Legendre transforms connecting them: Exercises — Hinglish deep dive

2.4.2 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Legendre transforms connecting them

Yahan sab kuch parent note par based hai. Agar koi symbol unfamiliar lage, woh note usse zero se build karta hai.

Level 1 — Recognition

Yeh test karte hain ki kya aap ek differential padh sakte ho aur machinery ka naam le sakte ho. Koi heavy algebra nahi.

Exercise 1.1 — Natural variables padho

Problem. Aapko yeh differential diya gaya hai: $d H = T d S + V d P + μ d N .$ (a) $H$ ke natural (independent) variables kya hain? (b) Is line se seedha nikalne wale teen slope formulas likho.

Recall Solution

HUM KYA KARTE HAIN: ek differential $d Φ = (\dots) d x + (\dots) d y + \dots$ mein, jis variables ko differentiate kiya ja raha hai ( $d x$ , $d y$ , …) woh natural variables hain, aur har ek ke aage ka coefficient corresponding slope hai.

(a) Jo differentials dikh rahe hain woh hain $d S$ , $d P$ , $d N$ . Toh natural variables hain: $H = H (S, P, N) .$ (b) Har coefficient ek partial derivative hai jisme baaki natural variables fixed rakhe jaate hain: $T = (\frac{\partial H}{\partial S})_{P, N}, V = (\frac{\partial H}{\partial P})_{S, N}, μ = (\frac{\partial H}{\partial N})_{S, P} .$ Yeh kyon kaam karta hai: ek differential bas yeh bookkeeping hai ki "jab har independent variable thoda sa nudge kare toh $Φ$ kitna hilta hai." $d P$ ka coefficient by definition $P$ ke respect mein partial hai. (Yaad karo $H = U + P V$ enthalpy hai — energy plus $P V$ room-making term.)

Exercise 1.2 — Conjugate pairs pehchano

Problem. $d U = T d S - P d V + μ d N$ se conjugate variable pairs list karo aur har ek ke liye batao ki kaun sa member extensive hai (system size ke saath scale karta hai) aur kaun sa intensive hai (size se independent).

Recall Solution

Conjugate pair hai (differentiate hone wala variable, uska coefficient). $d U$ (internal energy) se padh ke:

Pair	Extensive	Intensive
$(S, T)$	$S$	$T$
$(V, - P)$	$V$	$- P$
$(N, μ)$	$N$	$μ$

Split kyun matter karta hai: aap directly ek extensive knob nahi dial kar sakte (koi "entropy dial" nahi hota), lekin aap ek intensive wala reservoir se set kar sakte ho (thermostat $T$ set karta hai, piston/atmosphere $P$ set karta hai). Legendre transform exist karta hi isliye hai taaki awkward extensive member ko uske convenient intensive partner se swap kar sake. Dekho Internal Energy and the First Law.

Exercise 1.3 — Kis lab ke liye kaun sa potential?

Problem. Har experimental setup ko us potential se match karo jiske natural variables wahi hain jo tum control karte ho: (a) thermostat mein sealed rigid box; (b) room temperature par bench par rakha open beaker; (c) ek adiabatic, insulated cylinder jise ek movable lekin gas-tight piston seal karta hai, jiski bahar wali face atmosphere ke liye open hai.

Recall Solution

Controlled variables identify karo, phir woh potential chuno jiske natural variables match karen.

(a) Fixed volume ( $V$ ) + thermostat ( $T$ ) + sealed ( $N$ ) $\Rightarrow$ controls $(T, V, N)$ $\Rightarrow$ Helmholtz $F = U - T S$ , $F (T, V, N)$ (fixed $T$ par usable work).
(b) Atmosphere $P$ set karta hai, bench $T$ set karta hai, matter conserved $N$ $\Rightarrow$ controls $(T, P, N)$ $\Rightarrow$ Gibbs $G = U - T S + P V$ , $G (T, P, N)$ .
(c) Insulated matlab koi heat exchange nahi, isliye entropy $S$ conserved natural slot hai; piston atmosphere ke liye open hai, isliye woh pressure transmit karta hai aur $P$ fix karta hai. $N$ kyun abhi bhi fixed hai: ek piston ek solid, gas-tight wall hai — yeh pressure equalize karne ke liye slide karta hai lekin koi molecules iske paar nahi jaate, isliye cylinder ke andar gas count $N$ nahi badal sakta. (Ise ek permeable membrane se compare karo, jo particles ko through jaane deti hai aur instead chemical potential $μ$ fix karti hai — yeh grand-potential setup hai Exercise 5.3 ka.) Toh hum control karte hain $(S, P, N)$ $\Rightarrow$ Enthalpy $H = U + P V$ , $H (S, P, N)$ .

Kyun: equilibrium un variables ke fixed rahne par potential ka minimum hota hai. Sahi potential choose karne se lab conditions independent variables ban jaate hain, toh minimization clean hoti hai.

Level 2 — Application

Ab tum concrete functions par machine chalate ho. Har step mein, jab bhi $g = f - p x$ dikhe, picture yaad karo: yeh woh height hai jahan tangent (slope $p$ ) vertical axis ko hit karti hai.

Exercise 2.1 — Ek cubic ka Legendre transform

Problem. Maano $f (x) = x^{3}$ for $x > 0$ . Iska Legendre transform $g (p)$ compute karo aur slope relation $d g / d p = - x$ verify karo.

Recall Solution

Step 1 (KYA & KYUN): slope nayi variable ban jaati hai, toh $p = f^{'} (x) = 3 x^{2} .$ Yeh legal kyun hai: $x > 0$ ke liye slope $3 x^{2}$ strictly increasing hai, isliye har $p$ exactly ek $x$ naam deta hai — transform well-defined hai (Exercise 3.2 mein existence note dekho). Step 2 — invert (hamesha $g$ ko sirf $p$ ke terms mein chahiye; yahan $x > 0$ toh positive root lete hain): $x = (\frac{p}{3})^{1/2} .$ Step 3 — transform banao $g = f - p x$ (geometrically: tangent line ko axis tak giraao): $g (p) = x^{3} - p x = x^{3} - 3 x^{2} \cdot x = - 2 x^{3} = - 2 (\frac{p}{3})^{3/2} .$ Step 4 — defining slope relation check karo: $\frac{d g}{d p} = - 2 \cdot \frac{3}{2} (\frac{p}{3})^{1/2} \cdot \frac{1}{3} = - (\frac{p}{3})^{1/2} = - x . ✓$ Check confirm karta hai ki koi information lose nahi hui — $g$ apni khud ki slope ke through $x$ jaanta hai.

Exercise 2.2 — $F$ banao aur uski slopes nikalo

Problem. Ek system ki internal energy implicitly $d U = T d S - P d V$ ke through di gayi hai (fixed $N$ ). Uski Helmholtz free energy ( $F = U - T S$ , fixed temperature par usable work) yeh nikli: $F (T, V) = - a T^{2} - N k_{B} T ln V,$ jahan $a > 0$ constant hai. $P (T, V)$ aur $S (T, V)$ nikalo.

Recall Solution

Pehle, $d F = - S d T - P d V$ KYUN aata hai? Hum ise derive karte hain, isliye kuch assume nahi hota. Definition $F = U - T S$ se shuru karo aur dono sides ka differential lo, $T S$ term par product rule use karo: $d F = d U - d (T S) = d U - (T d S + S d T) .$ Ab first law $d U = T d S - P d V$ substitute karo: $d F = (T d S - P d V) - T d S - S d T = - S d T - P d V .$ Do $T d S$ terms cancel ho jaate hain — yahi cancellation Legendre subtraction $- T S$ ka poora point hai: yeh awkward $d S$ ko khatam karta hai aur $d T$ ko driver ke roop mein install karta hai. Kaun si slopes? $d F = - S d T - P d V$ se hum do rules padh lete hain: $S = - (\frac{\partial F}{\partial T})_{V}, P = - (\frac{\partial F}{\partial V})_{T} .$ Entropy ( $T$ mein differentiate karo, $V$ hold karke): $\frac{\partial F}{\partial T} = - 2 a T - N k_{B} ln V \Rightarrow S = 2 a T + N k_{B} ln V .$ Pressure ( $V$ mein differentiate karo, $T$ hold karke): $\frac{\partial F}{\partial V} = - \frac{N k _{B} T}{V} \Rightarrow P = \frac{N k _{B} T}{V} .$ Ideal-gas law $P V = N k_{B} T$ appear hoti hai — $F$ ki ek slope ke roop mein recover hoti hai, exactly jaisa parent note ne promise kiya tha. Dekho Maxwell Relations ki yeh do derivatives secretly kaise linked hain.

Exercise 2.3 — Enthalpy sign, sahi tarike se

Problem. $d U = T d S - P d V$ se shuru karke woh potential construct karo jiska doosre slot mein natural variable $V$ ki jagah $P$ ho. Sign carefully dikhao.

Recall Solution

Step 1 — slope ka sign padho (yahi poora game hai): $d U$ mein $d V$ ka coefficient $- P$ hai, toh $(\frac{\partial U}{\partial V})_{S} = - P .$ Step 2 — Legendre subtract (slope $\times$ variable): transform subtract karta hai $slope \times var = (- P) V$ (phir se: yeh tangent-line intercept hai, ab $V$ direction mein): $H = U - (- P) V = U + P V .$ Step 3 — differential check karo: $d H = d U + P d V + V d P = (T d S - P d V) + P d V + V d P = T d S + V d P .$ $P d V$ terms cancel ho jaate hain, $d P$ nayi driver ban jaati hai. Natural variables: $H = H (S, P)$ . ✓ ( $H = U + P V$ enthalpy hai.)

Level 3 — Analysis

Structure ko tod ke dekho aur samjho ki kyun hold karta hai.

Exercise 3.1 — $G$ se Maxwell relation derive karo

Problem. $d G = - S d T + V d P$ se woh Maxwell relation derive karo jo $(\partial S / \partial P)_{T}$ ko ek volume derivative se connect kare. Batao kaun sa quantity measurable hai aur kaun sa nahi.

Recall Solution

Key fact: $d G$ ek exact differential hai, isliye uske mixed second partials differentiation ke order se independent hain: $\frac{\partial}{\partial P} (\frac{\partial G}{\partial T})_{P} = \frac{\partial}{\partial T} (\frac{\partial G}{\partial P})_{T} .$ Ab $d G$ se padhe first-order slopes substitute karo ( $G = U - T S + P V$ Gibbs energy hai): $(\frac{\partial G}{\partial T})_{P} = - S, (\frac{\partial G}{\partial P})_{T} = V .$ Har ek ko differentiate karo: $(\frac{\partial ( - S )}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P} \Rightarrow - (\frac{\partial S}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P} .$ Measurability: $(\partial V / \partial T)_{P}$ directly measurable hai (thermal expansion — constant pressure par fluid ko heat karo aur dekho woh expand hota hai). $(\partial S / \partial P)_{T}$ directly measurable nahi hai (koi entropy meter nahi hota). Maxwell measurable ko unmeasurable se free mein trade karta hai. Dekho Maxwell Relations.

Exercise 3.2 — Convexity concavity mein flip hoti hai (aur transform exist kyun karta hai)

Problem. Convex parabola $f (x) = \frac{1}{2} a x^{2}$ (with $a > 0$ ) ka transform $g (p) = - \frac{p ^{2}}{2 a}$ hai, jo concave hai. (a) $g^{''} (p)$ compute karke concavity confirm karo. (b) Geometrically explain karo ki ek convex function ka Legendre transform concave kyun hota hai. (c) Woh condition batao $f$ par jo guarantee karti hai ki transform exist hi kare.

Recall Solution

(a) $g (p) = - \frac{p ^{2}}{2 a}$ , toh $g^{'} (p) = - \frac{p}{a}, g^{''} (p) = - \frac{1}{a} < 0.$ Kyunki $g^{''} < 0$ , $g$ concave hai. Compare karo $f^{''} (x) = a > 0$ (convex) se. Fact mein $g^{''} (p) \cdot f^{''} (x) = (- 1/ a) (a) = - 1$ : curvatures reciprocal-and-negated hain. (b) Geometric reason: transform variable slope $p = f^{'} (x)$ hai. Ek convex $f$ ke liye, jab $x$ badhta hai toh slope $p$ badhti hai (yahi convex ka matlab hai). Transform ki apni slope hai $d g / d p = - x$ , jo decrease karti hai jab $p$ (hence $x$ ) badhta hai. Woh quantity jiski slope decrease hoti hai woh concave hoti hai. Toh convex $\leftrightarrow$ concave transform ke under. Isliye $U$ (convex in $S$ ) $F$ (concave in $T$ ) ban jaata hai — dekho Thermodynamic Stability and Convexity. (c) Existence condition (woh point jo aksar skip ho jaata hai). $g (p)$ banane ke liye hum invert karte hain $p = f^{'} (x)$ ko taaki har $p$ ke liye ek single $x$ mile. Woh inversion guaranteed hai exactly jab $f^{'}$ strictly monotonic ho, yaani jab $f$ strictly convex ( $f^{''} > 0$ ) ya strictly concave ( $f^{''} < 0$ ) ho — kabhi flat nahi, kabhi wiggling nahi. Agar $f^{'}$ turn around kare (koi $p$ do alag $x$ 's se hit ho), toh "kaun sa $x$ ?" ambiguous hoga aur $g (p)$ ill-defined. Isliye thermodynamics insist karti hai ki potentials convex/concave hon: stability ( $C_{V} > 0$ , etc.) exactly woh condition hai jo $U, H, F, G$ ke beech har Legendre transform ko well-defined banati hai. Neeche ki figure dono curvatures side by side dikhati hai.

Figure (above). Blue hai convex parent $f (x) = \frac{1}{2} a x^{2}$ ("upar curve karta hai", $f^{''} > 0$ ); red hai iska Legendre transform $g (p) = - \frac{p ^{2}}{2 a}$ ("neeche curve karta hai", $g^{''} < 0$ ). Ise parts (a)–(b) ka visual statement samjho: transform ek bowl ko ek dome mein reflect karta hai. Iske draw ho paane ki wajah part (c) hai — convex bowl ki slope strictly increasing hai, toh har slope value $p$ ek unique point naam deti hai, aur red dome unambiguous hai.

Exercise 3.3 — Involution, ek clean convention

Problem. Dikhao ki $g (p) = - \frac{p ^{2}}{2 a}$ ko doosri baar transform karne par original $f (x) = \frac{1}{2} a x^{2}$ wapas milta hai. Poore time ek hi convention use karo: function $ϕ (u)$ ka Legendre transform hai $ψ (w) = ϕ - w u$ jahan $w \equiv ϕ^{'} (u)$ .

Recall Solution

Hum wahi same rule apply karte hain jo hamesha use karte hain — koi nayi sign convention nahi — $g (p)$ ko transform hone wala function treat karke. Uski apni slope variable, ise $w$ kahte hain, hai $w \equiv g^{'} (p)$ .

Step 1 — nayi slope. $w = \frac{d g}{d p} = - \frac{p}{a} .$ Step 2 — invert ( $w$ ke terms mein $p$ solve karo): $p = - a w .$ Step 3 — transform apply karo $ψ (w) = g (p) - w p$ : $ψ (w) = - \frac{p ^{2}}{2 a} - w p .$ $p = - a w$ term by term substitute karo: $- \frac{p ^{2}}{2 a} = - \frac{( - a w ) ^{2}}{2 a} = - \frac{a w ^{2}}{2}, - w p = - w (- a w) = + a w^{2} .$ Step 4 — do pieces add karo: $ψ (w) = - \frac{a w ^{2}}{2} + a w^{2} = + \frac{a w ^{2}}{2} .$ Step 5 — result padho aur variables match karo. Hamein mila $ψ (w) = \frac{1}{2} a w^{2}$ , jo exactly original $f$ hai jahan $w$ ki jagah $x$ ka role hai. Confirm karne ke liye ki variable sach mein $x$ hai: pehle transform ne obey kiya tha $d g / d p = - x$ , aur yahan $w = d g / d p$ hai, toh $w = - x$ . $ψ$ mein $w = - x$ substitute karo: $ψ = \frac{1}{2} a (- x)^{2} = \frac{1}{2} a x^{2} = f (x) . ✓$ Takeaway: ek single consistent rule ke saath, double transform $f$ par wapas land karta hai; single subtlety yeh hai ki intermediate slope variable $- x$ ke equal hai, jo precisely wahi hai jo relation $d g / d p = - x$ poore time track kar raha tha. Do applications = identity: Legendre transform ek involution hai.

Level 4 — Synthesis

Transform + physics ko ek single argument mein combine karo.

Exercise 4.1 — $F$ se ideal gas ki Gibbs energy

Problem. Ek gas ke liye $F (T, V, N)$ satisfy karta hai $d F = - S d T - P d V$ aur $P = N k_{B} T / V$ . $V$ -slot ko Legendre transform karke $G (T, P, N)$ construct karo, aur confirm karo ki $G$ sirf $P$ par depend karta hai ( $V$ par nahi).

Recall Solution

Yaad karo $F = U - T S$ (Helmholtz) hai aur hum $G = U - T S + P V$ (Gibbs) ki taraf ja rahe hain. Step 1 — kaun si slope swap karni hai. Hum $V$ ko uske conjugate se trade karna chahte hain. $V$ mein $F$ ki slope hai: $(\frac{\partial F}{\partial V})_{T} = - P .$ Step 2 — $V$ -slot par Legendre transform (slope $\times$ var $= (- P) V$ subtract karo, yaani $V$ direction mein tangent intercept): $G = F - (- P) V = F + P V .$ Step 3 — differential check karo: $d G = d F + P d V + V d P = (- S d T - P d V) + P d V + V d P = - S d T + V d P .$ $d V$ cancel ho jaata hai; driver ab $d P$ hai. Toh $G = G (T, P, N)$ . ✓ Step 4 — ideal gas ke saath sanity check: $P V = N k_{B} T$ , toh added term $P V = N k_{B} T$ finite aur well-defined hai; $V = N k_{B} T / P$ substitute karne par $G$ se $V$ ka har trace remove ho jaata hai. Transform ne exactly apna kaam kiya: $F (T, V)$ ko $G (T, P)$ mein losslessly convert kiya.

Exercise 4.2 — Fixed $T, P$ par equilibrium criterion

Problem. Ek beaker mein fixed $T$ aur $P$ par reaction ho rahi hai. $d G = - S d T + V d P + \sum_{i} μ_{i} d N_{i}$ (kaafi species $i$ wala Gibbs differential, har ek ka chemical potential $μ_{i}$ ) use karke explain karo ki equilibrium $\sum_{i} μ_{i} d N_{i} = 0$ kyun hota hai, aur kyun koi doosra potential itni cleanly yeh nahi deta.

Recall Solution

Yaad karo $G = U - T S + P V$ Gibbs energy hai, aur $μ_{i}$ species $i$ ka ek particle add karne ki energy cost hai. Step 1 — fixed conditions impose karo. Fixed $T$ aur $P$ matlab $d T = 0$ aur $d P = 0$ . Tab: $d G = - S 0 d T + V 0 d P + \sum_{i} μ_{i} d N_{i} = \sum_{i} μ_{i} d N_{i} .$ Step 2 — spontaneity aur equilibrium. Second law, fixed $(T, P)$ ke liye specialize karke, kehta hai ki ek spontaneous change sirf $G$ lower kar sakta hai, yaani $d G \leq 0$ . System $G$ mein downhill slide karta hai jab tak aur neeche nahi ja sakta — woh resting point minimum hai, jahan har allowed change $d G = 0$ deta hai. Step 1 ke saath combine karo: $i \sum μ_{i} d N_{i} = 0.$ Shabd mein: equilibrium par chemical potentials itne balanced hain ki particles ko species mein shuffle karna (reaction ki stoichiometry se tied allowed $d N_{i}$ 's) $G$ bilkul nahi change karta. Step 3 — $G$ specifically kyun. Sirf $G$ ke liye controlled lab variables $(T, P)$ natural variables hain. $U$ , $H$ , ya $F$ ke liye, kam se kam ek held-fixed lab variable natural nahi hai, toh differential clean $\sum_{i} μ_{i} d N_{i}$ form mein collapse nahi hota — aapke paas leftover $S d T$ ya $P d V$ terms rahenge jo in lab conditions ke under zero nahi hain, criterion ko muddy karte hain. Isliye hi constant- $T, P$ chemist $G$ ke liye pahuncha karta hai. Dekho Chemical Potential and Phase Equilibrium.

Level 5 — Mastery

Prove karo ya create karo.

Exercise 5.1 — Mechanics mein Legendre thermodynamics se match karta hai

Problem. Mechanics mein Hamiltonian hai $H = p \overset{q}{˙} - L$ , jahan $L (\overset{q}{˙})$ Lagrangian hai aur $p = \partial L / \partial \overset{q}{˙}$ . Yeh velocity $\overset{q}{˙}$ ko momentum $p$ ke liye swap karta ek Legendre transform hai. (a) Dikhao ki iska structure $F = U - T S$ jaisa hi hai. (b) Slope relation ka analogue verify karo.

Recall Solution

(a) Structures match karo. $S \to T$ swap karne wala thermodynamic transform hai: $F = U - T S, T = \frac{\partial U}{\partial S} .$ Mechanical transform hai: $H = p \overset{q}{˙} - L = - (L - p \overset{q}{˙}), p = \frac{\partial L}{\partial q ˙} .$ Identify karo $L \leftrightarrow U$ , $\overset{q}{˙} \leftrightarrow S$ (trade-away hone wala variable), $p \leftrightarrow T$ (uska conjugate slope). Ek overall sign convention ke alawa (mechanics likhta hai $p \overset{q}{˙} - L$ ; thermo likhta hai $U - T S$ ), yeh same operation hai: independent variable ko $\overset{q}{˙}$ se $p$ mein badalne ke liye slope $\times$ variable subtract karo. Dekho Lagrangian to Hamiltonian (Legendre in Mechanics). (b) Slope relation. $H = p \overset{q}{˙} - L$ differentiate karo: $d H = \overset{q}{˙} d p + p d \overset{q}{˙} - \frac{\partial L}{\partial q ˙} d \overset{q}{˙} = \overset{q}{˙} d p + = 0 (p - \frac{\partial L}{\partial q ˙}) d \overset{q}{˙} = \overset{q}{˙} d p .$ Toh $(\frac{\partial H}{\partial p}) = \overset{q}{˙}$ — exact analogue $d g / d p = - x$ ka (yahan sign convention $+ \overset{q}{˙}$ deta hai). Bilkul thermodynamics jaisa, awkward variable ( $\overset{q}{˙}$ ) eliminate ho jaata hai aur uska conjugate ( $p$ ) driver ban jaata hai.

Exercise 5.2 — Non-convex $f$ transform ko tod deta hai

Problem. Maano $f (x) = - \frac{1}{2} a x^{2}$ with $a > 0$ over $x \in R$ . (a) $p = f^{'} (x)$ compute karo aur dikhao ki naive substitution "eliminate $x$ " se $g (p) = + \frac{p ^{2}}{2 a}$ milta hai. (b) Explain karo ki concave-where-it-should-be-convex function thermodynamic language mein kaun si physical pathology correspond karti hai, aur kyun real stable systems isse avoid karte hain.

Recall Solution

(a) $p = f^{'} (x) = - a x$ , invert $x = - p / a$ . Tab: $g (p) = f - p x = - \frac{1}{2} a (\frac{p}{a})^{2} - p (- \frac{p}{a}) = - \frac{p ^{2}}{2 a} + \frac{p ^{2}}{a} = + \frac{p ^{2}}{2 a} .$ Algebra chal jaata hai, aur inverse $x (p) = - p / a$ yahan single-valued hai kyunki $f^{'} = - a x$ strictly monotonic hai (yeh $f$ strictly concave hai). Danger sirf tab aata hai jab $f^{'}$ non-monotonic ho (ek bump): tab koi $p$ do alag $x$ 's se hit hota hai aur transform ambiguous ho jaata hai — exactly Exercise 3.2(c) ki existence condition. (b) Physical matlab. Agar $U (S)$ concave hoti, toh $\partial^{2} U / \partial S^{2} = T / C_{V} < 0$ negative heat capacity $C_{V} < 0$ force karta — ek aisa system jo heat add karne par thanda ho jaata hai. Yeh ek runaway instability hai: koi bhi hot spot aur hotter hota jaata hai, koi bhi cold spot aur colder, jab tak phase separate na ho jaaye. Real stable equilibria require karte hain ki $U$ convex in $S$ ho (aur $V$ mein bhi), aur isliye hi unke liye Legendre transform hamesha well-defined hota hai. Dekho Thermodynamic Stability and Convexity.

Figure (above). Green hai stable $U (S)$ : woh upar curve karta hai (convex), isliye uski slope $T = \partial U / \partial S$ steadily badhti hai aur $C_{V} > 0$ . Red-dashed ek unstable concave stretch hai, jahan entropy add karne par slope lower hoti — negative heat capacity, part (b) ka runaway. Yellow line ek tangent hai, jiski slope us point par temperature hai: convexity exactly "every such tangent lies below the curve" hai, jo Legendre transform ko single-valued rakhta hai.

Exercise 5.3 — Ek potential design karo

Problem. Ek aisa "grand potential" $Ω$ invent karo ek system ke liye jo reservoir ke saath particles exchange karta hai (toh $μ$ controlled hai, $N$ nahi). $F (T, V, N)$ se shuru karo jahan $d F = - S d T - P d V + μ d N$ , $N$ -slot ko Legendre-transform karo aur $Ω$ , uska differential, aur uske natural variables do.

Recall Solution

Yaad karo $F = U - T S$ Helmholtz hai aur $μ$ chemical potential hai (ek particle add karne ki energy). Step 1 — kaun si slope swap karni hai. Hum $μ$ (intensive, reservoir-controlled) ko independent variable banana chahte hain $N$ ki jagah. $N$ mein $F$ ki slope hai: $(\frac{\partial F}{\partial N})_{T, V} = μ .$ Step 2 — $N$ -slot par transform (slope $\times$ var $= μ N$ subtract karo): $Ω = F - μ N .$ Step 3 — differential: $d Ω = d F - μ d N - N d μ = (- S d T - P d V + μ d N) - μ d N - N d μ = - S d T - P d V - N d μ .$ Step 4 — natural variables: drivers hain $d T$ , $d V$ , $d μ$ , toh $Ω = Ω (T, V, μ),$ yeh grand potential hai — grand canonical ensemble ke liye natural object. Yeh exactly woh potential hai jo grand partition function ke saath $Ω = - k_{B} T ln Z$ ke zariye pair karta hai.

Quick self-check

Recall Fixed

$(T, P, N)$ par kaun sa potential minimize hota hai? Gibbs free energy $G = U - T S + P V$ , kyunki $(T, P, N)$ uske natural variables hain — Exercise 4.2 dekho.

Recall Enthalpy vs grand potential ke liye Legendre term ka sign?

Enthalpy ke liye slope $\partial U / \partial V = - P$ negative hai, toh subtract karna $+ P V$ deta hai: $H = U + P V$ . Grand potential ke liye slope $\partial F / \partial N = + μ$ positive hai, toh subtract karna $- μ N$ deta hai: $Ω = F - μ N$ .