Energy Fluctuations and Heat Capacity

Derivation

Energy Fluctuations and Heat Capacity

D-157 Home PU-203 Threads chance · energy Depends on The Partition Function and F = -kT ln Z

Statement

For a system in the canonical ensemble at temperature T, the second derivative of ln Z with respect to β = 1/kT equals the variance of the energy, and this variance is fixed by the constant-volume heat capacity: ⟨(ΔE)²⟩ = ⟨E²⟩ − ⟨E⟩² = ∂²ln Z/∂β² = kT²C_V.

Why it matters

This is the simplest genuine fluctuation–dissipation relation. The left side, ⟨(ΔE)²⟩, is pure microscopic noise: how much the instantaneous energy jitters as the system trades quanta with its bath. The right side, C_V, is a macroscopic response you measure with a calorimeter. That two such different quantities are equal (up to kT²) is the first sign that spontaneous fluctuations and driven response are the same physics viewed from two sides.

It also explains why thermodynamics works at all. Because C_V ∝ N while ⟨E⟩ ∝ N, the relative fluctuation ΔE/⟨E⟩ ∝ 1/√N vanishes for macroscopic N — the energy of a large body is effectively sharp, and the ensemble average becomes the observed value.

Assumptions

Canonical ensemble at fixed T.The system exchanges only energy with a large bath, so state i has probability p_i = e^−βE_i/Z. Drop it and the Boltzmann weights vanish, Z is undefined, and moments of E have no meaning.

Fixed N and V (only energy fluctuates).Particle number and volume are held constant so the only fluctuating extensive variable is E. Drop it and particle exchange contributes, forcing the grand-canonical ensemble with its own ⟨ΔN²⟩ terms.

Z is twice differentiable in β; the energy spectrum is bounded below and the moment sums converge.Interchanging ∂/∂β with the sum requires Σ E_i² e^−βE_i to converge. Drop it (e.g. an unbounded density of states rising too fast) and ⟨E²⟩ diverges; the variance is not finite and the identity is empty.

The heat capacity is taken at constant volume.C_V = (∂U/∂T)_V uses U = ⟨E⟩ with no mechanical work. Drop constant-V and you differentiate along a path that does p dV work, giving C_p, which is not what ∂²ln Z/∂β² returns.

Derivation

β ≡ 1/(kT), Z = Σ_i e^−βE_i, p_i = e^−βE_i/Z

Definitions from the canonical ensemble; Z is the partition function of the assumed prior result. A

∂ln Z/∂β = (1/Z) ∂Z/∂β = −(1/Z) Σ_i E_i e^−βE_i = −⟨E⟩

Chain rule on ln Z; each ∂/∂β of e^−βE_i brings down −E_i. This reproduces the prior result ⟨E⟩ = −∂ln Z/∂β. A

∂²ln Z/∂β² = ∂/∂β [ (1/Z) ∂Z/∂β ] = (1/Z) ∂²Z/∂β² − [ (1/Z) ∂Z/∂β ]²

Product/quotient rule: differentiating Z⁻¹Z′ gives Z⁻¹Z″ − (Z′/Z)². Purely algebraic; no ensemble input yet. B

∂²Z/∂β² = Σ_i E_i² e^−βE_i = Z⟨E²⟩ ⇒ (1/Z)∂²Z/∂β² = ⟨E²⟩, [(1/Z)∂Z/∂β]² = ⟨E⟩²

Two derivatives bring down E_i²; dividing by Z re-weights into the ensemble average. Legality needs the moment sum to converge (see assumptions). B

∂²ln Z/∂β² = ⟨E²⟩ − ⟨E⟩² ≡ ⟨(ΔE)²⟩ ≥ 0

This is the definition of the variance, ΔE ≡ E − ⟨E⟩. Non-negativity is automatic and forces C_V ≥ 0 below. B

⟨(ΔE)²⟩ = ∂²ln Z/∂β² = −∂⟨E⟩/∂β

Combine step 5 with step 2: differentiating ⟨E⟩ = −∂ln Z/∂β once more in β gives the variance. Symbols only, no numbers yet. A

β = 1/(kT) ⇒ dβ/dT = −1/(kT²) ⇒ ∂/∂β = (dT/dβ) ∂/∂T = −kT² (∂/∂T)

Change the independent variable from β to T by the chain rule; dT/dβ = −kT² is the reciprocal of dβ/dT. B

−∂⟨E⟩/∂β = −(−kT²) ∂⟨E⟩/∂T = kT² (∂U/∂T)_V = kT² C_V

Insert step 7, then identify U = ⟨E⟩ and C_V = (∂U/∂T)_V at fixed volume. A

∴ ⟨(ΔE)²⟩ = kT² C_V

Equate the two expressions for the variance from steps 6 and 8. A

Result

⟨(ΔE)²⟩ = ⟨E²⟩ − ⟨E⟩² = ∂²ln Z/∂β² = k_BT² C_V

Reading. The mean-square spread of a system's instantaneous energy about its average is set entirely by how much its average energy changes with temperature. A body that soaks up heat readily (large C_V) is also a body whose energy visibly trembles; a body that barely responds to temperature is one whose energy is nearly frozen. Since ⟨(ΔE)²⟩ ≥ 0 always, the relation also proves C_V ≥ 0: a stable system cannot have negative heat capacity.

Units check. [k_B] = J K⁻¹, [T²] = K², [C_V] = J K⁻¹, so [k_BT²C_V] = (J K⁻¹)(K²)(J K⁻¹) = J². The left side ⟨(ΔE)²⟩ is an energy squared, J². Both sides are J². ✓

Limiting cases

Monatomic ideal gas: U = (3/2)Nk_BT, C_V = (3/2)Nk_B, so ⟨(ΔE)²⟩ = (3/2)N k_B²T² and ΔE/⟨E⟩ = √(2/3N) — vanishingly small for macroscopic N.
Large N (thermodynamic limit): ΔE/⟨E⟩ ∝ N^−1/2 → 0; the energy becomes sharp and the canonical and microcanonical descriptions agree.
Approaching a critical point: C_V diverges, so ⟨(ΔE)²⟩ diverges — giant fluctuations, the microscopic origin of critical opalescence.
T → 0 (third law): C_V → 0, so ⟨(ΔE)²⟩ → 0 faster than T²; the ground state is occupied with certainty and the energy stops fluctuating.

Breaks when

The system is thermally isolated (microcanonical). With fixed energy, ⟨(ΔE)²⟩ = 0 by construction; there is no bath to exchange energy with, and C_V must be obtained from entropy curvature, not from energy variance.
Particle number fluctuates (open system). In the grand-canonical ensemble energy and number fluctuate together; the total energy variance picks up cross terms and no longer equals kT²C_V alone.
The moment sum diverges. If the density of states grows so fast that Σ E_i² e^−βE_i does not converge (e.g. certain gravitating or Hagedorn-type spectra), ⟨E²⟩ is infinite and the identity is meaningless.
Long-range/non-additive systems. Self-gravitating systems can show negative microcanonical heat capacity; there the canonical ensemble is not equivalent and ⟨(ΔE)²⟩ = kT²C_V cannot represent a negative C_V, since the left side is non-negative.

Failure modes

Sign slip in the change of variable. Forgetting that dT/dβ = −kT² (both a factor kT² and a minus) turns −∂⟨E⟩/∂β into −kT²C_V and produces a spurious negative variance.
Dropping the cross term. Writing ∂²ln Z/∂β² = ⟨E²⟩ instead of ⟨E²⟩ − ⟨E⟩² — i.e. forgetting the −(Z′/Z)² piece from the quotient rule — gives a raw second moment, not the variance.
Using C_p instead of C_V. The identity holds at constant volume; substituting the measured C_p for a gas overestimates the fluctuation by the factor C_p/C_V = γ.
Confusing ΔE (per-system RMS) with ⟨E⟩ (mean). Reporting √⟨(ΔE)²⟩ as "the energy" rather than the spread of the energy; the two differ by the factor √N.
Assuming C_V is constant when integrating. For real systems C_V(T) varies, so ⟨(ΔE)²⟩ is kT²C_V(T) evaluated at the actual T, not a temperature-independent number.

Discussion

The deep content of this result is that noise measures response. The variance ⟨(ΔE)²⟩ describes spontaneous, undriven fluctuations of an equilibrium system; C_V describes how the system responds when you drive it by raising the temperature. Their proportionality is the thermodynamic ancestor of the general fluctuation–dissipation theorem, of which the Johnson–Nyquist relation (voltage noise ∝ RT) and the Einstein relation (diffusion ∝ mobility × kT) are close cousins. In every case a susceptibility equals the strength of an equilibrium fluctuation divided by a power of kT.

Practically, the identity is run backwards constantly. In molecular-dynamics and Monte-Carlo simulations you never differentiate the energy against temperature to get the heat capacity — that would require many runs at different T. Instead you sit at one temperature, record the energy time series, and compute C_V = ⟨(ΔE)²⟩/(k_BT²) from a single trajectory's fluctuations. The same logic lets calorimetry of a nanoscale object be inferred from its energy jitter.

The non-negativity ⟨(ΔE)²⟩ ≥ 0 is a thermodynamic stability statement in disguise: C_V ≥ 0 means adding heat cannot lower the temperature of a canonical system. When people quote "negative heat capacity" for a star or a gravitationally bound cluster, they are working microcanonically, where energy is fixed and this canonical identity simply does not apply — a reminder that ensemble equivalence can fail for long-range forces.

More formally, ln Z is the cumulant generating function for E in the variable −β: its first derivative gives the mean, its second the variance, its third the skewness, and so on. Because ln Z is convex in β (its second derivative, a variance, is non-negative), the free energy F = −kT ln Z inherits convexity properties that guarantee thermodynamic stability. Phase transitions are precisely the points where this generating function loses analyticity in the thermodynamic limit, and the divergence of its second cumulant is what makes C_V and the fluctuations blow up together.

Common misconceptions. (i) "Bigger systems fluctuate less" — no, the absolute fluctuation √⟨(ΔE)²⟩ ∝ √N grows; it is the relative fluctuation ∝ 1/√N that shrinks. (ii) "The heat capacity causes the fluctuations" — neither causes the other; both are consequences of the same underlying energy distribution, and the equality is an identity, not a mechanism.

Worked examples

Example 1 — Energy fluctuations of one mole of argon (monatomic ideal gas).

C_V = (3/2) N k_B = (3/2) R, ⟨(ΔE)²⟩ = k_BT² C_V = (3/2) N k_B² T²

Equipartition gives U = (3/2)Nk_BT; substitute into the boxed result. Symbols first. A

N = N_A = 6.022×10²³, k_B = 1.381×10⁻²³ J K⁻¹, T = 300 K

State the numbers and units before computing. A

C_V = (3/2)(8.314) = 12.47 J K⁻¹

R = N_Ak_B = 8.314 J K⁻¹ mol⁻¹. A

⟨(ΔE)²⟩ = (1.381×10⁻²³)(300)²(12.47) = 1.55×10⁻¹⁷ J²

Multiply k_BT²C_V: (1.381×10⁻²³)(9.00×10⁴)(12.47). A

√⟨(ΔE)²⟩ = 3.94×10⁻⁹ J, U = (3/2)RT = 3741 J, ΔE/U = 1.05×10⁻¹²

RMS spread over the mean; compare with √(2/3N) = 1.05×10⁻¹². B

√⟨(ΔE)²⟩ ≈ 3.9×10⁻⁹ J, ΔE/U ≈ 1.0×10⁻¹²

Reading. The gas holds 3741 J, and that energy wanders by only about 4 nanojoules — one part in 10¹². For any macroscopic body the energy is, for all practical purposes, perfectly sharp; this is why a mole of gas has a well-defined internal energy at all.

Example 2 — Heat capacity of a two-level system from its fluctuations. A single defect has two states of energy 0 and ε = 1.00×10⁻²⁰ J at T = 300 K. Find ⟨(ΔE)²⟩ directly, then extract C.

Z = 1 + e^−βε, p ≡ P(E=ε) = 1/(e^βε+1), ⟨(ΔE)²⟩ = ε² p(1−p)

A Bernoulli energy takes value ε with probability p, so its variance is ε²p(1−p). Symbols first. B

βε = ε/(k_BT) = (1.00×10⁻²⁰)/[(1.381×10⁻²³)(300)] = 2.414

Dimensionless ratio of level gap to thermal energy. A

e^βε = 11.18 ⇒ p = 1/12.18 = 0.0821, 1−p = 0.9179

Boltzmann occupation of the upper level. A

⟨(ΔE)²⟩ = (1.00×10⁻²⁰)²(0.0821)(0.9179) = 7.53×10⁻⁴² J²

Insert numbers into ε²p(1−p). A

C = ⟨(ΔE)²⟩/(k_BT²) = (7.53×10⁻⁴²)/[(1.381×10⁻²³)(9.00×10⁴)] = 6.06×10⁻²⁴ J K⁻¹

Invert the boxed result to read the heat capacity off the fluctuation. B

Check: C = k_B(βε)² e^βε/(e^βε+1)² = k_B(5.827)(11.18)/(12.18)² = 0.439 k_B

The closed-form Schottky heat capacity; 0.439 k_B = 6.06×10⁻²⁴ J K⁻¹ — agrees. C

⟨(ΔE)²⟩ ≈ 7.5×10⁻⁴² J², C ≈ 6.1×10⁻²⁴ J K⁻¹ = 0.44 k_B

Reading. A lone two-level system has a heat capacity that peaks near k_BT ≈ ε — the Schottky anomaly — precisely because that is where the energy fluctuates most (occupation nearest 50/50). Here k_BT is below ε, so we sit on the low-T flank at 0.44 k_B.

Problems

(A) A monatomic ideal gas contains N = 2.0×10²² atoms at T = 350 K. Find C_V, ⟨(ΔE)²⟩, the RMS fluctuation, and the relative fluctuation.
Solution
C_V = (3/2)Nk_B = 1.5(2.0×10²²)(1.381×10⁻²³) = 0.414 J K⁻¹. ⟨(ΔE)²⟩ = k_BT²C_V = (1.381×10⁻²³)(350²)(0.414) = 7.00×10⁻¹⁹ J². RMS = 8.4×10⁻¹⁰ J. U = (3/2)Nk_BT = 145 J, so ΔE/U = 8.4×10⁻¹⁰/145 = 5.8×10⁻¹², matching √(2/3N) = √(1/3.0×10²²) = 5.8×10⁻¹².
(B) One mole of an ideal diatomic gas (5 active degrees of freedom) is at T = 300 K. Find ⟨(ΔE)²⟩ and the RMS energy fluctuation.
Solution
C_V = (5/2)R = 2.5(8.314) = 20.79 J K⁻¹. ⟨(ΔE)²⟩ = (1.381×10⁻²³)(300²)(20.79) = 2.58×10⁻¹⁷ J². RMS = √(2.58×10⁻¹⁷) = 5.08×10⁻⁹ J. (Compare U = (5/2)RT = 6236 J, giving ΔE/U = 8.1×10⁻¹³.)
(B) A two-level defect with gap ε = 2.0×10⁻²¹ J sits at T = 150 K. Compute βε, the upper-level occupation p, ⟨(ΔE)²⟩, and the heat capacity C.
Solution
βε = (2.0×10⁻²¹)/[(1.381×10⁻²³)(150)] = 0.966. e^βε = 2.627, p = 1/3.627 = 0.2757, 1−p = 0.7243. ⟨(ΔE)²⟩ = ε²p(1−p) = (2.0×10⁻²¹)²(0.2757)(0.7243) = 7.99×10⁻⁴³ J². C = ⟨(ΔE)²⟩/(k_BT²) = (7.99×10⁻⁴³)/[(1.381×10⁻²³)(2.25×10⁴)] = 2.57×10⁻²⁴ J K⁻¹ = 0.186 k_B. (Since βε ≈ 1 we are near the Schottky peak region.)
(C) Starting from ⟨(ΔE)²⟩ = kT²C_V, show that for a monatomic ideal gas the relative fluctuation is ΔE/⟨E⟩ = √(2/3N), and evaluate it for N = 1.0×10²⁰.
Solution
⟨E⟩ = (3/2)Nk_BT and C_V = (3/2)Nk_B, so ⟨(ΔE)²⟩ = kT²·(3/2)Nk_B = (3/2)Nk_B²T². Then ΔE/⟨E⟩ = √[(3/2)Nk_B²T²] / [(3/2)Nk_BT] = √[(3/2)N] / [(3/2)N] = √(2/(3N)). For N = 1.0×10²⁰: ΔE/⟨E⟩ = √(2/3.0×10²⁰) = 8.2×10⁻¹¹ — about one part in 10¹⁰.
(C) A calorimetric sample has C_V = 2.00 J K⁻¹ at T = 300 K. (a) Find the RMS energy fluctuation. (b) A colleague claims a nanoscale flake of the same material, with C_V = 3.0×10⁻²¹ J K⁻¹, should show a relatively larger fluctuation. Quantify by comparing √⟨(ΔE)²⟩/(C_VT) (fluctuation as a fraction of a rough energy scale C_VT) for both.
Solution
(a) ⟨(ΔE)²⟩ = k_BT²C_V = (1.381×10⁻²³)(9.00×10⁴)(2.00) = 2.49×10⁻¹⁸ J²; RMS = 1.58×10⁻⁹ J. Fraction = 1.58×10⁻⁹/(2.00×300) = 2.6×10⁻¹². (b) Flake: ⟨(ΔE)²⟩ = (1.381×10⁻²³)(9.00×10⁴)(3.0×10⁻²¹) = 3.73×10⁻³⁹ J²; RMS = 6.1×10⁻²⁰ J. Fraction = 6.1×10⁻²⁰/(3.0×10⁻²¹×300) = 0.068. The relative fluctuation is ≈ 2.6×10¹⁰ times larger for the flake — the colleague is right: √⟨(ΔE)²⟩/(C_VT) ∝ 1/√C_V ∝ 1/√N, so tiny systems are dominated by their thermal noise.