THE EQUIPARTITION THEOREM

At equilibrium, energy does not play favourites. Every square-law way a system can store it — each translation, each spring — is handed the same ration: half a kT. Count the quadratic degrees of freedom and you have the energy. Rendered, not quoted.

source J. C. Maxwell, “On Boltzmann’s Theorem on the average distribution of energy in a system of material points,” Transactions of the Cambridge Philosophical Society, vol. 12 (1879), pp. 547–570 — developed with L. Boltzmann. Scientific Papers, vol. 2 (archive.org) · no single stable article DOI — AMBER.

Blue Team · builds & defends

3 THE MODEL

Write the energy as a sum of independent square-law terms, H = Σ cₓ qₓ². Weight each configuration by Boltzmann, e⁻ᴸᵀᵀ. The generalized equipartition theorem says ⟨q ∂H/∂q⟩ = kT.

For a purely quadratic term c·q² that forces ⟨c q²⟩ = ½kT — the constant c and the mass drop out. The share depends only on the power, not the stiffness.

General law: a term ∝|q|ⁿ carries kT/n. Quadratic (n=2) → ½kT. That is the whole engine.

5 THE LINEAGE

Energy shared equally among the modes — this is why the-ideal-gas carries &frac32;kT per atom: three translational (quadratic) modes, each ½kT.

Add rotations and vibrations and the same accounting sets every classical heat capacity: Cᵛ = (f/2) N k for f quadratic freedoms. Monatomic → 3/2 R; rigid diatomic → 5/2 R. The counting IS the physics.

7 THE WITNESS

A live re-check, independent of the boot selfcheck. It recomputes the mode shares now and confirms the quartic mode reads T/4, the quadratic T/2. Trip the tamper (window 6) and this flips red.

WITNESS · initialising…
The Machine

4 DATA IN in ↓

A temperature T and a list of degrees of freedom, each labelled by the power of its coordinate in the energy. Natural units: k = 1 (Boltzmann constant set to one), so energies read directly in units of T.

↓ ↓ ↓

0 THE PANEL LIT

Pick a system and a temperature. The panel counts each mode’s power and awards it kT/n live — verified against a direct numerical average over the Boltzmann distribution.

modepower nshare kT/nenergy
↓ ↓ ↓

8 DATA OUT out ↓

A total mean energy, and the proven law behind it: ½kT per quadratic mode — kT per oscillator, &frac32;kT per monatomic atom, and kT/4 for a quartic (anharmonic) mode.

booting…
Red Team · attacks & breaks

1 THE ADVERSARY WALL

Equipartition is classical, and it lies at low T. It assumes every mode is continuously excitable — that kT dwarfs the quantum level spacing.

When kT ≪ ħω a mode freezes out: it stops taking its share. Vibrations are frozen at room temperature; solid heat capacities fall to zero as T→0 (Debye), not to 3R. Push equipartition to a field with infinite modes and it predicts the ultraviolet catastrophe. WALL: the classical high-T limit is assumed, not universal.

2 THE GRAVEYARD

  • Every degree of freedom gets ½kT. only quadratic ones do; a quartic term gets kT/4, a linear one kT.
  • A solid’s molar heat capacity is 3R at all temperatures. it collapses toward 0 as T→0 — modes freeze (Einstein/Debye).
  • A diatomic gas always carries 7/2·kT. vibration is frozen near 300 K, so Cᵛ sits at 5/2 R.
  • Heavier atoms store more thermal energy per mode. the share is mass-independent — c and m cancel in the Gaussian integral.

6 THE TAMPER

Disclosed planted void. Press below to mis-assign ½kT to the quartic mode — treating a term ∝x⁴ as if it were ∝x². Its true share is kT/4. The Witness (window 7) recomputes and catches the lie.