Statistical mechanics — microstate, macrostate: Exercises

2.4.8 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Statistical mechanics — microstate, macrostate

Sirf do formulas hain jinpe hum tike hain, ek quick reminder — taaki kuch bhi bina naam ke use na ho:

Figure — Statistical mechanics — microstate, macrostate

Upar wala bar chart humaara map hai: height = $Ω (n)$ , yaani har macrostate ke peeche microstates ki count. Dekho kaise middle edges par tower karta hai — yeh ek fact hi har neeche diye exercise ka asli raaz hai.

L1 — Recognition

Exercise 1.1 (L1)

$N = 3$ coins ke system ke liye, neeche diye gaye har ko microstate ya macrostate classify karo: (a) "HTH" (b) "two heads" (c) coins par apply kiya gaya full list $(q_{1}, p_{1}, \dots)$ idea, yaani har coin ka named face (d) " $n = 1$ ".

Recall Solution

Microstate hai complete ordered pattern (kaun sa specific coin kya dikhata hai). Macrostate hai ek bulk count jo tum individual coins jaane bina padh sako.

(a) "HTH" — teeno coins ka exact face → microstate.
(b) "two heads" — sirf count $n = 2$ → macrostate.
(c) har coin ka named face → microstate.
(d) " $n = 1$ " — ek count → macrostate.

Exercise 1.2 (L1)

Compute kiye bina, $N = 3$ coins ke kis macrostate mein sabse zyada microstates hain, aur kis mein sabse kum? Sirf bars ki shape dekh ke answer do.

Recall Solution

Sabse zyada: middle ke nearest count, $n = 1$ ya $n = 2$ (dono tie karte hain $(1 3) = (2 3) = 3$ par). Sabse kum: extremes $n = 0$ aur $n = 3$ , har ek mein exactly $(0 3) = (3 3) = 1$ microstate. Bar chart symmetric hai aur middle mein peak karta hai — answer recognize karne ke liye bas itna hi chahiye.

L2 — Application

Exercise 2.1 (L2)

$N = 6$ two-state objects ke liye, (a) microstates ki total count, aur (b) macrostate $n = 2$ ki multiplicity compute karo.

Recall Solution

(a) Total $= 2^{6} = 64$ . Kyun: 6 objects mein se har ek count ko double karta hai. (b) $Ω (2) = (2 6) = \frac{6 !}{2 ! 4 !} = \frac{720}{2 \times 24} = \frac{720}{48} = 15.$ Toh 64 microstates mein se 15 mein exactly do ups hain.

Exercise 2.2 (L2)

Same $N = 6$ use karke, $n = 0, \dots, 6$ ke liye full multiplicity table $Ω (n)$ likho aur verify karo ki sum $2^{6}$ ke equal hai.

Recall Solution

$(0 6), (1 6), \dots, (6 6) = 1, 6, 15, 20, 15, 6, 1.$ Sum $= 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^{6} . ✓$ Check kyun kaam karta hai: saare macrostates par multiplicities ka sum har microstate ko exactly ek baar rebuild karna chahiye — yeh binomial theorem hai jisme $1 + 1 = 2$ .

Exercise 2.3 (L2)

$N = 6$ ke liye, equilibrium mein macrostate $n = 3$ (three ups) observe karne ki probability kya hai? Fundamental postulate use karo.

Recall Solution

Fundamental postulate kehta hai har accessible microstate equally likely hai, toh $P (n) = \frac{Ω ( n )}{2 ^{N}} = \frac{( 3 6 )}{2 ^{6}} = \frac{20}{64} = 0.3125.$ Peak macrostate $n = 3$ probability ka $31.25%$ carry karta hai — kisi bhi doosre single macrostate se zyada.

L3 — Analysis

Exercise 3.1 (L3)

$N = 10$ ke liye, ratio $\frac{Ω ( 5 )}{Ω ( 0 )}$ compute karo aur words mein explain karo ki yeh ratio kya measure kar raha hai.

Recall Solution

$Ω (5) = (5 10) = 252$ aur $Ω (0) = (0 10) = 1$ , toh ratio hai $252$ . Kya measure karta hai: "balanced" macrostate ( $n = 5$ ) 252 times zyada ways mein realize hota hai "all down" extreme se. Equal-microstate postulate ke under, balanced macrostate isliye 252× zyada likely hai. Multiplicities ka yeh ratio hi equilibrium ko overwhelming banata hai — aur entropy differences ka raw material hai, kyunki $S=k\ln\Omega$ $Ω$ ke ratio ko entropies ka difference bana deta hai.

Exercise 3.2 (L3)

Parent note claim karta hai ki peak ki relative width $1/ N$ ki tarah scale karti hai. Fair binomial ( $p = \frac{1}{2}$ ) ke liye, $n$ ka standard deviation $σ = \frac{1}{2} N$ hai aur mean $μ = N /2$ hai. $N = 100$ aur $N = 1 0^{4}$ ke liye relative width $σ / μ$ compute karo, aur batao ki $n \approx N /2$ ke alawa kuch observe karne ke liye yeh kya imply karta hai.

Recall Solution

Relative width $\frac{σ}{μ} = \frac{\frac{1}{2} N}{N /2} = \frac{1}{N}$ .

$N = 100$ : $1/ 100 = 0.1$ → up-count typically half ke $\pm 10%$ mein girta hai.
$N = 1 0^{4}$ : $1/ 1 0^{4} = 0.01$ → $\pm 1%$ ke andar. Implication: jaise-jaise $N$ badhta hai, fractional wobble $1/ N$ ki tarah shrink hota hai. Lab-sized $N \sim 1 0^{23}$ ke liye fluctuations $\sim 1 0^{- 11.5}$ hain — bilkul unmeasurable. Tum hamesha peak macrostate dekhte ho: yahi equilibrium hai. (Yeh binomial ka Gaussian limit hai.)

Neeche wali figure yeh shrinking relative width ko $N$ ke against log–log axes par plot karti hai. Cyan curve trace karo: yeh steadily neeche slide karta hai, aur do amber dots exactly woh $N = 100$ aur $N = 1 0^{4}$ answers mark karte hain jo humne abhi compute kiye. Log axes par straight-line shape $1/ N$ power law ka visual signature hai — $N$ mein har tenfold rise relative wobble ko roughly factor of three se girata hai.

Exercise 3.3 (L3)

$N = 4$ ke liye, direct listing se verify karo ki $Ω (1) = 4$ hai aur har microstate identify karo. Phir explain karo kyun "kaun sa coin up hai" macrostate ko invisible hai.

Recall Solution

Exactly ek up wale chaar microstates (up $↑$ , down $↓$ maan lo): $↑↓↓↓, ↓↑↓↓, ↓↓↑↓, ↓↓↓↑ .$ Yeh hai $(1 4) = 4$ . Kyun order macrostate ko invisible hai: macrostate sirf number $n = 1$ record karta hai; yeh kabhi nahi poochta kaun sa slot. Chaar patterns microscopically alag hain lekin macroscopically identical hain — "many micros, one macro" ka ek perfect miniature.

L4 — Synthesis

Exercise 4.1 (L4)

Do isolated spin systems, A ke paas $N_{A} = 4$ aur B ke paas $N_{B} = 4$ , contact mein rakhe gaye hain taaki ups ki total count $n_{tot} = 4$ fixed rahe (energy exchange = ups exchange). Joint macrostate hai " $n_{A}$ ups in A" (tab B mein $4 - n_{A}$ hain). Har joint macrostate ki multiplicity $Ω_{A} (n_{A}) \cdot Ω_{B} (4 - n_{A})$ compute karo, sabse probable split dhundho, aur Vandermonde identity $\sum_{n_{A}} (n _{A} 4) (4 - n _{A} 4) = (4 8)$ se totals confirm karo.

Recall Solution

Joint multiplicity $= (n _{A} 4) (4 - n _{A} 4)$ :

$n_{A}$	$(n _{A} 4)$	$(4 - n _{A} 4)$	product
0	1	1	1
1	4	4	16
2	6	6	36
3	4	4	16
4	1	1	1

Sabse probable split: $n_{A} = 2$ (product $= 36$ ) — energy equally share hui, exactly wahi jo "thermal equilibrium" ko hona chahiye. Check: $1 + 16 + 36 + 16 + 1 = 70 = (4 8)$ . $✓$ Synthesis point: joint system, 4 ups wale 8 spins ke ek pool ki tarah treat kiya gaya, mein $(4 8) = 70$ microstates hain; "A mein kitne hain" ke basis par split karna un 70 ko sirf partition karta hai. Equilibrium split $Ω_{A} Ω_{B}$ maximize karta hai, jo $S = k ln Ω$ ke zariye $S_{A} + S_{B}$ maximize karta hai — pure counting se nikalta hua second law. Yeh partitioned counting exactly microcanonical logic hai.

Exercise 4.2 (L4)

Ek ideal gas ka $Ω \propto V^{N} E^{3 N /2}$ hai. Agar volume fixed $E, N$ par double kar diya jaaye, toh $Ω$ kis factor se change hoga, aur entropy kitni change hogi? $N = 1$ mole $= 6.022 \times 1 0^{23}$ lo; $Δ S$ J/K mein do.

Recall Solution

Pehle, hume jo constant chahiye. Boltzmann's constant $k$ (kabhi $k_{B}$ likha jaata hai) woh fixed conversion factor hai jo "microstates ki count" ko thermodynamic entropy mein $S=k\ln\Omega$ mein convert karta hai. Iska value hai $k = 1.381 \times 1 0^{- 23} J/K .$ Yeh tiny hai kyunki ek single microstate mein sirf ek minuscule amount of entropy hoti hai; everyday J/K tak pahunchne ke liye tumhe Avogadro-many ki zaroorat hai.

Ab counting. $Ω \propto V^{N}$ , toh $V$ double karne par $Ω$ $2^{N}$ se multiply hota hai (har $N$ particle independently do guna room pata hai, aur choices multiply hoti hain): $\frac{Ω _{2}}{Ω _{1}} = (\frac{2 V}{V})^{N} = 2^{N} .$ Entropy change: $Δ S = k ln \frac{Ω _{2}}{Ω _{1}} = k ln 2^{N} = N k ln 2.$ Numerically $Δ S = (6.022 \times 1 0^{23}) (1.381 \times 1 0^{- 23}) (0.6931) \approx 5.76 J/K .$ Yeh free expansion ki textbook entropy hai — aur yeh seedha position cells $V^{N}$ count karne se nikli. Counting thermodynamics predict karta hai.

L5 — Mastery

Exercise 5.1 (L5)

Ek macroscopic sample ke liye 1% relative fluctuation ki probability estimate karo, taaki equilibrium ki crushing sharpness feel ho sake. $N = 1 0^{20}$ spins ke liye, binomial ka Gaussian approximation use karo: $P (n) \approx \frac{1}{2 π σ} exp [- \frac{( n - μ ) ^{2}}{2 σ ^{2}}]$ jisme $μ = N /2$ , $σ = \frac{1}{2} N$ . $n - μ = 0.01 μ$ (yaani mean ka 1%) ke fluctuation ke liye exponent compute karo aur interpret karo.

Recall Solution

1% fluctuation ka matlab hai $n - μ = 0.01 μ = 0.01 \cdot \frac{N}{2} = 0.005 N$ . Isse $σ = \frac{1}{2} N$ ke units mein measure karo: standard deviations ki count hai $z = \frac{n - μ}{σ} = \frac{0.005 N}{\frac{1}{2} N} = 0.01 N .$ $N = 1 0^{20}$ ke liye: $z = 0.01 \times 1 0^{10} = 1 0^{8}$ . Exponent hai $- z^{2} /2 = - \frac{1}{2} (1 0^{8})^{2} = - 5 \times 1 0^{15}$ . Toh $P \sim e^{- 5 \times 1 0^{15}}$ — itni chhoti probability ki iska koi physical meaning nahi. Ek macroscopic sample mein 1% fluctuation universe ki age mein kabhi nahi hota. Yeh irreversibility ka statistical bedrock hai.

Exercise 5.2 (L5)

Peak-height scaling. Stirling's approximation $ln N! \approx N ln N - N + \frac{1}{2} ln (2 π N)$ use karke, dikhao ki even $N$ ke liye most probable multiplicity satisfy karti hai $ln (N /2 N) \approx N ln 2 - \frac{1}{2} ln (\frac{π N}{2})$ , aur isliye peak probability $P (N /2) = (N /2 N) / 2^{N} \approx \frac{2}{π N}$ . $N = 100$ ke liye evaluate karo aur exact value se compare karo.

Recall Solution

Step 1 — central binomial ka log likho. Kyunki $(N /2 N) = \frac{N !}{[( N /2 )! ] ^{2}}$ , $ln (N /2 N) = ln N! - 2 ln (\frac{N}{2})! .$ Step 2 — Stirling ko uske $\frac{1}{2} ln (2 π N)$ tail ke saath insert karo (hum yeh tail rakhte hain kyunki yahi end mein $π N$ factor produce karta hai). Numerator ke liye, $ln N! \approx N ln N - N + \frac{1}{2} ln (2 π N) .$ Denominator mein har factorial ke liye, $N \to N /2$ replace karo: $ln (\frac{N}{2})! \approx \frac{N}{2} ln \frac{N}{2} - \frac{N}{2} + \frac{1}{2} ln (2 π \frac{N}{2}) = \frac{N}{2} ln \frac{N}{2} - \frac{N}{2} + \frac{1}{2} ln (π N) .$ Step 3 — subtract karo. Denominator line ko 2 se multiply karo aur numerator line se subtract karo: $ln (N /2 N) \approx [N ln N - N + \frac{1}{2} ln (2 π N)] - [N ln \frac{N}{2} - N + ln (π N)] .$ $- N$ aur $+ N$ cancel ho jaate hain. "Big" logs group karo: $N ln N - N ln \frac{N}{2} = N ln \frac{N}{N /2} = N ln 2$ . Yahi se $N ln 2$ ( $2^{N}$ ke same scale ka) aata hai. Step 4 — bacha hua $\frac{1}{2} ln$ terms collect karo, jo $π N /2$ produce karte hain: $\frac{1}{2} ln (2 π N) - ln (π N) = \frac{1}{2} ln (2 π N) - \frac{1}{2} ln ((π N)^{2}) = \frac{1}{2} ln \frac{2 π N}{( π N ) ^{2}} = \frac{1}{2} ln \frac{2}{π N} = - \frac{1}{2} ln \frac{π N}{2} .$ Steps 3–4 ko mila kar: $ln (N /2 N) \approx N ln 2 - \frac{1}{2} ln (\frac{π N}{2}) .$ Step 5 — $2^{N}$ se divide karo ( $N ln 2$ subtract karo) peak probability paane ke liye: $ln P (N /2) \approx - \frac{1}{2} ln (\frac{π N}{2}) ⟹ P (N /2) \approx \frac{2}{π N} .$ Step 6 — evaluate karo. $N = 100$ ke liye: $2/ (100 π) = 0.006366 = 0.07979 \approx 0.0798.$ Exact: $(50 100) / 2^{100} = 0.07959$ . $\sim 0.2%$ tak agreement — Stirling nail kar deta hai. Mastery insight: peak probability $1/ N$ ki tarah girta hai, phir bhi peak dominate karta hai kyunki har doosra macrostate aur bhi tezi se girta hai — distribution overall $1/ N$ relative width ki spike mein collapse ho jaati hai (Exercise 3.2). Yeh action mein binomial → Gaussian crossover hai.

Exercise 5.3 (L5)

Conceptual capstone. Ek paragraph mein, apne L1 wale self ko explain karo ki "equilibrium sabse probable macrostate hai" koi physical law nahi hai jo tumhe postulate karna pade, balki yeh counting se forced near-certainty hai — aur ek maatra genuine postulate kahan enter hota hai.

Recall Solution

Microstates constantly change hote rehte hain (ups ka microscopic pattern kabhi still nahi hota). Ek maatra postulate yeh hai ki equilibrium mein har accessible microstate equally likely hai — koi pattern favoured nahi. Yeh even-handedness dene par, macrostates equal nahi hote: ek macrostate ka chance sirf uska microstates ka share hai, $Ω (n) / 2^{N}$ . Kyunki $Ω$ balanced count par astronomically peaked hai ( $(n N)$ tower), almost har microstate usi ek macrostate ka hissa hai. Toh tum isse decree se nahi balki sheer weight of numbers se dekhte ho — relative width $1/ N$ ki tarah shrink karti hai jab tak koi bhi deviation observe karna impossible na ho jaaye. "Equilibrium" simply hai "jahan microstates overwhelmingly hain." Ek akela assumption hai equal microstate probability; baaki har conclusion — peak, uski sharpness, irreversibility — pure arithmetic hai jo us ek even-handed rule par sawaar hai.

Recall Self-test summary (sirf finish karne ke baad kholo)

L1: microstate = ordered pattern; macrostate = ek count. Peak middle mein, valleys extremes par.
L2: total $= 2^{N}$ ; $Ω (n) = (n N)$ ; $P = Ω/ 2^{N}$ .
L3: $Ω$ ke ratios probability drive karte hain; relative width $= 1/ N$ shrink hoti hai.
L4: independent systems → $Ω$ multiply karo, $S = k ln Ω$ add karo; gas $Ω \propto V^{N} E^{3 N /2}$ $V$ double karne par $Δ S = N k ln 2$ deta hai.
L5: Gaussian/Stirling spike quantify karte hain; ek maatra postulate hai equal microstate probability.

Connections

Entropy and Boltzmann's relation S = k ln Ω
Fundamental postulate of statistical mechanics
Binomial and Gaussian distributions
Microcanonical ensemble
Second law and irreversibility