Legendre transforms connecting them — Hinglish notes

2.4.2 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

1. Hum KAUNSI problem solve kar rahe hain?

Shuru karte hain internal energy se. Iski fundamental relation hai $U = U (S, V, N) .$

Dikkat yeh hai: $S$ ek lab variable nahi hai. Tum "entropy knob" nahi ghuma sakte. Tum $T$ ko heat bath se set kar sakte ho. Toh hum chahte hain ek nayi function jiska independent variable $S$ ki jagah $T$ ho, lekin phir bhi usme same physics ho.

Ek naive idea — "bas $S$ ko $T$ ke terms mein likho aur $U$ mein plug karo" — information kho deta hai (neeche steel-man kiya gaya hai). Legendre transform iski fix hai.

2. Scratch se derivation: Legendre transform HOTA kya hai?

KAISE — intercept derive karo. Point $x$ par tangent line ka equation hai $y = f (x) + p (X - x), p = f^{'} (x) .$ Iska $Y$ -intercept ( $X = 0$ set karo) hai $g = f (x) - p x .$

Hum $g$ ko slope $p$ ki function ke roop mein chahte hain, toh define karo

Check karo ki yeh information-preserving hai. $g = f - p x$ differentiate karo: $d g = df - p d x - x d p = = 0 (f^{'} - p) d x - x d p = - x d p .$

3. Char potentials banana

Hum transform ko $U (S, V, N)$ par apply karte hain, ek waqt mein ek conjugate pair swap karte hue. Conjugate pairs hain $(S, T)$ aur $(V, - P)$ .

(a) $S \to T$ swap karo (conjugate slope $T = \partial U / \partial S$ ). $S$ -slot par transform karo: $F = U - T S .$ Differentiate karo, $d U = T d S - P d V + μ d N$ use karke: $d F = d U - T d S - S d T = - S d T - P d V + μ d N .$ $T d S$ cancel ho gaya — $F$ ab $(T, V, N)$ par rehta hai. Yeh Helmholtz free energy hai.

(b) $V \to - P$ swap karo (slope $- P = \partial U / \partial V$ ). $V$ -slot par transform karo. Legendre subtract hai $(slope) \times (var) = (- P) V$ , toh hum $- P V$ subtract karte hain, yaani $P V$ add karte hain: $H = U + P V, d H = T d S + V d P + μ d N .$ $H$ $(S, P, N)$ par rehta hai — yeh enthalpy hai.

(c) Dono swap karo $S \to T$ aur $V \to - P$ : $G = U - T S + P V, d G = - S d T + V d P + μ d N .$ $G$ $(T, P, N)$ par rehta hai — yeh Gibbs free energy hai (lab ka favourite: $T$ aur $P$ dono controllable hain).

Figure — Legendre transforms connecting them

4. YEH sab kuch kyun power karta hai: Maxwell relations

Kyunki har $d Φ$ ek exact differential hai, mixed second partials commute karte hain. $d F = - S d T - P d V$ se: $(\frac{\partial}{\partial V} \frac{\partial F}{\partial T}) = (\frac{\partial}{\partial T} \frac{\partial F}{\partial V}) ⟹ (\frac{\partial S}{\partial V})_{T} = (\frac{\partial P}{\partial T})_{V} .$

5. Worked examples

6. Steel-manned mistakes

7. Flashcards

Legendre transform geometrically kya swap karta hai?

Ek description (point, value) se ek (slope, intercept) wali description mein — curve ko uski tangent-line envelope ke roop mein encode karta hai, koi info nahi jaati.

f (x)

ke Legendre transform ki definition?

g (p) = f - p x

jahan

p = f^{'} (x)

x

eliminate karne ke liye invert kiya jaata hai.

g (p) = f - p x

ka key slope relation?

d g / d p = - x

(aur transform ek involution hai).

S (T)

U

mein substitute kyun nahi kar sakte?

Yeh information kho deta hai; sirf

F = U - T S

use

- T S

term ke zariye rakhta hai, aur

S = - (\partial F / \partial T)_{V}

se recover karta hai.

F

ke natural variables aur definition?

F = U - T S

F (T, V, N)

d F = - S d T - P d V + μ d N

H

ke natural variables aur definition?

H = U + P V

H (S, P, N)

d H = T d S + V d P + μ d N

G

ke natural variables aur definition?

G = U - T S + P V

G (T, P, N)

d G = - S d T + V d P + μ d N

H = U + P V

(plus, minus nahi) kyun?

Kyunki slope

\partial U / \partial V = - P

hai; slope×V subtract karne par

- (- P) V = + P V

milta hai.

Maxwell relations kahaan se aate hain?

Har potential ke mixed second partials ki equality se (exact differentials).

F

se Maxwell relation?

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

Chemistry

G

kyun use karta hai?

Iske natural variables

(T, P)

exactly wahi hain jo lab mein control hote hain; equilibrium =

G

minimized.

Recall Feynman: 12-saal ke bachche ko samjhao

Ek pahaad imagine karo jo kaagaz par bana hai. Tum pahaad ko describe kar sakte ho har point ki height list karke, YA yeh list karke ki har jagah yeh kitna steep hai plus har slope-line ground se kahaan cross karti hai — dono descriptions tumhe ek hi pahaad batati hain. Physics mein, "height" energy hai. Kabhi kabhi jo cheez hum normally use karte hain (jaise chupi hui disorder, entropy) haath se pakadna impossible hai, lekin uske saath energy ki steepness (temperature!) kuch aisi hai jo thermostat aasaani se control karta hai. Legendre transform yeh trick hai ki energy pahaad ko us steepness ki madad se re-describe karo jo hum control kar sakte hain, bina koi information khoye — taaki real lab calculations ho sakein.