%% CAS_F0029_VERIFY.m — Partition function linkage (Z → F, U, S, Cv) % Pillar : Thermodynamics % CalRef : Math Appendix §5A, Thermodynamics Calibration §6A % Chain : Z(β) → F = −lnZ/β → U = −∂lnZ/∂β → S = kB(lnZ+βU) → Cv % Hardening: isAlways(..., 'Unknown', 'false') on every symbolic assertion % symvar-based presence checks; solver and has-function avoided clear; clc; fprintf('\n=== CAS F0029 : Partition function linkage ===\n\n'); %% ── Symbols ────────────────────────────────────────────────── syms beta kB T real positive syms lnZ(beta) % ln Z as symfun of beta %% ── A. Inputs ──────────────────────────────────────────────── % β = 1/(kB T) % F = −(1/β)·ln Z = −kB T ln Z % U = −∂(ln Z)/∂β F_free = -lnZ / beta; U_int = -diff(lnZ, beta); %% ── Step 1: F = −(1/β)·ln Z structure ────────────────────── % Verify F = −kB T ln Z when β = 1/(kB T) → 1/β = kB T F_kBT = -kB * T * lnZ; % Substitute β = 1/(kB T) into F_free: F_sub = subs(F_free, beta, 1/(kB*T)); % Need lnZ(1/(kBT)) — compare structurally by checking coefficient % Instead: verify −lnZ/β = −kBT·lnZ when β = 1/(kBT) res1 = simplify(F_sub + kB*T*lnZ(1/(kB*T))); assert(isAlways(res1 == 0, 'Unknown', 'false'), ... 'FAIL Step 1: F ≠ −kBT·lnZ'); fprintf('Step 1 PASS — F = −(1/β)·lnZ = −kBT·lnZ\n'); %% ── Step 2: U = −∂lnZ/∂β ─────────────────────────────────── % This is the definition. Verify structure: U depends on derivative of lnZ. % For a concrete Z: Z = exp(−β·E₀) → lnZ = −β·E₀ → U = E₀ syms E0 real positive lnZ_concrete = -beta * E0; U_concrete = -diff(lnZ_concrete, beta); res2 = simplify(U_concrete - E0); assert(isAlways(res2 == 0, 'Unknown', 'false'), ... 'FAIL Step 2: single-state U ≠ E₀'); fprintf('Step 2 PASS — Single state: Z=e^{−βE₀} → U = E₀\n'); %% ── Step 3: F from concrete Z ────────────────────────────── % F = −lnZ/β = −(−βE₀)/β = E₀ F_concrete = -lnZ_concrete / beta; res3 = simplify(F_concrete - E0); assert(isAlways(res3 == 0, 'Unknown', 'false'), ... 'FAIL Step 3: single-state F ≠ E₀'); fprintf('Step 3 PASS — Single state: F = E₀ (= U, zero entropy)\n'); %% ── Step 4: S = kB(lnZ + βU) ─────────────────────────────── % Derive: S = −∂F/∂T = (U − F)/T % From F = −lnZ/β and U = −∂lnZ/∂β: % S = (U − F)/T = (−∂lnZ/∂β + lnZ/β) / T % With β = 1/(kBT) → T = 1/(kBβ): % S = kBβ · (−∂lnZ/∂β + lnZ/β) = kB(lnZ + βU) % Verify algebraically: S_formula = kB * (lnZ + beta * U_int); % Alternative: S = (U − F)/T with T = 1/(kBβ) S_thermo = (U_int - F_free) * kB * beta; % (U−F)/T = (U−F)·kBβ res4 = simplify(S_formula - S_thermo); assert(isAlways(res4 == 0, 'Unknown', 'false'), ... 'FAIL Step 4: S = kB(lnZ+βU) ≠ (U−F)/T'); fprintf('Step 4 PASS — S = kB(lnZ + βU) = (U−F)/T\n'); %% ── Step 5: S for single state = 0 ───────────────────────── % Single state: lnZ = −βE₀, U = E₀ % S = kB(−βE₀ + β·E₀) = kB·0 = 0 S_single = kB * (lnZ_concrete + beta * E0); res5 = simplify(S_single); assert(isAlways(res5 == 0, 'Unknown', 'false'), ... 'FAIL Step 5: single-state S ≠ 0'); fprintf('Step 5 PASS — Single state: S = 0 (no degeneracy → zero entropy)\n'); %% ── Step 6: Two-level system ──────────────────────────────── % E₁ = 0, E₂ = ε: Z = 1 + e^{−βε} % lnZ = ln(1 + e^{−βε}) % U = −∂lnZ/∂β = ε·e^{−βε}/(1 + e^{−βε}) syms epsilon real positive Z_two = 1 + exp(-beta*epsilon); lnZ_two = log(Z_two); U_two = simplify(-diff(lnZ_two, beta)); U_two_expected = epsilon * exp(-beta*epsilon) / (1 + exp(-beta*epsilon)); res6 = simplify(U_two - U_two_expected); assert(isAlways(res6 == 0, 'Unknown', 'false'), ... 'FAIL Step 6: two-level U mismatch'); fprintf('Step 6 PASS — Two-level: U = ε·e^{−βε}/(1+e^{−βε})\n'); %% ── Step 7: Two-level high-T limit ───────────────────────── % β→0: Z→2, U→ε/2, S→kB·ln2 % Check U(β→0): ε·1/(1+1) = ε/2 U_highT = limit(U_two, beta, 0); res7 = simplify(U_highT - epsilon/2); assert(isAlways(res7 == 0, 'Unknown', 'false'), ... 'FAIL Step 7: high-T U ≠ ε/2'); fprintf('Step 7 PASS — Two-level high-T: U → ε/2\n'); %% ── Step 8: F = U − TS consistency ───────────────────────── % Verify: F = U − TS → F − U + TS = 0 % Using F = −lnZ/β, S = kB(lnZ + βU): % F − U + T·S = −lnZ/β − U + (1/(kBβ))·kB(lnZ + βU) % = −lnZ/β − U + lnZ/β + U = 0 FmUTS = F_free - U_int + (1/(kB*beta)) * S_formula; res8 = simplify(FmUTS); assert(isAlways(res8 == 0, 'Unknown', 'false'), ... 'FAIL Step 8: F ≠ U − TS'); fprintf('Step 8 PASS — F = U − TS (Legendre consistency)\n'); %% ── Step 9: Cv = −kBβ² ∂U/∂β ─────────────────────────────── % Cv = ∂U/∂T = (∂U/∂β)(∂β/∂T) = (∂U/∂β)(−β²kB) [since ∂β/∂T = −1/(kBT²) = −kBβ²] % Wait: ∂β/∂T = −1/(kBT²) and β² = 1/(kBT)² → kBβ² = 1/(kBT²) → ∂β/∂T = −kBβ² % Actually: β = 1/(kBT) → ∂β/∂T = −1/(kBT²) = −β²·kB % So Cv = (∂U/∂β)·(−β²·kB) % Verify for two-level system: dU_dbeta_two = diff(U_two, beta); Cv_two = simplify(-kB * beta^2 * dU_dbeta_two); % Cv should be kBβ²ε² · e^{−βε}/(1+e^{−βε})² Cv_expected = kB * beta^2 * epsilon^2 * exp(-beta*epsilon) / (1 + exp(-beta*epsilon))^2; res9 = simplify(Cv_two - Cv_expected); assert(isAlways(res9 == 0, 'Unknown', 'false'), ... 'FAIL Step 9: two-level Cv mismatch'); fprintf('Step 9 PASS — Cv = kBβ²ε²·e^{−βε}/(1+e^{−βε})² (Schottky anomaly)\n'); %% ── Step 10: Numerical — two-level at kBT = ε ────────────── % β = 1/ε (setting kB=1 for numerical check) % U = ε·e⁻¹/(1+e⁻¹) ≈ ε·0.2689 % Cv = β²ε²·e⁻¹/(1+e⁻¹)² ≈ 0.1966 (in units of kB) e_inv = exp(-1); U_num = e_inv / (1 + e_inv); % in units of ε Cv_num = e_inv / (1 + e_inv)^2; % in units of kB U_expected_num = 0.2689; Cv_expected_num = 0.1966; % e^{-1}/(1+e^{-1})^2 at βε=1 assert(abs(U_num - U_expected_num) < 0.001, 'FAIL Step 10a: U numerical'); assert(abs(Cv_num - Cv_expected_num) < 0.001, 'FAIL Step 10b: Cv numerical'); fprintf('Step 10 PASS — kBT=ε: U/ε=%.4f, Cv/kB=%.4f (Schottky at βε=1)\n', U_num, Cv_num); %% ── Step 11: Self-test — wrong sign in F ──────────────────── % Wrong: F = +(1/β)·lnZ (positive instead of negative) F_wrong = lnZ / beta; % Then U−F+TS ≠ 0 S_from_wrong = kB * (lnZ + beta * U_int); % S formula unchanged FmUTS_wrong = F_wrong - U_int + (1/(kB*beta)) * S_from_wrong; res11 = simplify(FmUTS_wrong); assert(~isAlways(res11 == 0, 'Unknown', 'false'), ... 'FAIL Step 11: wrong F sign not detected'); fprintf('Step 11 PASS — Wrong F = +lnZ/β breaks Legendre relation\n'); %% ── Summary ───────────────────────────────────────────────── fprintf('\n✓ ALL 11 ASSERTIONS PASSED — F0029 audit complete.\n'); fprintf(' Chain: Z(β) → F=−lnZ/β → U=−∂lnZ/∂β → S=kB(lnZ+βU) → Cv\n'); fprintf(' Cross-ref: F0016 (entropy production), F0017 (Cp−Cv)\n');