why a 1000 K poker is dim, and the violet ones are frozen far harder. Putting both wavelengths into $u_\lambda$,

$$\begin{aligned}\frac{u_\lambda(400\ \text{nm})}{u_\lambda(700\ \text{nm})} &= \left(\frac{700}{400}\right)^{5}\frac{e^{20.6}-1}{e^{36.0}-1}\\[4pt] &\approx 3\times10^{-6}.\end{aligned}$$

Violet is about three hundred thousand times weaker than red. Wien's law puts the peak at $\lambda_{\max} = 2.898\times10^{-3}/1000$ m $= 2.9$ μm, far in the infrared: most of what the poker emits is heat you feel, and what you see is the thin red edge of the tail.

The Sun's surface at 5800 K. Now $k_BT = 0.500$ eV, and $h\nu/k_BT$ is $3.5$ for red and $6.2$ for violet. The same ratio comes out as $1.1$: the visible band is filled almost evenly, and an even mixture of visible wavelengths is what the eye calls white. The peak is at

$$\begin{aligned}\lambda_{\max} &= \frac{2.898\times10^{-3}\ \text{m K}}{5800\ \text{K}}\\[4pt] &= 5.00\times10^{-7}\ \text{m} = 500\ \text{nm},\end{aligned}$$

in the blue-green, the middle of the visible band.

Check: can anything glow violet? Push the temperature up without limit. Every visible mode then has shallow steps, Planck's law goes over to the Rayleigh–Jeans form $u_\lambda = 8\pi k_BT/\lambda^4$, and the violet-to-red ratio approaches $(700/400)^4 = 9.4$ and stops growing. Red never drops out; the hottest stars look blue-white, not violet. The poker's sequence of red, orange, white is the visible band being filled from its low-energy end as successive modes thaw. There is no temperature at which the spectrum is a narrow band of green or violet, because a blackbody spectrum is never narrow.

What Planck had, and had not, shown

The formula works, and the constant $h$ it introduced will appear in every essay that follows. But notice how narrow the assumption was. Planck applied his energy steps to the oscillating charges in the walls of the box as they exchange energy with the radiation; we applied them directly to the modes of the radiation, which gives the same average and is how the argument is usually run today. For Planck himself the steps were a device that produced the right formula, and for years he hoped to see them explained away by ordinary physics. Nothing in the argument says why energy should come in units of $h\nu$, or what is actually discrete.

Optional: where the mode count $8\pi\nu^2/c^3$ comes from

Take a cubical box of side $L$. A standing wave must fit a whole number of half-wavelengths along each edge, so it is labelled by three positive whole numbers $(n_x, n_y, n_z)$, and its frequency is $\nu = (c/2L)\sqrt{n_x^2+n_y^2+n_z^2}$.

Picture each allowed triple as a point on a grid, one point per unit volume. Modes with frequency below $\nu$ are the grid points inside a sphere of radius $R = 2L\nu/c$, restricted to the one-eighth of the sphere where all three numbers are positive. Their number is $\tfrac18\cdot\tfrac43\pi R^3$, and since $R^3 = 8L^3\nu^3/c^3$ this equals $4\pi L^3\nu^3/3c^3$. Light has two independent polarisations, which doubles the count to $8\pi L^3\nu^3/3c^3$.

Differentiating with respect to $\nu$ gives the number between $\nu$ and $\nu+d\nu$, and dividing by the volume $L^3$ gives $8\pi\nu^2/c^3$ per cubic metre. The shape of the box drops out once the box is much larger than the wavelength.

Two questions

1. Your own glow. Skin is at about 310 K. At what wavelength does its thermal radiation peak, and how does the energy of one quantum at that wavelength compare with $k_BT$? Compare the ratio with the same ratio for the Sun.

Worked answer

$\lambda_{\max} = 2.898\times10^{-3}\ \text{m K}/310\ \text{K} = 9.35\times10^{-6}$ m, about 9 μm, in the infrared band that thermal cameras are built for. One quantum there has $h\nu = hc/\lambda = 2.12\times10^{-20}$ J $= 0.133$ eV, while $k_BT = 0.0267$ eV. The ratio is $4.97$.

For the Sun, the quantum at the 500 nm peak is $2.48$ eV and $k_BT = 0.500$ eV: the ratio is again $4.97$. This is no coincidence. It is the 4.965 in Wien's law. At any temperature, the wavelength peak sits where one quantum costs about five times the thermal energy, which is exactly the statement that the turnover is set by $h\nu/k_BT$ and nothing else.

2. What did the fix change? A friend summarises this essay as: "Planck cured the catastrophe by showing there are fewer high-frequency modes than the classical count says." What is wrong with that, and what is the correct statement? Support it with the average energy of a mode, as a fraction of $k_BT$, when $h\nu = 3k_BT$ and when $h\nu = 10k_BT$.

Worked answer

The mode count $8\pi\nu^2/c^3$ is identical in the Rayleigh–Jeans law and in Planck's law; it is geometry, and no one changed it. What changed is the energy each mode holds. With $x = h\nu/k_BT$, the fraction is $\langle E\rangle/k_BT = x/(e^x-1)$: for $x = 3$ it is $3/19.1 = 0.16$, and for $x = 10$ it is $10/22025 = 4.5\times10^{-4}$. The high-frequency modes exist but are almost always empty.

The tempting wrong turn is to picture quantisation as removing something from the box. It removes nothing. It makes the first step of a high-frequency mode too expensive for thermal motion to afford, and the Boltzmann factor does the rest. The same mechanism will return whenever a quantum system is colder than its level spacing.

Where this leaves us

We can now answer the opening question. A hot object's glow is a contest between the number of ways radiation can oscillate, which favours short wavelengths, and the energy each oscillation can afford, which cuts off once a quantum $h\nu$ costs more than a few $k_BT$. At 1000 K the cut-off sits in the infrared and only the red edge of the visible band survives; by 5800 K the whole band is filled and the glow is white; beyond that the spectrum stays broad, so it turns blue-white and never violet.

What we cannot yet say is what the steps belong to. Planck's argument concerns only how matter and radiation trade energy in a box. It is silent on whether light itself, travelling freely and far from any wall, is divided into quanta of $h\nu$. That is a question an experiment can settle, by letting light knock electrons out of a metal one at a time, and it is where the next essay begins.

Quantum Mechanics, Explained