As anyone who has taken vector calculus (read: most of you) knows, if a vector field is conservative, then it is the gradient of a potential function. In other words, if the vector field is two dimensional, the potential function is a surface and as such some things become really nice, like the line integral of a closed curve being zero. However, there are some very simple vector fields that are not conservative, which leads me to wonder what the corresponding surface would look like and in particular what characteristics of the surface disallow the nice formulations. For instance, the surface corresponding to the vector field $F = y \hat{i} – x \hat{j}$ is continuous but periodic, spiraling along the z-axis. I envision it to look kinda like this:
My question is thus: do all non-conservative vector fields (in 2-space) have corresponding surfaces that are periodic or discontinuous? If not, what other cases are there?
Best Answer
No. Non-conservative vector fields can be produced through many other vector potentials. By Helmholtz decomposition, a smooth vector field $F$ can be decomposition into a conservative vector field plus a rotation of some other conservative field: $$ F = \nabla \phi + \nabla^{\perp} \psi, $$ where $\nabla^{\perp}$ is like embedding the the 3D curl operator for scalar function in 2D: $$ \boldsymbol{C}^{1}(\mathbb{R}^2) \hookrightarrow \boldsymbol{C}^{1}(\mathbb{R}^3), \\ \nabla^{\perp} \psi(x,y) : = \left(\frac{\partial \psi}{\partial y},-\frac{\partial \psi}{\partial x}\right)\mapsto \left(\frac{\partial \psi}{\partial y},-\frac{\partial \psi}{\partial x},0\right) = \nabla\times (0,0,\psi). $$ Ignoring the conservative part of $F$, we can produce all sorts of non-conservative part of $F$ in $\mathbb{R}^2$ using very "smooth" potential $\psi$, neither periodic nor discontinuous. For example: let $\psi = e^{-x^2-y^2}/2$ $$ F = \nabla^{\perp}\psi = (- y\psi, x\psi). $$ You can easily check the field you gave is $\nabla^{\perp} xy$, a rotation of the conservative vector field $(x,y)$.
In fact, a $90^{\circ}$ degree rotation of any conservative vector field in $\mathbb{R}^2$ will make it non-conservative.
As joriki pointed out in the comments, the vector field generated by the spiral you gave is similar to "the gradient field of the polar angle" $$ F = \nabla \arctan \left(\frac{y}{x}\right) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right). \tag{1} $$ If the domain contains a curve winding around the origin, then this is not conservative. Otherwise, it is conservative indeed. Like you did there, we let $z$ be parametrized so that we glue different branches of $\arg (x+iy)$ together, the gradient flow is non-conservative. For a more detailed discussion you could refer to my answer here. Roughly the summary is: $$ \text{zero curl} + \text{simply-connectedness of the domain} \implies \text{conservative} \\ \text{gradient} + \text{no singularities in the domain} \implies \text{conservative} $$ Notice "curl zero" means $0$ everywhere, not like (1), if you include $\{0\}$ to make the domain simply-connected, then the curl is zero except this very point.
Lastly, to address your question again, the non-conservative field you found using that potential ("spiral") is actually a special kind among other non-conservative fields. It is a gradient, but it is not conservative (integral around a close path is non-zero).