Why the Maxwell and Planck distributions have the shapes they do

1

Two familiar distributions

The Maxwell–Boltzmann distribution for gas molecules and Planck's law for cavity radiation are taught in different chapters, yet they share one construction: each is a product of the states available at a given energy multiplied by the fraction of those states actually occupied. This page traces those two factors—and their separate physical origins—through both distributions.

(1)

(2)

Both distributions at the same temperature (1000 K). Each curve is rescaled to its own peak so the two maxima line up—only the shapes are compared.

Here is the plan:

1. put both on one energy axis → 2. split each into two factors → 3. trace each factor to its physical cause

Question 1

2

One coordinate, x = E / k_BT

Write both distributions as functions of the dimensionless energy . Temperature only sets the scale — against x, the shape is the same at every T. But a change of variable is not a relabelling: the density picks up a Jacobian factor. Where does that factor come from, and which part of the shape does it carry across?

The rule: densities, not functions A probability is an area under a density, so the density transforms — not the function itself. therefore gives . The absolute value appears because v(E) is monotonic: density flows only one way between the two axes.

Molecule, v → E. With , the Jacobian is . The Maxwell prefactor becomes , and the two powers multiply: . Collecting constants gives , (3). The is not assumed — it is from times from the Jacobian.

Light, λ → E. With , . Equation (2) carries in the wavelength coordinate; in energy that is . The Jacobian then multiplies by , so , which normalises to , (5). The is the formula’s own power in the λ coordinate — the Jacobian merely turns it into .

Photon number density. The row is already written in E — there is no substitution, only the rescaling . It is itself a density of states (, slides 4–5) times a Bose occupation (, slides 6–8); slides 4–7 build it from scratch. Normalised, it is , (4).

Start from	The step	Result
Speed distribution , molecule (1)	v → E: ; Jacobian	(3) Molecular energy distribution
Wavelength spectrum , cavity (2)	λ → E: ; Jacobian	(5) Energy density of the radiation in the cavity
Cavity energy density , (5)	The cavity energy density is the photon number density times one photon’s energy . Dividing by gives — slides 4–7 build that .	(4) Single-photon energy distribution

g(x) and f_ph(x) are single-particle energy probability densities (molecule, photon) and are directly comparable. p(x) is the cavity radiation energy density — the same information weighted by photon energy: .

Question 2

3

One construction, two shapes

Both equations are the product of the same two factors. Only the particle differs — and just two of its properties matter here.

the molecule — g(x)

geometry × dispersion:
E = p²/2m in d = 3
(slides 4–5)

×

statistics: identical particles
in the dilute limit
(slides 6–8)

=

equation (3)

the photon — f_ph(x)

geometry × dispersion:
E = pc in d = 3
(slides 4–5)

×

statistics: bosons, μ = 0
(slides 6–8)

=

equation (4)

Two properties of the particle decide everything Its mass sets the dispersion relation E(p), which fixes the rising factor. Its identity (indistinguishable or not, integer or half-integer spin) fixes the falling factor. Change either property and the curve changes in a predictable way — slide 10 puts both controls in your hands.

Question 3

4

I. The rising factor: counting states

A quantum state occupies one cell of volume h³ in momentum space — h along each of the three momentum axes. So counting states up to a fixed energy E is geometry: draw the momentum sphere of radius p(E) set by the dispersion relation, then divide its volume by h³.

The momentum sphere of radius p has volume , and one state occupies , so the number of states up to E is . The density of states is this count per unit energy, . Since p is itself a function of E, differentiate through the chain rule: . The first factor, , is the area of the shell at radius p — the number of momentum directions that carry energy E; the second, , is how far that shell stretches in momentum for each unit of energy. Their product is boxed as (6).

energy E (same for both particles) 1.00

(6)

The two factors in (6) have separate jobs: p² counts the momentum directions on the shell, while dp/dE converts a unit of energy into the momentum interval over which those directions are spread. In d dimensions a ball of volume replaces the 3-D sphere, so and ; photons add a factor of 2 for their two independent polarisations. This count is still pure geometry; it becomes a power of E only when the dispersion relation fixes how p grows with E — the next slide.

Question 4

5

II. The dispersion relation fixes the power

The previous slide counted the cells inside the momentum sphere, but that count has no energy dependence on its own — it is just . It becomes a power of E the moment the dispersion relation tells us how p grows with E. Write it as : α = 2 for a massive particle () and α = 1 for a massless one (). Feeding that into (6) gives the power of D(E).

The cancellation Massive (α = 2): , so the shell area grows as while the shell thins as . The two factors in (6) oppose each other — area up, thickness down — and cancel to . Massless (α = 1): , so and is constant; with nothing thinning against the growing area, . At the same E the photon’s momentum sphere is far larger — more states available, and more at every higher energy.

(7)

Here is the power, step by step: gives and , so substituting into (6), . The exponent is the only thing the dispersion fixes; the mass m lives only in the constant C. The three panels show the whole chain: A the dispersion E(p) — a parabola for α = 2, a straight line for α = 1; B the momentum shell that one energy E cuts out, its radius set by ; C log D against log E, a straight line of slope (1/2 for the massive particle, 2 for the photon). In d dimensions replace 3 by d, — slide 10 lets you vary both d and α.

Question 5

6

I. The falling factor: the occupation number

How many particles sit in each state? This factor contains no geometry. It is decided by what happens to the wavefunction when two identical particles are exchanged — and by nothing else.

Wavefunction under exchange	Particles	May two share a state?
symmetric	bosons, integer spin (0, 1, 2 …)	yes — and the count favours it
antisymmetric	fermions, half-integer spin (½, &frac32; …)	no — the wavefunction would vanish
no symmetry required	classical, distinguishable	the question does not arise

Do not read the third row as “only labelled particles give e^−x”: the same exponential emerges from identical particles once the gas is dilute — slide 8 fixes when that happens and why.

Where the ±1 comes from One formula, three levels of “identical-ness”; only the sign of the 1 moves. A symmetric wavefunction prefers to share a state (occupation may exceed 1); an antisymmetric one forbids sharing (occupation stays below 1); particles with labels get no correction at all. For photons μ′ = 0 (slide 8), so the boson row applies exactly as written. These rows were postulated from symmetry alone — the next slide derives the very same three by counting arrangements, and the ±1 falls out by itself.

Question 6

7

II. Exchange symmetry, counted directly

Slide 6 postulated three occupation rules from wavefunction symmetry. Here we derive the very same three rows by counting: two particles, two states, and the number of arrangements decides which row applies.

Rule	Ways to put n particles into g states	Occupation after maximising ln W
distinguishable	gⁿ	e^−x
identical bosons	(n+g−1)! / [n!(g−1)!]	1/(e^x−1)
identical fermions	g! / [n!(g−n)!]	1/(e^x+1)

Maximising the logarithm of each count, holding total energy and particle number fixed, reproduces exactly the three rows of slide 6. The ±1 appears because identical particles change the ways to fill a state, not the ways to choose which particle goes in — that single distinction is the whole of quantum statistics. Counting has now fixed the occupation numbers. But air molecules are identical quantum particles, so why do they follow the classical row? The answer is dilution — next slide.

Question 7

8

III. Why a dilute gas is Maxwell after all

Counting fixed the quantum occupations, and molecules are identical quantum particles — so why does a gas follow the classical e^−x? Because quantum corrections matter only when states are crowded. When x − μ′ ≫ 1, both quantum occupations collapse onto e^μ′−x — the classical shape e^−x times the constant e^μ′.

chemical potential μ′ = μ/k_BT −3.0

molecules μ′ ≪ 0 — at 300 K and 1 atm, μ′ ≈ −16 (of order −15 to −20, deepening as the gas thins). The Bose curve and its exponential asymptote are indistinguishable: the classical shape e^−x.

photons μ ≡ 0 — photon number is not conserved (emission and absorption are free). The −1 never vanishes: Planck’s spectrum is the genuinely quantum curve.

Question 8

9

Multiplying the two factors

The count of available states multiplied by the number filled gives the curve itself. The peak sits where the rising and falling factors balance.

Where the peak sits At the peak the two factors balance: . Classically the peak is x* = s, the power that rises at low energy. The Bose −1 pulls the peak slightly inward (bunching at low x), the Fermi +1 pushes it outward. Molecule (s = ½): x* = 0.500 · photon number (s = 2): x* ≈ 1.594 · photon energy (s = 3): x* ≈ 2.821.

Question 9

10

One law, all particles

Everything so far reduces to three knobs: the dispersion exponent α, the dimension d, and the statistics. Turn them and every distribution in the family appears — Maxwell and Planck are just two of the points.

dispersion exponent α (E ∝ p^α) 2.0

dimension d

statistics

density of states — x^d/α−1

occupation — decided by statistics

their product — the distribution

The rising factor is pure geometry: , the shell of one energy in d dimensions. The falling factor is pure identity: which statistics the particle obeys. Nothing else enters — change any knob and the curve responds in the way slides 4–8 predicted.

Question 10

11

Summary

	molecule	photon	comes from
dispersion relation	E = p²/2m	E = pc	whether the particle has mass
states available	√x	x²	the shell that this dispersion defines, in d = 3 — in general
states filled	e^−x	1/(e^x−1)	the symmetry of the wavefunction under exchange — bosons, fermions, or dilute
why e^−x for molecules	dilute gas, μ′ ≪ 0: quantum forms merge into the exponential		occupation ≪ 1 per state
most probable energy	0.500	1.594	the photon has far more states at high energy

One construction gives both shapes: states available multiplied by states filled.
The power is a consequence of , not an assumption — it is in general.
The in the Bose factor is a consequence of exchange symmetry, not an assumption.
The molecule’s is the dilute limit of quantum statistics — the same identity, different crowding.

Questions 11 and 12

Appendix

Equations and symbols

The numbering used throughout the deck, and a link to the slide where each equation is used.

Eq.	What it gives	Slide
(1)	Maxwell speed distribution	1
(2)	Planck spectrum, per unit wavelength	1
(3)	Molecular energy distribution	2, 9
(4)	Single-photon energy distribution	2, 9
(5)	Energy density of the radiation in the cavity	2
(6)	The count of states	4
(7)	Power of the density of states	5, 10

Symbol	Meaning
x = E/k_BT	dimensionless energy; x = 1 means the energy equals k_BT
α	the exponent in the dispersion relation : 2 massive, 1 massless
d	spatial dimension; in general
μ′ = μ/k_BT	chemical potential in units of k_BT; 0 for photons, ≪ 0 for a dilute gas
g(x)	energy probability density of one molecule — equation (3)
f_ph(x)	energy probability density of one photon — equation (4)
p(x)	energy density of the radiation in the cavity — equation (5)

Question 12