Material derivative of circulation, Kelvin's theorem

Question

i'm trying to work with the material derivative $$\frac{D}{Dt}=\frac{\partial}{\partial t}+(\vec{u}\cdot \vec{\nabla}) $$ of the circulation $\Gamma=\oint_{C=\partial S}\vec{u}\cdot d\vec{l}$, that can be expressed like $\Gamma=\iint_{S}\vec{\omega}\cdot d\vec{A}$. In Kelvins theorem, the material Derivative conmutes with the line integral around a closed contour, that it seems not vary over time (the closed curve). If i assume that it conmutes with the surface integral, this will lead me to $$\frac{D\Gamma}{Dt}=\iint_{S}\frac{D\vec{\omega}}{Dt}\cdot d\vec{A}+\iint_{S}\vec{\omega}\cdot \frac{D(d\vec{A})}{Dt}.$$

My problem is that i do not know how to treat the second term. The surface is arbitrary (with the restriction that it's boundary has to be C), and the variation of the normal to the surface is a priori unknown.

How can i solve this?

Answer 1

I'm afraid that this does not exactly address your problem, but I find it easiest to get Kelvin's theorem by the following route:

Euler's equation for a compressible fluid is $$ \frac{\partial {\bf v}}{\partial t}+ ({\bf v}\cdot \nabla) {\bf v}=- \frac{1}{\rho} \nabla P. $$ Let $$ \Gamma = \oint_{\gamma(t)}{\bf v}\cdot d{\bf r} $$ be the circulation around a contour $\gamma(t)$ that it being carried with the fluid. Then

$$ \frac{d\Gamma}{dt}= \frac{d}{dt}\oint_{\gamma(t)}{\bf v}\cdot d{\bf r}\nonumber\\ = \oint_{\gamma(t)}\left(\frac{\partial {\bf v}}{\partial t}+ ({\bf v}\cdot \nabla ){\bf v}\right)\cdot d{\bf r} + \oint_{\gamma(t)} {\bf v}\cdot ((d{\bf r}\cdot \nabla){\bf v})\nonumber\\ = \oint_{\gamma(t)}\left(- \frac{1}{\rho} \nabla P\right) \cdot d{\bf r} + \oint_{\gamma(t)} \frac 12 \nabla |{\bf v}|^2\cdot d{\bf r}\nonumber\\ = \int_\Omega \nabla\times \left(- \frac{1}{\rho} \nabla P\right)\cdot d{\bf A}+ \oint_{\gamma(t)} \frac 12 \nabla |{\bf v}|^2\cdot d{\bf r}.\nonumber $$

In the first term we have used Stokes' theorem with $\partial \Omega=\gamma$. Now the last term is clearly zero as $|{\bf v}|^2$ is single-valued. In the first term $$ \nabla\times \left(- \frac{1}{\rho} \nabla P\right)= \nabla P\times \nabla \left(\frac{1}{\rho}\right) $$ because the curl of gradiant vanishes. Now suppose that the flow is barotropic in that $P$ is a function of $\rho$ only. Then $\nabla P$ and $\nabla ({1}/{\rho})$ are parallel, and so this is term is zero also and $d\Gamma/dt$ is zero. This is Kelvin's circulation theorem.