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I’m following Anderson’s Fundamentals Of Aerodynamics. However, the proof for the above theorem is not provided. Every time I try it, things get really lengthy and cumbersome. Is there some nice and short method to getting around to prove the time rate of change of circulation over a chose closed loop 0 in an inviscid, incompressible flow ?


1 Answer 1


The circulation is given by

$$\Gamma = \oint_{C(t)}\boldsymbol{u}d\boldsymbol{r}.$$

Now, take the material derivative of this expression (remember to use the product rule because our contour is time-dependent)

$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\dfrac{D \boldsymbol{u}}{Dt}d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\left[\dfrac{D\boldsymbol{r}}{Dt}\right].$$

Now, we assume that the flow in inviscid and all volume forces are conservative (have a potential $\Psi$). Such that we can use Euler's equation for the first integral

$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\left[-\dfrac{\nabla p}{\rho}-\nabla \Psi \right]d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\boldsymbol{u}.$$

The term on the right-hand side is integrable and has an anti-derivative of $\boldsymbol{u}^2$. As we are integrating on a circle (starting point = endpoint) it must be zero. If we assume that $\rho(p)$, which is called barotropic, then we can conclude that the closed contour integral on the left-side is also zero. Note, that the closed contour integral for conservative fields $-\nabla \Psi$ is zero, also called path independence. Hence, we conclude

$$\dfrac{D \Gamma}{Dt} =0.$$

This means that for inviscid, barotropic flow's and conservative volume forces the circulation $\Gamma$ is conserved.


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