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Derivation

Boltzmann Entropy and Statistical Temperature

D-155 Home PU-203 Threads chance · energy Depends on The Clausius Inequality and Entropy as a State Function, stirling-approximation
Statement

For an isolated composite of two macroscopic systems exchanging energy at fixed total U = U1 + U2, the equilibrium energy partition is the one that maximises the total number of accessible microstates Ω = Ω1Ω2. Requiring this maximum forces the entropy to take the form S = k ln Ω (additive, monotonic in Ω) and identifies the statistical temperature through 1/T = (∂S/∂U)V,N, reproducing the thermodynamic entropy of the Clausius relation dS = δQ/T.

Why it matters

This is the bridge between the microscopic bookkeeping of statistical mechanics and the macroscopic quantities of thermodynamics. It explains why entropy exists as a state function, why it is additive, and why heat flows from hot to cold: equilibrium is simply the overwhelmingly most probable macrostate, and temperature is the parameter that governs how sharply the microstate count rises with energy.

It also fixes Boltzmann's constant k as the sole free multiplier, tying the abstract logarithm of a state count to units of joules per kelvin. Everything from the ideal-gas law to the Sackur–Tetrode equation and the Fermi–Dirac distribution descends from the two results proved here.

Assumptions
The composite is isolated and energy is the only exchanged extensive quantity (fixed V, N for each part).If particles or volume also flow, additional Lagrange conditions appear and equilibrium equalises chemical potential μ/T and pressure P/T, not temperature alone; the single equation dS = dU/T is incomplete. The systems are macroscopic, so Ω(U) is astronomically large and sharply peaked.If the systems have few degrees of freedom, the "most probable" partition is not overwhelmingly likely; fluctuations are large and temperature is not sharply defined. Weak coupling: the interaction energy is negligible compared with U1 and U2, so states factorise as Ω = Ω1Ω2.If coupling energy is not negligible (small systems, long-range forces), Ω does not factorise, entropy loses strict additivity, and 1/T can no longer be assigned to each subsystem independently. The microcanonical postulate: all accessible microstates at fixed energy are equally probable.If a priori probabilities are unequal (non-ergodic, glassy, or externally driven systems), Ω is not the correct weight and S = k ln Ω must be replaced by the Gibbs form S = −k Σ pi ln pi.
Derivation
1
Ω(U) = Ω1(U1) · Ω2(U2),   U1 + U2 = U = const
Under weak coupling the two systems are statistically independent at fixed energies, so accessible states multiply; the isolation constraint links the two energies. A
2
Equilibrium ⇔ Ω(U1) is maximal  ⇒  dΩ/dU1 = 0
The macrostate realised by the most microstates is overwhelmingly the most probable; with Ω sharply peaked the equilibrium partition sits at the maximum of the count. A
3
d/dU1 [Ω1(U1) Ω2(U − U1)] = Ω2 (dΩ1/dU1) + Ω1 (dΩ2/dU2)(dU2/dU1) = 0
Product rule, with the chain rule on Ω2 and dU2/dU1 = −1 from the constraint of Step 1. B
4
Ω2 (∂Ω1/∂U1) = Ω1 (∂Ω2/∂U2)
Rearranged the vanishing derivative, moving the second term across and cancelling the sign. Symbols only, no numbers yet. B
5
(1/Ω1)(∂Ω1/∂U1) = (1/Ω2)(∂Ω2/∂U2)  ⇔  (∂ ln Ω1/∂U1) = (∂ ln Ω2/∂U2)
Divide both sides by Ω1Ω2; each side is the logarithmic derivative of the microstate count, since (∂ ln Ω)/∂U = (1/Ω)(∂Ω/∂U). B
6
Define β ≡ (∂ ln Ω/∂U)V,N.   Equilibrium ⇒ β1 = β2
Step 5 says the quantity β is equal across two systems in thermal equilibrium — exactly the defining property of empirical temperature (zeroth law). β must therefore be a function of T alone. B
7
Require entropy additive: S = S1 + S2 while Ω = Ω1Ω2
Thermodynamic entropy is extensive; a product law for Ω plus an additive law for S forces S to depend on Ω through a function f obeying f(Ω1Ω2) = f(Ω1) + f(Ω2). C
8
f(x y) = f(x) + f(y)  ⇒  f(Ω) = k ln Ω
The only (continuous) solution of the Cauchy logarithmic functional equation is a constant times the logarithm; differentiate f(xy) w.r.t. x and set y=1 to get x f′(x) = f′(1) = k, then integrate. C
9
∂S/∂U = k (∂ ln Ω/∂U) = k β
Differentiate S = k ln Ω at fixed V, N using the β of Step 6. A
10
Clausius (prior result): dS = δQ/T = dU/T  (at fixed V, N)  ⇒  (∂S/∂U)V,N = 1/T
From the entropy of the Clausius inequality, with no work done (dV = 0) the first law gives δQ = dU. This is the thermodynamic definition of temperature. A
11
k β = 1/T  ⇒  β = 1/(k T)
Equate Step 9 with Step 10. This fixes the statistical β and pins k as the constant that converts a dimensionless log-count into thermodynamic entropy. A
Result
S = k ln Ω   and   1/T = (∂S/∂U)V,N = k (∂ ln Ω/∂U)V,N

Reading. Entropy is k times the logarithm of the number of microstates consistent with the macrostate. Two bodies are in equilibrium when the logarithmic slope of their microstate counts with energy is equal; that common slope is 1/(k T). A system whose state count rises steeply with added energy is "cold" (large β, small T) and greedily absorbs heat; one whose count is already flattening is "hot" and readily gives energy up. Heat therefore flows from hot to cold because that raises the total log-count.

Units check. Ω is a pure number, so ln Ω is dimensionless and [S] = [k] = J K⁻¹. For 1/T: [k ∂lnΩ/∂U] = (J K⁻¹)·(1/J) = K⁻¹ = [1/T]. ✓   With k = 1.380649×10⁻²³ J K⁻¹.

Limiting cases
  • Single microstate (Ω = 1): S = k ln 1 = 0 — the third-law ground state; a perfectly ordered non-degenerate system has zero entropy.
  • High temperature (β → 0): ln Ω is nearly flat in U; adding energy barely changes the accessible state count, so heat capacity governs how T responds.
  • Two identical systems: β1 = β2 is met with U1 = U2 = U/2 by symmetry — the intuitive "equal split" of energy.
  • Negative temperature (bounded spectrum): if Ω decreases with U near the top of a bounded energy band, ∂lnΩ/∂U < 0 and T < 0 — hotter than any positive temperature.
  • Classical continuum: Ω becomes a phase-space volume divided by h3NN! (indistinguishability), which cancels the Gibbs paradox and makes S extensive.
Breaks when
  • Small systems / strong coupling: when interaction energy is comparable to U1, U2, states no longer factorise (Step 1 fails), Ω ≠ Ω1Ω2, and entropy stops being additive — a single temperature cannot be assigned to each part.
  • Non-equilibrium / non-ergodic systems: glasses, driven systems, and systems that do not explore all accessible states violate the equal-a-priori-probability postulate, so ln Ω is not the correct entropy and the microcanonical β loses meaning.
  • Few degrees of freedom: for handfuls of particles Ω is not sharply peaked; the "most probable partition" is only weakly preferred, fluctuations dominate, and (∂S/∂U) does not define a stable temperature.
  • Long-range / gravitating systems: energy is non-additive and heat capacity can be negative; the maximisation of Ω1Ω2 need not have a stable interior maximum, so the derivation's equilibrium condition breaks.
Failure modes
  • Adding entropies of dependent systems. Writing S = S1 + S2 when the subsystems are correlated (strong coupling); additivity requires Ω = Ω1Ω2.
  • Maximising Ω instead of ln Ω. These share the same maximum, but students then try to add Ω rather than S, breaking extensivity.
  • Dropping N! (Gibbs paradox). Forgetting the indistinguishability factor gives a non-extensive entropy and a spurious entropy of mixing for identical gases.
  • Confusing β and T. Treating β as "the temperature" and forgetting the factor k; β has units J⁻¹, T has units K.
  • Assuming T > 0 always. Insisting Ω increases with U; for bounded spectra (spin systems) it can decrease, giving legitimate T < 0.
  • Holding the wrong variables fixed. Computing (∂S/∂U) without fixing V, N, so that δQ = dU no longer holds and the identification with 1/T is invalid.
Discussion

The derivation reframes the second law as a statement of probability. Equilibrium is not a state the system is forced into by a law of motion; it is simply the macrostate compatible with the vast majority of microstates. Because Ω for a mole of gas is of order e10²³, the peak in Ω(U1) is so sharp that deviations from the most-probable partition are never observed. The "arrow of time" is the overwhelming statistical drift toward larger ln Ω.

The logarithm is not a convenience but a necessity. Independent systems multiply their state counts while their entropies must add — only the logarithm turns products into sums (Steps 7–8). This is why the additivity of entropy and the multiplicativity of probability are two faces of the same fact, and why k is the unique dimensional bridge between them. Setting k = 1 recovers "natural" entropy as pure information (nats); the Boltzmann constant is a historical artifact of defining temperature in kelvin before its microscopic meaning was known.

The identity 1/T = ∂S/∂U also inverts the usual intuition about temperature. Temperature is not "how much energy" a system holds but "how reluctant it is to accept more" — the marginal entropy per unit energy. This is why a small hot object can heat a large cold one only until their slopes match, not until their energies match, and it is the seed of the canonical ensemble, where the Boltzmann factor e−E/kT arises from Taylor-expanding ln Ωreservoir(U − E) to first order in the small system's energy.

At the deepest level, the microcanonical S = k ln Ω is a special case of the Gibbs–Shannon entropy S = −k Σ pi ln pi evaluated on the uniform distribution pi = 1/Ω, which is itself the distribution that maximises S subject only to normalisation. The canonical and grand-canonical ensembles follow by adding constraints ⟨E⟩ and ⟨N⟩ with Lagrange multipliers β and −βμ. In this Jaynesian view, β = 1/(kT) is precisely the Lagrange multiplier conjugate to energy, and the entire structure of thermodynamics is the calculus of constrained entropy maximisation. The equality of β across systems (Step 6) is then the statement that multipliers conjugate to a shared, exchangeable quantity equalise at the joint maximum.

Common misconceptions. Entropy is not "disorder" in any everyday visual sense — a crystal can have high entropy if its vibrational state count is large. Entropy measures the number of microscopic arrangements, and a "tidy-looking" macrostate can be the overwhelmingly probable one. Likewise, the second law is not a strict prohibition: entropy can momentarily decrease, but for macroscopic N the probability is e−(order N), indistinguishable from zero.

Worked examples

Example 1 — Statistical temperature of a two-state paramagnet.

1
N non-interacting spins in field B; energy U = (n↑ − n↓)(−μB) with n↑ + n↓ = N. Ω = N! / (n↑! n↓!)
Microstate count is the number of ways to choose which spins are up (binomial). A
2
ln Ω ≈ N ln N − n↑ ln n↑ − n↓ ln n↓
Stirling's approximation (prior result), ln x! ≈ x ln x − x; the −x terms cancel by n↑+n↓=N. B
3
1/T = k (∂lnΩ/∂U) = k/(2μB) · ln(n↓/n↑)
Chain rule: dU = −2μB dn↑; differentiate the Stirling form and simplify. Symbols before numbers. C
4
N = 10²⁰, μB = 9.27×10⁻²⁴ J, take n↑ = 0.4N, n↓ = 0.6N
Insert numbers only now; a majority in the higher-energy (↑, since −μB lower for aligned) — here ↑ is the excited fraction. A
5
1/T = (1.381×10⁻²³)/(2·9.27×10⁻²⁴) · ln(0.6/0.4) = 0.745 · 0.405 = 0.302 K⁻¹
Evaluate the prefactor k/(2μB) = 0.745 K⁻¹ and multiply by ln 1.5 = 0.405. A
T = 1/0.302 ≈ 3.3 K

Reading. A 40:60 population imbalance corresponds to a positive temperature of a few kelvin. Push toward 50:50 and 1/T → 0 (infinite T); invert the majority (n↑ > n↓) and 1/T < 0 — a genuine negative temperature.

Units check. [k/μB] = (J K⁻¹)/J = K⁻¹; ln(ratio) dimensionless, so 1/T in K⁻¹. ✓

Example 2 — Heat flow direction from log-slope, Einstein solid.

1
Two Einstein solids, N oscillators each; Ω(q) = (q+N−1)! / [q! (N−1)!], energy U = q ħω
Number of ways to distribute q energy quanta among N oscillators (stars and bars). A
2
ln Ω ≈ N[(1+q/N) ln(1+q/N) − (q/N) ln(q/N)]  (q,N ≫ 1)
Stirling on all three factorials, grouping into the standard high-temperature Einstein-solid entropy. C
3
1/T = k ∂lnΩ/∂U = (k/ħω) ln(1 + N/q)
Differentiate ln Ω w.r.t. q (the ln terms combine neatly), then dU = ħω dq. Symbols first. C
4
Solid A: N=3000, qA=3000. Solid B: N=3000, qB=1000. ħω = 4.0×10⁻²¹ J
Choose numbers; A carries more energy per oscillator so should be hotter. A
5
TA = ħω/[k ln(1+3000/3000)] = 4.0×10⁻²¹/(1.381×10⁻²³·0.693) = 418 K;   TB = 4.0×10⁻²¹/(1.381×10⁻²³·ln 4) = 209 K
ln 2 = 0.693, ln 4 = 1.386; evaluate each temperature. A
TA ≈ 418 K > TB ≈ 209 K ⇒ heat flows A → B

Reading. B has the steeper log-slope (larger 1/T), so moving a quantum from A to B raises ln ΩB more than it lowers ln ΩA; total entropy increases until the slopes equalise at qA = qB = 2000, T ≈ 289 K.

Units check. [ħω/(k·dimensionless)] = J/(J K⁻¹) = K. ✓

Problems
  1. A system has Ω(U) = C UαN for constants C, α. Show that U = αN k T and hence identify the heat capacity.
    Solution S = k ln Ω = k(ln C + αN ln U). Then 1/T = ∂S/∂U = αNk/U, so U = αNkT. Heat capacity CV = ∂U/∂T = αNk. For a monatomic ideal gas α = 3/2, giving CV = (3/2)Nk — equipartition recovered.
  2. Two systems with heat capacities C1, C2 (constant) start at T1, T2 and reach a common final temperature. Find Tf and confirm total entropy rises.
    Solution Energy conservation: C1(Tf−T1) + C2(Tf−T2) = 0 ⇒ Tf = (C1T1 + C2T2)/(C1+C2). ΔS = C1 ln(Tf/T1) + C2 ln(Tf/T2). Since Tf is a weighted arithmetic mean and ln is concave, ΔS ≥ 0, with equality only if T1 = T2. E.g. C1=C2=C, T1=300, T2=400: Tf=350, ΔS=C ln(350²/(300·400)) = C ln(1.0208) = 0.0206C > 0.
  3. For the two-state paramagnet, show that T < 0 requires U > 0 (more spins in the high-energy state) and explain why a negative-temperature system is "hotter" than any positive-temperature one.
    Solution From Example 1, 1/T = (k/2μB) ln(n↓/n↑) where ↑ is the excited state. T < 0 needs the log negative, i.e. n↑ > n↓ (population inversion), which means U = (n↑−n↓)μB > 0. Placed in contact with any T > 0 body, the inverted system releases energy (lowering its ln Ω toward the 50:50 maximum from above), so energy always flows out of it — it is hotter than +∞ K. On the 1/T axis, negative temperatures sit above +∞.
  4. A reservoir at temperature T is in contact with a small system that can occupy states of energy Ei. By Taylor-expanding ln Ωres(U−Ei) to first order, derive the Boltzmann factor.
    Solution Probability pi ∝ Ωres(U−Ei) = exp[ln Ωres(U−Ei)]. Expand: ln Ωres(U−Ei) ≈ ln Ωres(U) − Ei (∂lnΩres/∂U) = const − βEi, with β = 1/(kT). Hence pi ∝ e−Ei/kT. Higher-order terms are suppressed because the reservoir is large (∂²lnΩ/∂U² ~ 1/(CVT) → 0). The statistical temperature β is exactly the expansion coefficient.
  5. Estimate the probability that 1 mole of gas, in a box, spontaneously has all molecules in the left half. Relate to entropy.
    Solution Each of N = 6.02×10²³ molecules independently in the left half with probability ½: P = (½)N = e−N ln 2. The entropy change for the gas to occupy half the volume is ΔS = Nk ln(½) = −Nk ln 2, and indeed P = eΔS/k = Ωhalf/Ωfull. Numerically ln P = −4.2×10²³, so P ~ 10−1.8×10²³ — utterly unobservable. This quantifies why the second law, though only statistical, is effectively absolute for macroscopic N.