# (Non-)Conservative Vector Fields

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?

• It's not clear to me what you mean by "the surface corresponding to the vector field" in the case of a non-conservative vector field. The surface corresponding to a conservative vector field is defined by a path integral, which is path-independent by definition. But for a non-conservative vector field, this is path-dependent. You seem to be assuming something like that the path-dependence only leads to integral multiples of some constant, but that's not the case. Your example has constant rotation, so the integral along a closed path is proportional to the enclosed area, which can be anything. – joriki May 11 '11 at 16:41
• @joriki: I've edited to put in a picture to show what I mean. – El'endia Starman May 11 '11 at 16:49
• @El'endia: That picture is exactly what I had in mind when I wrote my comment. It looks like you're assuming that for a given $(x,y)$ you can only change $z$ by integer multiples of some constant. That's not the case, since you can integrate along an arbitrary closed path and change the "potential" by the enclosed area. If that doesn't fit with what you mean by "the surface corresponding to the vector field", then I think you should explain what you mean by that. – joriki May 11 '11 at 16:57
• @El'endia: By the way, there is something that your picture is an appropriate representation of: This is the "potential" of the vector field $(y/r^2,-x/r^2)$ in the punctured plane, which has zero rotation. The reason for this not having a single-valued potential, however, is the puncture at the origin; the equivalence between a vector field being conservative, its rotation being zero, it being the gradient of a scalar potential and its path integral being path-independent only holds in simply connected domains. – joriki May 11 '11 at 17:04
• @El'endia: In that case, the quantization in the $z$ direction corresponds to the winding number around the origin. But this works only because the rotation of that vector field is zero everywhere except for the origin. In your example, the rotation is constant and non-zero, so you can get an arbitrary contribution from a closed path. – joriki May 11 '11 at 17:07

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.
The surface corresponding to the vector field $F= (y,-x)$ is continuous but periodic, spiraling along the $z$-axis.
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.