Multivariable calculus and vector analysis

Path-independence implies no circulation

If the vector field $\mathbf{F}$ is a path-independent vector field, then the path integral

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& & \hfill \end{array}$$

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

Suppose, for example, we have two paths $C_1$ and $C_2$ connecting point $\mathbf{p}$ with point $\mathbf{q}$.

Then we know that

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}& & \hfill \text{ (1)}\end{array}$$

because $\mathbf{F}$ is path-independent.

Now, what would happen is we turned the path $C_2$ around, so that it starts at point $\mathbf{q}$ and ends at point $\mathbf{p}$? (We will denote this path as $C_2^{-}$.) At any point, the tangent vector of $C_2^{-}$ will be the opposite of the tangent vector of $C_2$ because the path is going in the opposite direction. Consequently

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_2^{-}}\mathbf{F}\cdot d\mathbf{s}=-{\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}.& & \hfill \text{ (2)}\end{array}$$

If we define the path $C$ to be the path $C_1$ followed by the path $C_2^{-}$, the path $C$ will start at the point $\mathbf{p}$, go to the point $\mathbf{q}$ (via $C_1$) and then return to the point $\mathbf{p}$ (via $C_2$). In other words, the path $C$ will be a closed path.

Then path integral over $C$ is simply the path integral over $C_1$ plus the path integral over $C_2^{-}$. By combining equation (1) with equation (2), we see these last two integrals are opposites

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_2^{-}}\mathbf{F}\cdot d\mathbf{s}=-{\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}.& & \hfill \end{array}$$

We conclude that the path integral over $C$ is zero:

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_2^{-}}\mathbf{F}\cdot d\mathbf{s}=0.& & \hfill \end{array}$$

If $C$ is a closed path, we call the integral

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& & \hfill \end{array}$$

the circulation of $\mathbf{F}$ around $C$. If $\mathbf{F}$ 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 $\mathbf{F}$ 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

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}\ne 0,& & \hfill \end{array}$$

then we know that $\mathbf{F}$ is path-dependent. For the example vector field $\mathbf{F}\left(x,y\right)=\left(y,-x\right)$ 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 $\mathbf{F}$ is path-dependent.

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

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

5a. Green's theorem

5b. The theorem for conservative vector fields

• Path-independent or conservative vector fields
• An example of a conservative vector field
• Path-independence implies no circulation
• Understanding the conditions for path-independence
• Finding the potential function
• Path-independence example in three dimensions

5c. Stokes' theorem

5d. The divergence theorem

6. Review material