Boltzmann's entropy S = k_B ln(Ω): Worked examples

2.4.9 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Boltzmann's entropy S = k_B ln(Ω)

Intuition Yeh page kya hai

Parent note ne $S = k_{B} ln Ω$ ko scratch se banaya tha. Yahan hum use stress-test karte hain: hum formula par har tarah ki situation throw karte hain — chhote discrete systems, bade continuous wale, woh weird cases jahan entropy shrink karti hai, aur woh exam trick jo sabko pakadti hai. Agar koi bhi scenario exist karta hai, uske liye neeche ek worked cell hai.

Un symbols ki reminders jo hum use karte rahenge (sab parent mein define hain):

$Ω$ = microstates ki sankhya jo macrostate ke saath consistent hain (plain "tareekon" ki ginti).
$S = k_{B} ln Ω$ = entropy; $k_{B} = 1.380649 \times 1 0^{- 23} J/K$ .
$ln$ = natural logarithm — woh function jo products ko sums mein badalta hai (yahi pura reason hai ki yeh appear karta hai).
$N$ = particles ki sankhya; $n$ = moles ki sankhya.
$N_{A}$ = Avogadro's number $= 6.022 \times 1 0^{23}$ particles per mole — ek mole mein particles ki ginti; yeh ek per-particle result ko per-mole mein convert karta hai.
$R = N_{A} k_{B} = 8.314 J/(mol\cdotK)$ = gas constant (Boltzmann's constant ko per particle ki jagah per mole tak scale up kiya gaya).
$T$ = kelvin (K) mein absolute temperature. $Q$ = heat, woh energy jo thermally transfer hoti hai (joules, J mein).
$h$ = Planck's constant $= 6.626 \times 1 0^{- 34} J\cdots$ — phase space ka chhota "grain size." Hum ise Example 10 mein milte hain; yeh fix karta hai ki ek microstate kitna bada hota hai jab position aur momentum continuous hote hain.

The scenario matrix

$S = k_{B} ln Ω$ ke baare mein har problem inhi cells mein se kisi ek mein aati hai. Neeche ke examples mein se har ek ek tag carry karta hai jo batata hai ki woh kaun si cell discharge karta hai.

Cell	Kya ise alag banata hai	Chhupa hua khatara	Example
A. Discrete counting	Microstates haath se gino	Micro/macro ko confuse karna	Ex 1
B. Ratio of counts	Sirf $Ω_{f} / Ω_{i}$ matter karta hai	Absolute $Ω$ dhundhne ki koshish karna	Ex 2
C. Additivity check	Do systems ko join karo	$Ω$ add karna multiply karne ki jagah	Ex 3
D. Degenerate: $Ω = 1$	Ek akela microstate	Yeh sochna ki $ln 1 \neq = 0$	Ex 4
E. Large- $N$ / Stirling	$N$ astronomically bada hai	States list nahi kar sakte	Ex 5
F. Real-world word problem	Physical melting/mixing	Bhool jaana ki kya count hota hai	Ex 6
G. Sign flip: $S$ decreases	Capped energy, $\partial S / \partial E < 0$	"Zyada energy ⇒ zyada $S$ "	Ex 7
H. Exam twist: identical particles	$N!$ overcount	Gibbs paradox	Ex 8
I. Limiting behaviour	$E \to 0$ , $N \to \infty$ , $T \to 0$	Ill-defined limits	Ex 9
J. Continuous phase space	Integrals, $h^{3 N}$ chahiye	Non-dimensionless $Ω$	Ex 10

Example 1 — Cell A: Discrete counting from scratch

Worked example Chaar coins, macrostate = "heads ki sankhya"

Chaar fair coins hain. Woh macrostate jo tum dekh sakte ho woh hai "kitne heads hain." Macrostate "exactly 2 heads" ke liye $Ω$ aur $S$ dhundho. Kis macrostate ki entropy sabse zyada hai?

Forecast: abhi guess karo — kya "2 heads" "4 heads" se zyada ya kam likely hai? Kitna?

Steps

2 heads ke liye microstates list karo: ${H H T T, H T H T, H T T H, T H H T, T H T H, T T H H}$ → $Ω = 6$ . Yeh step kyun? Ek microstate har coin specify karta hai; macrostate sirf heads count karta hai. Ginti karna $Ω$ ka raw meaning hai.
Combination formula use karo $Ω = (k N) = \frac{N !}{k ! ( N - k )!} = \frac{4 !}{2 ! 2 !} = 6$ . Yeh step kyun? 4 coins ke liye listing kaam karti hai lekin 40 ke liye nahi — binomial coefficient hi count hai un tareekon ka jo choose karti hai ki kaun si positions heads hain.
Us macrostate ki entropy: $S = k_{B} ln 6 = 1.792 k_{B}$ . Yeh step kyun? $S = k_{B} ln Ω$ mein direct substitution.
"4 heads" compare karo: $Ω = (4 4) = 1$ , toh $S = k_{B} ln 1 = 0$ . Yeh step kyun? Dikhata hai ki extreme, ordered macrostate ki entropy zero hoti hai — yeh tidy-room case hai.

Verify: sab macrostates ka total $\sum_{k} (k 4) = 1 + 4 + 6 + 4 + 1 = 16 = 2^{4}$ hai ✓ (har ek $2^{4}$ raw microstates exactly ek baar count hota hai). "2 heads" ( $Ω = 6$ ) "4 heads" ( $Ω = 1$ ) ko beat karta hai — balanced macrostate ke sabse zyada ways hain, isliye highest $S$ hai.

Example 2 — Cell B: Sirf ratio survive karta hai

Worked example Volume triple hone tak free expansion

$N = 3.0 \times 1 0^{22}$ molecules ki gas freely expand karti hai (koi heat nahi, koi work nahi) toh uska accessible volume triple ho jaata hai. $Δ S$ dhundho.

Forecast: kya $Δ S$ microstates ki absolute sankhya par depend karta hai, ya sirf is par ki woh kaise bade?

Steps

Molecule ke per position-choices volume ke saath scale karti hain: $V$ triple karne se har molecule ke position options 3 se multiply hote hain, toh $Ω_{f} = 3^{N} Ω_{i}$ . Yeh step kyun? Har molecule independently choose karta hai → independent choices multiply hote hain (dice logic).
Entropy change: $Δ S = k_{B} ln Ω_{f} - k_{B} ln Ω_{i} = k_{B} ln \frac{Ω _{f}}{Ω _{i}} = k_{B} ln 3^{N} = N k_{B} ln 3$ . Yeh step kyun? Unknown absolute $Ω_{i}$ cancel ho jaata hai — ratio ka log sirf ratio chahiye. Isliye free expansion tractable hai.
Numbers: $Δ S = (3.0 \times 1 0^{22}) (1.380649 \times 1 0^{- 23}) (ln 3) = 0.455 J/K$ . Yeh step kyun? $ln 3 = 1.0986$ plug in karo.

Verify: thermodynamics $Δ S = n R ln (V_{f} / V_{i})$ deta hai. Yahan $n = N / N_{A} = 0.0498$ mol, toh $n R ln 3 = 0.0498 \times 8.314 \times 1.0986 = 0.455 J/K$ ✓ — counting picture classical thermo ko exactly reproduce karta hai.

Example 3 — Cell C: Additivity (counts multiply karo, entropies add karo)

Worked example Do boxes ko join karna

Box $A$ ke $Ω_{A} = 2^{10}$ microstates hain; box $B$ ke $Ω_{B} = 2^{15}$ hain. Woh independent hain. Combined entropy dhundho aur confirm karo ki woh $S_{A} + S_{B}$ ke barabar hai.

Forecast: kya $Ω_{A B}$ $2^{10} + 2^{15}$ ke barabar hai ya $2^{10} \times 2^{15}$ ?

Steps

Combined count: $Ω_{A B} = Ω_{A} Ω_{B} = 2^{10} \cdot 2^{15} = 2^{25}$ . Yeh step kyun? $A$ ke har ek state ke liye, $B$ apni kisi bhi state mein ho sakta hai → multiply karo, kabhi add nahi.
Combined entropy: $S_{A B} = k_{B} ln 2^{25} = 25 k_{B} ln 2$ . Yeh step kyun? Seedha substitution.
Alag-alag entropies: $S_{A} = k_{B} ln 2^{10} = 10 k_{B} ln 2$ , $S_{B} = 15 k_{B} ln 2$ ; sum $= 25 k_{B} ln 2$ . Yeh step kyun? Yahi payoff hai: $ln$ ne product $2^{25}$ ko sum $10 + 15$ mein badal diya.

Verify: $S_{A} + S_{B} = 25 k_{B} ln 2 = S_{A B}$ ✓. Additivity exactly logarithm ki wajah se hold karta hai — galat "counts add karo" rule nonsensical $ln (2^{10} + 2^{15})$ deta.

Example 4 — Cell D: Degenerate $Ω = 1$ case

Worked example Absolute zero par perfect crystal

Ek perfect crystal $T = 0$ par exactly ek lowest-energy arrangement mein hai. $S$ dhundho. Yeh kya predict karta hai?

Forecast: kya $ln 1$ zero hai, ek hai, ya undefined hai?

Steps

Ek microstate ⇒ $Ω = 1$ . Yeh step kyun? $T = 0$ par har particle unique ground configuration mein locked hai; sirf ek "tarika" hai.
$S = k_{B} ln 1 = 0$ , kyunki $ln 1 = 0$ (ek "ek tarike" ka log zero hai — koi chhupa hua multiplicity nahi, koi ignorance nahi). Yeh step kyun? Yeh boundary condition $f (1) = 0$ hai jisne parent ki derivation mein constant $C$ ko khatam kiya.

Verify: $ln 1 = 0$ exactly ✓. Yeh Third Law of Thermodynamics derive karta hai: perfect crystal ke liye entropy $\to 0$ jab $T \to 0$ . (Ek crystal jo residual disorder ke saath, jaise CO ice, $Ω > 1$ aur non-zero residual entropy rakhta hai — formula use bhi handle karta hai.)

Example 5 — Cell E: Large- $N$ , Stirling rescue karta hai

Worked example Two-state spins, most-probable macrostate

$N = 1 0^{24}$ non-interacting spins, har ek up ya down. Macrostate "exactly half up" ka $S$ dhundho aur confirm karo ki woh maximum $N k_{B} ln 2$ ke kareeb hai.

Forecast: $1 0^{24}$ spins ke saath tum states list nahi kar sakte. Kaun sa tool giant factorial ko kuch usable mein convert karta hai?

Steps

Count: $Ω = (N /2 N) = \frac{N !}{( N /2 )! ( N /2 )!}$ . Yeh step kyun? Same "choose which spins are up" logic jaise Ex 1 ke coins mein, ab enormous hai.
Log lo aur Stirling's approximation $ln M! \approx M ln M - M$ apply karo (huge $M$ ke liye valid). Yeh step kyun? $1 0^{24}$ ke factorials uncomputable hain; Stirling $ln (N!)$ ko simple arithmetic mein badal deta hai. Isliye statistical mechanics ko yeh chahiye — yeh counting se smooth thermodynamics tak ka bridge hai.
$ln Ω \approx N ln N - N - 2 [\frac{N}{2} ln \frac{N}{2} - \frac{N}{2}] = N ln N - N ln \frac{N}{2} = N ln 2$ . Yeh step kyun? $- N$ terms cancel hote hain; $ln N - ln (N /2) = ln 2$ .
Toh $S = k_{B} ln Ω \approx N k_{B} ln 2$ . Yeh step kyun? Yeh $N$ free two-state spins ki entropy ke barabar hai — matlab single "half-up" macrostate essentially sab microstates contain karta hai.

Verify (small- $N$ sanity): $N = 4$ ke liye, $(2 4) = 6$ aur $4 ln 2 \approx 2.77$ deta hai $e^{2.77} \approx 16 = 2^{4}$ ; Stirling chhote $N$ ke liye thoda overshoot karta hai (6 vs 16) lekin fraction $6/16$ log-scale dominance ki taraf approach karta hai jab $N$ badhta hai ✓. Approximation macroscopic $N$ ke liye log-mein-exact ban jaati hai.

Example 6 — Cell F: Real-world word problem (mixing / melting)

Worked example

$0^{\circ}$ C par ek mole ice ka melting Ice $T = 273.15$ K par latent heat $L_{f} = 6.01 kJ/mol$ ke saath water ban jaata hai. (a) Thermodynamics se $Δ S$ dhundho. (b) Ise per molecule microstate count $Ω_{water} / Ω_{ice}$ mein change ke roop mein interpret karo.

Forecast: liquid molecules ghoomte hain; crystal molecules pinned hain. Kya $Ω$ melting par upar ya neeche jaayega?

Steps

Constant $T$ par reversible melting: heat $Q$ (thermally absorb ki gayi energy, joules mein) latent heat $L_{f}$ ke barabar hai, toh $Δ S = \frac{Q}{T} = \frac{L _{f}}{T} = \frac{6010}{273.15} = 22.0 J/K$ (per mole). Yeh step kyun? Phase transition par heat fixed temperature $T$ par reversibly enter karti hai, toh $Δ S = Q / T$ directly (yeh $Q / T$ entropy change ki thermodynamic definition hai).
Counting se connect karo: $Δ S = k_{B} ln \frac{Ω _{water}}{Ω _{ice}}$ . Per molecule, $N_{A}$ (ek mole mein molecules ki sankhya) se divide karo: $\frac{Δ S}{N _{A}} = k_{B} ln r$ jahan $r$ per-molecule multiplicity ratio hai. Yeh step kyun? Boltzmann's formula macroscopic $Δ S$ ko log ke roop mein padhta hai ki ek molecule kitne zyada tareekon se arrange ho sakta hai.
$r$ ke liye solve karo: $ln r = \frac{Δ S}{N _{A} k _{B}} = \frac{Δ S}{R} = \frac{22.0}{8.314} = 2.646$ , toh $r = e^{2.646} = 14.1$ . Yeh step kyun? Kyunki $R = N_{A} k_{B}$ , per-mole $Δ S$ ko $R$ se divide karna hi per-molecule $ln r$ hai. Har molecule melting par accessible arrangements mein roughly $\sim 14$ ka factor gain karta hai — ek concrete, feel-able number.

Verify: $r$ se $Δ S$ rebuild karo: $R ln 14.1 = 8.314 \times 2.646 = 22.0 J/K$ ✓. Abstract count aur measured latent-heat entropy agree karte hain.

Example 7 — Cell G: Sign flip (entropy energy ke saath decreases karti hai)

Worked example Two-level spins jahan zyada energy ka matlab hai kam tarike

$N$ spins, har ek energy $0$ (down) ya $+ ε$ (up). $m$ = "up" spins ki sankhya, toh total energy $E = m ε$ . Dikhao ki jab $E$ apne maximum ke aadhe se zyada ho jaata hai, $\partial S / \partial E$ negative ho jaata hai — ek negative temperature.

Forecast: hum cheezein heat karte hain entropy raise karne ke liye. Kya energy add karna kabhi $S$ lower kar sakta hai?

Steps

Count: $Ω (m) = (m N)$ , toh $S = k_{B} ln (m N)$ . Yeh step kyun? $E$ fix karna $m$ fix karta hai; multiplicity sirf "choose which $m$ spins are up" hai.
$Ω$ $m = N /2$ par largest hai (half up), aur $m \to N$ ke roop mein shrink karta hai (sab up). Yeh step kyun? $m = N$ par sirf ek tarika hai (sab up), $Ω = 1$ , $S = 0$ — energy cap order force karta hai.
Kyunki $E = m ε$ $m$ ke saath increase karta hai lekin $S$ $m > N /2$ ke liye decrease karta hai, hume $(\frac{\partial S}{\partial E}) < 0$ milta hai. Yeh step kyun? $S$ $ln Ω$ track karta hai, $E$ nahi. Peak ke baad, zyada energy ka matlab hai kam arrangements.
Phir $\frac{1}{T} = \frac{\partial S}{\partial E} < 0 \Rightarrow T < 0$ — ek negative temperature, kisi bhi positive $T$ se hotter. Yeh step kyun? Yeh exactly woh Negative Temperature & Spin Systems regime hai jise parent ki fourth mistake-warning ne flag kiya.

Verify (numbers, $N = 4$ ): $Ω$ vs $m$ : ${1, 4, 6, 4, 1}$ . $m : 2 \to 3$ jaane par, $Ω : 6 \to 4$ gir jaata hai, toh $Δ S = k_{B} ln (4/6) = - 0.405 k_{B} < 0$ jabki $Δ E = + ε > 0$ ✓. $\partial S / \partial E$ ka sign negative hai — confirmed.

Example 8 — Cell H: Exam twist — indistinguishable particles (Gibbs)

Worked example Do identical gases mix karna — paradox

Do boxes, dono mein same gas at same $T, P$ hain, partition hatane se join ho jaate hain. Naïve counting ek "entropy of mixing" $Δ S = 2 N k_{B} ln 2$ predict karta hai. Sahi answer kya hai, aur kyun?

Forecast: jab aap ek identical gas ko identical gas se mix karte hain toh kuch physical nahi badalta. Kya $Δ S$ positive hona chahiye ya zero?

Steps

Naïve (distinguishable) count har half ke expansion ko Ex 2 ki tarah pure volume mein treat karta hai: $2 N$ molecules mein se har ek apna volume double karta hai → $Ω \to 2^{2 N} Ω$ , deta hai $Δ S = 2 N k_{B} ln 2$ . Yeh step kyun? Yeh trap hai: yeh kahega ki identical gases mix karna kuch nahi se entropy create karta hai.
Fix: identical particles indistinguishable hain, toh do molecules swap karna naya microstate nahi hai. Hume count ko har box ke liye $N!$ se divide karna hoga (Gibbs correction). Yeh step kyun? Permutations ka overcounting $Ω$ inflate karta hai; $1/ N!$ exactly fake states remove karta hai. Yahi woh $N!$ hai (aur $h^{3 N}$ grain size, Ex 10 dekho) jo parent ki third mistake-warning se aaya.
Sahi indistinguishable counting se, $Ω_{final}$ per molecule unchanged hai (same density, same $T$ ) → $Δ S = 0$ . Yeh step kyun? Physically kuch nahi hua, toh ek sahi formula ko zero dena chahiye — aur deta hai.

Verify: galat aur sahi mein fark $Δ S_{wrong} - Δ S_{right} = 2 N k_{B} ln 2 - 0 = 2 N k_{B} ln 2$ hai, jo precisely Stirling term $k_{B} ln [(2 N)! / (N! N!)] \approx 2 N k_{B} ln 2$ hai jise $N!$ factors cancel karte hain ✓. Gibbs Paradox & Indistinguishability dekho.

Example 9 — Cell I: Limiting behaviour ( $S$ vs $E$ ki geometry)

Worked example Spin system ke liye

$S (E)$ ki shape, poori tarah se $N$ -spin system ke liye $S$ ko energy $E = m ε$ ke function ke roop mein $E = 0$ se $E = N ε$ tak sketch karo aur interpret karo, dono endpoints aur peak cover karte hue. (Neeche di gayi figure ko apna sketch mano.)

Forecast: $S$ maximum kahan hai, aur dono ends par kya hota hai?

Steps

Endpoint $E = 0$ ( $m = 0$ , sab down): $Ω = 1$ , $S = 0$ . Slope $\partial S / \partial E > 0$ → ordinary positive $T$ , actually $T \to 0^{+}$ . Yeh step kyun? Lowest-energy state unique hai — degenerate ground, zero entropy (ek Third-Law-jaisa edge). Neeche figure mein far left par blue dot dekho.
Middle $E = N ε /2$ ( $m = N /2$ ): $Ω$ maximum, $S$ $N k_{B} ln 2$ par peak karta hai, slope $= 0$ → $1/ T = 0$ → $T = \pm \infty$ . Yeh step kyun? Peak par, energy add karna $S$ change karna band kar deta hai; yeh infinite-temperature crossover hai — figure mein dome ke top par pink dot.
Right endpoint $E = N ε$ ( $m = N$ , sab up): $Ω = 1$ phir se, toh $S = k_{B} ln 1 = 0$ , lekin curve yeh zero neeche aate hue reach karta hai slope $< 0$ ke saath → $T \to 0^{-}$ . Yeh step kyun? Energy cap top ko utna hi ordered banata hai jitna bottom ko — ek single "all up" microstate. Energy rakhne ki aur jagah nahi, toh count ek par wapas collapse ho jaata hai. Yeh figure mein far right par blue dot hai.
Conclusion: poore right half par ( $m > N /2$ ) slope $\partial S / \partial E < 0$ hai, toh $1/ T < 0$ aur $T < 0$ . Negative temperature "zero se thanda" nahi hai — yeh ek capped system ka high-energy aadha hai, sirf isliye reach kiya jaata hai kyunki $S$ wapas neeche aa jaati hai. Endpoint $E = N ε$ extreme negative-temperature state hai ( $T \to 0^{-}$ ), ordinary $T \to 0^{+}$ ground state ka mirror image. Yeh step kyun? Yeh Example 7 ke saath loop close karta hai: woh sign flip jo hum ne algebraically prove kiya tha woh exactly is dome ka falling right side hai, aur uska far end ek doosra $Ω = 1$ , $S = 0$ corner hai.

Verify (numbers, $N = 4$ , $k_{B}$ ke units mein): $S / k_{B} = ln {1, 4, 6, 4, 1} = {0, 1.386, 1.792, 1.386, 0}$ — symmetric, middle par peak $= ln 6 = 1.792$ , dono ends par zeros ✓. Drawn dome se exactly match karta hai.

Example 10 — Cell J: Continuous phase space aur grain size $h$

Worked example Classical gas ko

$h$ kyun chahiye: ek continuous system ko count karna Ek single classical particle of mass $M$ length $L$ ke 1-D box mein rehta hai jisme momentum magnitude $p_{m a x}$ tak hai. Uski "state" continuous phase space mein ek point $(x, p)$ hai (ek axis par position, doosre par momentum). Aap points count nahi kar sakte — infinitely many hain. Hum finite, dimensionless $Ω$ kaise paate hain?

Forecast: agar states ek smooth 2-D region banate hain, toh kaun cheez ek area ko microstates ki pure number mein badlegi?

Steps

Accessible region $(x, p)$ plane mein ek rectangle hai: $x \in [0, L]$ , $p \in [- p_{m a x}, p_{m a x}]$ . Iska area (ek phase-space integral $\int d x d p$ ) $A = L \cdot (2 p_{m a x})$ hai. Yeh step kyun? Continuous mechanics mein ek microstate ek point hai, toh "state ki maatra" ek phase-space area hai, allowed region par integrate karke prapt hoti hai (units: position × momentum = $m\cdotkg\cdotm/s = J\cdots$ ), plain count nahi.
Us area ke units $J\cdots$ hain — same units jaise Planck's constant $h$ . Pure number paane ke liye $h$ se divide karo: $Ω = \frac{A}{h} = \frac{2 L p _{m a x}}{h}$ . Yeh step kyun? $ln Ω$ ke andar $Ω$ dimensionless hona chahiye (aap "itne J·s" ka log nahi le sakte). $h$ ek cell ka natural quantum area hai — ek microstate per degree of freedom $h$ phase-space area occupy karta hai.
$N$ particles ke liye 3-D mein phase space mein $3 N$ position aur $3 N$ momentum axes hain, toh $h^{3 N}$ se divide karo; aur indistinguishable particles ke liye $N!$ se bhi divide karo (Ex 8): $Ω = \frac{1}{N ! h ^{3 N}} \int d^{3 N} x d^{3 N} p$ . Yeh step kyun? $h^{3 N}$ $Ω$ ko dimensionless banata hai ( $3 N$ degrees of freedom mein se har ek $h$ ka ek factor contribute karta hai); $N!$ permutation overcount remove karta hai. Dono milkar wahi hain jo parent ki third mistake-warning demand karti thi, aur dono hi Gibbs' paradox ko khatam karte hain.
Numbers (1-D single particle): $L = 1 m$ , $p_{m a x} = 1.0 \times 1 0^{- 24} kg\cdotm/s$ , $h = 6.626 \times 1 0^{- 34} J\cdots$ lo. Phir $Ω = \frac{2 ( 1 ) ( 1 0 ^{- 24} )}{6.626 \times 1 0 ^{- 34}} = 3.02 \times 1 0^{9}$ , aur $S = k_{B} ln Ω = k_{B} (21.83) = 3.01 \times 1 0^{- 22} J/K$ . Yeh step kyun? Ek concrete finite number nikalta hai — continuous system ab upar wale discrete systems ki tarah countable entropy rakhta hai.

Verify (units): $\frac{[ m ] [ kg\cdotm/s ]}{[ J\cdots ]} = \frac{kg\cdotm ^{2} / s}{kg\cdotm ^{2} / s} = 1$ (dimensionless) ✓. Numerically $ln (3.02 \times 1 0^{9}) = 21.83$ aur $k_{B} \times 21.83 = 3.01 \times 1 0^{- 22} J/K$ ✓. Microcanonical Ensemble se connect hota hai, jahan yeh $h^{3 N} N!$ counting standard recipe hai.

Recall Matrix ka one-line summary

Ways gino ⇒ unka log lo ⇒ ratios unknowns cancel karte hain, products sums ban jaate hain, $Ω = 1$ $0$ deta hai, huge $N$ ko Stirling chahiye, identical particles ko $1/ N!$ chahiye, continuous states ko dimensionless count banane ke liye $/ h^{3 N}$ chahiye, aur ek capped energy $S$ ko wapas neeche laati hai — negative temperature.

Active Recall

N

molecules ke liye

V

triple karne wala free expansion —

Δ S

N k_{B} ln 3 = n R ln 3

Independent systems

2^{10}

aur

2^{15}

ka combined weight?

2^{25}

; entropies add hokar

25 k_{B} ln 2

banti hain.

Ice melne par per-molecule multiplicity ratio (

Δ S = 22 J/K/mol

r = e^{Δ S / R} = e^{2.646} \approx 14

Spin system ka

S (E)

peak kahan hai?

m = N /2

par,

S = N k_{B} ln 2

, jahan

\partial S / \partial E = 0

(

T = \pm \infty

High-energy end

E = N ε

(sab spins up) par

S

kya hai?

Ω = 1

phir se toh

S = 0

; slope

< 0

ke saath reach hota hai, yaani

T \to 0^{-}

(extreme negative temperature).

Identical gases mix karne par

Δ S = 0

kyun?

Indistinguishability (

1/ N!

) naïve

2 N k_{B} ln 2

cancel kar deta hai; kuch physical nahi badla.

Continuous phase-space volume ko

h^{3 N}

se kyun divide karte hain?

Ek microstate per degree of freedom

h

phase-space area occupy karta hai; divide karne se

Ω

dimensionless ho jaata hai toh

ln Ω

well-defined hota hai.

Connections

2.4.09 Boltzmann's entropy S = k_B ln(Ω) (Hinglish)
Microcanonical Ensemble
Temperature as $\partial S/\partial E$
Negative Temperature & Spin Systems
Gibbs Paradox & Indistinguishability
Third Law of Thermodynamics
Shannon Information Entropy