Legendre transforms connecting them: Foundations — Hinglish deep dive

2.4.2 · D1 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Legendre transforms connecting them

Yeh page yeh assume karta hai ki aapne pehle kuch nahi dekha. parent page pe jo bhi symbol hain, sab yahan unpack kiye gaye hain, ek aisi order mein jahan har idea apne pehle waale par tika hua hai.

0. Woh ek prerequisite arrow jo kabhi skip nahi karte: ek slope

Kisi bhi thermodynamics se pehle, hum calculus ka ek idea chahte hain: kisi curve ka ek point par slope.

Ek smooth curve $y = f (x)$ imagine karo. Ek point par itna zoom in karo jab tak curve seedhi na lag jaaye — woh choti si seedhi piece ki ek steepness hoti hai. Wahi steepness slope hai, likha jaata hai $f^{'} (x)$ ya $\frac{df}{d x}$ .

Neeche wali figure slope ko define karti hai ek chote rise-over-run triangle ke zariye jo tangent line par baitha hai: orange line violet curve ko pink dot par sirf touch karti hai, aur uski steepness (rise ÷ run, dashed navy triangle) wahan $f^{'} (x)$ hai. Ise padhein: "tangent ki steepness hi number $f^{'} (x)$ hai."

Figure — Legendre transforms connecting them

1. Functions aur unke "natural variables"

Ek function $f (x)$ ek rule hai: $x$ daalo, number $f (x)$ nikalo. Jab hum $U (S, V, N)$ likhte hain, matlab "ek energy jo ek saath teen inputs par depend karti hai."

2. Chote changes: symbol $d$

Kisi quantity ke aage symbol $d$ ka matlab hai "isme infinitesimally small change." $d U$ = energy change ka ek whisker. $d S$ = entropy change ka ek whisker.

Curly $\partial$ aur uske subscripts. Symbol $\partial$ ("partial derivative") ek slope hai jo baaki sab inputs ko frozen rakhtey hue liya jaata hai. Chote subscript labels batate hain kaunse inputs frozen hain:

3. First law — jahan $T$ , $P$ , $μ$ paida hote hain

Energy conserved hoti hai: kisi closed chunk of stuff mein jo bhi energy daalo, woh iske andar kahin na kahin jaani chahiye. Energy daalne ke sirf do common tarike hain, plus matter add karne ke liye ek.

Intuition WHY heat

$T d S$ hai aur work $- P d V$ hai (physical origin)

Box ko squeeze karo (work). Piston ko ek tiny volume $d V < 0$ se pressure $P$ ke against andar push karo. Mechanics kehta hai gas par kiya gaya work force×distance $= P \times (area) \times (inward distance) = - P d V$ hai. Andar push karo ( $d V < 0$ ) aur yeh positive hai — tumne energy add ki. Yahi $- P d V$ term hai. (Yeh inexact $δ W = - P d V$ hai: transit mein work.)
Heat karo (heat). Temperature ki definition second law se hai $T = (\partial U / \partial S)$ : temperature exactly entropy ke ek unit ki energy price hai. Toh reversible whisker of heat $δ Q = T d S$ hai — disorder $d S$ add karo aur tum $T$ per unit pay karte ho. Yahan $δ Q$ §2 ka curly $δ$ carry karta hai kyunki heat path-dependent transfer hai; phir bhi uski reversible value exact combination $T d S$ ke barabar hoti hai. Yahi $T d S$ term hai.
Ek particle add karo. Ek aur molecule andar slide karna $μ$ per particle cost karta hai, $μ d N$ deta hai. Teen energy inflows add karo aur first law milta hai. Dono inexact transfers $δ Q + δ W$ milkar ek exact $d U$ banate hain — yahi first law ka content hai.

$d U = T d S - P d V + μ d N .$

Yeh thermodynamics ka first law hai (energy bookkeeping). Ab ise §2 ke differential ke saath term-by-term match karo. Jo bhi $d S$ ko multiply karta hai woh $\partial U / \partial S$ hona chahiye, wagairah:

Neeche wali figure $T$ ko $U$ -vs- $S$ curve ki actual slope ke roop mein dikhati hai: ek point (pink dot) chuno, tangent (orange) draw karo, uski steepness hi wahan temperature hai.

4. Extensive vs intensive, aur conjugate pairs

Variables ko pair karne se pehle, hume ek classification chahiye.

First law dobara dekho: $d U = T d S - P d V + μ d N$ . Har term ek extensive change aur ek intensive partner ka product hai:

extensive (ek variable)	uska intensive slope-partner
$S$	$T$
$V$	$- P$
$N$	$μ$

5. Legendre transform khud — exact operation

Upar sab kuch setup tha. Yahan machine hai, ek baar ke liye clearly bataya gaya.

Intercept derive karo. Point $x$ par tangent line ka slope $p = f^{'} (x)$ hai, toh uska equation hai $Y = f (x) + p (X - x)$ . Uska $y$ -intercept padhne ke liye $X = 0$ set karo, jise hum $b$ kehte hain: $b = f (x) - p x .$

Isse ek thermodynamic potential banao. $U (S, V, N)$ lo aur $S$ -slot ka Legendre-swap karo (jiska slope $T$ hai). Standard thermodynamic sign land karne ke liye slope×variable, $T \cdot S$ subtract karo: $F = U - T S .$ Ab $d F$ find karo. Hume product $T S$ ka change chahiye, aur yahan reason hai ki hum sirf " $T d S$ " nahi likh sakte: dono $T$ aur $S$ move kar sakte hain, isliye product ke change ke do contributions hain. Yeh calculus ka ordinary product rule hai, $d (T S) = T d S + S d T (bilkul jaise d (uv) = u d v + v d u),$ kyunki $T$ aur $S$ sides wale rectangle mein ek chota sa change dono sides par ek thin strip add karta hai. Isliye $d F = d U - d (T S) = d U - T d S - S d T .$ First law $d U = T d S - P d V + μ d N$ substitute karo, $T d S$ pieces cancel ho jaate hain: $d F = (T d S - P d V + μ d N) - T d S - S d T = - S d T - P d V + μ d N .$ Toh $F$ depend karta hai $(T, V, N)$ par. Yeh Helmholtz free energy hai.

Enthalpy $H = U + P V$ . $V$ -slot swap karo (slope $- P$ , isliye $- P \cdot V$ subtract karo, yani add $P V$ ). Same product rule, $d (P V) = P d V + V d P$ : $d H = d U + d (P V) = (T d S - P d V + μ d N) + (P d V + V d P) = T d S + V d P + μ d N .$ $- P d V$ aur $+ P d V$ cancel ho jaate hain, $H$ ko $(S, P, N)$ par chhod jaate hain.

Gibbs $G = U - T S + P V$ . Dono slots ek saath swap karo; dono products par product rule apply karo: $d G = d U - d (T S) + d (P V) = d U - (T d S + S d T) + (P d V + V d P) .$ First law insert karo aur $T d S$ aur $P d V$ pairs cancel karo: $d G = - S d T + V d P + μ d N,$ toh $G$ rehta hai $(T, P, N)$ par — dono lab-controllable.

6. Convexity — swap allowed kyun hai

"Point-description" ko "slope-description" se swap karne ke liye, har slope ko ek unique point naam dena chahiye. Iske liye curve ko consistently ek taraf bend karna hoga, kabhi back and forth nahi wiggle karna.

Neeche wali figure ek well-behaved convex curve (har slope use ek baar hit karta hai) ko ek wiggly curve se contrast karti hai jahan ek hi slope do points par touch karta hai — wahan transform ambiguous hai.

7. Exact differentials — Maxwell relations kyun free mein nikle aate hain

§2 mein hum exact-vs-inexact distinction se mile. Ek exact differential $d U$ hai jo ek genuine function $U (S, V)$ se aaya. Uski pehchaan: mixed second slopes order ki parwah nahi karte.

Kyunki $\partial U / \partial S = T$ aur $\partial U / \partial V = - P$ , ise $U$ par apply karne se milta hai $(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V}$ — ek Maxwell relation, gift-wrapped. Poora set dekho Maxwell Relations mein.

Prerequisite map — pieces aage kaise jaate hain

Diagram ko upar-se-neeche ek dependency chain ke roop mein padhein: har box ek symbol hai jo ab aapka hai, aur har arrow ka matlab hai "neeche waale box se pehle upar wala box chahiye." Calculus ka ek idea (slope, top-left) do independent streams mein fan out hota hai — physics stream (first law → $T, P, μ$ ke meanings → conjugate pairs) aur geometry stream (convexity). Legendre transform box se pehle dono streams aani chahiye; yahi poori baat hai — transform ko slope-meaning (physics) aur ek-slope-ek-point guarantee (geometry) chahiye. Phir woh, exact-differential facts ke saath, chaar potentials $U, H, F, G$ mein merge ho jaate hain.

§5 ne already $F = U - T S$ banaya; parent page saare chaar potentials same tarike se assemble karta hai.

Legendre transform sirf thermodynamics nahi hai — wahi slope-swap ek Lagrangian ko Hamiltonian mein badal deta hai mechanics mein, dekho Lagrangian to Hamiltonian (Legendre in Mechanics). Jab Partition Functions se miloge, tum dekkhoge $F = U - T S$ dobara aata hai $F = - k_{B} T ln Z$ ke roop mein.

Equipment checklist

Khud ko test karo — daayein side cover karo. Agar koi bhi answer fuzzy hai, parent page se pehle woh section dobara padhna.

f^{'} (x)

ka ek phrase mein matlab kya hai?

Tangent line ki steepness (slope)

y = f (x)

ke point

x

par.

Entropy

S

intuitively kya hai, aur ise seedhe set kyun nahi kar sakte?

Microscopic arrangements kitne spread-out hain uska count; koi lab "entropy knob" nahi hota, isliye hum ise

T

se swap karte hain.

Curly

\partial

ka matlab kya hai, aur uske subscripts kya batate hain?

Ek slope jisme ek input wiggle hoti hai jabki subscripted inputs constant values par held fixed rehti hain.

d

aur

δ

mein kya fark hai?

d

ek exact differential hai (genuine state function ka change, path-independent);

δ

inexact hai (ek path-dependent transfer jaise heat ya work, koi underlying state function nahi).

First law likhao aur har term ka physical origin batao.

d U = T d S - P d V + μ d N

;

- P d V

compression work hai,

T d S

reversible heat hai,

μ d N

added particles ki energy hai.

- P = \partial U / \partial V

kyun hai aur

+ P

kyun nahi?

First law

- P d V

term par minus carry karta hai, isliye squeeze karna (negative

d V

)

U

badhata hai.

Extensive vs intensive — ek-line test?

System clone karo: extensive quantities double ho jaati hain (

U, S, V, N

), intensive wahi rehti hain (

T, P, μ

g = p x - f

kyun hai (

f - p x

nahi)?

Flipped sign

d g / d p = + x

banata hai,

g

ko convex rakhta hai aur transform ko clean symmetric involution; dono signs otherwise valid hain.

F = U - T S

differentiate karte waqt,

T S

term kaunsa rule handle karta hai aur kyun?

Product rule

d (T S) = T d S + S d T

, kyunki dono

T

aur

S

change kar sakte hain, isliye rectangle

T S

dono taraf badhta hai.

Convex vs concave, aur convexity physically kahan fail hoti hai?

Convex upar curve karta hai (

f^{''} \geq 0

), concave neeche (

f^{''} \leq 0

); dono unique slopes dete hain, lekin phase transition curve ko flat kar deta hai isliye convexity fail hoti hai.

Exact differential ki kaunsi signature Maxwell relations deti hai, aur kyun?

Mixed second partials equal hote hain (Clairaut's theorem), kyunki genuine smooth function ke cross-slopes order par depend nahi karte.