Conservative vector fields — potential functions: Visual walkthrough

4.4.26 · D2 · Maths › Multivariable Calculus › Conservative vector fields — potential functions

Yeh page parent note par boxed result ki picture-by-picture derivation hai, Conservative vector fields — potential functions: $\int_{C} \nabla f \cdot d r = f (B) - f (A) .$

Hum us formula ka har piece kamayenge.

Step 1 — Vector field kya hota hai? (har point par ek arrow)

KYA HAI: Hum plane ko ghaans ke maidan ki tarah sochte hain, aur har tainde par ek chhota arrow chipkaate hain.

KYU HAI: Baaki sab — force, work, potential — in arrows se padhte hain. Pehle inhe dekhna zaroori hai.

PICTURE: Figure mein kaale arrows $F$ hain. Ek laal arrow dekho: uski horizontal shadow $P$ hai, uski vertical shadow $Q$ hai. $P$ aur $Q$ ka matlab bas itna hi hai — ek arrow ki do shadows.

Figure — Conservative vector fields — potential functions

Step 2 — Chhupa hua hill: ek scalar function $f$

KYA HAI: Har point par arrow ki jagah, $f$ har point par ek height rakhta hai.

KYU HAI: Conservative field woh hai jo aisi hill se aayi ho. "Arrows neeche ki taraf point karte hain" kehne se pehle, hill ka drawn hona zaroori hai.

PICTURE: Nestedlaal loops $f$ ki contour lines hain — har loop ek fixed altitude hai. Jahan loops paas paas hain, hill steep hai; jahan door hain, wahan flat hai.

Step 3 — Gradient $\nabla f$ : sabse steep uphill arrow

KYA HAI: Hum hill ki do directional slopes se har point par ek naya arrow banate hain.

KYU do partial slopes, ek nahi? Ek dimension mein hill ki ek hi slope hoti hai. Do dimensions mein steepness depend karta hai aap kis taraf dekhte ho, isliye hume do numbers chahiye — east mein rise aur north mein rise — aur inhe vector mein stack karna ek sabse steep-uphill direction deta hai. Yeh exactly woh tool hai jo Gradient and directional derivative provide karta hai.

PICTURE: Marked point par laal gradient arrow seedha uphill point karta hai, us contour line ke perpendicular jis par woh baitha hai. Conservative field $F = \nabla f$ matlab hai kaale field arrows in laal gradient arrows hain.

Step 4 — Path ke saath kaam: line integral

KYA HAI: Har instant par hum field arrow aur velocity arrow ka dot product lete hain, phir saare instants ko add karte hain.

KYU dot product? Dot product $a \cdot b$ measure karta hai kitna $a$ , $b$ ke saath point karta hai. Work sirf force ka woh part count karta hai jo aapki motion ke saath push karta hai — sideways force koi kaam nahi karti. Dot product exactly woh tool hai jo along-part rakhta hai aur sideways-part discard karta hai. (Zyada Line integrals of vector fields mein.)

PICTURE: Curve ke saath, har dot par laal velocity arrow aur kaala field arrow ek angle par milte hain; dot product sirf unka aligned overlap leta hai.

Step 5 — Ek magic move: chain rule integrand ko pure derivative mein badal deta hai

Ab $F = \nabla f$ ko work integral mein substitute karo: $\int_{a}^{b} \nabla f (r (t)) \cdot r^{'} (t) d t .$

Integrand ko term by term dekho: $(f_{x}, f_{y}) \nabla f \cdot (x^{'} (t), y^{'} (t)) r^{'} = f_{x} x^{'} (t) + f_{y} y^{'} (t) .$

KYA HAI: Hum claim karte hain yeh sum exactly woh time-rate-of-change hai jis par height $f$ badlati hai jab hum curve ke saath ride karte hain: $\frac{d}{d t} f (x (t), y (t)) = \frac{\partial f}{\partial x} x^{'} (t) + \frac{\partial f}{\partial y} y^{'} (t) .$

YEH KYU SACH HAI? Yeh multivariable chain rule hai. Ek chhote time step $d t$ mein dot right ki taraf $x^{'} d t$ aur upar $y^{'} d t$ move karta hai. Sideways motion ka har bit height ko (us direction mein slope) × (distance) se change karta hai: east part $f_{x} x^{'} d t$ contribute karta hai, north part $f_{y} y^{'} d t$ contribute karta hai. Unhe add karo — yeh total height change $df$ hai. $d t$ se divide karo aur formula mil jaata hai. Toh integrand hai altitude-along-the-curve ka derivative.

PICTURE: Do chhote right-triangle steps stacked: ek horizontal nudge $x^{'} d t$ jo $f_{x} x^{'} d t$ climb karta hai, aur ek vertical nudge $y^{'} d t$ jo $f_{y} y^{'} d t$ climb karta hai. Unki heights ek climb $df$ mein add ho jaati hain (laal).

Step 6 — Height changes add karo: endpoints tak telescoping

Kyunki integrand ek perfect derivative hai, Fundamental Theorem of Calculus apply hota hai. Yeh kehta hai: kisi quantity ke saare chhote changes ko add karna uska net change start se end tak deta hai.

$\int_{a}^{b} \frac{d}{d t} [f (r (t))] d t = end B par height f (r (b)) - start A par height f (r (a)) .$

KYA HAI: Saare chhote climbs $df$ collapse ho jaate hain ("telescope") sirf final height minus starting height mein.

KYU: Har intermediate height ek baar add hoti hai aur agle step mein ek baar subtract hoti hai, toh saara beech cancel ho jaata hai — sirf do endpoints bachte hain. Isliye tedha-medha path irrelevant hota hai.

PICTURE: Chhote laal rises ki ek staircase; jab aap unhe sum karte ho, har internal step cancel ho jaata hai aur sirf $A$ aur $B$ ke beech ka total vertical gap bachta hai.

Step 7 — Do paths, same answer (path-independence visible roop mein)

KYA HAI: Usi $A$ se usi $B$ tak do bilkul alag curves $C_{1}$ aur $C_{2}$ lo. Step 6 kehta hai dono integrals $f (B) - f (A)$ ke barabar hain, toh woh ek doosre ke barabar hain.

KYU IMPORTANT HAI: Yeh "conservative" ka physical meaning hai — aap ek hoshiyaar route chunke alag amount ka kaam nahi pa sakte.

PICTURE: Ek chhota seedha laal path aur ek lamba looping kaala path endpoints $A, B$ share karte hain. Dono same net height climb karte hain, toh dono same kaam karte hain.

Step 8 — Closed loop: net work exactly zero hota hai (degenerate case $A = B$ )

KYA HAI: Ab path ko wahan waapas aane do jahan se shuru hua, toh $B = A$ . Toh: $\oint_{C} \nabla f \cdot d r = f (A) - f (A) = 0.$

KYU YEH CRUCIAL EDGE CASE HAI: Ek closed loop "free-energy" test hai. Agar aap hamesha ke liye circle karte rehne se energy gain kar sakte, toh field conservative nahi hogi. Formula ise forbid karta hai: end height = start height ⟹ net work $= 0$ . Yeh Conservation of energy in physics ek line mein hai, aur yeh Green's theorem aur Curl of a vector field se "andar koi swirl nahi" condition ke roop mein juda hua hai.

PICTURE: Contour lines ke upar padi ek closed laal loop: woh har contour ko upar jaate waqt utni hi baar cross karta hai jitni baar neeche aate waqt, toh heights zero mein cancel ho jaate hain. Usके saath, ek hole ke aas paas ek loop jahan woh cancel nahi hote.

Ek-picture summary

Upar sab kuch ek sentence mein ek picture ke saath hai: field arrows kisi hill ke neeche ki taraf arrows hain, toh work = height climbed = end height − start height.

Recall Feynman retelling — saari walkthrough plain words mein

Ek hilly park ki picture karo jahan, har jagah, ek arrow dikhata hai ball kis taraf roll karegi: seedha neeche. Woh arrow-map hamaari field $F$ hai, aur hill ka height-map $f$ hai. Ab bench se swings tak koi bhi tedha-medha path chalo. Har chhote step mein, "slope se jo push milti hai" times jitni doori chale woh exactly kitna aapka altitude badla — yeh chain rule hai, aur yeh ek akal mand move hai. Kyunki har step ka kaam sirf ek chhoti height-change hai, inhe sab add karne par sirf total height change milta hai: swings ki height minus bench ki height. Beech ke zig-zags sab cancel ho jaate hain. Toh koi bhi path lo, koi farq nahi, sirf shuru aur khatam matter karta hai. Aur agar bench waapas loop chaloge, toh aap usi height par khatam karte ho — total work zero hai. Aap circle karte rehne se free energy kabhi nahi milegi. Ek warning: iske liye andar koi holes ke bina ek proper single hill chahiye; agar ground mein ek puncture hai (jaise ek whirlpool ka center), toh "height" har lap mein $2 π$ climb kar sakti hai aur trick toot jaata hai.

Recall Quick self-test

Conservative field ke liye path ka shape kyun matter nahi karta? ::: Work $f (B) - f (A)$ ke barabar hota hai, jo sirf endpoints use karta hai. Kaunsa ek theorem $\nabla f \cdot r^{'}$ ko $\frac{d}{d t} f (r (t))$ mein convert karta hai? ::: Multivariable chain rule. Work integral mein dot product kya throw away karta hai? ::: Force ka woh part jo motion ke perpendicular hai (sirf aligned part kaam karta hai). Closed-loop integral zero kyun hota hai? ::: Start aur end heights barabar hoti hain, toh $f (A) - f (A) = 0$ . Loop integral zero failkyun ho sakta hai chahe $F = \nabla f$ locally ho? ::: Jab domain mein hole ho (simply connected nahi), toh koi single-valued hill exist nahi karti.