Question bank — Entropy and disorder — Boltzmann S = k·ln(W)

True. Yeh statistical mechanics ka fundamental postulate hai; equilibrium koi preferred microstate nahi hai balki ek aisa macrostate hai jiske paas in equally-likely microstates ka sabse zyada hissa hota hai.

Almost, lekin sahi tarike se kaho. Nature us macrostate ki taraf drift karti hai jahan $W$ sabse bada ho kyunki woh almost saare microstates apne paas rakhta hai; $N\sim 10^{23}$ particles wale system ke liye woh peak itni sharp hoti hai ki effectively certain lagta hai — lekin yeh overwhelming probability ka statement hai, koi force ka law nahi.

False. $S=k\ln W$, toh $W$ ko double karne se sirf $k\ln 2$ add hota hai; entropy $W$ ke saath nahi balki $W$ ke logarithm ke saath badhti hai.

True. $S_{AB}=S_A+S_B$ kyunki $W_{AB}=W_A{W}_B$ aur $\ln$ us product ko sum mein badal deta hai — yahi additivity ek exact wajah hai ki logarithm kyun aata hai.

True. $S=k\ln 1=0$; ek akela unique arrangement (jaise ek perfect flawless crystal jo $T=0$ tak cool ho gaya ho, jahan saari jiggling ruk jaati hai) mein count karne ke liye kuch nahi hota.

False. Ek isolated system ki entropy kabhi decrease nahi hoti; ek sub-system (fridge ka interior, ek freezing puddle) entropy kho sakta hai jab tak surroundings zyada gain karein, toh total phir bhi badhta hai.

False. "Disorder" ek fuzzy visual label hai; entropy precisely $k\ln (\text{accessible microstates})$ hai. Mixed isotopes se bana ek neat-looking crystal isotope arrangements ki wajah se badi entropy carry kar sakta hai jo tum dekh nahi sakte.

False. $k$ sirf units set karta hai (J/K) taaki $S=k\ln W$ purani thermodynamic (Clausius) entropy se match kare; macrostates ki ranking sirf $W$ pe depend karti hai, jise $k$ kabhi reorder nahi karta.

False (subtle). Vacuum mein free expansion irreversible hai aur $Q=0$ hota hai; relation $Q=T\Delta S$ (from $dS=d{Q}_{\text{rev}}/T$) sirf ek reversible path pe hold karta hai. Entropy phir bhi badhti hai kyunki yeh ek state function hai — tum ise ek imaginary reversible route pe compute karte ho, actual route pe nahi.

False in general. Ordinary systems ke liye haan, lekin ek bounded, finite energy ladder wala system (jaise spins jo sirf upar ya neeche point kar sakte hain) ek aisi state tak pahunch sakta hai jahan energy add karne se zyada spins higher level pe jaate hain, jo actually $W$ ko reduce karta hai. Wahan $S$ energy badhne ke saath girti hai — precisely yahi "negative absolute temperature" ka matlab hai (hotter-than-infinity, not colder-than-zero).

Error hai ek force invent karna. Nature blind hai — har microstate equally likely hai; disordered macrostate sirf unhe contain karta hai vastly more numbers mein, isliye tum wahan sirf counting se pahunchte ho.

Galat operation. $\ln (3W)=\ln 3+\ln W$, isliye $S$ $k\ln 3$ se increase hota hai (ek addition), 3 ke factor se multiply nahi hota.

Microstate counts multiply hote hain ($W_{AB}=W_A{W}_B$), lekin entropies add hoti hain: $S_{AB}=S_A+S_B=2S$. Tum $W$ multiply karte ho, $S$ add karte ho — exactly yahi $\ln$ enforce karta hai.

Tum kaun se coins heads hain yeh count karna hoga: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ alag choices hain. $W$ arrangements ki count hai, heads ki number nahi.

Zero entropy ka matlab $W=1$ hai, yaani exactly ek microstate — yeh perfectly possible hai, sirf middle macrostates ki bahut-si-ways ki tulna mein bahut improbable hai. Rare $\ne$ forbidden.

Heat nayi configurations ko enable karti hai, lekin entropy rise $W_{\text{liquid}}\gg W_{\text{solid}}$ se aati hai — molecules ko bahut zyada positions/orientations ka access milta hai. Count cause hai; heat enabler hai.

Same core idea. Shannon entropy bits mein measured $\ln (\text{count})$ of possibilities hai; Boltzmann ki wahi log-of-multiplicity hai jo joules per kelvin mein factor $k$ ke zariye measure ki jaati hai.

Error — yeh Gibbs paradox hai. Agar dono sides ke molecules identical aur indistinguishable hain, toh ek left molecule ko right wale se swap karna koi naya microstate nahi banata, isliye $W$ actually badhta nahi aur $\Delta S=0$ hota hai. $Nk\ln 2$ ka jump sirf tabhi aata hai jab do gases alag (distinguishable) species hoon.

Sirf $f(xy)=f(x)+f(y)$ entropy ko additive banata hai jabki multiplicities multiply hoti hain, aur us equation ka unique continuous solution $f=\text{const}\cdot \ln$ hai. Square root products ko sums mein nahi badhata.

Kyunki $W$ exponentially $N$ mein grow karta hai (roughly $W\sim a^N$), toh $\ln W=N\ln a$ — ek exponential ka log linear hota hai. Dono effects cancel ho jaate hain ek bulk, extensive quantity dene ke liye.

$N$ two-sided choices ke liye counts binomial $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ follow karte hain, jiska peak $n=N/2$ pe hota hai. Iska width (appreciable probability ka spread) lagbhag $\sqrt{N}$ particles ka hai, jabki peak position $N/2$ hai, toh relative width $\sqrt{N}/N=1/\sqrt{N}$ hai — fluctuations ki Central-Limit shrinking. $N\sim 10^{23}$ pe, $1/\sqrt{N}\sim 10^{-11}$, ek unimaginably thin spike, toh essentially har microstate central macrostate ka hissa hai.

Absolute zero pe ek system apni unique ground configuration mein settle ho jaata hai, toh $W\to 1$ aur $S=k\ln 1=0$ — yeh third law of thermodynamics ka statistical content hai.

"Preferred" macrostate ko refer karta hai, kisi single microstate ko nahi. Equilibrium macrostate jeetatа hai kyunki woh equally-weighted microstates ka overwhelming share rakhta hai.

Har species achanak poore doubled volume ko access kar leti hai — position choices free expansion ki tarah multiply hote hain, toh $W$ jump karta hai aur $S$ purely counting se badhti hai, koi energy transfer required nahi. (Crucial yeh hai ki iske liye gases distinguishable honi chahiye; Gibbs-paradox trap upar dekhein.)

Ek akela low-energy state per state zyada probable hai, lekin aksar bahut zyada high-energy states hote hain; observed population "probability per state" aur "number of states" ke beech balance karta hai — phir se ek multiplicity argument.

$W=1$ aur $S=k\ln 1=0$ — entropy ka degenerate floor, woh reference point jo third law fix karta hai.

Koi bhi physically realizable macrostate esa nahi ho sakta; $W=0$ ka matlab hoga ki woh kisi microstate se correspond nahi karta, yaani woh simply exist hi nahi karta. Har observable macrostate ka $W\ge 1$ hota hai.

Entropy macrostate ki property hai, kisi ek microstate ki nahi. Expansion ke baad accessible position-space double ho gayi, toh $W$ (aur $S$) badh gaya regardless of specific microstate jis par tum nazar daalo.

Formally haan, $\Delta S=1\cdot k\ln 2$, lekin ek molecule ke saath fluctuations enormous hain aur "Second Law" sirf statistical hai — molecule original half mein wapas kaafi baar ja sakta hai. Thermodynamic certainty ke liye large $N$ chahiye.

Haan — yahi Gibbs correction ka point hai. Kyunki indistinguishable particles ko permute karna wahi microstate deta hai, honest count un $N!$ fake rearrangements ko divide out karta hai, $W$ ko lower karta hai aur entropy ko properly extensive rakhta hai.

Haan, ek capped energy ladder wale system ke liye (spin systems, certain lasers). Jab zyada than half particles top level mein aa jaate hain, extra energy population ko "all-up" configuration ki taraf shove karti hai, toh $W$ shrink karta hai aur $S$ girta hai — yeh negative absolute temperature ka regime hai, jo actually kisi bhi positive temperature se hotter hota hai.

Haan, locally — yeh neater dikhta hai. Lekin yeh surroundings mein itni entropy dump karta hai (usually heat ke roop mein) ki system-plus-surroundings ka total $W$ phir bhi badhta hai, second law ko honour karta hai.

Haan — central macrostate ($n\approx N/2$) ke paas sabse bada $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ hai, isliye sabse bada $\ln W$ hai, isliye un fixed coins ke liye available maximum entropy hai.