Worked examples — Statistical mechanics — microstate, macrostate

2.4.8 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Statistical mechanics — microstate, macrostate

Setup ke liye parent topic dekho, aur peak ki maths ke liye Binomial and Gaussian distributions dekho.

---

Scenario matrix

Is topic ka har counting problem in cells mein se kisi ek mein aata hai. Hum har ek ko cover karenge.

Cell	Kya special hai	Example jo cover karta hai
A — Edge macrostate ($n=0$ ya $n=N$)	Sirf $\Omega =1$ microstate; picture ka ek "corner"	Ex 1
B — Peak macrostate ($n=N/2$)	Sabse bada $\Omega$; yahi equilibrium hai	Ex 2
C — Off-peak, generic $n$	Ordinary combination; peak ke saath ratios	Ex 3
D — Degenerate input ($N=0$, $N=1$, empty)	Formula toot na jaaye; $0!=1$ check karo	Ex 4
E — Large-$N$ limiting behaviour	Peak sharpen hota hai $\sim 1/\sqrt{N}$; Gaussian shape	Ex 5
F — Biased two-state (uneven weights)	Microstates a priori equally likely nahi — subtle!	Ex 6
G — Continuous system (gas: $V,E$ scaling)	Lists nahi, phase space mein volumes count karo	Ex 7
H — Real-world word problem	Everyday shaking ko $\Omega$ mein translate karo	Ex 8
I — Exam twist (composite / multi-state)	Do subsystems combine karo; multiplicities multiply hote hain	Ex 9

Kuch bhi shuru hone se pehle, ek figure jo coin/spin problem ki puri geography fix kar deta hai.

---

Ex 1 — Cell A: corner macrostate

Forecast: Padhne se pehle guess karo — kya yeh $6$ hai? $36$? Kuch aur bada?

1. Multiplicity likho. Macrostate "$n$ up out of $N$" ka $\Omega (n)=\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ hai. Yeh step kyun? Yeh wohi counting rule hai jo parent ne derive ki thi: choose karo ki $N$ slots mein se kaun se $n$ up hain.
2. $n=N=6$ plug karo. $\Omega (6)=\left(\genfrac{}{}{0ex}{}{6}{6}\right)=\frac{6!}{6!0!}=\frac{720}{720\cdot 1}=1.$ Yeh step kyun? Har slot force hoke up hai — literally ek hi sequence hai $\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow$.
3. $n=0$ plug karo. $\Omega (0)=\left(\genfrac{}{}{0ex}{}{6}{0}\right)=\frac{6!}{0!6!}=1.$ Yeh step kyun? "Sab down" bhi ek single sequence hai. Formula $1$ deta hai sirf isliye ki $0!=1$ — woh convention hi edges ko sahi banati hai.

Verify: $\left(\genfrac{}{}{0ex}{}{6}{6}\right)=\left(\genfrac{}{}{0ex}{}{6}{0}\right)=1$, aur dono row $N=6$ ki sabse choti entries hain (figure mein pyramid dekho). ✓

---

Ex 2 — Cell B: peak (equilibrium)

Forecast: Peak $n=N/2=3$ par hai. Compute karne se pehle $\Omega (3)$ guess karo.

1. Middle evaluate karo. $\Omega (3)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{720}{6\cdot 6}=\frac{720}{36}=20.$ Yeh step kyun? $n=3$ se $6$ slots evenly split hote hain; $3!$ ups ko order karta hai aur $3!$ downs ko, dono divide ho jaate hain.
2. Corner se compare karo. $\Omega (3)/\Omega (6)=20/1=20$. Yeh step kyun? Ek tiny system ke liye bhi, "half up" ke already 20× zyada tarike hain "all up" se. Yahi ratio equilibrium ka reason hai.
3. Peak ki probability. $P(3)=\Omega (3)/2^N=20/2^6=20/64=\mathrm{0.3125.}$ Yeh step kyun? Fundamental postulate of statistical mechanics ke mutabik, $2^6=64$ microstates mein se har ek equally likely hai, isliye macrostate ki probability uska share $\Omega /2^N$ hai.

Verify: Pascal's triangle ki row-6 hai $1,6,15,20,15,6,1$; sum $=64=2^6$, aur $20$ sach mein max hai. ✓

---

Ex 3 — Cell C: ek off-peak macrostate aur uske odds

Forecast: Peak $\Omega (5)$ se bada ya chota? Kitna?

1. $\Omega (2)$ compute karo. $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=\frac{10!}{2!8!}=\frac{10\cdot 9}{2}=45.$ Yeh step kyun? Choose karo ki 10 mein se kaun se 2 slots heads hain; $10\cdot 9$ ordered picks aadhe ho jaate hain kyunki 2 chosen slots interchangeable hain.
2. Peak $\Omega (5)$ compute karo. $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=\frac{10!}{5!5!}=252.$ Yeh step kyun? Compare karne ke liye kuch chahiye — equilibrium value.
3. Probability. $P(2)=45/2^{10}=45/1024\approx \mathrm{0.0439.}$ Yeh step kyun? Wahi $\Omega /2^N$ rule. Off-peak macrostates rare hain par impossible nahi.
4. Peak se ratio. $45/252\approx 0.179$ — sirf $N=10$ ke liye, half-heads se already lagbhag 6× kam likely hai. Yeh step kyun? Dikhata hai ki modest $N$ par bhi peak dominate karta hai; Ex 5 ka preview hai.

Verify: $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=45$, $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, $45/1024\approx 0.04395$. ✓

---

Ex 4 — Cell D: degenerate tiny systems

Forecast: Ek-coin system — kitne macrostates, kitne total microstates?

1. $N=1$: macrostates list karo. $n=0$ (down) aur $n=1$ (up). Yeh step kyun? Ek object ke saath sirf do "kitne up" values hoti hain.
2. Unki multiplicities. $\left(\genfrac{}{}{0ex}{}{1}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{1}{1}\right)=1$; total $=1+1=2=2^1$. ✓ Yeh step kyun? Confirm karta hai ki ${\sum }_n\left(\genfrac{}{}{0ex}{}{N}{n}\right)=2^N$ sabse chote nontrivial size par bhi hold karta hai.
3. $N=0$: empty system. Ek macrostate hai ($n=0$) jiska $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=\frac{0!}{0!0!}=1$ hai, aur $2^0=1$ total microstate. Yeh step kyun? "Empty configuration" ek legitimate single microstate hai. Multiplicity yahan kabhi $0$ nahi — machine gracefully degrade karta hai.

Verify: $\left(\genfrac{}{}{0ex}{}{1}{0}\right)+\left(\genfrac{}{}{0ex}{}{1}{1}\right)=2=2^1$ aur $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1=2^0$. ✓

---

Ex 5 — Cell E: large-$N$ limit aur sharp peak

Forecast: Kya $P(50)$ $1$ ke paas hoga, ya chota? (Trap: yeh actually chota hai — padhte raho.)

1. Peak probability. $P(50)=\left(\genfrac{}{}{0ex}{}{100}{50}\right)/2^{100}\approx \mathrm{0.0796.}$ Yeh step kyun? Sabse zyada likely single macrostate bhi 8% se neeche hai. Peak $n$ mein broad hai par $N$ ke relative narrow hai — yeh do alag baatein hain.
2. Relative width. Fair two-state system ke liye $n$ ka standard deviation $\sigma =\frac{1}{2}\sqrt{N}$ hai. Yahan $\sigma =\frac{1}{2}\sqrt{100}=5$. Yeh step kyun? Binomial spread $\sqrt{N}$ ki tarah badhta hai jabki centre $n=N/2$ $N$ ki tarah badhta hai.
3. Fractional spread. $\frac{\sigma }{N/2}=\frac{5}{50}=0.10$, yaani $10\%$. Yeh step kyun? Yeh $1/\sqrt{N}=1/10$ hai. $N=10^{23}$ ke liye yeh $\sim 3\times 10^{-12}$ ho jaata hai — tum sirf aur sirf peak measure karte ho. Yahi equilibrium hai.

Verify: $\sigma =\frac{1}{2}\sqrt{100}=5$, relative spread $=1/\sqrt{100}=0.1$, aur $\left(\genfrac{}{}{0ex}{}{100}{50}\right)/2^{100}\approx 0.0796$. ✓

---

Ex 6 — Cell F: biased objects (microstates equally likely NAHI)

Forecast: Kya $\Omega (3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ phir bhi aata hai? Haan — par yeh poori kahani nahi hai.

1. Sequences count karo. 3 heads wale $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ ordered patterns phir bhi hain. Yeh step kyun? Arrangement ki combinatorics unchanged hai — bias se patterns ki sankhya nahi badalti.
2. Har pattern ko weight do. Har specific "3H,2T" sequence ki ab probability $p^3{q}^2=0.7^3\cdot 0.3^2=0.343\cdot 0.09=0.03087$ hai. Yeh step kyun? Jab outcomes biased hoon, microstates equally likely nahi hote, isliye hum sirf $2^N$ se divide nahi kar sakte. Hum count ko per-sequence weight se multiply karte hain.
3. Combine karo. $P(3\text{ H})=\left(\genfrac{}{}{0ex}{}{5}{3}\right)p^3{q}^2=10\times 0.03087=\mathrm{0.3087.}$ Yeh step kyun? Yeh general binomial distribution hai; fair coin special case hai $p=q=\frac{1}{2}$, jahan $p^n{q}^{N-n}=2^{-N}$ har sequence ke liye hota hai aur hum $\Omega /2^N$ recover karte hain.

Verify: $10\cdot 0.7^3\cdot 0.3^2=0.3087$. ✓

---

Ex 7 — Cell G: ek continuous system (ideal gas)

Forecast: Multiplicity $2^{\text{(kuch)}}$ se badhti hai — kya exponent $N$, $3N/2$, ya $2N$ hai?

1. $V$-dependence isolate karo. Sirf $V^N$ badalta hai. $\frac{{\Omega }_{2V}}{{\Omega }_V}=\frac{(2V{)}^N}{V^N}=2^N.$ Yeh step kyun? $N$ particles mein se har ek independently doubled position space access karta hai — factors multiply hote hain, har particle ke liye $2$ ka ek factor milta hai.
2. $N=100$ plug karo. Factor $=2^{100}\approx 1.27\times 10^{30}$. Yeh step kyun? Concrete size: sirf 100 particles box double karne se $10^{30}$ aur microstates gain karte hain. Isisi wajah se gas available volume fill karne dauda hai.
3. Log lo (entropy language mein). $\ln (2^{100})=100\ln 2\approx \mathrm{69.3.}$ Yeh step kyun? Entropy $S=k\ln \Omega$ hai (Entropy and Boltzmann's relation S = k ln Ω), isliye $\Delta S=k\cdot 100\ln 2$ — ek positive jump, second law se match karta hai.

Verify: $(2V{)}^{100}/V^{100}=2^{100}$ aur $\ln 2^{100}=100\ln 2\approx 69.31$. ✓

---

Ex 8 — Cell H: real-world word problem

Forecast: Coin-flip ke 50% se zyada ya kam? Sum karne se pehle guess karo.

1. Model banao. Har marble ek fair two-state object hai (L/R). "Left par number" $=n$ macrostate hai; total microstates $=2^{20}=1,048,576$. Yeh step kyun? Wahi coin machine — "marbles" sirf costume hai.
2. Favourable multiplicities sum karo. Balanced matlab $n\in \{8,9,10,11,12\}$: ${\Omega }_{\text{bal}}=\left(\genfrac{}{}{0ex}{}{20}{8}\right)+\left(\genfrac{}{}{0ex}{}{20}{9}\right)+\left(\genfrac{}{}{0ex}{}{20}{10}\right)+\left(\genfrac{}{}{0ex}{}{20}{11}\right)+\left(\genfrac{}{}{0ex}{}{20}{12}\right).$ Yeh step kyun? Macrostate probabilities add hoti hain kyunki macrostates mutually exclusive hain; har ek apna microstate count contribute karta hai.
3. Numbers. $125970+167960+184756+167960+125970=772616.$ Yeh step kyun? Symmetry note karo $\left(\genfrac{}{}{0ex}{}{20}{8}\right)=\left(\genfrac{}{}{0ex}{}{20}{12}\right)$ — distribution $n=10$ ke aas-paas mirror-symmetric hai.
4. Probability. $P=772616/1048576\approx \mathrm{0.737.}$ Yeh step kyun? Nearly 74% — balanced band probability ka bada hissa capture karta hai, theek isliye kyunki peak ke paas ke macrostates dominate karte hain (Ex 5 ka lesson disguise mein).

Verify: $\left(\genfrac{}{}{0ex}{}{20}{8}\right)=125970$, sum $=772616$, aur $772616/2^{20}\approx 0.7368$. ✓

---

Ex 9 — Cell I: exam twist (composite system)

Forecast: Do counts multiply hote hain ya add? Aur kya "joined" count "separate" count se bada ya chota hai?

1. Alag multiplicities. ${\Omega }_A=\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3,{\Omega }_B=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6.$ Yeh step kyun? Har subsystem wahi rule se independently count hota hai.
2. Combine karo — multiply karo. ${\Omega }_{A+B}^{\text{(fixed each)}}={\Omega }_A\cdot {\Omega }_B=3\times 6=18.$ Yeh step kyun? Independent choices multiply hoti hain (fundamental counting principle): $A$ ke 3 mein se koi bhi pattern $B$ ke 6 mein se kisi ke saath bhi pair ho sakta hai.
3. Ab constraint relax karo. Agar sirf total $n=3$ out of $N=7$ fixed hai, toh ${\Omega }_{\text{total}}=\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35.$ Yeh step kyun? Internal partition hatane se naye microstates khulte hain (jaise $n_A=0,n_B=3$) jo rigid split ne forbid kiye the.
4. Interpret karo. $35>18$: systems ko share karne dene se microstate count badhta hai. Isliye heat flow karta hai — combined system us arrangement ki taraf slide karta hai jahan sabse zyada microstates hain (Temperature as ∂S/∂E aur Microcanonical ensemble dekho). Yeh step kyun? Pure counting ko spontaneous change ki direction se connect karta hai.

Verify: $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)=3\cdot 6=18$; $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$; aur identity ${\sum }_k\left(\genfrac{}{}{0ex}{}{3}{k}\right)\left(\genfrac{}{}{0ex}{}{4}{3-k}\right)=\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ (Vandermonde) confirm karta hai ki relaxed count saari internal splits collect karta hai. ✓

---

Recall Kaun sa cell kaun sa tha?

Edge $n=0,N$ ::: Ex 1 (Cell A) Peak / equilibrium ::: Ex 2 (Cell B) Generic off-peak ratio ::: Ex 3 (Cell C) Degenerate $N=0,1$ ::: Ex 4 (Cell D) Large-$N$ sharpening $1/\sqrt{N}$ ::: Ex 5 (Cell E) Biased (weighted) microstates ::: Ex 6 (Cell F) Continuous gas $V^N{E}^{3N/2}$ ::: Ex 7 (Cell G) Word problem ::: Ex 8 (Cell H) Composite / multiplied multiplicities ::: Ex 9 (Cell I)

---

Connections

• Parent — microstate & macrostate
• Binomial and Gaussian distributions
• Fundamental postulate of statistical mechanics
• Entropy and Boltzmann's relation S = k ln Ω
• Second law and irreversibility
• Microcanonical ensemble
• Temperature as ∂S/∂E
• Phase space and Liouville's theorem