Question bank — Statistical mechanics — microstate, macrostate

Sahi — yeh hi Fundamental postulate of statistical mechanics hai; har accessible microstate equal weight carry karta hai, aur baaki sab (kaunsa macrostate dikhta hai) sirf unhe count karne se aata hai.

Galat — ek macrostate ki probability $\Omega /{\Omega }_{\text{total}}$ hoti hai (uski apni count total $2^N$ ke upar), isliye jo macrostates ke peeche zyada microstates hote hain woh bahut zyada likely hote hain; "equal" wala flat rule sirf microstates pe apply hota hai.

Galat — iska multiplicity $\Omega =1$ hai, isliye yeh allowed hai; yeh sirf fantastically improbable hai ($1/2^N$), jo impossible se alag statement hai.

Galat — ek microstate har coordinate fix karta hai, isliye har bulk average bhi fix ho jaata hai, toh woh exactly ek macrostate ka hota hai; many-to-one map doosri taraf jaata hai ("bahut micros, ek macro").

$\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ jaisi discrete counting ke liye Sahi — tum distinct configurations count kar rahe ho; continuous phase space ke liye $\Omega$ ek volume ban jaata hai aur use pure number banane ke liye cell size $h^{3N}$ se divide karna padta hai (jahan $h$ Planck's constant hai).

Galat — microstate wildly churn karta rehta hai aur kabhi repeat nahi hota; sirf macrostate frozen hota hai kyunki system jo bhi microstate explore karta hai uski almost har state wahi most-probable macrostate share karti hai.

Microstates ke roop mein Sahi (har specific ordered sequence ki probability $1/2^N$ hoti hai); confusion yeh hai ki "roughly aadhe heads" ek macrostate hai jo aisi bahut saari sequences bundle karta hai, isliye woh jeetta hai.

Galat — kyunki $\Omega \propto V^N$ hai ($N$ particles mein se har ek ko independently zyada room $V$ milta hai), $V$ double karne se $\Omega$ $2^N$ se multiply hoti hai, jo astronomically bada factor hai; yeh violent growth exactly wahi wajah hai kyun gas apne container ko fill karne ke liye expand karti hai (Second law and irreversibility).

Absolute terms mein Sahi (width $\sim \sqrt{N}$), lekin relative width $\sqrt{N}/N=1/\sqrt{N}$ shrink hoti hai — isliye bade $N$ ke liye distribution ek razor-sharp spike hoti hai (Binomial and Gaussian distributions).

Ulta hai — microstate woh single complete arrangement hai; macrostate woh bulk description hai jo bahut saare microstates share karte hain.

Choices multiply hote hain, add nahi: $N$ objects mein se har ek ke paas 2 options hain jo $2\times 2\times \cdots =2^N$ dete hain, fundamental counting principle se, $2N$ nahi.

Galat count — "all up" ek specific microstate hai ($\Omega =1$), isliye $P=1/2^N$; $N$-fold reasoning "exactly one up" describe karta hai, jo $\left(\genfrac{}{}{0ex}{}{N}{1}\right)=N$ hai.

Macrostate label $n$ pe sum $2^N$ hai (binomial theorem with $1+1$); yeh har microstate ko ek baar recover karta hai, aur $2^N\ne N!$ sirf tiny $N$ ko chhod kar.

Yeh $E^{3N/2}$ ki tarah scale karta hai — energy $3N$-dimensional momentum sphere ka radius $\sqrt{2mE}$ fix karti hai, aur uski surface $E^{(3N-2)/2}\approx E^{3N/2}$ ki tarah scale karti hai, har particle ke liye linearly nahi.

$\ln 1=0$, isliye us macrostate ki entropy zero hai — perfectly ordered "all up" state low-entropy extreme hai, high wala nahi (Entropy and Boltzmann's relation S = k ln Ω).

$\sim 2^{10^{23}}$ items list karna impossible hai, lekin combinatorics ($\left(\genfrac{}{}{0ex}{}{N}{n}\right)$, phase-space volume) $\Omega$ closed form mein deta hai — is method ki poori power enumeration ke bina count hai.

Uski multiplicity baaki sab ko dwarf karti hai aur relative fluctuation $1/\sqrt{N}$ ki tarah shrink hoti hai; $N\sim 10^{23}$ pe woh spread $10^{-11}$ ke aas paas hai, isliye measured bulk variables perfectly steady lagte hain.

Sirf tab jab har microstate equally weighted ho toh $P(\text{macrostate})=\Omega /{\Omega }_{\text{total}}$ ho sakta hai; iske bina hame probabilities assign karne ka koi principled tarika nahi hoga aur "counting" meaningless hoti.

"Phail gayi" macrostate ke paas "ek corner mein clustered" wali se overwhelmingly zyada microstates hain, isliye system almost certainly uski taraf evolve karti hai — irreversibility sirf multiplicity ka ek landslide hai.

$N!$ saare objects ko order karta hai, lekin $n$ ups ko apne beech mein reorder karna ya $N-n$ downs ko apne beech mein reorder karna same macrostate deta hai, isliye un internal permutations ko divide out karna padta hai.

Zyada energy ek bada momentum sphere kholta hai, isliye $\Omega$ badhta hai ($\propto E^{3N/2}$); kyunki $S=k\ln \Omega$, $E$ ke saath badhta hai, yahi wajah hai ki $\partial S/\partial E$ positive hota hai, aur by definition $1/T=\partial S/\partial E$ (Temperature as ∂S/∂E).

$\left(\genfrac{}{}{0ex}{}{N}{0}\right)=1$ — sabko neeche rakhne ka exactly ek tarika hai, "all up" jaisi hi akeli count, isliye dono extremes equally (un)likely hain.

Do microstates (up, down) aur do macrostates ($n=0,n=1$) — ek degenerate case jahan map one-to-one hai aur "bahut micros per macro" wali picture abhi switch on nahi hui.

Yeh $1/\sqrt{N}$ ki tarah zero ho jaata hai, isliye macrostate effectively deterministic ho jaata hai — yahi limit wajah hai ki bulk thermodynamics sharp aur reproducible hoti hai.

Sirf weakly: $\Omega (1)=2$ versus $\Omega (0)=\Omega (2)=1$, isliye $P(n=1)=1/2$ — "sharp peak" wali story large-$N$ phenomenon hai aur tiny systems ke liye simply exist nahi karti.

Haan agar woh koi constraint violate kare (jaise $n=N+1$ ups impossible hai), jo $\Omega =0$ aur $P=0$ deta hai — aisa macrostates simply accessible nahi hote aur counting se drop out ho jaate hain.

$E$ fix karna $3N$-dimensional momentum sphere ka radius pin karta hai, isliye allowed microstates ek thin shell (ek surface) pe hote hain, single point pe nahi — yeh shell microcanonical counting region hai (Microcanonical ensemble).