Blackbody Radiation and the Quantum

Beginner

Look into a campfire after the flames die down: the embers glow orange, and a coal fanned by the wind brightens to yellow. A potter judges the heat of a kiln by the colour inside it. Hot things glow, and their colour tells how hot they are.

Step 1. Everything gives off this thermal radiation, even you and a cup of tea. At everyday temperatures it is all infrared, invisible to our eyes; thermal cameras turn it into pictures. Only above about 800 K (roughly 500 °C) does a faint red glow appear.

Step 2. As an object heats up, two things change. It gives off far more energy: double its temperature in kelvin and it radiates 16 times as much. And its colour shifts: dull red, orange, yellow, then white. The hottest stars look blue-white, the coolest red.

Step 3. The glow is a mixture of many wavelengths. In the animation the purple curve shows how the light is shared among them. As the knob turns up, the hump slides towards the rainbow band of visible light and the gold rays grow fast. Even white hot, most of the curve stays in the infrared, which is why filament bulbs wasted so much energy as heat.

Step 4. Around 1900 the best theory of light predicted the wrong curve: it said hot objects should pour out endless violet and ultraviolet light. Max Planck fixed it with a bold idea: light is given off in small packets of energy, and a packet of blue light carries more energy than a packet of red. A warm object can rarely spare enough for the big blue packets, so its glow fades out on the blue side.

Check yourself: one star looks orange, another blue-white. Which is hotter? Answer: the blue-white one.

Intermediate

What a blackbody is

It is tempting to think that only glowing objects radiate. In fact everything radiates simply because it is warm: its charged particles jostle with thermal motion, and accelerated charges emit electromagnetic waves. Near room temperature this thermal radiation is almost all infrared, around 10 µm; only above about 800 K does enough of it reach the visible band to show a dull red glow.

A blackbody absorbs all the radiation falling on it, at every wavelength, so the light it gives off depends only on its temperature $T$, not on its material or shape. The best laboratory version is a small hole in a closed oven: light entering it is absorbed after bouncing around inside, so the hole looks black when cold and glows with the pure thermal spectrum when hot. Stars and glowing coals come close, and the animation treats its bar as one.

Emissivity, $\epsilon$, is the fraction of a blackbody's radiation that a real surface emits at the same temperature. By Kirchhoff's law this equals the fraction of incoming radiation the surface absorbs: good absorbers are good emitters. Soot has $\epsilon \approx 0.95$, polished silver about 0.02.

Two laws: the peak and the total power

Wien's displacement law gives the wavelength at which the spectrum peaks:
${\lambda }_{\max }T=b=2.898\times 10^{-3}\mathrm{m}\cdot \mathrm{K}$
Double $T$ and ${\lambda }_{\max }$ halves. With $b=2.898\times 10^6\mathrm{nm}\cdot \mathrm{K}$, a person at 310 K peaks near 9300 nm, deep in the infrared, while the Sun, whose spectrum peaks near 500 nm, must be about 5800 K at its surface: Wien's law is a thermometer that works at any distance.

The Stefan–Boltzmann law gives the total power radiated, over all wavelengths, by a surface of area $A$:
$P=\epsilon \sigma A{T}^4$
$\sigma =5.67\times 10^{-8}W/({\mathrm{m}}^2\cdot {\mathrm{K}}^4)$
A blackbody, with $\epsilon =1$, radiates $\sigma T^4$ per square metre. Doubling $T$ multiplies this by $2^4=16$, and the bar in the animation, heated from 300 K to 3500 K, ends up radiating $(3500/300{)}^4\approx 18500$ times more.

A body also absorbs radiation from its surroundings; if they are at $T_{\mathrm{env}}$, the net power it loses is
$P_{\mathrm{net}}=\epsilon \sigma A(T^4-T_{\mathrm{env}}^4)$
When the two temperatures are equal, both sides keep radiating but the net flow is zero: thermal equilibrium.

Planck's quanta: why the blue side fades

Around 1900 the accepted theory predicted radiation growing without limit towards short wavelengths: any warm object would pour out ultraviolet light. Max Planck explained the measured curve by assuming that matter exchanges energy with radiation of frequency $f$ only in whole quanta:
$E=hf=\frac{hc}{\lambda }$
where $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is Planck's constant. With $hc\approx 1240\mathrm{eV}\cdot \mathrm{nm}$, a red photon at 620 nm carries 2.0 eV and a violet one at 413 nm 3.0 eV.

Each vibration of the hot material has a thermal energy of order $kT$, where $k=1.381\times 10^{-23}J/K$ is Boltzmann's constant: 0.30 eV at 3500 K. A quantum much bigger than $kT$ is rarely emitted, its chance falling roughly as $e^{-hf/kT}$. So the long-wavelength side follows the old theory, while on the short side the quanta become too costly and the curve plunges; the peak sits where $hf\approx 5kT$.

Worked example: the bar in the animation

At the end of the animation the bar is at $T=3500\mathrm{K}$. The two laws give
${\lambda }_{\max }=\frac{b}{T}=\frac{2.898\times 10^{-3}}{3500}=8.28\times 10^{-7}\mathrm{m}=828\mathrm{nm}$
$\sigma T^4=5.67\times 10^{-8}\cdot 3500^4\approx 8.51\times 10^{6}W/{\mathrm{m}}^2$

These are the final values on the ${\lambda }_{\max }$ and $\sigma T^4$ readouts, 828 nm and 8509 kW/m²: each square centimetre radiates about 850 W. The peak lies just beyond the red end of the visible band, which is at 750 nm, so the dashed blue line ends just right of the rainbow strip, and only about 18 % of the power is visible light.

At 1750 K, half that temperature, the peak was twice as far out, at 1656 nm, and the power 16 times smaller. At the start, 300 K, the peak was at 9659 nm, far off the axis. The gold rays, whose length grows as $T^4$, appear only at 2000 K, when the power is still 11 % of its final value. The room at 300 K sends back only $\sigma \cdot 300^4\approx 0.46kW/{\mathrm{m}}^2$, negligible for the glowing bar.

Worked example: the Sun and the Earth

How hot is the Sun's surface, judged from its power alone? Above the atmosphere, sunlight delivers 1361 W/m². That power left the Sun's surface, of radius $R=6.96\times 10^8\mathrm{m}$, and has spread over a sphere of radius $d=1.496\times 10^{11}\mathrm{m}$, so at the surface the flux was $(d/R{)}^2\approx 46200$ times larger:
$\sigma T^4=1361\cdot 46200\approx 6.29\times 10^{7}W/{\mathrm{m}}^2$
$T={\left(\frac{6.29\times 10^7}{5.67\times 10^{-8}}\right)}^{1/4}\approx 5770\mathrm{K}$
close to the 5800 K that Wien's law gave from the 500 nm peak: the total power and the shape of the spectrum agree.

The Earth intercepts sunlight on its cross-section $\pi R_E^2$ and reflects about 30 % of it, but it radiates from its whole surface, $4\pi R_E^2$. In balance,
$\sigma T_E^4=\frac{0.70\cdot 1361}{4}\approx 238W/{\mathrm{m}}^2\Rightarrow T_E\approx 255\mathrm{K}$
that is −18 °C, with a peak near $b/T_E\approx 11\mu \mathrm{m}$. The measured average is about 288 K: the 33 K difference is the greenhouse effect, as water vapour and carbon dioxide absorb part of that infrared and send it back down.

Reading the spectrum and the colours

The purple curve plots $B_{\lambda }$, the power radiated per unit wavelength, against $\lambda$; the area under it is the total power. Here its peak is always drawn at the same height (“peak scaled to 1”), so the curve shows where the light lies and the $\sigma T^4$ readout says how much. It is lopsided, falling steeply on the short-wavelength side and trailing a long infrared tail: at every temperature only a quarter of the power lies below ${\lambda }_{\max }$.

The colour comes from the whole visible slice, 380 to 750 nm. At 1000 K only the far red tail reaches it: dull red. Orange and yellow join in as $T$ rises, and at 5000 to 6000 K the slice is nearly flat and looks white; hotter stars look bluish white. No blackbody looks green: when the peak sits in the green, the curve is so broad that the mix is white.

Common mistakes

• “Red hot” is the coolest visible glow, not the hottest: red stars are near 3000 K, the Sun 5800 K, blue-white stars 10 000 K or more.
• The peak is not the colour: the bar at 3500 K peaks in the infrared, at 828 nm, yet glows white hot.
• Cold objects radiate too: a wall at 300 K emits about 460 W per square metre, but absorbs about as much from its surroundings.
• A hotter body is brighter at every wavelength, red included: at 3500 K the bar emits 4600 times more at 650 nm than at 1500 K, and 195 000 times more at 450 nm.
• Use kelvins: warming a hotplate from 20 °C to 40 °C multiplies its radiation by $(313/293{)}^4\approx 1.30$, a rise of 30 %, not 16 times.

Where you meet it

Thermal cameras and ear thermometers read infrared, and astronomers read star temperatures from their colours. Filament lamps at 2700 to 3000 K put only about 7 to 11 % of their power into visible light, which is why LEDs replaced them. Planck's quanta lead next to the Photoelectric Effect and to Atomic Models: the Bohr Atom.

Advanced

Radiation in a box: counting the modes

Every body glows, and the colour of the glow depends on its temperature. To find the spectrum from first principles, take the simplest glowing object: a closed cavity whose walls are held at the temperature $T$. The radiation inside comes into equilibrium with the walls, and its spectrum cannot depend on what the walls are made of. If it did, two cavities at the same temperature, joined through a filter that passes only one colour, would trade energy at that colour, and heat would flow between bodies at equal temperature, which the second law forbids.

So the radiation is universal, and we may study the easiest cavity of all: a cube of side $L$ with reflecting walls. Classically the field inside is a sum of standing waves, and the first job is to count them.

Legend

$n_x$	Whole number of half-wavelengths of a standing wave along $x$.
$n_y$	Whole number of half-wavelengths along $y$.
$n_z$	Whole number of half-wavelengths along $z$.
$R$	Distance of a mode's point from the origin, $\sqrt{n_x^2+n_y^2+n_z^2}$; modes longer than $\lambda$ have $R<2L/\lambda$.
$dR$	Thickness of the shell holding the modes between $\lambda$ and $\lambda +d\lambda$: $dR=2Ld\lambda /{\lambda }^2$.

Figure 1 is the counting device. A standing wave must fit the cube along each axis: along $x$ exactly $n_x$ half-wavelengths fit into $L$, and likewise $n_y$ along $y$ and $n_z$ along $z$, each a positive whole number. So every mode is one triple $(n_x,n_y,n_z)$, that is one point of a lattice with spacing 1 in the space spanned by the axes $n_x$, $n_y$, $n_z$. Each lattice point owns a unit cube of this space, so counting modes amounts to measuring volume.

The wavelength of a mode is set by the distance of its point from the origin. Measured along the $x$ axis, the wave repeats every $2L/n_x$, and likewise along $y$ and $z$. The reciprocals of these three axis wavelengths add like the components of a vector, which gives
$\frac{1}{{\lambda }^2}=\frac{n_x^2+n_y^2+n_z^2}{4L^2}\Rightarrow \lambda =\frac{2L}{R}$
where $R=\sqrt{n_x^2+n_y^2+n_z^2}$ is that distance. Points far from the origin are short waves.

Now count. The modes with wavelengths longer than $\lambda$ are the points closer to the origin than $R=2L/\lambda$. Because all three numbers are positive, these points fill one eighth of a ball, the octant outlined in blue in Figure 1. Its volume is $\frac{1}{8}\cdot \frac{4\pi }{3}{R}^3$, and every point stands for two modes, because light is a transverse wave whose electric field can oscillate in two independent directions (two polarisations). The number of modes longer than $\lambda$ is therefore
$N=2\cdot \frac{1}{8}\cdot \frac{4\pi }{3}{R}^3=\frac{8\pi L^3}{3{\lambda }^3}$

The spectrum needs the number of modes in a narrow band of wavelengths. Lengthening the wavelength from $\lambda$ to $\lambda +d\lambda$ shrinks $R=2L/\lambda$ by $dR=2Ld\lambda /{\lambda }^2$, so these modes lie in the gold shell of thickness $dR$ just inside the blue surface. The shell's volume is $\frac{1}{8}4\pi R^{2}dR$; with two polarisations it holds $8\pi L^{3}d\lambda /{\lambda }^4$ modes. Dividing by the volume $L^3$ of the cube and by $d\lambda$ gives the number of modes per unit volume and unit wavelength:
$n(\lambda )=\frac{8\pi }{{\lambda }^4}$
The size of the box has dropped out, as it must. The steep $1/{\lambda }^4$ drives everything that follows: short waves fit into a given space in far greater numbers than long ones.

Equipartition and the ultraviolet catastrophe

Classical physics also says how much energy each mode holds. The field of one standing wave swings back and forth in time exactly like a mass on a spring, its energy passing between the electric and the magnetic field. Equipartition of energy gives such an oscillator, in equilibrium at the temperature $T$, the mean energy $kT$, where $k$ is Boltzmann's constant: $\frac{1}{2}kT$ for each of its two quadratic energy terms. The energy per unit volume and unit wavelength is then $u_{\lambda }=n(\lambda )kT=8\pi kT/{\lambda }^4$.

What a detector or an eye receives is the radiance $B_{\lambda }$: the power leaving a surface per unit area, per unit solid angle and per unit wavelength. Radiation of energy density $u_{\lambda }$ travels at the speed of light $c$, spread evenly over all $4\pi$ steradians of direction, so $B_{\lambda }=c{u}_{\lambda }/(4\pi )$, and
$B_{\lambda }^{\mathrm{RJ}}=\frac{2ckT}{{\lambda }^4}$
This is the Rayleigh–Jeans law. It matches measurements at long wavelengths, but it grows without limit as $\lambda \to 0$, so the total energy ${\int }_0^{\infty }{u}_{\lambda }d\lambda$ is infinite. Every cavity would hold an infinite energy, piled up in the shortest waves: the ultraviolet catastrophe.

The dashed curve in this level's animation is this law at 3500 K. At the right edge of the graph, 3000 nm, it is already 2.1 times Planck's curve; at the 828 nm peak it is 29 times too high; it comes within 11 % only beyond about 20 µm. The failure is not a detail of the model. The mode count is sound geometry; what fails is equipartition itself. The same failure shows up in the specific heats of gases, where molecular vibrations that equipartition says should share the heat stay frozen at low temperature.

Planck's quantised oscillators

In 1900 Planck found the way out. He kept the mode count and changed a single assumption: an oscillator cannot hold any energy it likes, only whole multiples of a fixed step.

Legend

$E$	Energy of the oscillator.	$\mathrm{J}$
$hf$	Spacing of the energy levels: one quantum of an oscillator of frequency $f$.	$\mathrm{J}$
$e^{-nx}$	Boltzmann weight of level $n$, with $x=hf/kT$: each level weighs $e^{-x}$ times as much as the one below.
$f$	Frequency of the oscillator, that is of the mode.	$\mathrm{Hz}$
$kT$	Mean energy of an oscillator according to classical equipartition; the dashed line.	$\mathrm{J}$
$kT/h$	The frequency at which one step $hf$ equals $kT$.	$\mathrm{Hz}$
$\bar{E}$	Mean energy of the quantised oscillator, $hf/(e^{hf/kT}-1)$.	$\mathrm{J}$

The left half of Figure 2 draws this rule as a ladder. An oscillator of frequency $f$ may hold only the energies $E_n=nhf$, with $n=0,1,2,\dots$; the rungs are a step $hf$ apart, where $h$ is a new constant of nature, Planck's constant. At the temperature $T$, statistical physics weights rung $n$ by its Boltzmann factor $e^{-E_n/kT}=e^{-nx}$, with the shorthand $x=hf/kT$: the size of one step measured in units of $kT$. The green bars show these weights for $hf=kT$, that is $x=1$. Each bar is $e^{-1}\approx 0.37$ times the one below it, so most oscillators sit on the lowest rungs.

The mean energy is the weighted average of the rungs. The weights form a geometric series with ratio $e^{-x}$,
${\sum }_{n=0}^{\infty }{e}^{-nx}=\frac{1}{1-e^{-x}}$
and the energy-weighted sum is $hf$ times minus the derivative of this series with respect to $x$:
${\sum }_{n=0}^{\infty }nhf{e}^{-nx}=\frac{hf{e}^{-x}}{(1-e^{-x}{)}^2}$
Dividing the second sum by the first gives the mean energy:
$\bar{E}=\frac{hf{e}^{-x}}{1-e^{-x}}=\frac{hf}{e^{hf/kT}-1}$

The purple curve on the right of Figure 2 is this $\bar{E}$, plotted against the frequency $f$; the dashed line is the classical $kT$. Follow the curve from left to right. When the step is small, $hf\ll kT$, the approximation $e^x\approx 1+x$ gives $\bar{E}\approx kT$, so the curve starts on the classical line: low frequencies, that is long wavelengths, still obey equipartition. At the marked point $f=kT/h$ one step costs exactly $kT$, and the curve has already fallen to $kT/(e-1)\approx 0.58kT$.

When $hf\gg kT$, $\bar{E}\approx hf{e}^{-hf/kT}$ is exponentially small. Even the first rung is then too expensive, nearly every oscillator stays on rung 0, and the short-wavelength modes are frozen out. Their huge number no longer matters, and the catastrophe is gone.

Replacing $kT$ by $\bar{E}$, with $f=c/\lambda$, turns the Rayleigh–Jeans law into Planck's law:
$B_{\lambda }=\frac{2h{c}^2}{{\lambda }^5}\frac{1}{e^{hc/\lambda kT}-1}$
Planck fitted it to the precise measurements made in Berlin in 1900 and obtained values of $h$ and $k$ within a few per cent of today's, $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ and $k=1.381\times 10^{-23}J/K$.

Photons and Einstein's coefficients

Planck quantised the oscillators that emit the light. In 1905 Einstein went further and proposed that light itself comes in quanta of energy $hf$, later called photons. In 1917 he derived Planck's law once more, from nothing but atoms trading photons with radiation, and in doing so found a process nobody had suspected.

Legend

$N_1$	Number of atoms in the lower level.
$N_2$	Number of atoms in the upper level.
$hf$	Gap between the levels, equal to the energy of each photon drawn.	$\mathrm{J}$
$B_{12}{u}_f$	Rate at which one atom in the lower level absorbs a photon; $u_f$ is the energy density of the radiation per unit frequency.	${\mathrm{s}}^{-1}$
$A_{21}$	Rate at which one excited atom emits a photon spontaneously.	${\mathrm{s}}^{-1}$
$B_{21}{u}_f$	Rate at which one excited atom is stimulated to emit by the radiation.	${\mathrm{s}}^{-1}$

Figure 3 shows atoms with two energy levels a gap $hf$ apart, bathed in radiation whose spectral energy density, the energy per unit volume and unit frequency, is $u_f$. There are $N_1$ atoms in the lower level and $N_2$ in the upper one. Each panel is one process: the filled dot marks where the atom starts, the open circle where it ends, the straight arrow is its jump, and the wavy arrows are photons of energy $hf$.

On the left is absorption: a photon disappears and the atom jumps up. An atom in the lower level does this at the rate $B_{12}{u}_f$, proportional to the amount of radiation present. In the middle is spontaneous emission: an excited atom drops on its own, at the rate $A_{21}$, and sends out a photon in a random direction. On the right is stimulated emission: a passing photon makes an excited atom drop, at the rate $B_{21}{u}_f$. One photon goes in and two come out.

In equilibrium the number of upward jumps per second equals the number of downward ones, and the populations follow Boltzmann's law, $N_2/N_1=e^{-hf/kT}$. Dividing the balance by $N_2$ and collecting the terms in $u_f$ gives
$N_1{B}_{12}{u}_f=N_2(A_{21}+B_{21}{u}_f)\Rightarrow u_f=\frac{A_{21}}{B_{12}{e}^{hf/kT}-B_{21}}$

Compare this with Planck's law written per unit frequency, $u_f=(8\pi h{f}^3/c^3)/(e^{hf/kT}-1)$. As $T\to \infty$, $e^{hf/kT}\to 1$ and Planck's $u_f$ grows without limit, so the denominator above must vanish there: $B_{12}=B_{21}$. What remains then matches only if $A_{21}/B_{21}=8\pi h{f}^3/c^3$. Absorption and stimulated emission are equally strong, and they fix the rate of spontaneous emission.

Stimulated emission is essential. Without it the balance would give $u_f=A_{21}/(B_{12}{e}^{hf/kT})$, Wien's exponential form, which fails at long wavelengths. The stimulated photon is an exact copy of the one that triggered it, with the same frequency, direction and phase: that is the principle of the laser.

Why, then, is a laser hard to make? The ratio of stimulated to spontaneous emission, $B_{21}{u}_f/A_{21}$, works out to $1/(e^{hf/kT}-1)$, the mean number of photons per mode. For the glowing bar of the animation, at 3500 K and at its 828 nm peak, it is only 0.007: almost all the light of a hot body is spontaneous. A laser therefore needs a population inversion, $N_2>N_1$, which no body in thermal equilibrium can provide. In 1924 Bose derived the whole law by counting photons as indistinguishable particles, the start of Bose–Einstein statistics.

Wien's and Stefan's laws from Planck's law

Two rules of thermal radiation were known before Planck: the peak wavelength falls as $1/T$, and the total power rises as $T^4$. Both now follow from Planck's formula, with their constants built from $h$, $c$ and $k$.

First the peak. Use the dimensionless variable $x=hc/\lambda kT$, the photon energy $hc/\lambda$ in units of $kT$. Substituting $\lambda =hc/(xkT)$ gives
$B_{\lambda }=\frac{2k^5{T}^5}{h^4{c}^3}\frac{x^5}{e^x-1}$
so at a fixed $T$ the peak lies where $x^5/(e^x-1)$ is largest. Setting its derivative to zero gives $5x^4(e^x-1)=x^5{e}^x$, that is $x=5(1-e^{-x})$. Start from the guess $x=5$ and put it into the right-hand side: it returns 4.966, and one more round settles on the root $x=4.965$. Hence

${\lambda }_{\max }T=\frac{hc}{4.965k}=2.898\times 10^{-3}\mathrm{m}\cdot \mathrm{K}$

Wien's constant is not a new constant of nature but a combination of $h$, $c$ and $k$.

Next the total power. The power leaving unit area of the surface, $M$, is $\pi$ times the integral of $B_{\lambda }$ over all wavelengths. The factor is $\pi$ rather than $2\pi$ because the radiance is added over all directions of the hemisphere, each weighted by the cosine of its angle from the normal: a surface seen at a slant looks smaller. Changing the variable to $x$, with $d\lambda =-hcdx/(x^{2}kT)$, turns the integral into a pure number times $T^4$, and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$:
$M=\pi {\int }_0^{\infty }{B}_{\lambda }d\lambda =\frac{2\pi k^4{T}^4}{h^3{c}^2}{\int }_0^{\infty }\frac{x^{3}dx}{e^x-1}$
$M=\sigma T^4,\sigma =\frac{2{\pi }^5{k}^4}{15h^3{c}^2}$
which gives the Stefan–Boltzmann constant $\sigma =5.670\times 10^{-8}W/({\mathrm{m}}^2\cdot {\mathrm{K}}^4)$.

The fourth power has a simple reading. Modes with $hf$ up to a few $kT$ are active and the rest are frozen out, so the active modes reach up to a frequency proportional to $T$. Their number per unit volume grows as the cube of that frequency, so as $T^3$, and each holds an energy of order $kT$: the energy density, and with it the power emitted, grows as $T^4$.

Both laws were found before Planck. Stefan read the fourth power from measurements in 1879; Boltzmann derived it in 1884 from thermodynamics and the radiation pressure $u/3$, where $u$ is the energy density; and in 1893 Wien showed that thermodynamics forces the form $B_{\lambda }=T^{5}g(\lambda T)$, with some function $g$ of the single variable $\lambda T$. Thermodynamics could not find $g$; that took the quantum.

Reading the family of Planck curves

The scaling $B_{\lambda }=T^{5}g(\lambda T)$ is what this level's animation shows as the bar heats up. Every curve has the same shape, squeezed horizontally by the factor $1/T$ and stretched vertically by $T^5$. From 2000 K to 3500 K, a factor 1.75 in temperature, the peak moves from 1449 nm to 828 nm and rises $1.75^5\approx 16.4$ times, while the area under the curve, the total power, grows only $1.75^4\approx 9.4$ times, from 907 to 8509 kW/m². The height outgrows the area because the curve also gets narrower.

As the bar heats, the purple curve rises past the grey curves it leaves behind and lies above each of them at every wavelength, because $1/(e^{hc/\lambda kT}-1)$ increases with $T$ at every $\lambda$: a hotter body is brighter in every colour. The small graph is the quartic $\sigma T^4$. It reaches half its final value only at $3500/2^{1/4}\approx 2943\mathrm{K}$, after more than four fifths of the temperature rise.

A caution about peaks. $B_{\lambda }$ is a density per unit wavelength. Plotted per unit frequency, the same radiation peaks where $x=3(1-e^{-x})$, at $x=2.821$, which at 3500 K corresponds to 1457 nm, not 828 nm. The peaks differ because equal steps in wavelength are not equal steps in frequency: $|df|=cd\lambda /{\lambda }^2$, so converting one density into the other multiplies it by a factor that depends on $\lambda$, and the maximum moves.

For the Sun the two peaks lie at 502 nm and 883 nm. The power in any band is the same in both plots; only the position of the maximum of a density depends on the variable chosen. “The Sun peaks in the green” describes a graph, not the light.

Worked problem: a glowing rod cools

Planck's law also tells us how fast a hot body cools when radiation alone carries its heat away. Switch off the current when the bar is at $T_0=3500\mathrm{K}$. Model it as a tungsten rod of diameter $d=1.0\mathrm{mm}$ in a vacuum, as inside a lamp bulb, so that radiation is the only way out for heat. Take the density $\rho =19300kg/{\mathrm{m}}^3$, a mean specific heat $c_p=170J/(\mathrm{kg}\cdot \mathrm{K})$ and a mean emissivity $\epsilon =0.30$, the fraction of the blackbody power $\sigma T^4$ that the real surface emits. How long does the rod take to reach 1750 K, where it radiates $2^4=16$ times less?

Set up an energy balance for a piece of length $\ell$ away from the ends. Its heat content changes at the rate $m{c}_{p}dT/dt$, with the mass $m=\rho \pi d^2\ell /4$, and it loses $\epsilon \sigma T^4$ through each square metre of its surface $\pi d\ell$. Dividing both by the surface gives, per unit area,
$\frac{\rho d{c}_p}{4}\frac{dT}{dt}=-\epsilon \sigma T^4$
The length has dropped out; only the ratio of volume to surface, $d/4$, matters. Separate the variables, $-dT/T^4=4\epsilon \sigma dt/(\rho d{c}_p)$, and integrate from $T_0$ at $t=0$. Since the integral of $-T^{-4}$ is $\frac{1}{3}{T}^{-3}$,
$\frac{1}{T^3}-\frac{1}{T_0^3}=\frac{12\epsilon \sigma }{\rho d{c}_p}t\Rightarrow T=\frac{T_0}{(1+t/\tau {)}^{1/3}}$
with the time constant $\tau =\rho d{c}_p/(12\epsilon \sigma T_0^3)\approx 0.375\mathrm{s}$.

The cooling curve

The solution has a shape worth remembering: a fast start and a very long tail.

Legend

$T$	Temperature of the rod.	$\mathrm{K}$
$t$	Time since the current was switched off.	$\mathrm{s}$
$T_0$	Starting temperature.	$\mathrm{K}$
$T_0/2$	Half the starting temperature.	$\mathrm{K}$
$T_0/4$	A quarter of the starting temperature.	$\mathrm{K}$
$\tau$	Time constant of the rod, $\tau =\rho d{c}_p/(12\epsilon \sigma T_0^3)$.	$\mathrm{s}$

Figure 4 plots this solution, the temperature $T$ against the time $t$. The curve leaves $T_0$ almost vertically and then flattens into a long tail. The dots mark the two halvings of the temperature. $T=T_0/2$ when $(1+t/\tau {)}^{1/3}=2$, so $1+t/\tau =8$ and $t=7\tau$; $T=T_0/4$ when $1+t/\tau =64$, at $t=63\tau$. The gold bars under the time axis show the two intervals: $7\tau$ for the first halving and $63\tau -7\tau =56\tau$ for the second, eight times as long.

Now the numbers. Differentiating $T$ at $t=0$ gives the initial cooling rate $T_0/(3\tau )\approx 3100K/s$, the steep start of the curve. The first halving, $7\tau \approx 2.6\mathrm{s}$, takes the bar from white heat to a yellow-orange glow at 1750 K. The next halving, to 875 K, where the dull red is almost gone, takes eight times as long, about 21 s. The reason: at each stage of the second halving the bar is at half the temperature of the matching stage of the first, so it radiates $2^4=16$ times more slowly, while it has only half as much heat to lose, 875 K of cooling instead of 1750 K; and $16/2=8$. The radiation the bar receives from a room at 300 K is $(300/875{)}^4\approx 1.4\%$ of what it emits even at 875 K, so ignoring it is fair.

In reality the emissivity of tungsten falls to about 0.1 near 1000 K, so the red stage lasts longer still. A lamp filament a few hundredths of a millimetre thick has a $\tau$ dozens of times shorter, since $\tau$ is proportional to $d$, and fades in a fraction of a second.

Limits of validity

Planck's law describes radiation in thermal equilibrium with matter at a single temperature. Real surfaces emit less, $\epsilon (\lambda )B_{\lambda }$, with an emissivity between 0 and 1 that depends on wavelength. Tungsten's is higher in the visible than in the infrared, so its light looks a little hotter than the metal really is. A thin gas emits and absorbs only narrow spectral lines, and the Sun's spectrum is a near-blackbody crossed by dark absorption lines. Lasers, LEDs and sodium lamps are not thermal sources at all.

The classical laws survive as limits: Wien's exponential for $hf\gg kT$, and Rayleigh–Jeans for $hf\ll kT$. Radio astronomy lives in the second: a 1 GHz signal has $hf/kT\approx 0.016$ even at 3 K. The law also assumes bodies and gaps much larger than the thermal wavelength, the wavelength near the emission peak, about 10 µm at room temperature. A dust grain much smaller than that radiates less than $\sigma T^4$ per unit area, while two surfaces a fraction of a micrometre apart can exchange heat many times faster than two blackbodies, through evanescent waves that tunnel across the gap.

Within its domain the law is extraordinarily exact. The cosmic microwave background, the afterglow of the hot early universe, follows a Planck curve at 2.725 K, peaking near 1.06 mm, to within about 50 parts per million of its peak.

Where this leads

Planck treated quantisation as a property of the emitters; Einstein's light quanta took it literally, and the Photoelectric Effect is where you meet them next. Discrete energy levels and their spectral lines are the subject of Atomic Models: the Bohr Atom, and the double nature of the photon returns in Wave–Particle Duality. Light made without heat, across a band gap, appears in Semiconductors, Diodes and LEDs. The failure of equipartition links back to Kinetic Theory of Gases, and the entropy of radiation, which guided Planck in 1900, to The Second Law of Thermodynamics and Entropy.