Path-independence implies no circulation

If the vector field F is a path-independent vector field, then the path integral

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

does not depend on the actual path C; the path integral depends only on the beginning point (call it p) and end point (call it q) of the path C.

Suppose, for example, we have two paths C1 and C2 connecting point p with point q.

\includegraphics[width=2.5in]{pathindcirc3.eps}
Then we know that

$\displaystyle \int_{{C_1}}^{}$F . ds = $\displaystyle \int_{{C_2}}^{}$F . ds (1)

because F is path-independent.

Now, what would happen is we turned the path C2 around, so that it starts at point q and ends at point p? (We will denote this path as C2-.) At any point, the tangent vector of C2- will be the opposite of the tangent vector of C2 because the path is going in the opposite direction. Consequently

$\displaystyle \int_{{C_2^-}}^{}$F . ds = - $\displaystyle \int_{{C_2}}^{}$F . ds. (2)

If we define the path C to be the path C1 followed by the path C2-, the path C will start at the point p, go to the point q (via C1) and then return to the point p (via C2). In other words, the path C will be a closed path.

\includegraphics[width=2.5in]{pathindcirc2.eps}

Then path integral over C is simply the path integral over C1 plus the path integral over C2-. By combining equation (1) with equation (2), we see these last two integrals are opposites

$\displaystyle \int_{{C_2^-}}^{}$F . ds = - $\displaystyle \int_{{C_1}}^{}$F . ds.    

We conclude that the path integral over C is zero:

$\displaystyle \int_{{C}}^{}$F . ds = $\displaystyle \int_{{C_1}}^{}$F . ds + $\displaystyle \int_{{C_2^-}}^{}$F . ds = 0.    

If C is a closed path, we call the integral

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

the circulation of F around C. If F represents fluid flow, this integral indicates the tendancy for the fluid to circulate around the curve C. We could use the above argument to show that F is path-independent if and only if the circulation around any closed curve is zero.

We can use this result as a test for path-dependence. If we can find a single path C where

$\displaystyle \int_{{C}}^{}$F . ds$\displaystyle \ne$0,    

then we know that F is path-dependent. For the example vector field F(x, y) = (y, - x) shown at the end of the previous reading, one can see the nonzero circulation around any circular path centered at the origin. This observation is enough to conclude that F is path-dependent.

The key point to remember is that path-independence means there is no circulation around any curve.



Duane Nykamp
nykamp@math.umn.edu
2005-11-02