Determining from its graph whether a vector field is conservative

Question

Given the graph of a vector field, how can I tell whether it is conservative or non-conservative?

Answer 1

One cannot always make this determination visually, but one can apply some ad hoc tests that guarantee one or the other. For example (here we assume the given vector field $\bf F$ is $C^1$):

• If one can find, for example, an smooth, oriented loop $\gamma$ such that at every point of $\gamma$ the unit tangent vector makes a zero, acute, or right angle, and at least at one point where the vector field is nonzero makes a zero or acute angle with the vector field, then the vector field cannot be conservative: This condition guarantees that $\int_{\gamma} {\bf F} \cdot d{\bf s} > 0$, but for a conservative vector field that integral is zero for all loops $\gamma$.
• If the vector field is invariant under rotation about some point, then it is conservative: By translating we may take the distinguished point to be the origin, and by construction $\bf F$ has potential $f\left(\sqrt{x^2 + y^2}\right)$, where $f(r) := \int_a^r {\bf F}(x, 0) \cdot d{\bf x}$, where $d{\bf x}$ is the infinitesimal vector pointing in the positive $x$-direction and $(a, 0)$ is some point in the domain of $\bf F$ on the positive $x$--half-axis. (Strictly speaking, this construction assumes that the domain of $\bf F$ is connected.)

Answer 2

If it is conservative then $\vec F = \nabla \phi $ for some potential $\phi$, using a very useful identity, $\nabla \times \vec F = \nabla \times \nabla\phi = 0$, this mean that if the field is conservative, it won't curl around any point it will be straight lines, something that looks like electric or gravitationnal field !