Kelvin’s Circulation Theorem

Prove Kelvin’s Circulation Theorem, assuming that the fluid in question is barotropic and there is no rotation. So

$$\frac{D\Gamma}{Dt} = \frac{D}{Dt} \oint \vec{v} \cdot d\vec{l} = 0$$

where \(\vec{v}\) is the velocity vector and \(\Gamma\) is the circulation.

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We can start with the momentum equations for barotropic flow in 3D:

\begin{align*}\frac{D \vec{v}}{D t} &= -\frac{1}{\rho}\nabla p - g\hat{k} \\ &= -\nabla \left(\frac{p}{\rho} + gz \right)\end{align*}

Note that this takes the form of a gradient. We will come back to this later.

Let's write the definition of the circulation:

$$ \Gamma = \oint\vec{v} \cdot d\vec{l} $$

If we take the time derivative of this, we get:

\begin{align*} \frac{D \Gamma}{D t} &=\oint \frac{D \vec{v}}{D t} \cdot\vec{dl} \\ &+\oint \vec{v} \cdot \frac{D\vec{dl}}{Dt}\end{align*}

The material line element \(\vec{dl}\) moves in the direction of changing velocity, so

$$\frac{D\vec{dl}}{Dt} = \nabla \vec{v} \cdot \vec{dl}$$

So:

\begin{align*} \frac{D \Gamma}{D t} &=\oint \frac{D \vec{v}}{D t} \cdot\vec{dl} \\&+ \oint (\vec{v} \cdot \nabla \vec{v} )\cdot \vec{dl} \\ &=\oint \frac{D \vec{v}}{D t} \cdot\vec{dl} \\ &+\oint \frac{1}{2}\nabla \|\vec{v}\|^2 \cdot \vec{dl}\end{align*}

The last thing we must do is convert these line integrals to surface integrals via Stokes' theorem:

$$ \oint \vec{f}\cdot \vec{dl} = \iint\limits_S \nabla \times \vec{f} \cdot \vec{dS} $$

Both of our line integrals disappear because the curl of a gradient is zero (and now we finally get to use the momentum equation!)

\begin{align*} \require{cancel}&-\cancel{\iint\nabla \times \nabla \left(\frac{p}{\rho} + gz \right) \cdot\vec{dS}} \\&+\cancel{\frac{1}{2}\iint \nabla \times\nabla \|\vec{v}\|^2 \cdot \vec{dS}} \\ &= 0 \end{align*}

Thus,

$$\frac{D \Gamma}{D t} = 0$$