Visual walkthrough — Stream function, velocity potential

2.2.11 · D2 · Physics › Fluid Mechanics › Stream function, velocity potential

Hum ek nadi se milenge, arrows count karenge, aur koi bhi aisa symbol nahi likhenge jo pehle draw na kiya ho.

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Step 0 — Flow kya hoti hai, pictures mein?

Figure dekho: har grid point par ek blue arrow baitha hai. Uska horizontal shadow $u$ hai, uska vertical shadow $v$ hai. Toh flow describe karne ka matlab hai har point par do numbers jaanna — do poori functions $u(x,y)$ aur $v(x,y)$. Yeh bahut saara bookkeeping hai. Is page ka baaki hissa ek aise tarike ki talash hai jisse ek number carry kiya ja sake do ki jagah.

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Step 1 — Un-squishable rule (incompressibility)

Picture mein kya dikhaya gaya hai. $\Delta x$ by $\Delta y$ side wala ek tiny square. Paani left face se andar aata hai aur right face se bahar jaata hai (yeh hai $u$ traffic); yeh bottom se andar aata hai aur top se bahar jaata hai ($v$ traffic). Net inflow count karo aur use zero set karo.

• Left/right imbalance per unit area $=\frac{\partial u}{\partial x}$ — "rightward speed kitna change hota hai jab tum right chalte ho." Agar $u$ right jaate jaate speed up karta hai, toh un faces se zyada bahar jaata hai jitna andar aata hai.
• Bottom/top imbalance $=\frac{\partial v}{\partial y}$ — "upward speed kitna change hota hai jab tum upar chalte ho."

Total change ko zero set karne se continuity equation milti hai (jo Continuity Equation mein derive ki gayi hai):

$$\underset{\text{net right-face outflow}}{\underbrace{\frac{\partial u}{\partial x}}}+\underset{\text{net top-face outflow}}{\underbrace{\frac{\partial v}{\partial y}}}=0.$$

Yeh ek equation hai jo $u$ aur $v$ ko baandh ke rakhti hai. Yeh woh leash hai jo hum Step 2 mein use karenge do functions ko ek mein shrink karne ke liye.

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Step 2 — $(u,v)$ ko ek scalar $\psi$ mein pack karna

Hum ek aisi single function chahte hain jo automatically Step 1 ke rule ko follow kare, taaki incompressibility free mein aa jaye. Trick yeh hai: $u$ aur $v$ ko ek hill $\psi (x,y)$ ke slopes ki tarah build karo.

Hum kya karte hain — check. Yeh do definitions continuity equation mein daalo:

$$\frac{\partial }{\partial x}\left(\underset{=u}{\underbrace{\frac{\partial \psi }{\partial y}}}\right)+\frac{\partial }{\partial y}\left(\underset{=v}{\underbrace{-\frac{\partial \psi }{\partial x}}}\right)={\psi }_{xy}-{\psi }_{yx}.$$

Kyun yeh zero ho jaata hai. Do partial derivatives lene ka order smooth fields ke liye matter nahi karta: ${\psi }_{xy}={\psi }_{yx}$ (yeh Clairaut's theorem hai, picture mein: ek smooth hill ko east-then-north differentiate karna = north-then-east). Toh do terms equal aur opposite hain — woh cancel ho ke $0$ dete hain chahe $\psi$ kuch bhi ho.

Humne do functions ko ek se replace kar liya, aur un-squishable rule ab by construction hold karta hai.

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Step 3 — Kyun $\psi$ ki constant lines streamlines hain

Tangent condition kya kehti hai. Agar ek tiny step $(dx,dy)$ ek streamline ke saath lie karta hai, toh usse $(u,v)$ ke parallel hona chahiye. Do vectors parallel hote hain jab unka "cross" combination zero ho:

$$udy-vdx=0.$$

Use aise padho: step ka sideways part flow ke tilt se match karta hai.

Usi step par $\psi$ mein kya change aata hai. Do slope contributions ko jodkar:

$$d\psi =\underset{=-v}{\underbrace{\frac{\partial \psi }{\partial x}}}dx+\underset{=u}{\underbrace{\frac{\partial \psi }{\partial y}}}dy=-vdx+udy=udy-vdx.$$

Kyun yeh punchline hai. Right side exactly tangent condition hai. Toh ek streamline ke saath $udy-vdx=0$, jiska matlab hai $d\psi =0$ — $\psi$ streamline ke saath chalte waqt change nahi hota. Isliye:

$$\boxed{\psi =\text{const}⟺\text{a streamline}}$$

Figure mein green curves $\psi$-hill ki level curves hain, aur blue arrows unke tangent hain — kabhi cross nahi karte.

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Step 4 — $\psi$ paani count karta hai (flux = labels ka difference)

Hum kya compute karte hain. Streamline ${\psi }_1$ ke ek point aur streamline ${\psi }_2$ ke ek point ko join karne wali koi bhi line draw karo. Us line ko per second cross karne wala paani hai

$$Q=\int (udy-vdx)=\int d\psi ={\psi }_2-{\psi }_1.$$

Kyun yeh gorgeous hai. Kyunki $udy-vdx$ hi $d\psi$ hai (Step 3), integral sirf $\psi$ mein tiny changes add karta hai aur endpoints par telescope karta hai — crossing line ki shape matter nahi karti, sirf woh do streamlines matter karti hain jo wo connect karti hai. Toh $\psi$ literally "mere left mein kitna paani hai" ka meter hai. Do units apart do streamlines hamesha unke beech do units flow carry karti hain.

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Step 5 — Non-spinning rule ek doosra scalar $\varphi$ paida karta hai

Incompressibility ne humein $\psi$ diya. Ek alag physical rule ek alag scalar deta hai.

Yeh humein kya deta hai. Calculus ka ek theorem kehta hai: agar ek field ka zero curl ho (simply-connected patch mein), toh woh kisi scalar hill $\varphi$ ka gradient hai. Toh hum likh sakte hain

$$u=\frac{\partial \varphi }{\partial x},v=\frac{\partial \varphi }{\partial y}.$$

Kyun iss $\varphi$ mein automatically koi spin nahi hai. Kisi bhi gradient का curl identically zero hota hai — wahi Clairaut cancellation jaise Step 2 mein, ab hamare liye kaam karta hai: ${\partial }_x{\varphi }_y-{\partial }_y{\varphi }_x=0$. Toh velocity ko gradient ki tarah likhne se irrotationality free milti hai, exactly jaise use $\psi$ ke zariye likhne se incompressibility free milti thi.

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Step 6 — Jab DONO exist karte hain: Cauchy–Riemann aur Laplace

Agar flow dono un-squishable aur non-spinning hai ("ideal flow"), $\psi$ aur $\varphi$ saath-saath exist karte hain. Ab $u$ aur $v$ mein se har ek ke do naam hain — ek har scalar se. Naam barabar set karo:

Yeh Cauchy-Riemann Equations hain — wahi jo ek analytic complex function define karte hain.

Kyun har scalar Laplace's equation solve karta hai (dekho Laplace Equation):

• $\varphi$: incompressibility kehti hai $u_x+v_y=0$; $u={\varphi }_x,v={\varphi }_y$ substitute karo:

$$\underset{u_x}{\underbrace{{\varphi }_{xx}}}+\underset{v_y}{\underbrace{{\varphi }_{yy}}}=0\Rightarrow {\nabla }^2\varphi =0.$$

• $\psi$: irrotationality kehti hai $v_x-u_y=0$; $v=-{\psi }_x,u={\psi }_y$ substitute karo:

$$\underset{v_x}{\underbrace{-{\psi }_{xx}}}-\underset{u_y}{\underbrace{{\psi }_{yy}}}=0\Rightarrow {\nabla }^2\psi =0.$$

Har rule doosre scalar ko Laplace mein feed karta hai. Yeh beautiful crossover notice karo: incompressibility $\varphi$ ko tame karta hai, irrotationality $\psi$ ko tame karta hai. Isliye har ideal 2D flow ek pair of harmonic hills hai.

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Step 7 — Kyun do maps right angles par cross karte hain

Hum kya compare karte hain. Kisi bhi point par do gradient arrows:

$$\nabla \varphi =({\varphi }_x,{\varphi }_y)=(u,v),\nabla \psi =({\psi }_x,{\psi }_y)=(-v,u).$$

Pehla $\varphi$-level-curves (equipotentials) ke across point karta hai; doosra $\psi$-level-curves (streamlines) ke across point karta hai.

Kyun perpendicular. Unka dot product lo — right angle ka test (zero matlab $90^{\circ }$):

$$\nabla \varphi \cdot \nabla \psi =(u)(-v)+(v)(u)=-uv+uv=0.$$

Zero, hamesha. Toh equipotentials aur streamlines har jagah $90^{\circ }$ par milte hain: woh flow net bunate hain, curvy graph paper ki ek sheet. Yeh parent note ka final claim hai, ab dekha gaya.

Recall Edge cases jo tumhe skip nahi karne chahiye

Stagnation points ($u=v=0$): dono gradients vanish ho jaate hain, toh dot-product test trivially $0=0$ padhta hai — orthogonality statement wahan khali hai aur grid lines fork ho sakti hain (jaise ek body ki naak par). Kuch bhi toot nahi raha; net mein bas ek crossing point hai. Reveal ::: Ek stagnation point par $\nabla \varphi =\nabla \psi =\vec{0}$; flow net mein X-shaped junction ho sakta hai. Yeh degenerate hai, contradiction nahi.

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Worked example pictures ke saath — ek point source

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Ek-picture summary

Recall Feynman: poori walkthrough plain words mein

Upar se dekhi gayi ek shant nadi ki picture karo. Har jagah paani ka ek arrow hai — yeh do numbers hain, sideways speed aur upward speed, aur har jagah do numbers juggle karna thaka deta hai.

Pehli trick. Paani squish nahi ho sakta, toh kisi bhi tiny box se utna hi baahaar jaata hai jitna andar aata hai. Woh single rule humein nadi ko "lane numbers" se paint karne deta hai — ise $\psi$ kaho. Paani kabhi lanes nahi badlta, toh ek equal lane number ki line ek aisi path hai jo paani actually follow karta hai (ek streamline), aur do lane numbers ke beech ka gap exactly utna paani hai jitna us channel mein bahhta hai. Ek number ne do ko replace kar diya.

Doosri trick. Agar paani mein koi little whirlpools bhi nahi hain, toh hum ise "heights" se paint kar sakte hain — ise $\varphi$ kaho — aur paani hamesha higher ground ki taraf slide karta hai. Phir se do ki jagah ek number.

Jab dono sach hon, har painting ek special tarike se smooth hoti hai (Laplace's equation), aur — magic finale — lane-lines aur height-lines har jagah perfect right angles par cross karti hain, jaise bent graph paper. Ise ek source par test karo: equal height ke circles, equal lane ke spokes, $90^{\circ }$ par milte hue, source se nikla hua paani exactly iske strength ke barabar add hota hai.

Do scalars, do physical rules se born, ek orthogonal grid mein bunay. Yeh ek saanson mein poora chapter hai.

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