Circulation per unit area

When discussing the idea behind Green's theorem, I claimed that the "microscopic" circulation of a two-dimensional vector field F was

"microscopic circulation" = $\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$.    

We can prove that the quantity

$\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$.    

does represent "circulation per unit area," which is probably a better term for it than "microscopic circulation."

Remeber that the circulation of F around the closed curve C is

$\displaystyle \int_{C}^{}$F . ds.    

so the "circulation per unit area" is simply the above integral divided by the area inside C:

$\displaystyle {\frac{{\int_C \mathbf{F} \cdot d\mathbf{s}}}{{\text{Area inside $C$}}}}$.    

It turns out if you let C shrink down to a point, this ratio becomes

$\displaystyle {\frac{{\int_C \mathbf{F} \cdot d\mathbf{s}}}{{\text{Area inside $C$}}}}$ $\displaystyle \rightarrow$ $\displaystyle {\frac{{\partial F_2}}{{\partial x}}}$ - $\displaystyle {\frac{{\partial F_1}}{{\partial y}}}$,    

assuming C is oriented counterclockwise.

We will sketch a proof of this for a rectangular curve C, oriented counterclockwise. Let the lower-left point of C be (a, b), its width be $ \Delta$x, and its height be $ \Delta$y. Label the edges of the rectangle by C1, C2, C3, and C4.

\includegraphics[width=3in]{circsmall2.eps}.

We assume box small enough to approximate F as constant along each edge.

Along the curve C1, the value of y is constantly b, but the value of x changes from a to a + $ \Delta$x. We assume we can ignore that change in x (since the box is small), and simply approximate F(x, y) as F(a, b) all along the segment C1.

Along the curve C2, the value of x is constantly a + $ \Delta$x, but y changes from b to b + $ \Delta$y. Since we assume we can ignore the change in y, we approximate F(x, y) as F(a + $ \Delta$x, b) all along segment C2.

Similiarly, along C3, y = b + $ \Delta$y, and we approximate x as a (even though it ranges between a and a + $ \Delta$x). We approximate F(x, y) as F(a, b + $ \Delta$y) all along segment C3.

Along C4, x = a, and we approximage y as b + $ \Delta$y (even though it ranges between b and b + $ \Delta$y). We approximate F(x, y) as F(a, b) all along segment C4.

We summarize these approximations in the following figure.

\includegraphics[width=3in]{circsmall7.eps}.

Next, to compute the integral $ \int_{C}^{}$F . ds, we remember that it is the same thing as integrating the scalar-valued function F . T, where T is the unit tangent vector along C:

$\displaystyle \int_{C}^{}$F . ds = $\displaystyle \int_{C}^{}$F . T ds.    

We need to compute F . T along each segment of C. The tangent vector T is constant along each segment, and we are approximating F as constant along each segment, so the dot product F . T will be constant along each segment of C.

Along C1, the path is directed in the positive x direction, so the unit tangent vector is T = (1, 0). The dot product F . T is simply the first component of F(a, b):

F . T = F1(a, b).    

Along C2, the path is directed in the positive y direction, so T = (0, 1), and F . T is simply the second component of F(a + $ \Delta$x, b):

F . T = F2(a + $\displaystyle \Delta$x, b).    

Along C3, the path is directed in the negative x direction, so the unit tangent vector is T = (- 1, 0). The dot product F . T is minus the first component of F(a, b + $ \Delta$y):

F . T = - F1(a, b + $\displaystyle \Delta$y).    

Along C4, the path is directed in the negative y direction, so T = (0, - 1), and F . T is minus the second component of F(a, b):

F . T = - F2(a, b).    

We summarize these findings in the following figure.

\includegraphics[width=3in]{circsmall6.eps}.

The integrals are now easy to compute. Along C1, F . T is constant, so the integral is simply its value F1(a, b) times the length of the segment $ \Delta$x:

$\displaystyle \int_{{C_1}}^{}$F . ds = $\displaystyle \int_{{C_1}}^{}$F . T ds = $\displaystyle \int_{{C_1}}^{}$F1(a, b) ds = F1(a, b)$\displaystyle \Delta$x.    

Along C2, the integral is F2(a + $ \Delta$x, b) times its length $ \Delta$y:

$\displaystyle \int_{{C_2}}^{}$F . ds = F2(a + $\displaystyle \Delta$x, b)$\displaystyle \Delta$y.    

Similarly,

$\displaystyle \int_{{C_3}}^{}$F . ds = - F1(a, b + $\displaystyle \Delta$y)$\displaystyle \Delta$x,    

and

$\displaystyle \int_{{C_4}}^{}$F . ds = - F2(a, b)$\displaystyle \Delta$y.    

The integral around all of C is just the sum along the four segments

$\displaystyle \int_{C}^{}$F . ds = $\displaystyle \int_{{C_1}}^{}$F . ds + $\displaystyle \int_{{C_2}}^{}$F . ds + $\displaystyle \int_{{C_3}}^{}$F . ds + $\displaystyle \int_{{C_4}}^{}$F . ds    
  = F1(a, b)$\displaystyle \Delta$x + F2(a + $\displaystyle \Delta$x, b)$\displaystyle \Delta$y - F1(a, b + $\displaystyle \Delta$y)$\displaystyle \Delta$x - F2(a, b)$\displaystyle \Delta$y.    

The circulation per unit area is the integral divided by the area of the rectangle, which is $ \Delta$x$ \Delta$y

$\displaystyle {\frac{{\int_C \mathbf{F} \cdot d\mathbf{s}}}{{\Delta x \Delta y}}}$ = $\displaystyle {\frac{{F_2(a+\Delta x,b) \Delta y - F_2(a,b) \Delta y - (F_1(a,b+\Delta y)\Delta x - F_1(a,b) \Delta x)}}{{\Delta x \Delta y}}}$,    

where I simply rearranged the terms in the numerator.

Half of the numerator is multiplied by $ \Delta$y and half is multiplied by$ \Delta$x. If we separate these into two fractions, we can cancel the $ \Delta$y in the first fraction with the $ \Delta$y in the demoninator

$\displaystyle {\frac{{F_2(a+\Delta x,b) \Delta y - F_2(a,b) \Delta y}}{{\Delta x \Delta y}}}$ = $\displaystyle {\frac{{F_2(a+\Delta x,b) -F_2(a,b)}}{{\Delta x}}}$.    

In the second fraction, we can cancel the $ \Delta$x,

$\displaystyle {\frac{{F_1(a,b+\Delta y)\Delta x - F_1(a,b) \Delta x}}{{\Delta x \Delta y}}}$ = $\displaystyle {\frac{{F_1(a,b+\Delta y) - F_1(a,b)}}{{\Delta y}}}$.    

Putting these back together, we have

$\displaystyle {\frac{{\int_C \mathbf{F} \cdot d\mathbf{s}}}{{\Delta x \Delta y}}}$ = $\displaystyle {\frac{{F_2(a+\Delta x,b) -F_2(a,b)}}{{\Delta x}}}$ - $\displaystyle {\frac{{F_1(a,b+\Delta y) - F_1(a,b)}}{{\Delta y}}}$.    

Now, we let the curve C shrink down to a point. This means that $ \Delta$x$ \to$ 0 and $ \Delta$y$ \to$ 0. In this limit, the two fractions become something familar: partial derivatives of F.

$\displaystyle \lim_{{\Delta x, \Delta y \to 0}}^{}$$\displaystyle {\frac{{\int_C \mathbf{F} \cdot d\mathbf{s}}}{{\Delta x \Delta y}}}$ = $\displaystyle \lim_{{\Delta x \to 0}}^{}$$\displaystyle {\frac{{F_2(a+\Delta x,b) -F_2(a,b)}}{{\Delta x}}}$ - $\displaystyle \lim_{{\Delta y \to 0}}^{}$$\displaystyle {\frac{{F_1(a,b+\Delta y) - F_1(a,b)}}{{\Delta y}}}$    
  = $\displaystyle {\frac{{\partial F}}{{\partial x}}}$(a, b) - $\displaystyle {\frac{{\partial F}}{{\partial y}}}$(a, b).    

We have shown the the circulation per unit area around the point (x, y) = (a, b) is

$\displaystyle {\frac{{\partial F}}{{\partial x}}}$(a, b) - $\displaystyle {\frac{{\partial F}}{{\partial y}}}$(a, b).    

This is exactly what we integrate over a region D to obtain the total circulation around the border of D, according to Green's theorem

$\displaystyle \int$$\displaystyle \int_{D}^{}$$\displaystyle \left(\vphantom{\frac{\partial F}{\partial x}(x,y) - \frac{\partial F}{\partial y}(x,y)}\right.$$\displaystyle {\frac{{\partial F}}{{\partial x}}}$(x, y) - $\displaystyle {\frac{{\partial F}}{{\partial y}}}$(x, y)$\displaystyle \left.\vphantom{\frac{\partial F}{\partial x}(x,y) - \frac{\partial F}{\partial y}(x,y)}\right)$dA = $\displaystyle \int_{{\partial D}}^{}$F . ds.    

where C = $ \partial$D is the path going counterclockwise around D.


Duane Nykamp
nykamp@math.umn.edu
2005-02-28