My understanding of the conservative field is that it is any vector field that satisfies any of these three equivalent conditions: $$\oint_C\vec{F}.d\vec{s}=0$$for any closed path $C$ in the domain,$$\vec{F}=\vec{\nabla}\phi$$for some scalar field $\phi$ defined over the domain, and$$\vec{\nabla}\times\vec{F}=\vec{0}$$ at every point in the domain.

However, our teacher told us today that a conservative field and a field derived from a potential are not the same thing. In my research on the issue I found this wolfram page that states that the last condition is not equivalent to the others if the domain $D$ is not simply connected.

Can anyone provide me with an example on the case ?

And in this case, what becomes the definition of a conservative field ?


1 Answer 1


Consider $\mathbf{F} = (-y/\sqrt{x^2 + y^2}) \mathbf{e}_x + (x/\sqrt{x^2 + y^2}) \mathbf{e}_y = (1/r) \mathbf{e}_\theta$ on the multiply connected domain $D =\mathbb{R}^2 \setminus (0,0)$.

Note that $\mathbf{F}$ is the gradient of a function $\phi$ in $\hat{D} =D \setminus \{(r,\theta): \theta = 0 \}$ but not throughout $D$.

In $\hat{D}$ we have for $\phi(r,\theta) = \theta$

$$\mathbf{F}(r,\theta) = \nabla \phi = \frac{\partial \theta}{\partial r}\mathbf{e}_r + \frac{1}{r}\frac{\partial \theta}{\partial \theta} \mathbf{e}_\theta + \frac{\partial \theta}{\partial z} \mathbf{e}_z =\frac{1}{r}\mathbf{e}_\theta .$$

This field has zero curl throughout $D$, i.e.,

$$ \nabla \times \mathbf{F} = \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{1}{r} \right)\mathbf{e_z} = 0,$$

but it is non-conservative. Around any circular contour $C$ centered at the origin, we have

$$\oint_C \mathbf{F} \cdot d\mathbf{s} = \int_0^{2\pi} \frac{1}r r\, d\theta = 2 \pi \neq 0.$$

It is impossible to satisfy both $\mathbf{F} = \nabla \phi$ where $\phi$ is continuous and differentiable and $\oint_C \mathbf{F} \cdot \, d \mathbf{s} \neq 0.$

  • $\begingroup$ However there's a line where $\phi$ is discontinuous (or else it would not be a function on $\mathbb R^2\setminus\{(0,0)\}$), and on that line, obviously we cannot have $\mathbf F=\nabla\phi$ as the latter isn't defined there. $\endgroup$
    – celtschk
    Feb 25, 2017 at 13:47
  • $\begingroup$ @celtschk we need the domain to be not simply connected. The issue now is : this answer shows that the first condition is not equivalent to the others, not the last one. $\endgroup$
    – Tofi
    Feb 25, 2017 at 13:55
  • $\begingroup$ @VIP: The answer shows that the first condition is not equivalent to the third. It doesn't show anything about the second condition, because the provided $\phi$ does not have the properties claimed. $\endgroup$
    – celtschk
    Feb 25, 2017 at 14:05
  • $\begingroup$ @celtschk now I got you, you mean the function is discontinuous on $\theta=0$. $\endgroup$
    – Tofi
    Feb 25, 2017 at 16:19
  • $\begingroup$ @VIP: If for $\theta$ you use the range $\theta\in [0,2\pi)$, then the discontinuity is at the line with $\theta=0$. With the more common choice of $\theta\in(-\pi,\pi]$, the discontinuity is on the line with $\theta=\pi$. But the important point is, no matter how you do it, you always get a discontinuity on some line. $\endgroup$
    – celtschk
    Feb 25, 2017 at 16:37

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