Visual walkthrough — Blackbody radiation from statistical mechanics — Planck distribution

Before we write the destination formula, we need the names of two constants of nature that will appear inside it. We meet them properly (with pictures) in Steps 6 and 7 — here we only give their plain-word meaning so no symbol is unexplained.

Our destination — the thing every step is building toward:

$u(\nu ,T)d\nu =\underset{\text{how many modes}}{\underbrace{\frac{8\pi {\nu }^2}{c^3}}}\times \underset{\text{energy per mode}}{\underbrace{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}}d\nu$

Here $u(\nu ,T)$ is the energy per unit volume per unit frequency — how much light-energy sits in the box between colour $\nu$ and colour $\nu +d\nu$. The little $d\nu$ is a thin slice of frequency. $\nu$ (Greek "nu") is frequency, $T$ is temperature, $c$ the speed of light, and $h,k_B$ the two constants just introduced. Everything else we define as we go.

Step 1 — Put light in a box

WHAT. Imagine a hollow metal cube, side length $L$. Heat its walls to temperature $T$. The walls glow, filling the inside with electromagnetic (light) waves. In equilibrium the light bouncing around must form standing waves — patterns that don't drift, like a plucked guitar string.

WHY. A wave that isn't a standing wave would slosh energy around forever and never settle. Only the standing patterns are stable, so those are the only ones we count. A standing wave must be pinned to zero at both walls (the metal shorts out the field there), exactly like a rope tied at both ends.

PICTURE. Look at the figure. The first pattern fits half a wavelength across the box, the next fits one full wavelength, then one-and-a-half, and so on. Only whole numbers of half-wavelengths fit.

Because a whole number of half-wavelengths must fit along each axis, the wave's "tightness" along $x$ can only take special values:

$k_x=\frac{n_x\pi }{L}$

• $k_x$ — the wavenumber along $x$: how many radians of wave per metre. Bigger $k_x$ = tighter ripples.
• $n_x$ — a counting integer $1,2,3,\dots$: which harmonic (how many bumps).
• $\pi /L$ — the smallest step; each extra bump adds this much to $k_x$.

The same holds for $y$ and $z$ with their own integers $n_y,n_z$.

Step 2 — Every mode is a dot in "mode-space"

WHAT. A mode is fixed by three integers $(n_x,n_y,n_z)$. Plot each triple as a point on a 3D grid. Every mode = one dot, and the dots sit exactly one unit apart in each direction.

WHY. Counting waves is hard; counting evenly-spaced dots is easy. By turning "how many modes" into "how many grid points," we can use volume to count instead of adding one-by-one.

PICTURE. The figure shows the grid of dots in the first octant (all three integers positive). One dot occupies exactly one unit cube of space, so the number of dots inside a region ≈ the volume of that region.

We measure how far a dot sits from the corner:

$R=\sqrt{n_x^2+n_y^2+n_z^2}$

• $R$ — the radius in mode-space, the straight-line distance from the origin to the dot. (We use $R$, not $n$, so it never collides with the counting integer $n$ we will meet in Step 7.)
• The square-root-of-sum-of-squares is just Pythagoras in 3D: it turns three counts into one distance.

All dots with the same $R$ lie on a sphere of radius $R$. That's the key: modes of the same "size" form a shell.

Step 3 — Count modes with a sphere (careful: one-eighth, and only when $R\gg 1$)

WHAT. To count all modes out to radius $R$, take the volume of a sphere of radius $R$ — but keep only the corner where all three integers are positive.

WHY. Since each dot fills one unit cube, the count is the volume. But $n_x,n_y,n_z$ must be positive (there's no "minus-first harmonic"), so only $\frac{1}{8}$ of the sphere counts — the single octant nearest the corner.

PICTURE. The figure shows the full sphere faded, and the solid amber wedge is the one-eighth we keep.

$N(R)=\underset{\text{keep one octant}}{\underbrace{\frac{1}{8}}}\times \underset{\text{full sphere volume}}{\underbrace{\frac{4}{3}\pi R^3}}=\frac{\pi R^3}{6}$

• $N(R)$ — total number of modes with radius up to $R$.
• $\frac{1}{8}$ — the positive-integer restriction; drop it and you'd overcount eightfold.
• $\frac{4}{3}\pi R^3$ — ordinary volume of a ball of radius $R$.

Step 4 — Translate mode-count into frequency

WHAT. Physicists want colour (frequency $\nu$), not the abstract radius $R$. Convert.

WHY. Two facts link $R$ to $\nu$. The wavevector magnitude is $|\vec{k}|=R\pi /L$ (Pythagoras on the $k$'s from Step 1). And any light wave satisfies $|\vec{k}|=2\pi \nu /c$ (wave speed = frequency × wavelength). Set them equal.

PICTURE. The figure is a "dial converter": turn the $R$ knob, read off $\nu$ — a straight-line (linear) relationship.

$\frac{R\pi }{L}=\frac{2\pi \nu }{c}⟹R=\frac{2L\nu }{c}$

• Left side: wave tightness from the box geometry.
• Right side: wave tightness from being light at frequency $\nu$.
• The $\pi$'s cancel; $L$ and $c$ set the scale. Bigger box or higher $\nu$ → bigger $R$.

Substitute into $N=\pi R^3/6$:

$N(\nu )=\frac{\pi }{6}{\left(\frac{2L\nu }{c}\right)}^3=\frac{4\pi L^3{\nu }^3}{3c^3}$

Step 5 — Two polarizations, then differentiate → density of states

WHAT. Double the count (light has two independent polarizations per direction), divide by the box volume $L^3$ to get a per-volume number, then differentiate to get the count in a thin frequency slice.

WHY.

• ×2: for every wavevector, light can wiggle in two perpendicular directions — two distinct modes.
• ÷ $L^3$: we want a property of the material, not of our particular box size.
• Differentiate: we don't want "all modes up to $\nu$," we want "how many modes at $\nu$ within a sliver $d\nu$." The derivative is exactly "how fast the count grows per unit frequency."

PICTURE. The figure shows $N(\nu )$ as a rising curve; its slope at each point is $g(\nu )$, the number living in one thin vertical strip.

$g(\nu )=\frac{1}{L^3}\frac{d}{d\nu }[2N(\nu )]=\frac{1}{L^3}\frac{d}{d\nu }\left[\frac{8\pi L^3{\nu }^3}{3c^3}\right]=\boxed{\frac{8\pi {\nu }^2}{c^3}}$

• $g(\nu )$ — density of states: modes per unit volume per unit frequency. See Density of states.
• The $L^3$ cancels — good, the answer shouldn't care about box size.
• The derivative of ${\nu }^3$ is $3{\nu }^2$; the 3 cancels the 3 in the denominator.

Step 6 — The classical energy per mode (and why it explodes)

WHAT. Classical physics (Equipartition theorem) says every oscillator holds the same average energy $k_{B}T$, whatever its frequency. Here $k_B$ is Boltzmann's constant — the exchange rate between temperature and energy — so $k_{B}T$ is the typical thermal energy at temperature $T$.

WHY it's a catastrophe. Multiply "how many modes" by "energy each":

$u(\nu )=g(\nu )\cdot k_{B}T=\frac{8\pi {\nu }^2}{c^3}{k}_{B}T$

This is the Rayleigh-Jeans law. As $\nu \to \infty$ the ${\nu }^2$ climbs forever — the box would hold infinite energy. That impossible result is the ultraviolet catastrophe.

PICTURE. The figure overlays the true (measured) spectrum against the Rayleigh–Jeans curve rocketing to infinity. They agree only at low $\nu$; the classical curve then runs off the top of the page.

• $k_B$ — Boltzmann's constant, the conversion between temperature and energy.
• $k_{B}T$ — the classical energy per oscillator; the same for every mode — that's the fatal assumption.

We need to replace the flat $k_{B}T$ with something that shrinks at high $\nu$.

Step 7 — Quantize the energy: the staircase, not the ramp

WHAT. Planck's leap: a mode of frequency $\nu$ can only hold energy in whole lumps of size $h\nu$, where $h$ is Planck's constant:

${ϵ}_m=mh\nu ,m=0,1,2,\dots$

WHY. If energy came in a smooth ramp, every mode could accept an arbitrarily tiny sip and equipartition would win. Force energy onto a staircase and a high-$\nu$ mode faces a first step $h\nu$ so tall that a thermal kick of size $k_{B}T$ can't climb it. So it stays on step 0, holding zero energy.

PICTURE. The figure contrasts the classical smooth ramp with the quantum staircase; the amber arrow shows a thermal kick of height $k_{B}T$ — it easily reaches many low steps but falls short of a single tall step.

• $h$ — Planck's constant, the size of one energy lump per unit frequency.
• $h\nu$ — the step height; grows with frequency, so high-$\nu$ steps are unaffordable.
• $m$ — how many lumps this mode currently holds (an occupation number, a different integer from the geometric radius $R$ of Steps 2–4).

Now compute the true average. First we need a way to turn "list of allowed energies" into "average energy at temperature $T$." That machinery is the Partition function.

Each energy level ${ϵ}_m$ carries a Boltzmann weight $e^{-\beta {ϵ}_m}$ — a relative probability that says "how likely is the mode to sit on step $m$." The partition function is just the sum of all those weights (a normaliser):

$Z={\sum }_{m=0}^{\infty }{e}^{-\beta mh\nu }={\sum }_{m=0}^{\infty }(e^{-\beta h\nu }{)}^m=\frac{1}{1-e^{-\beta h\nu }}$

• The sum is a geometric series with ratio $e^{-\beta h\nu }<1$, so it closes to $1/(1-e^{-\beta h\nu })$.

Apply it:

$⟨ϵ⟩=-\frac{\partial }{\partial \beta }\ln \left[\frac{1}{1-e^{-\beta h\nu }}\right]=\frac{h\nu e^{-\beta h\nu }}{1-e^{-\beta h\nu }}=\boxed{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}$

(last step: multiply top and bottom by $e^{\beta h\nu }$ and put $\beta =1/k_{B}T$ back). This is $h\nu$ times the Bose–Einstein occupation $\bar{n}=1/(e^{h\nu /k_{B}T}-1)$ — the average number of lumps in the mode.

Step 8 — Assemble, and watch the catastrophe die

WHAT. Multiply Step 5 (count) by Step 7 (energy):

$u(\nu ,T)=\frac{8\pi {\nu }^2}{c^3}\cdot \frac{h\nu }{e^{h\nu /k_{B}T}-1}=\frac{8\pi h{\nu }^3}{c^3}\frac{1}{e^{h\nu /k_{B}T}-1}$

WHY it now converges. The ${\nu }^3$ up top still wants to explode — but at high $\nu$ the denominator $e^{h\nu /k_{B}T}$ grows exponentially, and exponential decay beats any power. So $u$ rises, peaks, and falls to zero.

PICTURE. The figure plots the finished Planck curve: rising as ${\nu }^3$ on the left, bending over at the peak, dying exponentially on the right. The disaster is contained.

Edge cases — check both ends:

• Low frequency ($x\to 0$, where $x=h\nu /k_{B}T$): use $e^x-1\approx x$, so $⟨ϵ⟩\to k_{B}T$ and we recover Rayleigh–Jeans. Classical physics is the cheap-photon corner of Planck's law.
• High frequency ($x\to \infty$): $e^x-1\approx e^x$, so $u\to \frac{8\pi h{\nu }^3}{c^3}{e}^{-h\nu /k_{B}T}$ — Wien's exponential tail.
• Degenerate $T\to 0$: every step is infinitely expensive, $⟨ϵ⟩\to 0$ at all $\nu$ — a cold body doesn't glow. ✓
• Degenerate $\nu \to 0$: $u\to 0$ because $g(\nu )\to 0$ (no modes at zero frequency), even though each holds $k_{B}T$. ✓

The one-picture summary

Everything on one canvas: count grows as ${\nu }^2$ (green, climbing), energy per mode falls off (cyan, dropping via the exponential), and their product is the Planck bump (amber). "${\nu }^3$ up, $e^x$ down."

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