Kelvin's Circulation Theorem

Definition (Circulation)

For incompressible fluids, we can define the circulation of the fluid. Imagine drawing a closed loop $C_0$ in the fluid and connecting a specific set of fluid particles. As the fluid flows, those specific particles move and deform into a new loop, denoted by $C_t={\varphi }_t(C_0)$ where ${\varphi }_t$ is the flow map ($\dot{{\varphi }_t}={\mathbf{u}}_t\circ {\varphi }_t$).

The circulation of the velocity along the curve is the line integral of the velocity field ${\mathbf{u}}_t$ around that moving loop:

$${\Gamma }_{C_t}:={\oint }_{C_t}{\mathbf{u}}_t\cdot d\ell ={\oint }_{C_t}{\mathbf{u}}_t^{♭}$$

where $♭$ is the metric dual operator that creates the covector from the original vector field.

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Theorem (Kelvin’s Circulation Theorem)

Suppose the flow velocity ${\mathbf{u}}_t$ satisfies the Euler equation:

$$\frac{\partial }{\partial t}{\mathbf{u}}_t+{\nabla }_{{\mathbf{u}}_t}{\mathbf{u}}_t=-\nabla p_t$$

Kelvin’s Circulation theorem states that if you follow a closed loop of fluid particles as they flow, the circulation around that specific loop will remain perfectly constant over time. That is,

$$\frac{\partial }{\partial t}{\Gamma }_{C_t}=\frac{d}{dt}{\oint }_{C_t}{\mathbf{u}}_t\cdot d\ell =0$$

Abstract Formulation & Proof

We can further abstract this. By converting the velocity vector field into a covector field via $\mathbf{u}\cdot dl={\mathbf{u}}^{♭}$, we can take the pullback on the domain and get:

$$\frac{d}{dt}{\oint }_{C_t}\mathbf{u}\cdot dl=\frac{d}{dt}{\oint }_{C_t}{\mathbf{u}}^{♭}=\frac{d}{dt}{\oint }_{C_0}\left({\varphi }_t^{\ast }{\mathbf{u}}^{♭}\right)={\oint }_{C_0}\frac{d}{dt}\left({\varphi }_t^{\ast }{\mathbf{u}}^{♭}\right)=0$$

for all closed curves $C_0$.

Proof:

We compute the integrand.

$$\begin{array}{rl}\frac{d}{dt}\left({\varphi }_t^{\ast }{\mathbf{u}}^{♭}\right)& =\frac{d}{dt}\left[(d\varphi {)}^{\top }\mathbf{u}\right]\\ & =\underset{d(u\circ \varphi )}{\underbrace{(d\dot{\varphi })}}\mathbf{u}+(d\varphi {)}^{\top }\dot{\mathbf{u}}\\ & ={\left[(\nabla \mathbf{u})d\varphi \right]}^{\top }\mathbf{u}-(d\varphi {)}^{\top }\left({\nabla }_{\mathbf{u}}\mathbf{u}+\nabla p\right)\\ & =(d\varphi {)}^{\top }\left[(\nabla \mathbf{u}{)}^{\top }\mathbf{u}-{\nabla }_{\mathbf{u}}\mathbf{u}+\nabla p\right]\end{array}$$

via

1. product rule
2. substitution with the flow map and the Euler Equation.
3. factorization and simplification

Then, let

$$-\nabla \tilde{p}=(\nabla \mathbf{u}{)}^{\top }\mathbf{u}-{\nabla }_{\mathbf{u}}\mathbf{u}-\nabla p$$

be a new spatial gradient to absorb those inner terms. Thus,

$$\frac{d}{dt}\left({\varphi }_t^{\ast }{\mathbf{u}}^{♭}\right)=(d\varphi {)}^{\top }[-\nabla \tilde{p}]=-{\varphi }_t^{\ast }(d\tilde{p})$$

Therefore

$$\begin{array}{rl}\frac{d}{dt}{\oint }_{C_t}\mathbf{u}\cdot dl& ={\oint }_{C_0}\frac{d}{dt}\left({\varphi }_t^{\ast }{\mathbf{u}}^{♭}\right)\\ & ={\oint }_{C_0}-{\varphi }_t^{\ast }(d\tilde{p})\\ & ={\oint }_{C_t}d\tilde{p}\\ & =0\end{array}$$

where the last line is via Stokes Theorem, since the curl of a gradient field is $0$.