Free energy from partition function: Exercises — Hinglish deep dive

2.4.12 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Free energy from partition function

Definition Glossary — is page pe use hone wale har symbol ka matlab

$Z$ = partition function: ek akela number jo har microstate ka Boltzmann weight sum karta hai. "Kitne tarike, properly weighted, system apni energy hold kar sakta hai."
$E_{i}$ = microstate $i$ ki energy (system ki ek specific configuration).
$ε$ = ek two-level system ki excitation energy: upper state ki energy cost ground state ke relative (jo $0$ pe baith ta hai).
$T$ = absolute temperature (kelvin mein); system ke aas-paas heat bath kitna garam hai.
$k_{B}$ = Boltzmann's constant $\approx 1.38 \times 1 0^{- 23} J/K$ — temperature aur energy ke beech conversion factor.
$β$ = inverse temperature, shorthand $β \equiv \frac{1}{k _{B} T}$ ke liye. Bada $β$ = thanda, chhota $β$ = garam.
$e^{- β E_{i}}$ = kisi state ka Boltzmann weight: $0$ aur $1$ ke beech ek number jo jitna zyada energy state cost kare utna chhota ho jaata hai.
$p_{i} = e^{- β E_{i}} / Z$ = system ke microstate $i$ mein hone ki probability.
$U = ⟨ E ⟩$ = internal energy, sabhi states pe average energy.
$S$ = entropy, ek measure ki kitne states effectively kaam mein hain ("messiness").
$C$ = heat capacity $= \partial U / \partial T$ : temperature ek degree badhne pe absorb ki gayi energy.
$F = U - T S$ = Helmholtz free energy; fixed $T, V, N$ pe minimize hone wala potential.
$V$ = volume, $N$ = particles ki sankhya, $p$ = pressure, $μ$ = chemical potential (ek particle add karne ki energy cost).
$Z$ = grand partition function; $Φ = - k_{B} T ln Z$ = grand potential; $⟨ n ⟩$ = ek site pe average particle number.
$h$ = Planck's constant; $ℏ \equiv h /2 π$ ("h-bar"). Yeh phase space mein ek cell ka size set karte hain taaki states count ho sakein.
$ω$ = ek oscillator ki angular frequency; $m$ = mass; $x, p$ = position aur momentum; $λ$ = thermal wavelength.

Shuru karne se pehle, ek figure us picture ko fix karta hai jis par hum baar baar laute hain: two-level system, hamara workhorse. Isse neeche Figure 1 label kiya gaya hai aur solutions mein usi naam se refer kiya jaata hai.

Figure — Free energy from partition function

Figure 1 — The two-level system.

Do horizontal levels dekho: energy $0$ (ground) aur energy $ε$ (excited). Kam temperature pe system neeche baithta hai; jaise-jaise heat daali jaati hai, upper level bharne lagta hai. Neeche ka har exercise ya toh yahi picture hai, iska copy hai, ya ek smooth (oscillator/gas) analogue hai.

Level 1 — Recognition

Goal: kya aap sahi formula shelf se utha ke numbers plug in kar sakte ho?

Recall Solution

KYA: master bridge $F = - k_{B} T ln Z$ directly apply karo. KYUN: kuch derive nahi karna — $F$ define hi hota hai $- k_{B} T ln Z$ se jab $Z$ mil jaaye. $F = - k_{B} T ln 4 = - k_{B} T (1.3863) \approx - 1.386 k_{B} T .$ Sign negative hai kyunki $Z > 1$ matlab $ln Z > 0$ hai, aur aage ek minus hai. Bahut accessible states wale system ki free energy low (bahut negative) hoti hai — nature ise pasand karta hai.

Recall Solution

KYA: do states pe Boltzmann weights $e^{- β E_{i}}$ ka sum karo. KYUN: definition $Z = \sum_{i} e^{- β E_{i}}$ — har state ek exponential contribute karta hai. $Z = e^{- β \cdot 0} + e^{- β ε} = 1 + e^{- β ε} .$ $β ε = 1$ pe: $Z = 1 + e^{- 1} = 1 + 0.3679 = 1.3679$ . Picture (Figure 1): " $1$ " ground level hai jo hamesha full weight se count hota hai; $e^{- β ε}$ excited level ka shrunk weight hai — shrunk isliye kyunki isme energy $ε$ lagti hai.

Level 2 — Application

Goal: $ln Z$ differentiate karo aur real thermodynamics nikalo.

Recall Solution

KYA: $ln Z$ ka $β$ -derivative lo. KYUN tool: $U$ average energy $\sum_{i} p_{i} E_{i}$ hai; parent note ne dikhaya ki yeh weighted sum $- \partial_{β} ln Z$ ke barabar hai — derivative har exponent se ek $E_{i}$ "pull down" karta hai aur $Z$ se division (jo $ln$ ke andar chhupa hai) averaging automatically kar deta hai. $ln Z = ln (1 + e^{- β ε}), \frac{\partial l n Z}{\partial β} = \frac{- ε e ^{- β ε}}{1 + e ^{- β ε}} .$ $U = - \frac{\partial l n Z}{\partial β} = \frac{ε e ^{- β ε}}{1 + e ^{- β ε}} = \frac{ε}{e ^{β ε} + 1} .$ $β ε = 1$ pe: $U / ε = 1/ (e^{1} + 1) = 1/3.7183 = 0.2689$ . Cases mein sanity: $β \to \infty$ (thanda) $\Rightarrow U \to 0$ (ground state). $β \to 0$ (garam) $\Rightarrow U \to ε /2$ (dono states equally likely). Hamari value $0.269 ε$ beech mein hai, jaisa hona chahiye.

Recall Solution

KYA: parent derivation se assembled relation $S = k_{B} β U + k_{B} ln Z$ use karo. KYUN: yeh $U$ (jo pehle se mila) aur $Z$ (jo pehle se mila) reuse karne deta hai — koi nayi sums nahi.

$β U = β \cdot \frac{ε}{e ^{β ε} + 1} = \frac{β ε}{e ^{β ε} + 1} = \frac{1}{3.7183} = 0.2689$ .
$ln Z = ln 1.3679 = 0.3133$ . $S / k_{B} = 0.2689 + 0.3133 = 0.5822.$ Cross-check: ek 2-state system ki maximum possible entropy $ln 2 = 0.693$ hai (dono states equally likely). Hum usse neeche hain kyunki finite $T$ pe system abhi bhi thoda ground state favour karta hai. ✓

Level 3 — Analysis

Goal: extensivity, limits, aur systems ko combine karna.

Recall Solution

KYA: single-system partition functions multiply karo, phir log lo. KYUN tool: independent systems ke exponent mein energies additive hoti hain, aur exponent mein add karne ka matlab hai exponentials ko multiply karna — isliye $Z_{tot} = Z_{1}^{N}$ . Factorisation kyun hold karta hai iske liye Boltzmann distribution and microstates dekho. $Z_{tot} = Z_{1}^{N} = (1 + e^{- β ε})^{N} .$ $F = - k_{B} T ln Z_{1}^{N} = - N k_{B} T ln (1 + e^{- β ε}) .$ $β ε = 1$ pe: $F / (N k_{B} T) = - ln (1.3679) = - 0.3133$ . Lesson (extensivity): $F \propto N$ . Logarithm ne product $Z_{1}^{N}$ ko sum $N ln Z_{1}$ mein badal diya. Yahi reason hai ki free energy — $Z$ itself nahi — natural thermodynamic potential hai. Helmholtz vs Gibbs free energy dekho.

Recall Solution

KYA: har temperature extreme mein $ln (1 + e^{- β ε})$ examine karo. KYUN: limits reveal karti hain ki formula physically behave kar raha hai ya nahi — jo formula $0$ ya $\infty$ par misbehave kare woh suspect hai.

$T \to 0$ ( $β \to \infty$ ): $e^{- β ε} \to 0$ , toh $ln (1 + 0) = 0$ , deta hai $F \to 0$ . Physically system ground state mein freeze ho gaya hai (energy $0$ , entropy $0$ ), isliye $F = U - T S \to 0$ . ✓
$T \to \infty$ ( $β \to 0$ ): $e^{- β ε} \to 1$ , toh $ln 2$ , deta hai $F \to - k_{B} T ln 2 \to - \infty$ . Entropy term $- T S$ dominate karta hai ( $S \to k_{B} ln 2$ ), aur $F$ neeche girta jaata hai. ✓ Figure 2 mein $F / ε$ , $U / ε$ , aur $T S / ε$ ko $k_{B} T / ε$ ke against plot kiya gaya hai taaki aap dekh sako ki $F = U - T S$ $T$ badhne pe neeche kheenchta jaata hai.

Figure 2 — $F$ , $U$ aur $T S$ temperature ke against, ek two-level system ke liye.

Level 4 — Synthesis

Goal: ek continuous system ke liye puri thermodynamics derive karo.

Recall Solution

KYA: do Gaussian integrals karo, phir $ln Z$ differentiate karo. KYUN Gaussian tool: ek quadratic Hamiltonian ka Boltzmann weight $e^{- β (\dots)}$ $p$ aur $x$ mein Gaussian hai; standard result $\int_{- \infty}^{\infty} e^{- a x^{2}} d x = π / a$ is sawaal ka jawab deta hai ki "sabhi positions aur momenta pe total weight kya hai?" $Z = \frac{1}{h} \int e^{- β p^{2} /2 m} d p \int e^{- β m ω^{2} x^{2} /2} d x = \frac{1}{h} \frac{2 π m}{β} \frac{2 π}{β m ω ^{2}} = \frac{2 π}{h β ω} = \frac{1}{ℏ β ω} = \frac{k _{B} T}{ℏ ω} .$ ( $h = 2 π ℏ$ use kiya.) Energy: $ln Z = - ln β + const$ , isliye $U = - \frac{\partial l n Z}{\partial β} = - (- \frac{1}{β}) = \frac{1}{β} = k_{B} T .$ Yeh equipartition hai: do quadratic degrees of freedom ( $p$ aur $x$ ), har ek $\frac{1}{2} k_{B} T$ contribute karta hai. Heat capacity: $C_{V} = \frac{\partial U}{\partial T} = \frac{\partial ( k _{B} T )}{\partial T} = k_{B}$ . Free energy: $F = - k_{B} T ln (k_{B} T /ℏ ω)$ .

Recall Solution

KYA: $S = - \partial F / \partial T$ use karo. KYUN: yeh thermodynamic identity $d F = - S d T - p d V + μ d N$ hai jo fixed $V, N$ pe padhi jaati hai; Legendre transforms in thermodynamics dekho. Maano $x = k_{B} T /ℏ ω$ , toh $F = - k_{B} T ln x$ . Phir $S = - \frac{\partial F}{\partial T} = k_{B} ln (\frac{k _{B} T}{ℏ ω}) + k_{B} T \cdot \frac{1}{T} = k_{B} ln (\frac{k _{B} T}{ℏ ω}) + k_{B} .$ Check $F = U - T S$ : $U = k_{B} T$ ke saath, $U - T S = k_{B} T - T [k_{B} ln x + k_{B}] = k_{B} T - k_{B} T ln x - k_{B} T = - k_{B} T ln x = F . ✓$ $U$ ka $k_{B} T$ $- T S$ ke andar ke $+ k_{B} T$ ko cancel kar deta hai, sirf $- k_{B} T ln Z$ bachta hai — wahi "U cancels" magic jaisi parent derivation mein hai.

Level 5 — Mastery

Goal: naye results build karo aur frameworks ko connect karo.

Recall Solution

KYA: $U$ ko $T$ ke respect mein differentiate karo, $β = 1/ k_{B} T$ use karke. KYUN: $C = \partial U / \partial T$ measure karta hai ki system har degree mein kitni energy soak karta hai — ek two-level ("Schottky") system ka fingerprint. Likho $U = ε (e^{β ε} + 1)^{- 1}$ . Chain rule $β$ se: $\frac{d β}{d T} = - \frac{1}{k _{B} T ^{2}} = - k_{B} β^{2}$ . $\frac{\partial U}{\partial β} = ε \cdot (- \frac{ε e ^{β ε}}{( e ^{β ε} + 1 ) ^{2}}) = - \frac{ε ^{2} e ^{β ε}}{( e ^{β ε} + 1 ) ^{2}} .$ $C = \frac{\partial U}{\partial T} = \frac{\partial U}{\partial β} \frac{d β}{d T} = (- \frac{ε ^{2} e ^{β ε}}{( e ^{β ε} + 1 ) ^{2}}) (- k_{B} β^{2}) = k_{B} (β ε)^{2} \frac{e ^{β ε}}{( e ^{β ε} + 1 ) ^{2}} .$ $β ε = 2$ pe: $e^{2} = 7.389$ , denominator $(8.389)^{2} = 70.38$ . $C / k_{B} = 4 \cdot \frac{7.389}{70.38} = 4 \cdot 0.10498 = 0.4199.$ Physics: $C \to 0$ dono $T \to 0$ (frozen) aur $T \to \infty$ (aur absorb karne ki jagah nahi) par, beech mein peak karta hai — Schottky anomaly. Entropy — Gibbs and Boltzmann definitions dekho.

Recall Solution

KYA: har state ko $e^{- β E}$ ki jagah $e^{- β (E - μ N)}$ se weight karo. KYUN naya tool: ab particle number change ho sakta hai, isliye hum reservoir ko $μ$ per particle "pay" karne dete hain. Yeh grand canonical recipe hai. Step 1 — $Z$ likho. Do states hain: empty $(N = 0, E = 0)$ aur occupied $(N = 1, E = ε)$ : $Z = e^{- β (0 - μ \cdot 0)} + e^{- β (ε - μ \cdot 1)} = 1 + e^{- β (ε - μ)} .$ Step 2 — grand potential. Definition directly apply karo: $Φ = - k_{B} T ln (1 + e^{- β (ε - μ)}) .$ Step 3 — average occupation. Trick $⟨ n ⟩ = \frac{1}{β} \frac{\partial ln Z}{\partial μ}$ kaam karti hai kyunki $μ$ ke respect mein differentiate karne se har exponent se ek factor $β N$ neeche aata hai, exactly number-weighted average build karta hai: $⟨ n ⟩ = \frac{1}{β} \frac{\partial l n Z}{\partial μ} = \frac{e ^{- β (ε - μ)}}{1 + e ^{- β (ε - μ)}} = \frac{1}{e ^{β (ε - μ)} + 1} .$ Interpretation. Yeh Fermi–Dirac distribution hai — two-level energy jaisi same algebraic shape, lekin ab yeh particles count karta hai. $μ = ε$ set karne par (kisi bhi $T$ pe) $⟨ n ⟩ = \frac{1}{e ^{0} + 1} = \frac{1}{2}$ milta hai: chemical potential exactly woh energy hai jis par ek level half-filled hoti hai.

Recall Solution

KYA: $p = - \partial F / \partial V$ fixed $T, N$ pe use karo; Ideal gas from the partition function dekho. KYUN: pressure ke liye sirf $Z_{1}$ ki $V$ -dependence matter karti hai, aur $Z_{1} \propto V$ hai, isliye $ln Z_{1} = ln V + (V-independent terms)$ . $F = - N k_{B} T ln Z_{1} = - N k_{B} T (ln V - 3 ln λ) .$ $p = - \frac{\partial F}{\partial V} = N k_{B} T \frac{\partial l n V}{\partial V} = N k_{B} T \cdot \frac{1}{V} = \frac{N k _{B} T}{V} .$ Dono sides ko volume $V$ se multiply karne par ideal-gas law milta hai: $p V = N k_{B} T . ✓$ Poori ideal-gas law sirf $ln V$ ka derivative hai. ( $1/ N!$ Gibbs correction $F$ ko ek $V$ -independent amount se shift karta hai, isliye $p$ unchanged rehta hai — yeh entropy extensivity fix karta hai, pressure nahi.)

Recall Ladder ka ek-line summary

L1 $F = - k_{B} T ln Z$ mein plug karo ::: L2 $U, S$ ke liye $ln Z$ differentiate karo ::: L3 $Z$ 's multiply karo, log $F$ ko extensive banata hai ::: L4 oscillator ke liye Gaussian integrals + equipartition ::: L5 heat-capacity peak, grand ensemble, ideal-gas law — sab ek $ln Z$ se.