Free energy from partition function: Visual walkthrough

2.4.12 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Free energy from partition function

Hum sirf yeh maante hain ki aap add, multiply, aur ek bar chart padh sakte hain. Baaki sab — Boltzmann weights, partition function, entropy, logarithm trick — yahan aapke saamne banaya jaayega.

Step 1 — Ek system energy levels ki ek list hai

KYA. Ek chhote se system ki picture banao. Woh apni marzi ki koi bhi energy nahi rakh sakta — woh sirf ek fixed set of microstates mein se kisi ek mein baith sakta hai, aur har ek ki apni energy hoti hai. Unhe $i = 1, 2, 3, \dots$ se label karo aur unki energies ko $E_{1}, E_{2}, E_{3}, \dots$ kaho. Hum unhe ek ladder ki rungs ki tarah draw karte hain: neechi rung = kam energy, oopar ki rung = zyada energy.

KYUN. Kuch bhi count ya weight karne se pehle, hume woh cheez chahiye jo count ki jaayegi. Levels ki yeh ladder hi poori statistical mechanics ka raw material hai.

PICTURE. Neeche di gayi ladder dekho. Har rung ki height uski energy $E_{i}$ hai. Abhi kuch bhi weight nahi kiya gaya hai — yeh sirf menu hai.

Step 2 — Thande systems ko neeche ki rungs pasand hain: the Boltzmann weight

KYA. System ek bade heat bath ko temperature $T$ par touch karta hai. Bath system ko energy udhaarne ya dump karne deta hai. System ke rung $i$ par baithne ki probability Boltzmann weight ke proportional hoti hai: $w_{i} = e^{- β E_{i}}, β \equiv \frac{1}{k _{B} T} .$ Term by term: $e$ exponential hai (yeh exponent mein add karne ko multiply mein badal deta hai — hum ise baad mein use karenge); $E_{i}$ rung ki energy hai; $β = 1/ k_{B} T$ "thandak" hai (bada $β$ = thanda, chhota $β$ = garam); $k_{B}$ Boltzmann's constant hai, woh chhota sa number jo temperature ko energy mein convert karta hai.

$e^{- β E}$ kyun use karein, na ki, say, $1/ E$ ? Kyunki hum ek aisi rule chahte hain jahan energies add hon lekin weights multiply hon. Agar do independent pieces ki energies $E_{a}$ aur $E_{b}$ hain, toh joint energy $E_{a} + E_{b}$ hai, aur $e^{- β (E_{a} + E_{b})} = e^{- β E_{a}} e^{- β E_{b}}$ cleanly ek product mein split ho jaata hai. Sirf exponential hi aisa karta hai. Yeh ek property poori derivation ko drive karti hai.

PICTURE. Weight $e^{- β E_{i}}$ ek girta hua curve hai: neechi rungs ke liye lamba, upar jaane par chhhota hota jaata hai. Thanda (bada $β$ ) = steep fall, sirf bottom rungs matter karti hain. Garam (chhota $β$ ) = gentle fall, saari rungs roughly equal.

Step 3 — Weights ko add karo: woh sum hi $Z$ hai

KYA. Har rung ka weight ek number mein add karo, partition function: $Z = \sum_{i} e^{- β E_{i}} .$ Yahan $\sum_{i}$ ka matlab hai "saari rungs par sum karo." $Z$ ek single number hai jo $β$ par depend karta hai (aur energies ke through $V, N$ par bhi).

KYUN. Do reasons hain. (1) Weights ko honest probabilities mein badalne ke liye hume unke total se divide karna hoga — woh total $Z$ hai. (2) $Z$ secretly sab kuch store karta hai: ek baar $Z$ mil jaaye toh hum individual rungs ko kabhi nahi choote. Yeh woh "master number" hai jo parent note ne promise kiya tha.

PICTURE. Weight-bars ko end to end stack karo. Stack ki poori length $Z$ hai. Stack ka woh fraction jo rung $i$ ka hai, woh uski probability hai: $p_{i} = \frac{e ^{- β E_{i}}}{Z}, \sum_{i} p_{i} = 1.$ $p_{i}$ ka matlab hai "stack ka kitna hissa rung $i$ ka hai" — aur shares hamesha $1$ add hote hain.

Yeh Boltzmann distribution and microstates se connect hota hai, jahan $p_{i}$ ki shape entropy maximise karke derive ki jaati hai.

Step 4 — Average energy $U$ , $ln Z$ ka slope hai

KYA. Internal energy $U$ average rung height hai, har rung ko uske share se weight karke: $U = ⟨ E ⟩ = \sum_{i} p_{i} E_{i} = \frac{1}{Z} \sum_{i} E_{i} e^{- β E_{i}} .$

Ab calculus ka ek chhota piece. Ek derivative $\partial / \partial β$ ka jawab deta hai: "agar main thandak $β$ ko thoda nudge karun, toh yeh quantity kitni tezi se badlegi?" Hum ise isliye use karte hain kyunki exponential ko differentiate karne se energy neeche aa jaati hai: $\frac{\partial}{\partial β} e^{- β E_{i}} = - E_{i} e^{- β E_{i}} .$ Yahi toh woh $E_{i}$ hai jise hum chahte the! $i$ par sum karne se: $\sum_{i} E_{i} e^{- β E_{i}} = - \frac{\partial Z}{\partial β} .$ $Z$ se divide karo aur $\frac{1}{Z} \frac{\partial Z}{\partial β} = \frac{\partial l n Z}{\partial β}$ pehchano (chain rule: $ln Z$ ka slope $Z$ ke slope ka $\frac{1}{Z}$ guna hota hai): $U = - \frac{\partial ln Z}{\partial β}$

KYUN. Humne saari rungs par ek messy weighted sum ko ek function ke ek derivative se replace kar diya. Yahi poori statistical mechanics ka payoff hai: $ln Z$ differentiate karo, microstates dubara kabhi count mat karo.

PICTURE. $ln Z$ ko $β$ ke against plot karo. Yeh neeche ki taraf slope karta hai (zyada thandak $\to$ kam accessible rungs $\to$ chhota $Z$ ). Us downhill slope ki steepness, minus sign se positive bana ke, $U$ hai.

Step 5 — Entropy $S$ hai "shares kitni spread out hain"

KYA. Entropy measure karta hai ki probability kaafi saari rungs par kitni spread hai. Gibbs definition: $S = - k_{B} \sum_{i} p_{i} ln p_{i} .$ Term by term: $p_{i}$ rung $i$ ka share hai; $ln p_{i}$ ek chhote share ke liye ek bada negative number hai; overall minus aur $k_{B}$ $S$ ko ek positive energy-per-temperature banate hain.

Jo hum jaante hain woh substitute karo: Step 3 se, $p_{i} = e^{- β E_{i}} / Z$ , isliye $ln p_{i} = - β E_{i} - ln Z .$ Plug in karo: $S = - k_{B} \sum_{i} p_{i} (- β E_{i} - ln Z) = k_{B} β = U i \sum p_{i} E_{i} + k_{B} ln Z = 1 i \sum p_{i} .$ Pehla bracket $U$ hai (Step 4), doosra $1$ hai (shares ka sum ek hota hai). $k_{B} β = 1/ T$ use karke: $S = \frac{U}{T} + k_{B} ln Z$

KYUN. Humne deliberately $ln p_{i}$ use kiya taaki $ln Z$ term alag ho jaaye — woh loose $k_{B} ln Z$ piece hi agले step mein free energy ban jaayega.

PICTURE. Ek share ek tall spike ki tarah draw ki gayi hai (ordered, kam $S$ ); kaafi saare equal shares ek flat plateau ki tarah (spread out, zyada $S$ ). Har ek ke neeche $- \sum p_{i} ln p_{i}$ ka size dikhaya gaya hai.

Dekho Entropy — Gibbs and Boltzmann definitions ki $- k_{B} \sum p_{i} ln p_{i}$ $k_{B} ln Ω$ mein kyun collapse hoti hai jab saare shares equal hote hain.

Step 6 — $F = U - T S$ assemble karo aur $U$ ko gayab hote dekho

KYA. Helmholtz free energy defined hai $F = U - T S$ (iski natural variables $T, V, N$ hain; dekho Helmholtz vs Gibbs free energy). Step 5 ka $S$ ka result substitute karo: $F = U - T (\frac{U}{T} + k_{B} ln Z) = U - U - k_{B} T ln Z .$ $U$ terms equal aur opposite hain — woh ek doosre ko cancel kar dete hain: $F = - k_{B} T ln Z$

KYUN cancel hona zaroori hai. $F$ ko $T, V, N$ par sirf $Z$ ke through depend karna chahiye; bacha hua average energy $U$ wahan hone ki koi wajah nahi. Cancellation yeh theory bata rahi hai ki bookkeeping consistent hai.

PICTURE. Do stacked bars, $+ U$ internal energy se aur $- U$ jo $- T S$ ke andar chhupi thi, cancel ho jaate hain; jo surviving strip bachi woh lavender $- k_{B} T ln Z$ hai.

Step 7 — Logarithm kyun: systems ko join karna

KYA. Do independent systems lo, $a$ aur $b$ . Ek joint microstate hai " $a$ rung $α$ par AUR $b$ rung $γ$ par", energy $E_{α} + E_{γ}$ ke saath. Tab $Z_{ab} = \sum_{α, γ} e^{- β (E_{α} + E_{γ})} = (\sum_{α} e^{- β E_{α}}) (\sum_{γ} e^{- β E_{γ}}) = Z_{a} Z_{b} .$ Partition functions multiply karte hain. Lekin real thermodynamic quantities jaise energy aur free energy extensive hone chahiye — do identical tanks of gas jodo toh $F$ ko double hona chahiye, square nahi. Ilaaj hai logarithm: $F_{ab} = - k_{B} T ln (Z_{a} Z_{b}) = F_{a} - k_{B} T ln Z_{a} + F_{b} - k_{B} T ln Z_{b} .$ $ln$ product ko sum mein badal deta hai, isliye $F$ add hota hai. Yahi poora reason hai ki $F$ , $- k_{B} T ln Z$ hai aur $- k_{B} T Z$ nahi.

KYUN yahan. Yeh boxed formula se missing "log kyun" hai — woh property jo $F$ ko, na ki $Z$ ko, natural thermodynamic potential banati hai.

PICTURE. Left: do boxes side by side, weights multiply ho rahe hain ( $Z_{a} Z_{b}$ ). Right: $ln$ ke baad, do contributions end to end hain, add ho rahe hain ( $F_{a} + F_{b}$ ).

Edge & limiting cases — two-level system har corner check karta hai

KYA. Machine ko sabse simple system par test karo: do rungs $0$ aur $ε$ par. $Z = e^{0} + e^{- β ε} = 1 + e^{- β ε}, U = - \frac{\partial l n Z}{\partial β} = \frac{ε}{e ^{β ε} + 1} .$

KYUN. Jo formula aap trust karte ho use extremes se bachna chahiye. Do limits:

Thanda, $T \to 0$ ( $β \to \infty$ ): $e^{β ε} \to \infty$ , isliye $U \to 0$ . System ground rung mein freeze ho jaata hai. ✓
Garam, $T \to \infty$ ( $β \to 0$ ): $e^{β ε} \to 1$ , isliye $U \to ε /2$ — dono rungs equally likely, average height midpoint hai. ✓
Degenerate $ε \to 0$ : dono rungs merge ho jaate hain, $Z \to 2$ , $F \to - k_{B} T ln 2$ — pure entropy of a two-way coin flip, zero energy. ✓

PICTURE. Energy curve $U (T)$ smoothly $0$ se plateau $ε /2$ tak rise karti hai; do dashed asymptotes cold aur hot limits hain.

The one-picture summary

KYA. Poori derivation ek single flow mein compress ki gayi: levels ka ladder $\to$ Boltzmann weights $\to$ unhe $Z$ mein sum karo $\to$ $ln Z$ se $U$ (slope) aur $S$ (spread) nikalo $\to$ $F = U - T S$ assemble karo $\to$ $U$ cancel hota hai $\to$ $F = - k_{B} T ln Z$ . Side note ka logarithm extensivity explain karta hai.

Recall Poore walkthrough ki Feynman retelling

Ek ladder ki picture banao jahan har rung ek energy hai jo system rakh sakta hai. Bahar ek warm bath system ko neechi rungs comfortable banata hai aur oopar ki rungs unlikely — rung $i$ ki "comfiness" woh girta hua number $e^{- β E_{i}}$ hai. Saari comfinesses add karo aur ek grand total milta hai, $Z$ . Kisi bhi rung ki comfiness ko us total se divide karo aur uski honest probability milti hai — ladder par uska share. $Z$ ke andar do cheezein rehti hain. Pehli, agar aap poocho "jab main system ko thanda karta hun toh $ln Z$ kitni tezi se girta hai?", toh jawab hi average energy $U$ hai — rung by rung dobara add karne ki zaroorat nahi. Doosri, "shares kitni rungs par spread hain?" entropy $S$ hai, aur jab aap ise work out karte hain toh paate hain $S = U / T + k_{B} ln Z$ . Ab free energy banao $F = U - T S$ : energy ka $U$ aur $T S$ ke andar chhupi $U$ equal aur opposite hain, isliye woh ek doosre ko cancel kar dete hain, aur jo bacha woh sirf $F = - k_{B} T ln Z$ hai. Log kyun? Do systems glue karo aur unke comfiness-totals multiply karte hain; lekin do tanks ki free energy ko add karna chahiye. Logarithm woh machine hai jo multiply karne ko add karne mein badal deta hai — yahi poora reason hai ki nature hume $Z$ ki jagah $ln Z$ deta hai.

Active recall

Ek baar energy ladder milne ke baad, kaunsa single number saari thermodynamics store karta hai? ::: Partition function $Z = \sum_{i} e^{- β E_{i}}$ .
Rung by rung sum kiye bina $Z$ se $U$ kaise nikalte hain? ::: $U = - \partial (ln Z) / \partial β$ ( $β$ ke against $ln Z$ ka downhill slope).
$F = U - T S$ mein $U$ ka kya hota hai? ::: Yeh exactly $S$ ke andar $U / T$ piece se cancel hota hai, aur $F = - k_{B} T ln Z$ bachta hai.
Logarithm kyun aur sirf $Z$ kyun nahi? ::: Independent systems $Z$ ko multiply karte hain; $ln$ product ko sum mein badal deta hai taaki $F$ extensive (additive) ho.
Two-level system mein $T \to \infty$ par $U$ kya hai? ::: $ε /2$ — dono rungs equally likely, average midpoint hai.
Two-level system mein $T \to 0$ par $U$ kya hai? ::: $0$ — system ground rung mein freeze ho jaata hai.