Question bank — Kelvin's circulation theorem

False. Inviscid hona kaafi nahi — tumhe barotropic ($\rho =\rho (p)$) bhi chahiye taaki $\frac{1}{\rho }\nabla p$ ek pure gradient ban sake; agar $\rho$ temperature pe bhi depend kare toh pressure term survive karta hai aur circulation generate karta hai. Dekho Baroclinic vorticity generation.

False. Sirf ek material loop ke around jo flow ke saath convect kare. Space mein fixed ek loop generally $\Gamma$ change hote dekhta hai kyunki alag-alag swirl wala fluid usse cross karta hai.

True. Har material loop ka $\Gamma =0$ initially hai, Kelvin har ek ko hamesha ke liye $0$ pe rakhta hai, toh $\nabla \times \mathbf{u}=0$ persist karta hai — Lagrange/Cauchy corollary (dekho Vorticity and the vorticity equation).

False. Yeh ek material loop ke through vorticity ka flux conserve karta hai ($\Gamma ={\iint }_S\mathbf{\omega }\cdot d\mathbf{A}$ via Stokes' theorem), local $\mathbf{\omega }$ nahi. Vorticity stretching se intensify ho sakti hai jabki $\Gamma$ fixed rehti hai.

True. Agar $\rho =\text{const}$ toh trivially $\rho =\rho (p)$, toh pressure term $\oint \nabla P\cdot d\mathbf{\ell }=0$ pe reduce ho jaata hai. Lekin barotropic zyada general (weaker) condition hai.

False. Thin boundary layers se door, viscous effects chhote hote hain aur flow nearly ideal behave karti hai, toh $\Gamma$ bahut nearly conserved hoti hai — isliye airfoil theory (Kutta–Joukowski lift theorem) itni achhi tarah kaam karti hai.

False. Gravity conservative hai ($\mathbf{f}=-\nabla \Phi$), aur $\oint \nabla \Phi \cdot d\mathbf{\ell }=0$ kisi bhi closed loop ke around. Yeh loop par per unit mass zero net work karta hai, toh yeh $\Gamma$ nahi badal sakta.

Broadly true as an analogy. Conserved $\Gamma$ with $\Gamma =\omega A$ force karta hai $\omega$ ko upar jab area shrink ho — skater-pulling-in-arms effect. Yeh rotational inertia ke material tubes ke along preserve hone ka fluid statement hai.

Circulation constant rehti hai; vorticity chaar guni hoti hai ($\omega \propto 1/A$ kyunki $\Gamma =\omega A$ fixed hai). Speaker ne conserved quantity ko intensified wali se confuse kar liya.

Kuch bhi violate nahi hota — original material loop ka total $\Gamma$ $0$ rehta hai; shed vortex $-{\Gamma }_{\text{bound}}$ bound circulation $+{\Gamma }_{\text{bound}}$ ko exactly cancel karta hai.

Notation carefully dekho: circulation khud hai $\oint \mathbf{u}\cdot d\mathbf{\ell }$ (velocity dotted with ek length step $d\mathbf{\ell }$), jabki yeh term hai $\oint \mathbf{u}\cdot d\mathbf{u}$ (velocity dotted with ek velocity change $d\mathbf{u}$) — ek alag object jo loop ke stretching se janam leta hai. Yeh $\oint d(\frac{1}{2}|\mathbf{u}{|}^2)$ ke equal hai, ek closed loop ke around ek perfect differential, jo exactly $0$ hai; toh stretching kuch contribute nahi karti.

Hamesha nahi — sirf tab jab $\frac{1}{\rho }\nabla p=\nabla P$ ek pure gradient ho, jiske liye barotropicity chahiye. Ek baroclinic fluid mein yeh term genuinely nonzero hoti hai aur circulation create karti hai.

Element rigidly nahi move karta; neighbouring particles ki velocities thodi alag hoti hain, toh $D(d\mathbf{\ell })/Dt=d\mathbf{u}\ne 0$. Yeh stretch aur rotate karta hai — yahi vortex stretching ka poora point hai.

Woh alag hain. Bernoulli's principle pressure aur speed ko ek streamline ke along relate karta hai (ek energy statement); Kelvin ek material loop par circulation conserve karta hai (ek rotational/angular-momentum statement). Dono ko ideal-flow assumptions chahiye lekin alag cheezein kehte hain.

False consequence. Ek tube ke along conserved circulation ($\nabla \cdot \mathbf{\omega }=0$ ke saath) vortex lines ko close ya boundary tak reach karne par force karta hai — woh mid-fluid terminate nahi ho sakti. Dekho Helmholtz vortex theorems.

Kyunki $D/Dt$ (moving fluid follow karte hue change) wahi hai jo derivation ke do pieces ko — stretching term aur "Euler substitution" ($D\mathbf{u}/Dt$ ko $-\frac{1}{\rho }\nabla p-\nabla \Phi$ se swap karna) — perfect-differential loop integrals mein combine karta hai jo cancel ho jaate hain. Ek fixed loop $\partial /\partial t$ use karta hai instead, alag swirl wale fluid ko isse cross karne deta hai, toh uski $\Gamma$ freely change ho sakti hai. Top pe deforming-loop figure dekho.

Sirf tabhi $\frac{1}{\rho }\nabla p$ ko ek single-valued pressure function $P(p)=\int \frac{dp}{\rho (p)}$ ke gradient ke roop mein likha ja sakta hai — ek antiderivative sirf $p$ se bana. Kyunki $\nabla P=\frac{dP}{dp}\nabla p=\frac{1}{\rho }\nabla p$, uska loop integral $\oint \nabla P\cdot d\mathbf{\ell }=\oint dP=0$ apni start value pe waapas aata hai. Agar $\rho$ temperature pe bhi depend kare, toh koi aisa single-valued $P$ exist nahi karta aur term survive karta hai (baroclinic torque, $\nabla \rho \times \nabla p$ se).

Viscosity Euler equation mein ek $\nu {\nabla }^2\mathbf{u}$ term add karta hai (Euler equation for ideal fluids wahi hai jo Kelvin use karta hai); yeh ek gradient nahi hai, toh uska loop integral necessarily vanish nahi karta, aur yeh vorticity ko material loops ke across diffuse karta hai, jisse $D\Gamma /Dt\ne 0$ milta hai. Vorticity friction ke through leak karti hai.

Yeh wing enclosing ek loop ki total circulation ko $0$ rehne par force karta hai, toh lift ke liye needed bound circulation tabhi allowed hai jab ek equal-and-opposite starting vortex shed ho. Woh shedding lift ($L=\rho U{\Gamma }_{\text{bound}}$, Kutta–Joukowski lift theorem) ko physically possible banati hai.

Kyunki $\Gamma =\iint \mathbf{\omega }\cdot d\mathbf{A}$; jab tube ka cross-section thinner stretch hoti hai (area shrinks), $\mathbf{\omega }$ ko flux fixed rakhne ke liye badhna padta hai. $\Gamma$ ki conservation spin-up cause karti hai rather than rokne ke.

Kyunki ek conservative force $-\nabla \Phi$ hai, aur $\oint \nabla \Phi \cdot d\mathbf{\ell }=\oint d\Phi =0$ ek closed loop ke around — potential apni start value pe waapas aata hai, toh per unit mass net work zero hai.

Jab constant density aur constant pressure ki surfaces ek doosre ke relative tilt hoti hain, $\nabla \rho$ aur $\nabla p$ alag directions mein point karte hain; unka cross product $\nabla \rho \times \nabla p$ nonzero hota hai aur ek torque ki tarah act karta hai jo fluid ko spin karta hai (neeche tilted-surfaces figure dekho). Barotropic flow in surfaces ko parallel rakhti hai, toh torque vanish ho jaata hai.

Uski circulation $0$ approach karti hai, $\Gamma =\iint \mathbf{\omega }\cdot d\mathbf{A}\to 0$ ke saath jab area $\to 0$; theorem ab bhi trivially hold karta hai kyunki ek degenerate loop shuru se $0$ thi agar woh us tarah shuru hui thi.

Exactly $0$, kyunki $\oint \mathbf{u}\cdot d\mathbf{\ell }=\mathbf{u}\cdot \oint d\mathbf{\ell }=\mathbf{u}\cdot 0=0$ (loop ka total displacement close ho jaata hai). Kelvin phir ise $0$ pe rakhta hai — koi swirl appear nahi ho sakti.

Haan — stretching term aur conservative-force term ab bhi vanish karte hain; sirf pressure term $\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }$ survive karta hai, deta hai $D\Gamma /Dt=\oint \frac{1}{\rho }\nabla p\cdot d\mathbf{\ell }\ne 0$. Woh surviving term baroclinic source hai.

Cleanly nahi — theorem smooth, differentiable fields assume karta hai; ek sheet ke across velocity discontinuous hai aur $\mathbf{\omega }$ singular hai, toh material-derivative manipulation wahan invalid hai.

Nonzero. Ek paddle ek non-conservative, non-pressure force apply karta hai jo loop ke around net work karta hai, toh yeh ek external circulation source ki tarah act karta hai — exactly woh ingredient jo Kelvin ki idealisations exclude karti hain.

Haan. Har instant pe $\oint \nabla \Phi \cdot d\mathbf{\ell }=\oint d\Phi =0$ explicit time dependence se regardless, kyunki spatial gradient ab bhi ek closed loop ke around zero integrate karta hai.

Haan. Domain multiply-connected hai (cylinder ek hole hai jisse loop shrink hokar nahi ja sakta), toh ise wrap karne wala ek loop fixed nonzero $\Gamma$ hold kar sakta hai jo Kelvin conserve karta hai — yahi exactly lift ke peeche bound circulation hai. Multiply-connected figure neeche dekho.

Nahi. Obstacle enclosing na karne wala loop ek point tak shrink ho sakta hai (irrotational flow mein $\Gamma =0$ deta hai), lekin woh loop jo obstacle wrap karta hai nahi kar sakta — woh trapped bound circulation carry kar sakta hai. Loops ki do categories topologically alag hain, aur Kelvin har category ki value alag se conserve karta hai.