Microstate i ki Boltzmann probability hai
pi=Ze−βEi.Yeh step kyun? Bath T fix karta hai; fixed average energy par entropy maximize karne se exactly yahi exponential weight milta hai, aur Z se divide karne par ∑ipi=1 normalize ho jaata hai.
U=⟨E⟩=∑ipiEi=Z1∑iEie−βEi.
Dhyan do ∂β∂e−βEi=−Eie−βEi, toh
∑iEie−βEi=−∂β∂Z.
Isliye
U=−Z1∂β∂Z=−∂β∂lnZ.Yeh step kyun? Humne weighted sum ko Z ke derivative mein convert kar diya — yahi poora point hai: microstates sum karne ki jagah lnZ differentiate karo.
F=U−TS=U−T(TU+kBlnZ)=U−U−kBTlnZ.F=−kBTlnZYeh kyun beautiful hai:U exactly cancel ho jaata hai. Free energy purelnZ hai — yeh intuition confirm karta hai ki F multiplicative Z ka additive (extensive) partner hai.
Recall Feynman: ek 12-saal ke bache ko explain karo
Socho ek badi bag mein dice hain jo kisi bhi number par gir sakte hain, lekin thande dice "prefer" karte hain chhote numbers aur garam dice wild hote hain. Partition function Z ek aisa number hai jo secretly jaanta hai ki pure bag mein kitne likely outcomes hain. "Free energy" F sirf Z ka ek stretched-out version hai — hum Z ko logarithm se squish karte hain taaki jab aap do bags ko saath rakh do, unke free energies simply add ho sakein multiply karne ki jagah. Us ek stretched number se aap bag ka temperature behavior, uski energy, aur uski messiness (entropy) predict kar sakte ho — bina har ek dice ko ek-ek karke count kiye.
Canonical partition function define karo. ::: Z=∑ie−βEi, sabhi microstates par sum, β=1/kBT.
Z ke terms mein Helmholtz free energy? ::: F=−kBTlnZ.
F mein logarithm kyun hai (sirf Z nahi)? ::: Independent systems apne Z multiply karte hain; ln product ko sum mein convert karta hai toh F extensive (additive) hota hai.
Z se internal energy? ::: U=−∂(lnZ)/∂β=Z1(−∂βZ).
Z aur U se entropy? ::: S=U/T+kBlnZ=−∂F/∂T.
Z se pressure? ::: p=kBT∂(lnZ)/∂V=−∂F/∂V.
F=U−TS derive karte waqt kya cancel hota hai? ::: U exactly cancel ho jaata hai, −kBTlnZ bachta hai.
N identical independent two-level systems ke liye, F=? ::: F=−NkBTln(1+e−βε).