I was solving problem 2.4.13 from the book "George B Arfken, Hans J Weber - Mathematical Methods For Physicists- Sixth edition" and the problems was that:

Problem 2.4.13

A force is described by

$\vec{F} = -\hat{x}\frac{y}{x^2+y^2} + \hat{y}\frac{x}{x^2+y^2}$

(a) Express $F$ in circular cylindrical coordinates.

Operating entirely in circular cylindrical coordinates for (b) and (c),

(b) Calculate the curl of $F$ and

(c) Calculate the work done by $F$ in encircling the unit circle once counter-clockwise.

(d) How do you reconcile the results of (b) and (c)?

So, I found:

a) $-\hat{x} = -\hat{\rho}\cos{\varphi}+\hat{\varphi}\sin{\varphi}$; $\ \ \ \hat{y} = \hat{\rho}\sin{\varphi}+\hat{\varphi}\cos{\varphi}$

$y = \rho\sin{\varphi}$; $\ \ \ x = \rho\cos{\varphi}$

then $\vec{F}$ becomes:

$\vec{F} = \frac{1}{\rho}\hat{\varphi}$

b)$\nabla\times\vec{F} = \frac{1}{\rho} \begin{vmatrix} \hat{\rho} & \rho\hat{\varphi} & \hat{z} \\ \frac{\partial}{\partial\rho} & \frac{\partial}{\partial\varphi} & \frac{\partial}{\partial z} \\ 0 & \frac{\rho}{\rho} & 0 \\ \end{vmatrix} = 0$

c) $\int\vec{F}\cdot d\vec{r} = \int F_i h_i dq_i \implies \int_0^{2\pi} \frac{1}{\rho}\rho d\varphi = 2\pi$

So my question is:

Why does the work integral not vanish?

I took a look into a solution of this question in internet (the only that I found) and there stay that:

"$\nabla\times \vec{F}$ is not defined at the origin. A cut line from the origin out to infinity (in any direction) is needed to prevent one from encircling the origin. The scalar potential $\psi = \varphi$ is not single-valued."

But that wasn't clear enough for me (mainly the italicized part), could someone be more clear than that?

  • $\begingroup$ Perhaps someone can give you a concise answer, but that person is not me. I do not have Arfken's book, but I feel very confident that the answer is there. Look in the index for "Riemann surface" and "contour integration". $\endgroup$
    – garyp
    Jan 30, 2021 at 1:17

5 Answers 5


The problematic line in your reasoning is in assuming that $\nabla \times \vec F=0$ implies that $\vec F$ is conservative, i.e. that $\vec F = \nabla \phi$ for some $\phi$. If this were true, then it would imply that $\oint \vec F \cdot \mathrm d\vec r = 0$ via the fundamental theorem of vector calculus, but alas it is not.

We are only permitted to make the leap $\nabla \times \vec F = 0 \implies \vec F = \nabla\phi$ if $\vec F$ is differentiable on a simply-connected domain. Simple-connectedness means that every loop in the domain can be continuously shrunk to a point without leaving the domain. This is not the case here - $\vec F$ is not differentiable (or even defined) anywhere on the $z$-axis, so any loop encircling it cannot be shrunk to a point. On such domains, $\nabla \times \vec F = 0$ is necessary but not sufficient for $\vec F$ to be conservative.

An alternate viewpoint is provided by the given solution. If you try to find some $\phi$ such that $F = \nabla \phi$, you will only be able to do so on a domain which does not completely surround the origin. For example, $\phi(x,y,z) = \tan^{-1}(y,x)$ is defined for all of $\mathbb{R^3}$ except for any arbitrary slice (across which the angle jumps by $2\pi$), but that's not good enough, as $\vec F=\nabla \phi$ has to hold everywhere in order for $\vec F$ to be conservative.

  • $\begingroup$ Just to check that I've undertand, that also goes for the electrostatic field $\vec{E} = \frac{kq}{r^2} \hat{r}$, since the origin is not at the domain? We know that there is a electric potential, but say that $\nabla\times\vec{E} = 0 \implies \vec{E} = -\nabla\varphi$ (in eletrostatic case) should not be enough to prove that there is a electric potential ? Thanks for your answer ! $\endgroup$ Jan 30, 2021 at 3:27
  • 3
    $\begingroup$ @LucasSievers In fact, $\mathbb R^3$ without the origin is simply connected, so an irrotational field on that domain is automatically conservative. You would have to subtract an entire axis from $\mathbb R^3$ to change that. In $\mathbb R^2$ on the other hand, a single point is sufficient. $\endgroup$
    – J. Murray
    Jan 30, 2021 at 4:07
  • 2
    $\begingroup$ @LucasSievers I've updated my answer to clarify that in this case, the entire $z$-axis has been removed from the domain, which renders it non-simply connected. If only the origin is removed, then $\mathbb R^3$ is simply connected. $\endgroup$
    – J. Murray
    Jan 30, 2021 at 6:07
  • $\begingroup$ @Lucas Sievers You might be interested in looking up De Rham Cohomology $\endgroup$
    – Daniel
    Feb 6, 2021 at 20:42

The core idea in this problem is that integrating the curl over the region inside the unit circle is not a valid operation, because the curl is not defined at the origin.*

You can integrate over a region that looks like the unit circle, but has a wedge cut out of it so that the boundary goes most of the way along the unit circle, comes in along an almost-radial line, loops around the origin, and then goes back to the circle along another almost-radial line. The limit of this wedge being infinitesimally small is the "cut line" in the quote.

(Another approach to removing the origin is to have an infinitesimally-small circle around the origin, in which case you end up getting a curl integral over "the unit circle minus a point at the origin". This integral corresponds to the work done by going around the unit circle, minus the work required to go around the inner circle.)

* As pointed out by @NoLongerBreathedIn, it is possible to use a generalization of integrating over functions (integrating over distributions) to accommodate singularities like the one at the origin. This approach seems to be outside the scope of the question, which asks why Arken and Weber assert that they need to remove a "cut line" from the circle to invoke Stokes' theorem in the context of function integrals.


There is a subtle error in your computation. $\nabla \times \vec{F} = 0$ only holds when $\rho \neq 0$. This is evident from your computation: $$ \nabla \times \vec{F} = \frac{1}{\rho} | \dots | \text{,} $$ because this is undefined when $\rho = 0$. This "blow-up" at $\rho = 0$ is how the computation "warns" that something weird is going on when $\rho = 0$ and you should analyze the origin more carefully.

If you had, you would have discovered that the curl was not zero throughout the interior of a circular path enclosing the origin and would have been led to the arguments in the other answers.


It’s also useful to think about the force that you’ve calculated: $F={\hat\phi}/\rho$: the force points in the ${\hat\phi}$ direction, i.e., tangent to a circle around the origin.

If you travel around such a circle, the force is always either working against or pushing you along, and so you never regain any lost energy, or vice versa. It’s not like falling into a potential well and climbing out of it as would be the case for a conservative force.


The issue is that you have computed the curl of $f$ to be 0 everywhere except at the origin (where the force blows up, so the curl can be forgiven for blowing up as well).

I disagree with RLH's assertion that integrating the curl over the region inside the unit circle is invalid. It's perfectly valid, and has integral $2\pi$.

Of course, this requires integrating a distribution, which is a difficult thing to understand.

  • $\begingroup$ My assertion of validity was in the context of being limited to function integrals, based on the scope of the question and its source materials. I could also have discussed how this relates to cohomology of forms, but that's likewise out of scope. $\endgroup$
    – RLH
    Jan 30, 2021 at 19:14
  • $\begingroup$ You're not wrong. $\endgroup$ Feb 1, 2021 at 17:21

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