Test for a conservative vector field

Question

I have a vector $\mathbf{f}$ that satisfies \begin{align} \oint_C \mathbf{f} ds = 0, \end{align} for any smooth closed contour $C$. I believe this means that \begin{gather} \oint_C \mathbf{f} \cdot \mathbf{\hat{e}_x} ds = 0,\\ \oint_C \mathbf{f} \cdot \mathbf{\hat{e}_y} ds = 0. \end{gather} where $\mathbf{\hat{e}_x}$ and $\mathbf{\hat{e}_y}$ are unit vectors in the $x$ and $y$ direction. Does this mean that I can also say that \begin{align} \oint_C \mathbf{f} \cdot \mathbf{\hat{t}} ds = 0, \end{align} and that $\mathbf{f}$ is a conservative vector field? $\mathbf{\hat{t}}$ denotes the unit vector tangent to the curve $C$.

Answer 1

My claim is that if $\vec{f}$ is smooth then,

$$\oint_C \vec{f} \ ds = 0 \quad \text{for all closed smooth contours $C$,} $$ is possible only for $\vec{f}=\vec{0}$.

Consider a rectangular contour of width $L$ and height $e$. Give the contour rounded edges to ensure it is sufficiently smooth.

If $\vec{f}$ is smooth, then for sufficiently small $e$ the values of $\vec{f}$ on the top of the rectangle are close to the values of $\vec{f}$ on the bottom of the rectangle. We also have that for sufficiently small $e$ the contributions to the overall integral from the sides is negligible compared to the contribution from the top and bottom.

The result is that,

$$ \oint \vec{f} ds = 2\int_{top} \vec{f} ds + O(e). $$

Now lets suppose $\vec{f}$ satisfies the hypothesized condition. This would mean that,

$$ 2\int_{top} \vec{f} ds = 0 ,$$

for any value of the length $L$ and any orientation of the rectangle. Without loss of generality let the rectangle be parallel to the $x$-axis. Then we must be able to find smooth functions $f_x$ and $f_y$ such that,

$$ 2\int_{x_0}^{x_0+L} f_x(x) dx = 0 \qquad 2\int_{x_0}^{x_0+L} f_y(x) dx = 0$$

If $\vec{f}$ isn't identically equal to $\vec{0}$ then these can't be $0$ for all choices of $x_0$ and $L$. To see this suppose that there is any neighborhood in which $f_x(x)$ is positive. Then we could pick $x_0$ and $L$ so that only this neighborhood is involved in the integration, resulting in a positive value for the integral.

Therefore there can't be any neighborhood in which $f_x(x)$ is positive. Similarly there can't be any neighborhood in which $f_x(x)$ is negative. All that is left is for $f_x(x)$ to be zero everywhere. A similar argument applies to $f_y$.