# Line Integral as Circulation - But why?

In most vector calculus texts say that if if the vector field $\vec{F}$ is viewed as the velocity vector of a fluid, then the surface integral $\iint_{S} \vec{F} \cdot d\vec{S}$, called flux, could be viewed as the amount of fluid that cross the surface $S$ per unit of time. On the other hand, the line integral $\oint_{C} \vec{F} \cdot d\vec{s}$, called circulation, could be interpreted as the amount of fluid transported tangentially to the closed curve $C$ per unit of time. Well, while I find the proposition about the flux very intuitive (and dimesionally sound), I'm very confused about the interpretation of circulation. Supposedly, the dot product $\vec{F} \cdot \Delta\vec{s}$ gives the amount of fluid transported along the path $\Delta\vec{s}$. This doesn't make sense to me (It isn't even well defined what be transported along a path means...). More over, the units of this product are $[m^2/s]$, wich looks very weird to me. Can someone help me understand this interpretation of circulation?

Related question

• Note that, if $\vec{F}$ is velocity of a fluid ($m^3/s$ - volume per unit time), then our product $\vec{F}\cdot\delta\vec{s}$ has units $m^3$, not $m^2/s$. Jun 27 '14 at 16:28
• I think whoever said it was the amount of fluid spoke loosely. It's not a volume of fluid nor a mass of fluid. But I'm not sure what to call it instead. If you break the flow at a point on the boundary into a normal component and a tangential component, then this one is the tangential component. Jun 27 '14 at 16:52
• The local velocity of a fluid has units $m/s$. Volumetric rate or flux (velocity integrated over a surface) has units $m^3/s$.
– RRL
Jun 27 '14 at 17:54

## 1 Answer

Circulation does not represent an amount (or rate of amount) of transported fluid. That would be the flux of the velocity field through a surface:

$$Q = \int_{S} \mathbf{u} \cdot d\mathbf{S}.$$

By Stokes' theorem, circulation is the flux of vorticity $\nabla \times \mathbf{u}$ through a surface with the closed curve as its boundary

$$\int_{C}\mathbf{u} \cdot d \mathbf{l}=\int_{S} \nabla \times\mathbf{u} \cdot d\mathbf{S},$$

and is a measure of the angular momentum of the fluid per unit volume within the closed curve.

In the simplest case of rigid rotation, the azimuthal velocity component is $u_\theta=\omega r$ and the circulation is $2\pi \omega r^2.$ Multiplying by the fluid density $\rho$ you have a quantity $2\pi \rho \omega r^2.$with units $kg\cdot m^{-1}\cdot s^{-1}.$

In the case of rigid rotation, the vorticity is

$$\nabla \times \mathbf{u}= \frac1{r}\frac{\partial(ru_\theta)}{\partial r}\mathbf{e}_z=2\omega \mathbf{e}_z,$$

so locally a fluid element has angular velocity -- a small marker convected with fluid would rotate.

In contrast, a line vortex has a velocity field $u_\theta = O(1/r)$ and the vorticity is $0$ except at the singular point $r=0$ -- a small marker convected with fluid would not rotate. The circulation is non-zero only for a curve that encloses the axis of the vortex (due to the singularity).

In both cases, however, fluid particles follow circular paths and the entire mass of fluid has angular momentum. The circulation is non-zero in both cases for appropriate curves.