Maxwell relations — derivation from each potential: Visual walkthrough

2.4.3 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell relations — derivation from each potential

Step 1 — Do cheez'on par depend karne wala ek function ek landscape hai

KYA HAI. Maan lo ek number $z$ do knobs, $x$ aur $y$ , par depend karta hai. Likho $z = z (x, y)$ . Socho $x$ east ki taraf ja raha hai, $y$ north ki taraf, aur $z$ us east–north floor ke upar zameen ki height hai.

KYUN. Thermodynamics mein har quantity do doosri cheez'on par depend karti hai (energy depend karti hai kitna disorder hai aur kitni jagah hai par). Physics ko touch karne se pehle hum woh ek hi picture banate hain jo hamein chahiye: ek flat map ke upar baithi ek hill.

PICTURE. Figure dekho. Neeche ka flat grid $(x, y)$ floor hai. Uske upar curved sheet height $z$ hai. Floor par ek spot → sheet par ek height.

Figure — Maxwell relations — derivation from each potential

Step 2 — Ek chosen direction mein "steepness": partial derivative

KYA HAI. Hill par khade ho. East ki taraf munh karo ( $x$ badhao, $y$ frozen rakho). Har step east par aapki height kitni tezi se chadhti hai? Woh number partial derivative hai: $east ki taraf steepness (\frac{\partial z}{\partial x})_{y}$

KYUN. Humein ek aisa tool chahiye jo steepness ko sirf ek direction mein measure kare, kyunki ek hill ek hi spot par east mein steep aur north mein flat ho sakti hai. Subscript $y$ is tool ka poora point hai.

$\partial z / \partial x$ — "height mein change per change in $x$ "
subscript $_{y}$ — "jabki $y$ frozen rakha gaya hai" (aap ek seedhi east line par chalte ho, north mein kabhi drift nahi karte)

PICTURE. Blue arrow east-facing tangent slope hai; yellow arrow north-facing tangent slope hai. Same point, do alag steepnesses.

Step 3 — Differential: ek chhote step se total height change

KYA HAI. Ek tiny step lo: thoda east ( $d x$ ) aur thoda north ( $d y$ ). Tera total height change hai east-steepness × east-step plus north-steepness × north-step: $d z = ise M kaho (\frac{\partial z}{\partial x})_{y} d x + ise N kaho (\frac{\partial z}{\partial y})_{x} d y = M d x + N d y$

KYUN. Koi bhi chhota move "kuch east + kuch north" hota hai. Hum har leg par gained height add karte hain. Do slopes ko $M$ aur $N$ naam dena humein yeh bhulne deta hai ki yeh kisi height se aaye hain aur unhe sirf floor par do functions ki tarah treat karte hain.

$M$ — east slope, jo khud position $(x, y)$ ka function hai
$N$ — north slope, jo bhi $(x, y)$ ka function hai
$d z$ — chhote diagonal step se total rise

PICTURE. Chhota step ek red east leg aur ek green north leg mein split hota hai; har leg par charha hua height total $d z$ dene ke liye stack hota hai.

Step 4 — Jadui sawaal: kya climbing ka order matter karta hai?

KYA HAI. Ek tiny square ke diagonally-opposite corner tak pahunchne ke do tarike:

Path A: pehle east jao, phir north.
Path B: pehle north jao, phir east.

Ek asli hill par dono same corner par land karte hain, toh dono ek hi total height chadhte hain.

KYUN. Yeh har Maxwell relation ka geometric seed hai. Agar height real hai (ek state function), toh do paths ko agree karna hi hoga — aur unhe agree karne par majboor karna slopes $M$ aur $N$ ke beech ek relationship pin down karega.

PICTURE. Floor par square, do colored paths ke saath (east-then-north vs north-then-east) same top corner par milte hue.

Step 5 — "Order matter nahi karta" ko ek equation mein badalna

KYA HAI. Poochho: east slope $M$ kaise change hota hai jab main north ki taraf nudge karta hun? Woh hai $(\frac{\partial M}{\partial y})_{x}$ . Aur: north slope $N$ kaise change hota hai jab main east ki taraf nudge karta hun? Woh hai $(\frac{\partial N}{\partial x})_{y}$ . "Order matter nahi karta" fact inhe equal hone par majboor karta hai: $(\frac{\partial M}{\partial y})_{x} = (\frac{\partial N}{\partial x})_{y}$

KYUN. Dono sides literally ek hi second derivative $\frac{\partial ^{2} z}{\partial x \partial y}$ hain — hill ki cross-curvature — do possible orders mein measure ki gayi. Ek smooth hill ki ek cross-curvature hoti hai, toh orders match karte hain. Yeh Schwarz / equality-of-mixed-partials theorem hai (dekho Equality of mixed partial derivatives (Schwarz theorem)).

left side — "east slope $M$ lo, dekho yeh north jaate hue kaise tilt karta hai"
right side — "north slope $N$ lo, dekho yeh east jaate hue kaise tilt karta hai"
dono equal hain kyunki dono = $\partial^{2} z / \partial x \partial y$

PICTURE. Tiny square phir se, ab yeh dikhate hue ki "far corner par extra rise" ek hi number hai chahe aap ise $M$ ke northward steepen hone se attribute karo ya $N$ ke eastward steepen hone se.

Recall Kyun do second derivatives match karni chahiye

Far corner par extra height ::: yeh (east-slope mein change jaise tum north jaate ho) ke barabar hai aur saath hi (north-slope mein change jaise tum east jaate ho) ke bhi — dono same corner describe karte hain, toh yeh ek hi number hai.

Step 6 — Asli hill load karo: internal energy $U$

KYA HAI. Ab abstract pieces ko physics se naam do. Height internal energy $U$ hai; floor coordinates entropy $S$ (disorder) east ki taraf aur volume $V$ (jagah) north ki taraf hain. First+Second Laws dete hain (dekho First and Second Laws of Thermodynamics): $d U = M T d S N - P d V$

KYUN. Yeh hamara $d z = M d x + N d y$ hai ek specific hill ke saath. Term-by-term match karte hue:

$z \to U$ (height energy hai)
$x \to S$ , $y \to V$ (floor axes disorder aur jagah hain)
$M \to T$ (energy ka east slope disorder ke against temperature hai)
$N \to - P$ (energy ka north slope jagah ke against minus pressure hai)

$P$ par minus koi choice nahi hai — energy girti hai jab gas expand karke bahar push karti hai, toh north slope negative hai.

PICTURE. Step 1 jaisi hi landscape, ab re-labelled: axes $S$ aur $V$ , height $U$ , east slope $T$ tag kiya hua aur north slope $- P$ tag kiya hua.

Step 7 — Cross-differentiate karo: relation nikal ke aata hai

KYA HAI. Step-5 box mein $M = T$ aur $N = - P$ plug karo: $(\frac{\partial M}{\partial y})_{x} = (\frac{\partial N}{\partial x})_{y} ⟹ (\frac{\partial T}{\partial V})_{S} = (\frac{\partial ( - P )}{\partial S})_{V}$ Minus ko bahar nikalo: $(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V}$

KYUN. Humne yahan kuch physical nahi kiya — humne sirf Step 5 ki pure-calculus identity ko rename kiya. Physics Step 6 mein pehle se load ho chuki thi. Minus sign seedha $- P$ se inherit hua hai; humne ise kabhi guess nahi kiya.

$(\frac{\partial T}{\partial V})_{S}$ — fixed disorder par jab aap jagah badhate ho toh temperature kaise change hoti hai
$- (\frac{\partial P}{\partial S})_{V}$ — minus, fixed jagah par jab aap disorder badhate ho toh pressure kaise change hota hai
yeh do mushkil-se-picture karne wali quantities hill ki smoothness se equal hone par majboor hain

PICTURE. Step-5 square physics labels ke saath: east-slope $T$ north ( $V$ ) ki taraf move karne par tilt karta hai, jo equals karta hai north-slope $- P$ ko east ( $S$ ) ki taraf move karne par tilt karte hue.

Step 8 — Edge & degenerate cases: picture kab toot jaati hai?

KYA HAI. Equal-climb argument ko ek smooth hill chahiye. Yeh kahan fail ho sakta hai?

Ek kink / phase transition. Boiling ke dauran, fixed pressure par heat add karne se temperature stuck flat rehne ke saath volume badhta hai — surface par ek sharp crease develop ho jaati hai. Crease par slope single-valued nahi hota, toh mixed partials us exact edge par equal guaranteed nahi hain. Transition se door (pure phase) hill phir smooth hoti hai aur relation hold karta hai.
Ek flat direction (degenerate slope). Agar koi slope exactly zero hai — jaise ideal gas mein $(\frac{\partial U}{\partial V})_{T} = 0$ — relation phir bhi hold karta hai; zero ek perfectly good, smooth value hai. Zero koi break nahi hai.
Boundaries ( $V \to 0$ , $T \to 0$ ). Jaise $T \to 0$ par surface flat hoti hai (Third Law: $S$ involve karne wale slopes $0$ ki taraf tend karte hain); relation ek limit ke roop mein survive karta hai, dono sides zero ki taraf jaate hue.

KYUN. Reader ko koi aisa scenario nahi milna chahiye jo humne skip kiya. Rule simple hai: smooth ⇒ relation holds; kink ⇒ sirf kink par re-examine karo.

PICTURE. Do mini-surfaces side by side: ek smooth dome (relation valid) aur ek phase line par creased ridge (relation valid hai har smooth face par, exactly crease par ambiguous).

Ek-picture summary

Upar ki sab cheez ek diagram mein collapse hoti hai: ek smooth energy hill $U (S, V)$ , uske floor par ek tiny square, aur yeh single statement ki dono climbing orders same corner par pahunchte hain — jo, slopes $T$ aur $- P$ label karne ke baad, hai hi Maxwell relation.

Recall Feynman: poora walkthrough simple words mein

Ek smooth grassy hill socho. Teri altitude sirf is baat par depend karti hai ki tu kahan khada hai, is par nahi ki tu kaise chala. Toh agar tu thoda east phir thoda north jaaye, tu same height par pahunchega jaise north phir east jaate — hill routes mein favourites nahi khelti. Ab, "east slope kitna steep hai, aur kya woh steepness khud change hoti hai jaise main north ki taraf drift karta hun?" Yeh nikalta hai ki woh number match karna chahiye "north slope kitna steep hai, aur kya woh change hota hai jaise main east ki taraf drift karta hun?" — kyunki dono secretly far corner par same thoda extra rise measure kar rahe hain. Bas yahi trick hai. Phir hum kehte hain: hill ki height ko energy hone do, east ko disorder aur north ko room. Energy ka east slope temperature hai; north slope minus pressure hai. Woh names "orders must match" rule mein daalo, aur nikalta hai ek fact jo connect karta hai ki temperature zyada room par kaise react karti hai aur pressure zyada disorder par kaise react karta hai — do cheezein jo directly picture karna almost impossible hain, ek smooth hill ki shape se free mein bandh ki gayi hain.

Connections

Maxwell relations — derivation from each potential — parent, saari chaar relations
Equality of mixed partial derivatives (Schwarz theorem) — Step-5 engine
First and Second Laws of Thermodynamics — $d U = T d S - P d V$ ka source
Thermodynamic potentials & Legendre transforms — baaki teen hills
Joule expansion and internal energy — jahan $(\partial U / \partial V)_{T} = 0$ kaam aata hai