The definition of vorticity is $\boldsymbol{\omega} = \nabla \times \mathbf{v}$, where $\mathbf{v}$ is the velocity vector field.

Now, if I look at a rotating flow in cylindrical coordinates I find that: $$\nabla \times \mathbf{v} = \frac{1}{r}\frac{\partial (r v_{\theta})}{\partial r},$$ in case of a free vortex I also know that $v_\theta \propto 1/r$ and therefore the derivative in the above equation vanishes for a vortex centered at $r=0$. In other words, the vorticity is zero everywhere, $\boldsymbol{\omega} = \mathbf{0}$.

I can also look at the situation globally, and instead of the localised curl I take a line integral of the speed along a circular path distance $r$ from the centre. In this case I find that: $$C = \int_{\text{circular path}}{\mathbf{v}\cdot d\mathbf{l} = 2\pi ru_{\theta}},$$ a finite constant. But from the Stoke's theorem I know that: $$\int_{\text{enclosed area}}{\left(\nabla \times \mathbf{v}\right)\cdot d \mathbf{A}} = \int_{\text{enclosing curve}}{\mathbf{v}\cdot d\mathbf{l}},$$ but if the circulation is a finite non-zero constant, the curl must be also non-zero somewhere within the enclosed area! Thus the vorticity is non-zero somewhere in the velocity vector field!

These two findings seemingly contradict each other, where am I making a mistake?


2 Answers 2


There is a singularity at the origin: a delta function in the vorticity field. The vorticity is zero (irrotational flow) everywhere but at the origin, where it is infinite. The circulation around any path not enclosing the origin is zero. The circulation around any path enclosing the origin is a constant (non-zero).

  • $\begingroup$ I did not notice the simple pole at the origin. Is it then that only non-physical flows can have zero vorticity everywhere with non-zero circulation? i.e. that if I find something similar happening, the field must have a pole somewhere? $\endgroup$
    – Akerai
    Jan 20, 2018 at 3:34
  • $\begingroup$ I would say yes: Stoke's theorem works. But there are physical flows that approximate the vortex you described. Water draining out of a basin has all its vorticity concentrated right at the center of the vortex. It's just a large vorticity over a small area rather than an infinite one at a point. $\endgroup$
    – Ben51
    Jan 20, 2018 at 3:40

Stokes' theorem cannot be applied here. The reason is that Stokes' theorem can only be applied to a surface that is enclosed by a simple, closed, piecewise-smooth curve (see any decent textbook on vector analysis). As Ben51 points out there is a singularity at the origin, so that must be excluded from the flow domain, which creates a hole in the domain. This means that the surface (the 'enclosed area' of your integration domain) is not enclosed by a simple closed curve, but it now lies between two curves, one is the outer enclosing curve and one is the curve which cuts out the origin. This means that we cannot apply Stokes theorem to compute the circulation, but we can of course compute it directly using line integration, $\Gamma = \oint_C {\bf v} \cdot {\bf dr} $.

For a similar reason we cannot use Stokes' theorem to compute the circulation of 2D flow around an aerofoil (this is an important calculation since a lift force can only exist of we have non-zero circulation around the aerofoil). In this case in a 2D plane the aerofoil surface forms an inner boundary to the surface which is enclosed by the curve along which you wish to compute the circulation. The circulation can again be computed directly from the line integral though.


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