Free energy from partition function: Question bank

2.4.12 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Free energy from partition function

Shuru karne se pehle, symbols ki ek-line reminder taaki kuch bhi bina explanation ke use na ho:

Definition Symbol refresher

$k_{B}$ — Boltzmann's constant, woh fixed conversion factor ( $\approx 1.38 \times 1 0^{- 23} J/K$ ) jo temperature ko energy per microstate mein convert karta hai.
$T$ — us heat bath ki absolute temperature (kelvin mein) jisme system rakha hai; zyada $T$ matlab bath system mein zyada energy push karti hai.
$V$ — woh volume jo system occupy karta hai (canonical setup mein fixed rakha jaata hai).
$N$ — system mein number of particles (canonical setup mein yeh bhi fixed rakha jaata hai).
$Z = \sum_{i} e^{- β E_{i}}$ — partition function, ek single number jo har microstate $i$ ka Boltzmann weight $e^{- β E_{i}}$ sum karta hai.
$β = 1/ (k_{B} T)$ — "inverse temperature"; bada $β$ = thanda, chhota $β$ = garam.
$F = - k_{B} T ln Z$ — Helmholtz free energy (natural variables $T, V, N$ ).
$U = - \frac{\partial ( ln Z )}{\partial β}$ — internal (average) energy, yaani $ln Z$ ka $β$ ke saath derivative.
$S = - \frac{\partial F}{\partial T}$ — entropy.

Neeche ki saari cheezein do pictures pe tiki hain. Pehli dikhati hai kyun $Z$ 's ko multiply karna, $F$ 's ko add karna ban jaata hai (extensivity). Doosri dikhati hai $F (T)$ ki shape aur kyun uski slope $- S$ hai.

Figure — Free energy from partition function

True or false — justify

$Z$ kisi bhi system ke liye dimensionless hota hai. ::: Discrete sum $e^{- β E_{i}}$ ke liye yeh dimensionless hota hai, lekin classical phase-space $Z$ ko dimensionless banane ke liye har degree of freedom ke liye $h$ se divide karna padta hai (aur identical particles ke liye $N!$ se) — yeh factors chhod do toh $Z$ mein stray units aa jaate hain.
Har energy level mein ek constant $E_{0}$ add karne se saare measurable predictions unchanged rehte hain. ::: $Z$ aur $F$ ke liye False hai (dono shift hote hain: $Z \to e^{- β E_{0}} Z$ , $F \to F + E_{0}$ ), lekin un observables ke liye True hai jo derivatives jaise $p$ aur $S$ se aate hain; $U$ exactly $E_{0}$ se shift hota hai, jaisa physically expected hai.
$F = - k_{B} T ln Z$ sirf equilibrium par valid hai. ::: True — $Z$ Boltzmann-weighted microstates maanta hai, jo ki equilibrium distribution hai; out of equilibrium $F = U - T S$ defined hai lekin $- k_{B} T ln Z$ ke barabar nahi hota.
Kyunki independent subsystems ke liye $Z$ multiply hota hai, unke liye $F$ add hota hai. ::: True — logarithm ka yahi poora point hai (figure 1 dekho): $ln (Z_{1} Z_{2}) = ln Z_{1} + ln Z_{2}$ , toh $F$ extensive hai jabki $Z$ multiplicative hai.
$F$ hamesha negative hota hai. ::: False — $F$ ka sign energy ke zero aur $Z$ par depend karta hai; sirf $F$ mein changes physically meaningful hain, toh uska raw sign koi universal meaning nahi rakhta.
Temperature kam karna hamesha $F$ ko kam karta hai. ::: False — figure 2 mein slope $\partial F / \partial T = - S \leq 0$ hai, toh $T$ ghatne par $F$ badhta hai (kyunki $S \geq 0$ ); $F$ tabhi ghatata hai jab $T$ badhta hai.
Equilibrium state fixed $T, V, N$ par $F$ ko minimize karta hai. ::: True — yahi defining variational property hai: fixed $T, V, N$ par saari conceivable arrangements mein, asli equilibrium woh hai jo $F = U - T S$ ki sabse kam value par ho, sirf $U$ ki minimum par nahi.
Microstates ki sankhya double karne se hamesha $F$ double ho jaata hai. ::: False — $F \propto ln Z$ hai, toh $Z$ double karne se $F$ mein sirf ek fixed $k_{B} T ln 2$ subtract hota hai; sirf pure system ki independent copy replicate karne se $F$ double hota hai.

Spot the error

" $U = - \frac{\partial Z}{\partial β}$ ." ::: Normalization missing hai: $- \partial Z / \partial β = \sum_{i} E_{i} e^{- β E_{i}}$ ek unnormalized weighted sum hai; tumhe $Z$ se divide karna hoga, jo $U = - \frac{1}{Z} \frac{\partial Z}{\partial β} = - \frac{\partial ( ln Z )}{\partial β}$ deta hai.
" $S = + \partial F / \partial T$ ." ::: Wrong sign: $d F = - S d T - p d V + μ d N$ mein built-in minus hai, toh $S = - \partial F / \partial T$ ; plus sign entropy ko negative bana deta.
" $N$ independent particles ke liye $Z = N Z_{1}$ ." ::: Galat operation: energies exponent ke andar add hoti hain, jo exponentials ko multiply karti hain, toh $Z = Z_{1}^{N}$ (indistinguishability ke liye $1/ N!$ tak), $N Z_{1}$ nahi.
" $p = + \partial F / \partial V$ ." ::: Phir sign error: $d F = - p d V$ se fixed $T, N$ par, pressure $p = - \partial F / \partial V$ hai; result $p = k_{B} T \partial_{V} ln Z$ mein already correct sign baked in hai.
" $F = - k_{B} T Z$ (log drop kar diya)." ::: Log ke bina $F$ systems combine hone par add ki jagah multiply karta, extensivity todta; logarithm exactly wahi hai jo $Z$ ki product structure ko ek additive potential mein convert karta hai.
"Kyunki $p_{i} = e^{- β E_{i}} / Z$ , sabse probable microstate hamesha ground state hota hai." ::: Sabse probable single microstate lowest-energy wala hota hai, lekin sabse probable energy level ek high wala ho sakta hai agar uske bahut saare microstates hain — ek level ki probability $p_{i}$ times uski degeneracy hoti hai.

Why questions

$F$ mein $Z$ ki jagah logarithm kyun hai? ::: Independent systems apna $Z$ multiply karte hain; $ln$ lene se woh product ek sum ban jaata hai (figure 1), toh $F$ extensive ho jaata hai (system size ke saath scale karta hai) — ek thermodynamic potential ki zaroori requirement.
$F = U - T S$ mein $U$ exactly cancel kyun hota hai? ::: Kyunki $S = U / T + k_{B} ln Z$ hai, $T \cdot (U / T) = U$ term exactly $F = U - T S$ mein $U$ ko subtract karta hai, sirf pure $- k_{B} T ln Z$ bachta hai — yeh ek structural fact hai, coincidence nahi.
Entropy $- \partial F / \partial T$ kyun hai na ki directly $Z$ se counting wali koi cheez? ::: Dono same cheez hain: Gibbs entropy $S = - k_{B} \sum_{i} p_{i} ln p_{i}$ se shuru karo jahan $p_{i} = e^{- β E_{i}} / Z$ hai, toh $ln p_{i} = - β E_{i} - ln Z$ ; substitute karne par $S = k_{B} β U + k_{B} ln Z = U / T + k_{B} ln Z$ milta hai, aur $F = - k_{B} T ln Z$ ko differentiate karne par exactly yahi reproduce hota hai — thermodynamic derivative aur microscopic count agree karte hain kyunki $F$ inhi $p_{i}$ se banaya gaya tha.
Classical harmonic oscillator $ω$ se independent $U = k_{B} T$ kyun deta hai? ::: Kyunki $Z \propto 1/ β$ hai jo $ln Z = - ln β + const$ banata hai, aur $ω$ sirf constant mein hota hai; equipartition energy $k_{B} T$ (do quadratic degrees of freedom) stiffness se independent hai — Equipartition theorem dekho.
Hum microstates par wapis jaane ki jagah $Z$ se differentiation karke saari thermodynamics kyun bana sakte hain? ::: Har observable energy ka average ya fluctuation hai, aur woh $ln Z$ ke $β$ - aur $V$ -derivatives mein encoded hain; toh single function $ln Z$ woh saari microstate information store karta hai jo humein chahiye.
Fixed $T$ par $U$ nahi, $F$ minimize kyun hota hai? ::: Fixed $T$ par system bath se energy exchange kar sakta hai, toh woh internal energy ko entropy ke against trade karta hai; $F = U - T S$ minimize karna "low energy" aur "high disorder" ko sahi balance karta hai, jo sirf $U$ minimize karna ignore karta hai — Helmholtz vs Gibbs free energy dekho.

Edge cases

$T \to 0$ par $F$ kya hoga? ::: $Z$ mein sirf ground state bachta hai, toh $Z \to g_{0} e^{- β E_{0}}$ aur $F \to E_{0} - k_{B} T ln g_{0} \to E_{0}$ ; entropy term vanish ho jaata hai aur $F$ ground-state energy tak pahunch jaata hai.
Two-level system mein $T \to \infty$ par $U$ kya hoga? ::: $β \to 0$ dono states ko equally likely banata hai, toh $U \to ε /2$ ( $0$ aur $ε$ ka average), $ε$ nahi — high temperature matlab maximal ignorance hai, maximal energy nahi.
Kya $Z$ diverge ho sakta hai, aur iska matlab kya hai? ::: Haan — jaise infinite volume mein free particle ya hydrogen atom ko saare bound states par sum karo; divergent $Z$ signal karta hai ki koi normalizable equilibrium additional constraint (volume cutoff, regularization) ke bina exist nahi karta.
Two-level system mein entropy ki value $T \to 0$ versus $T \to \infty$ par kaisi behave karti hai? ::: $T \to 0$ par sirf ground state occupied hai toh $S \to 0$ (ek accessible state, zero disorder); $T \to \infty$ par dono states equally likely hain toh $S \to k_{B} ln 2$ (maximum, zero nahi) — dono limits physically opposite hain, toh $S$ high $T$ par vanish nahi karta.
Kya $F = - k_{B} T ln Z$ grand canonical ensemble mein valid hai? ::: Nahi — wahaan controlled potential grand potential $Φ = - k_{B} T ln Z$ hai jisme $Z$ particle number par bhi sum karta hai; Grand canonical ensemble and grand potential dekho.
Agar saare microstates ki same energy $E$ ho toh kya hoga? ::: Tab $Z = Ω e^{- β E}$ aur $F = E - k_{B} T ln Ω$ , toh $S = k_{B} ln Ω$ exactly Boltzmann entropy recover karta hai — canonical result microcanonical par collapse ho jaata hai.
Kya formula distinguishable particles assume karta hai? ::: Single-system $F = - k_{B} T ln Z$ aisi koi assumption nahi karta, lekin bahut saare identical particles ke liye $Z$ banane mein entropy-of-mixing paradox se bachne ke liye $1/ N!$ (Gibbs correction) chahiye — Ideal gas from the partition function dekho.

Recall Traps ka one-line summary

Lagbhag har error teen mein se ek hai: $ln$ drop karna (extensivity todta hai), $1/ Z$ normalization drop karna (unnormalized averages), ya ek minus sign drop karna ( $d F = - S d T - p d V + μ d N$ se).