Divergence theorem (Gauss's theorem) — statement, flux-divergence connection: Visual walkthrough

4.4.33 · D2 · Maths › Multivariable Calculus › Divergence theorem (Gauss's theorem) — statement, flux-diver

Recall Teen hypotheses (abhi plain words mein, symbols baad mein)

Yeh finished result tab valid hoga jab:

Region finite aur closed off ho — ek solid lump $V$ jise aap paint mein duba sako, jiske chaaron taraf ek definite skin ho (mathematicians ise compact kehte hain: closed aur bounded).
Skin finitely many smooth patches se bani ho — jaise ek soccer ball ya cube: smooth panels jo edges aur corners par milte hain. Ise piecewise-smooth kehte hain. Step 4 explain karta hai ki edges aur corners koi trouble kyun nahin karte.
Field kabhi andar blow up na kare — iska rate-of-change (woh derivatives jo hum Step 2 mein milte hain) $V$ mein har jagah finite rahe; yeh continuous first partial derivatives condition hai. Step 7 exactly dikhata hai kya galat hoga agar yeh fail hoti hai.

Hum yahan inka wording deliberately words mein kar rahe hain. Jis moment kisi symbol jaise $\hat{n}$ , "flux," ya "divergence" ki zaroorat padegi, uski apni definition box neeche pehle ek picture par earn karegi.

Step 0 — Vector field kya hai, aur "wall ke paas se flow" kya hota hai?

Ab woh key idea jo baar baar kaam aayegi: ek flat wall ke paas se per second kitna fluid cross karta hai?

Figure — Divergence theorem (Gauss's theorem) — statement, flux-divergence connection

Figure mein wall dekho. Uska area $A$ hai aur ek chosen arrow $\hat{n}$ (length 1) seedha usse bahar ki taraf point karta hai — outward normal (yeh bilkul wahi symbol hai jo finished theorem ko chahiye; hum ise abhi, picture par, use karne se pehle define kar rahe hain). Fluid arrow $F$ wall se ek angle par takraata hai.

Dot product kyun aur kuch nahin? Humein ek single number chahiye "do arrows kitne aligned hain," jahan perfectly aligned full value de, perpendicular zero de, opposite negative de (flow andar aa raha hai). Dot product exactly woh machine hai. Yeh ek quantity — flow-across — neeche sabka atom hai.

Step 1 — Ek tiny box ke do $x$ -faces se flux bahar

$x$ -direction face karne wale do walls se shuru karo (figure mein left aur right walls).

Right wall $x + Δ x$ par baitha hai. Iska outward normal $+ \hat{x} = (1, 0, 0)$ hai. Ise cross karne wala flow wahan evaluate kiya gaya $x$ -push $P$ hai, wall area $Δ y Δ z$ se multiply karke: $out through right = + P (x + Δ x, y, z) Δ y Δ z .$
Left wall $x$ par baitha hai. Iska outward normal doosri taraf point karta hai, $- \hat{x}$ , toh iska flow $F \cdot (- \hat{x}) = - P (x, y, z)$ hai: $out through left = - P (x, y, z) Δ y Δ z .$

Subtraction kyun: positive $P$ right par bahar push karta hai lekin left par andar. Net escape right-outflow minus left-inflow hai. Woh difference agले step ka star hai.

Step 2 — Difference ko slope mein convert karo (derivative appear hoti hai)

Hamare paas $P (x + Δ x) - P (x)$ hai: $P$ mein kitna change aaya jab hum $Δ x$ right mein step kiya. Yeh exactly woh hai jo ek derivative measure karta hai.

Ek tiny step ke liye, "change = slope × step": $P (x + Δ x, y, z) - P (x, y, z) = \frac{\partial P}{\partial x} Δ x + leftover error ε (Δ x) Δ x .$

Yahan har term apni jagah earn karta hai:

$\frac{\partial P}{\partial x} Δ x$ — straight-line prediction: slope times step.
$ε (Δ x) Δ x$ — leftover error, woh amount jitna true curve us step par us straight line se bend away karta hai. Number $ε (Δ x)$ $0$ ki taraf shrink hota hai jab $Δ x \to 0$ — yeh precisely derivative ki definition hai (slope woh value hai jis par ratio $(P (x + Δ x) - P (x)) /Δ x$ approach karta hai).

Toh, sirf surviving term rakhte hue, $net x -flux ⟶ \frac{\partial P}{\partial x} = Δ V, box ka volume Δ x Δ y Δ z = \frac{\partial P}{\partial x} Δ V .$

Humne abhi kya kiya: do face-flows ko ek clean expression se replace kiya — ek slope times box ka volume $Δ V$ , plus ek error jo humne abhi prove kiya negligible hai. Yahan surfaces quietly volumes ban jaate hain.

Step 3 — $y$ aur $z$ ke liye repeat karo: divergence paida hoti hai

$x$ -argument ne $x$ ke baare mein kuch special use nahin kiya. Ise $y$ ke do faces ke liye (using $Q$ ) aur $z$ ke do faces ke liye (using $R$ ) rerun karo — har ek ka apna leftover error jo same reason se vanish hota hai: $net y -flux \to \frac{\partial Q}{\partial y} Δ V, net z -flux \to \frac{\partial R}{\partial z} Δ V .$

Teeno pairs add karo — poore tiny box ka total flow bahar:

Yeh kyun matter karta hai: humne theorem ek infinitesimal box ke liye prove kar di hai. Baaki bookkeeping hai ise ek badi region tak grow karne ki.

Step 4 — Boxes stack karo: interior walls cancel (aur edges/corners harmless hain)

Poore solid region $V$ ko in tiny boxes ke grid se bharo.

Do neighbouring boxes ke beech ki koi bhi shared wall dekho (amber mein highlighted). Left box ke liye us wall ka outward normal right point karta hai; right box ke liye usi wall ka outward normal left point karta hai. Toh: $+ P Δ y Δ z (left box se flow out through wall) + - P Δ y Δ z (right box se flow out through wall) = 0.$

Kya bachta hai: sirf woh walls jinke doosri taraf koi neighbour nahin hai — outer walls. Milkr woh outer walls boundary surface $S = \partial V$ ki ek staircase-shaped approximation form karte hain.

Intuition Edges aur corners: "piecewise-smooth" kyun kaafi hai

Hamaari hypothesis ne kaha tha $S$ finitely many smooth patches se bani hai jo edges aur corners par milte hain. Kya yeh junctions argument spoil karte hain? Dekho yeh kya hain: ek edge ek 1-dimensional curve hai, ek corner ek single 0-dimensional point hai. Jab hum $V$ ko side $h$ ke boxes se tile karte hain, edge ko touch karne wale boxes ka total surface area $h$ ki tarah shrink karta hai (ek thin ribbon), aur corner par $h^{2}$ ki tarah — dono flat-patch areas se kaafi faster $0$ jaate hain (jo finite rehte hain). Toh junction boxes ka flux limit mein vanish ho jaata hai, aur surviving outer faces khud smooth patches par converge karte hain. Finite number of edges aur corners kuch contribute nahin karte — yeh exactly isliye hai ki "piecewise-smooth" (na ki "perfectly smooth everywhere") ek sufficient hypothesis hai.

Step 5 — Sab add karo aur boxes shrink karo (sums integrals kyun ban jaate hain)

Saare boxes par Step 3 ka box law add karo. Left par hum divergence×volume sum karte hain; right par hum face-flows sum karte hain, lekin Step 4 se sirf exterior faces bachte hain: $inside sum boxes \sum (\nabla \cdot F) Δ V = skin sum exterior faces \sum (F \cdot \hat{n}) Δ S + (errors \to 0) .$

Yahan $Δ S$ apne naam ka deserve karta hai.

Ab $Δ x, Δ y, Δ z \to 0$ jaane do — boxes points tak shrink hote hain, aur jagged outer surface true smooth surface $S$ ko hug karta hai (figure mein staircase sphere par close in hota dikhta hai).

Intuition (value × tiny piece) ka sum

kyun integral hai Yeh integral sign ka poora matlab hai, toh ise plainly kehna worth hai. Ek Riemann sum ek region ko bahut saare small pieces mein chop karta hai, ek sampled value ko har piece ke size se multiply karta hai, aur add karta hai. Ek integral defined hai as woh number jin par woh Riemann sums approach karte hain jab pieces zero size tak shrink hote hain. Kyunki $F$ ke continuous partials hain (hamaari hypothesis!), $\nabla \cdot F$ ek continuous function hai, aur continuous functions ka guarantee hai ki unke Riemann sums ek single limit mein converge karte hain — the integral. Toh:

$boxes \sum (\nabla \cdot F) Δ V$ solid par ek Riemann sum hai: value $\nabla \cdot F$ times piece-volume $Δ V$ . Iska limit hai triple integral $∭_{V} (\nabla \cdot F) d V$ .
$exterior faces \sum (F \cdot \hat{n}) Δ S$ skin par ek Riemann sum hai: value $F \cdot \hat{n}$ times piece-area $Δ S$ . Iska limit hai surface integral $\iint_{S} F \cdot \hat{n} d S$ . Aur Steps 2–4 ke leftover errors sab ek extra factor carry karte hain jo $0$ jaata hai, toh woh drop out ho jaate hain. Do convergent Riemann sums jo box-by-box equal hain unke equal limits hain.

Step 6 — Edge case: outward normal optional nahin hai

Upar sab kuch outward normals use kiya taaki outflow positive count ho.

Agar aapne accidentally inward normal $- \hat{n}$ use kar liya, har dot product sign flip karta hai: $\iint_{S} F \cdot (- \hat{n}) d S = - ∭_{V} (\nabla \cdot F) d V .$

Step 7 — Degenerate case: andar ek hidden source (singularity)

Box argument assume karta tha ki $F$ har interior point par smooth hai (finite slopes — "continuous first partials" hypothesis). Kya ho agar koi aise spot ho jahan $F$ blow up kare?

Point-source field lo $F = \frac{r}{∣ r ∣ ^{3}} = \frac{( x , y , z )}{( x ^{2} + y ^{2} + z ^{2} ) ^{3/2}} .$ Origin se door $\nabla \cdot F = 0$ hai, toh naive inside-sum $0$ deta. Lekin field origin par undefined hai — koi tiny box wahan baith nahin sakta, toh Step 3 us ek point par fail hoti hai ("continuous first partials on $V$ " hypothesis violated hai).

Ek-picture summary

Ek picture, poori kahani: tiny boxes region ko tile karte hain; har box ke andar divergence measure karti hai kitna fluid appear hota hai; shared walls cancel hote hain; sirf outer skin ka flux bachta hai; grid shrink karna sums ko boxed theorem ke do integrals mein convert karta hai.

Recall Walkthrough ki Feynman retelling

Humne ek single question se shuru kiya — ek wall ke paas se kitna fluid cross karta hai? — aur uska jawab $F \cdot \hat{n}$ times wall ka area se diya: sirf woh part jo seedha through point karta hai escape karta hai. Woh number flux hai. Phir humne smallest possible room banaya, ek tiny box, aur uske chhe walls ko do-do karke add kiya. Har pair ne field ka ek difference diya box ke paas, aur ek tiny step par difference sirf ek slope hai — ek derivative — plus ek tiny bend-error jo box se faster shrink karta hai, toh kabhi matter nahin karta. Teen slopes add karo aur aapko ek number milta hai, divergence, times box ka volume: itna box leak karta hai. Phir humne poori region ko in boxes se tile kiya aur kuch beautiful notice kiya: jahan bhi do boxes touch karte hain, jo ek se leak hota hai woh doosre mein enter karta hai, toh woh inner walls completely cancel ho jaate hain. Yahan tak ki ball ke edges aur corners par baithne wale boxes vanishingly little area carry karte hain, toh woh kuch spoil nahin karte. Sirf outer skin bachti hai. Saare boxes par "leak per box" add karo aur yeh "flow across the skin" ke barabar hai. Woh do adding-up Riemann sums hain, aur continuous cheezein ke Riemann sums integrals ban jaate hain jab pieces vanish hote hain — toh humein divergence ka volume integral aur surface par flux integral milta hai. Woh equality — inside production equals outside escape — divergence theorem hai. Do warnings: normals outward point karo ya har sign flip ho jaata hai, aur kabhi bhi kisi aise spot par box mat run karo jahan field explode kare — pehle ise cut out karo.

Recall Quick self-check

Interior faces cancel kyun hote hain? ::: Neighbouring boxes ek wall share karte hain opposite outward normals ke saath, toh unke flux contributions equal aur opposite hote hain. Ek tiny box kya deta hai? ::: Net flux out $\approx (\nabla \cdot F) Δ V$ , plus ek error jo $Δ V$ se faster shrink karta hai. Box sums integrals kyun ban jaate hain? ::: Yeh continuous functions ke Riemann sums hain; definition se aisi sums integral mein converge karti hain jab pieces zero tak shrink hote hain. $S$ ke edges aur corners proof spoil kyun nahin karte? ::: Edges/corners ke finite set ko touch karne wale boxes ka total area $h$ ya $h^{2}$ ki tarah shrink karta hai, toh unka flux limit mein vanish ho jaata hai. $d S$ kisse bana hai? ::: Tiling boxes ke little flat face areas ( $Δ y Δ z$ , etc.) se, $S$ ki ek infinitesimal tile tak shrunk. Derivation kya break karta hai, aur fix kya hai? ::: $V$ ke andar ek singularity; ise ek small sphere se excise karo aur uske flux ko alag handle karo.

Connections

Flux integrals — skin side $\iint_{S} F \cdot \hat{n} d S$ jo humne box faces se sum kiya.
Divergence and curl — divergence Step 3 ka flux-per-box number hai.
Stokes' theorem — boundary curve ke saath circulation ke liye same tile-and-cancel idea.
Green's theorem — is walkthrough ka flat 2D version.
Gauss's law (electromagnetism) — Step 7 ka $4 π$ point-source case.
Continuity equation — time ke saath motion mein local box law $\nabla \cdot F$ .