Third law of thermodynamics — S → 0 as T → 0: Visual walkthrough

1.7.25 · D2 · Physics › Thermodynamics › Third law of thermodynamics — S → 0 as T → 0

Step 1 — "Microstate" kya hota hai? Pehle box draw karo

KYA. Kisi bhi formula se pehle, humein us cheez ki ek picture chahiye jo hum count karne waale hain. Socho ek tiny crystal ko atoms ki ek grid ki tarah. Har atom ko kisi na kisi tarah arrange kiya ja sakta hai — yahan, har atom ke paas ek chhota arrow hai (use uski orientation ya wiggle direction kaho). Har arrow ka ek complete choice ek microstate hai: ek specific tarika jisme poora crystal dikh sakta hai.

KYU. Entropy na "heat" hai aur na "temperature" — yeh possibilities ka count hai. Count karne ke liye pehle yeh jaanna zaroori hai ki hum kya count kar rahe hain. Woh "kya" hi microstate hai. Jab tak yeh box exist nahi karta, is page ki koi bhi cheez samajh nahi aayegi.

PICTURE. Figure dekho: wohi crystal (same energy, same temperature — the macrostate) teen alag-alag tarike se draw ki gayi hai. Teeno allowed hain; har ek ek microstate hai. Aisi distinct pictures ki jo sankhya hogi, use hum $W$ kahenge.

Figure — Third law of thermodynamics — S → 0 as T → 0

Step 2 — Pictures count karo: $W$ introduce karo

KYA. Count ko ek naam do. Maano $W = distinct internal pictures number of microstates consistent with the macrostate.$ Yahan $W$ ek plain counting number hai: $W = 1, 2, 3, \dots$ . Agar crystal mein $N$ atoms hain aur har ek 2 taraf point kar sakta hai, to $W = 2^{N}$ (choices ko multiply karo).

KYU. Hum chahte hain ek single number jo measure kare "yeh kitne tarike se dikh sakta hai". Bada $W$ matlab bahut freedom = bahut disorder. Chhota $W$ matlab system pin down ho gaya hai. Yeh woh raw material hai jise Third Law $1$ tak squeeze kar dega.

PICTURE. Figure choices ko multiply hote dikhata hai: 1 atom → 2 pictures, 2 atoms → 4 pictures, 3 atoms → 8 pictures. Har naya atom count ko double kar deta hai, isliye powers aate hain.

Step 3 — Logarithm kyun? Multiplying ko adding mein badlo

KYA. Boltzmann entropy ko define karta hai is tarah: $S = J/K, sets the units k_{B} natural log ln count W .$ $ln$ natural logarithm hai: $ln W$ poochta hai "e ki kaun si power $W$ deti hai?"

KYU yeh tool aur $W$ itself kyun nahi? Do reasons hain, dono picture mein clearly dikhte hain:

Additivity. Do independent crystals ko side by side rakho. Unka combined count multiply hota hai: $W_{tot} = W_{1} W_{2}$ . Lekin hum chahte hain entropy add ho (system double karo → entropy double ho). Logarithm woh ek function hai jo multiplication ko addition mein badal deta hai: $ln (W_{1} W_{2}) = ln W_{1} + ln W_{2}$ . Koi aur simple tool yeh nahi karta.
Explosion ko kabu mein karna. $W = 2^{N}$ astronomically bada hota hai. $ln 2^{N} = N ln 2$ ek sane, size-proportional number hai.

PICTURE. Figure do boxes ko fuse hote dikhata hai: counts neeche multiply hote hain, lekin $ln$ values stack (add) hoti hain. Woh stacking hi poora reason hai ki $ln$ kyun choose kiya gaya.

Recall

$W$ ko directly entropy kyun nahi use karte? Kyunki do independent systems ki entropy add honi chahiye, jabki unke microstate counts multiply hote hain. ::: Logarithm $W_{1} W_{2}$ ko $ln W_{1} + ln W_{2}$ mein convert karta hai, entropy ko additive banata hai.

Step 4 — Temperature matlab "main kitni rungs climb kar sakta hoon?"

KYA. Ek crystal mein energy ladder of levels ki form mein aati hai: ek sabse neeche ki rung (the ground state) aur usse upar ki rungs (excited states). Temperature $T$ decide karta hai ki kitni thermal energy, roughly $k_{B} T$ , ladder pe upar jaane ke liye available hai.

KYU. Accessible microstates ki sankhya is baat par depend karti hai ki system kitni rungs tak pahunch sakta hai. Hot ( $k_{B} T$ bada) → bahut saari rungs reachable → bahut saare microstates → bada $W$ → bada $S$ . Yeh abstract count $W$ ko physical dial $T$ se jodta hai jo hum actually turn karte hain.

PICTURE. Energy ladder: high $T$ par amber glow bahut saari rungs tak pahunchta hai (bahut saare arrangements lit up); jaise-jaise $T$ girta hai, glow sirf bottom rung ki taraf shrink ho jaata hai.

Step 5 — Bottom tak cool karo: $T \to 0$ ground state force karta hai

KYA. $T \to 0$ lo. Phir $k_{B} T \to 0$ : koi bhi rung climb karne ke liye essentially koi thermal energy nahi hai. System single lowest rung, the ground state, par force ho jaata hai.

KYU. Koi bhi upar ki rung bottom se $Δ E$ energy gap pe baithi hai. Use occupy karne ka chance $e^{- Δ E / (k_{B} T)}$ ki tarah fade hota hai; jaise $T \to 0$ , woh exponential crash ho kar $0$ ho jaata hai. To har excited possibility switch off ho jaati hai aur sirf ground rung bachti hai.

PICTURE. Step 4 waali ladder, lekin ab amber glow poori tarah bottom rung par collapse ho gayi hai. Saari upar wali rungs dark hain.

Step 6 — PERFECT crystal ke liye bottom rung unique hoti hai: $W = 1$

KYA. Ek perfect crystal ke liye — har atom apni unique lattice site par locked, koi defects nahi, koi orientation ambiguity nahi — ground state ek single arrangement hoti hai. Usse count karo: $W = 1$ .

KYU. "Perfect" ka matlab hai ki sabse kam energy wale pattern mein saare atoms ko place karne ka literally sirf ek hi tarika hai. Koi choice nahi bachi, to ground-state pictures ka count exactly $1$ hai.

PICTURE. Ek clean lattice, har arrow identical aur aligned. Koi alternative picture draw karne ke liye nahi hai — figure ka point uski uniqueness hi hai.

Ab count ko Boltzmann mein plug karo: $S = k_{B} ln W = k_{B} ln 1 = k_{B} \cdot 0 = 0.$ Term by term: $W = 1$ (ek ground picture) → $ln 1 = 0$ (ek ka log zero hota hai) → $k_{B}$ se multiply karo phir bhi $0$ . Zero possibilities-beyond-one ka matlab zero entropy.

Step 7 — Degenerate case: jab bottom rung ke $g$ floors hon

KYA. Maano ground state unique nahi hai — woh $g$ -fold degenerate hai (system $g$ equally-low arrangements mein se kisi mein bhi freeze ho sakta hai). Phir $T = 0$ par bhi $W = g$ hai, aur $S (0) = k_{B} ln g .$

KYU. CO jaise real solids mein har molecule CO ya OC ki taraf pointing hoke freeze ho sakta hai. Cooling thermal wiggling ko hata deta hai lekin yeh decide nahi kar sakta ki har molecule ne kaun si frozen orientation li — woh choice trap ho jaati hai. $g > 1$ ke saath, $ln g > 0$ , to entropy zero nahi pahunchti. Yeh leftover residual entropy hai.

PICTURE. $T = 0$ par ek CO crystal: molecules randomly CO/OC ki tarah locked. $N$ molecules ke liye $g = 2^{N}$ — figure highlight karta hai ki har site pe ek stuck binary choice hai.

Dekho bhi Residual entropy of ice and CO.

Step 8 — Payoff: wapas upar climb karne se ABSOLUTE entropy milti hai

KYA. Kyunki Third Law $S (0) = 0$ pin kar deta hai (perfect crystal ke liye), hum bilkul bottom se entropy add kar sakte hain: $S (T) = = 0 S (0) + \int_{0}^{T} \frac{C _{p} ( T ^{'} )}{T ^{'}} d T^{'} .$ Term by term: $S (0) = 0$ woh anchor hai jo Third Law deta hai; $C_{p} (T^{'})$ heat capacity hai (har degree ke liye kitni heat chahiye); $T^{'}$ se divide karna $d S = δ Q_{r e v} / T$ se aata hai.

Integral bottom par blow up kyun nahi karta? $T = 0$ ke paas solids Debye law follow karte hain $C_{p} = a T^{3}$ (dekho Heat capacity and Debye T-cubed law). Phir integrand $a T^{'3} / T^{'} = a T^{'2} \to 0$ hai: yeh origin par smoothly vanish ho jaata hai, to area finite hai. Third Law aur $C_{p}$ ka vanish hona self-consistent hain.

PICTURE. $0$ se $T$ tak $C_{p} / T$ curve ke neeche ka area — shaded amber — origin par flat start karta hai (koi divergence nahi) aur $S (T)$ mein accumulate hota hai.

Ek picture mein poora summary

Ek single $S$ -vs- $T$ chart par poori journey: perfect crystal (cyan) $T = 0$ par exactly zero tak glide karti hai; imperfect crystal (amber) residual plateau $k_{B} ln g > 0$ par level off ho jaati hai; dono curves flatten ho jaati hain (slope $\to 0$ ) kyunki $C_{p} \to 0$ . Woh saath mein flattening yeh bhi reason hai ki tum $T = 0$ tak kabhi pahunch nahi sakte (unattainability — dekho Adiabatic demagnetization and reaching low temperatures).

Recall Feynman: poora walkthrough plain words mein

Humne atoms ka ek box draw kiya aur bola: ek exact arrangement ek microstate hai, aur $W$ unhe count karta hai. Independent atoms apni choices multiply karte hain, to $W$ powers ki tarah explode karta hai — isliye hum logarithm lete hain, jo "multiply" ko "add" mein badal deta hai aur entropy ko sane aur additive rakhta hai. Temperature bas energy ladder climb karne ka energy budget hai: hot matlab bahut saari rungs reachable, bahut saare arrangements, badi entropy. Cool karo aur budget khatam ho jaata hai — kisi bhi upper rung ki probability $e^{- Δ E / k_{B} T}$ ki tarah fade hoti hai — to sab bottom rung par gir jaate hain. Agar crystal perfect hai, to neeche exactly ek arrangement hai, $W = 1$ , aur $ln 1 = 0$ , to entropy zero ho jaati hai. Agar bottom par kai stuck configurations hain (CO is taraf ya us taraf pointing), to thoda sa mess freeze ho jaata hai: residual entropy $k_{B} ln g$ . Aakhir mein, kyunki zero pin down hai, hum bottom se heat capacity add kar ke absolute entropy nikal sakte hain — aur sum behave karta hai, kyunki $C_{p}$ zero ke paas $T^{3}$ ki tarah khatam ho jaata hai.

Connections

Boltzmann entropy S = k ln W — Steps 1–3 mein use hone wala counting engine.
Heat capacity and Debye T-cubed law — Step 8 ka integral kyun converge karta hai.
Absolute entropy and standard molar entropy — $S (0) = 0$ ka practical payoff.
Residual entropy of ice and CO — Step 7 ka $g > 1$ case.
Adiabatic demagnetization and reaching low temperatures — summary figure mein unattainability.
Second law of thermodynamics — $d S = δ Q_{r e v} / T$ provide karta hai.