I learned from my Physics textbooks that there is zero net work (W) done by the force when moving a particle through a trajectory that starts and ends in the same place i.e.


Now, I need to verify whether the force $$\vec{F}=\dfrac{x \hat{i} + y \hat{j}}{(x^2+y^2)^{\frac{3}{2}}}$$ is conservative or not.

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I substituted $x=r\cos(\theta)$ and $y=r\sin(\theta)$, in order to prove that if I move a body by applying the given force through a complete circle then the work done will be zero.

Now, suppose I start from $\theta=0$ and move anticlockwise then my unit vector for displacement at any angle $\theta$ should be $-\sin(\theta) \hat{i} + \cos(\theta) \hat {j}$.

So total work done in traversing the circular path is


Now is showing $W'=0$ sufficient to prove that $\vec{F}$ is conservative ? Also, is their any easier way ?


3 Answers 3


Proving that

$\oint \vec{F}\cdot d\vec{r}=0$

is sufficient to establish that the force is conservative if it is true for all possible paths. You only proved it for a single path, that is the one on a circle of radius $r$ centered at the origin.

There can be a more efficient way to prove the same result, depending on the context, using the same equation in the differential form instead of the integral form. The idea is to use Stokes theorem to write

$\oint \vec{F}\cdot d\vec{r}=\int\limits_\Omega (\nabla\times\vec{F})\cdot d\vec{A},$

where $\Omega$ is the surface enclosed by the closed path of the left hand side. Now, observe that the right hand side equation will always be zero if


everywhere. It is often much simpler to prove this instead.

For the sake of completeness, there is also a third option. A conservative force can be written as the gradient of a potential $\phi$, that is


This follows from the preceding condition as the curl of a gradient is always zero (as long as the function $\phi$ is well behaved, which you can suppose through much of an undergrad physics curriculum.)


Let $\mathbf F\colon \Omega \subseteq \mathbb R ^d \to \mathbb R ^d$ be a smooth vector field, defined on an open subset of $\mathbb R ^d$. Consider the three following conditions.

  1. For every closed path $\gamma$ in $\Omega$: $$\intop _{\gamma} \mathbf F \cdot \text d \mathbf s=0$$
  2. There exists $\phi \colon \Omega \to \mathbb R$ such that $\mathbf F = -\nabla \phi$.
  3. $$\nabla \times \mathbf F \equiv 0.$$

It is easy to show that $(2)\implies (1)$. Also, $(1)\implies (2)$ if $\Omega $ is connected, for in that case $$\phi (\mathbf r ) =-\intop _{\gamma(\mathbf r ,\mathbf r_0)} \mathbf F \cdot \text d \mathbf s=0,$$where $\gamma (\mathbf r ,\mathbf r _0)$ is any path from some fixed $\mathbf r_0$ to $\mathbb r$, defines a smooth function whose gradient is $\mathbf F$.

Let's investigate the relations of $(1),(2)$ with $(3)$. By direct computation, $(2)\implies (3)$ (if $\phi$ is sufficently regular, e.g. if the second derivatives are continuous). Also, if $\Omega$ is connected, since $(1)\implies (2)$, $(1)\implies (3)$. I want to show you that $(1)$ doesn't follow from $(3)$. This is relevant to your case. Let $\Omega = \mathbb R^2 - \lbrace 0 \rbrace$ and let $$\mathbf F = \frac{-y\mathbf e _1+x \mathbf e _2 }{\sqrt {x^2+y^2}}.$$ You can readily show that $\nabla \times \mathbf F = 0$, but $$\intop _{C} \mathbf F \cdot \text d \mathbf s=2\pi,$$ where $C$ is the unit circle oriented anticlockwise.

The problem here is that $\Omega$ is connected but not simply connected. If we were to use the rotor criterion here we would conclude, for example, that the magnetic field generated by a uniform current in an infinite wire is a gradient, which is wrong. However, a weaker version of $(3)\implies (1),(2)$ holds: if $\nabla \times \mathbf F=0$, then $\mathbf F $ is locally (say, in a solid sphere contained in $\Omega$) a gradient.

  • $\begingroup$ Note that $$\mathbf F \cdot \text d \mathbf s =\delta \theta$$ is the variation of the angle subtended by $\mathbf r$. $\endgroup$
    – pppqqq
    Dec 7, 2016 at 10:49
  • $\begingroup$ So to check my reasoning for why (3) doesn't imply (1) , the definition of $\vec{F}$ (a) requires we exclude {0} due to singularity and therefore (b) makes any interior surface to $C$ not-simply-connected, such that you cannot apply stoke's theorem in the regular fashion? $\endgroup$
    – forky40
    Jan 4, 2017 at 19:23
  • 1
    $\begingroup$ Exactly, if $F$ was well defined (and smooth) at the origin, you could apply Stokes' theorem no doubt. Loosely speaking, Stokes' theorem requires that $C=\partial S$ be the boundary of a two-dimensional surface $S$ where $\nabla \times \mathbf F$ is well defined: $$\intop _{\partial S}\mathbf F \cdot \text d \mathbf s=\intop _S (\nabla \times \mathbf F )\cdot \text d \mathbf S$$, which is $=0$ if $\nabla \times \mathbf F=0$ in the whole of $S$. (I say "loosely speaking" because there are different formulations of Stokes' theorem using different notions of "curve", "boundary", etc.) $\endgroup$
    – pppqqq
    Jan 5, 2017 at 9:30

There are three equivalent criteria for demonstration of a conservative force $\mathbf F$: The vanishing of the curl of $\mathbf F$, the line integral of $\mathbf F$ around a closed curve being zero and the existence of a scalar potential that $\mathbf F$ may be written as the gradient of. Any one of these imply the other two (and thus all are related by iff statements).

E.g the former two can be seen to be related through Stoke's theorem $$ \iint_S (\nabla \times \mathbf F) \cdot \mathbf{dS} = \oint_{\partial S} \mathbf F \cdot \mathbf{dr}$$ with $\partial S$ the boundary of an orientable surface $S$.

So, by simply computing the curl of $\mathbf F$ and seeing it vanishes you would also have demonstrated that $\mathbf F$ is conservative.

  • $\begingroup$ How to compute the curl ? I am new to this topic. BTW my question just asked whether my method is sufficient. Is my method correct and sufficient ? $\endgroup$
    – user135951
    Dec 7, 2016 at 10:03
  • $\begingroup$ See tutorial.math.lamar.edu/Classes/CalcIII/CurlDivergence.aspx and the contents therein - you compute a determinant formed with entries being partial derivatives and components of $\mathbf F$, c.f cross product. $\endgroup$
    – CAF
    Dec 7, 2016 at 10:06
  • 1
    $\begingroup$ Hi CAF, note that the theorem you claim is not strictly valid in the OP's case, since the domain of the force field (the punctured plane) is not simply connected. $\endgroup$
    – pppqqq
    Dec 7, 2016 at 10:12
  • $\begingroup$ It is not completely sufficient because you didn't demonstrate that the line integral around any closed curve is zero just for one in particular. Since there are an infinite number of closed curves, better just computing the curl of F ;) $\endgroup$
    – CAF
    Dec 7, 2016 at 10:13
  • $\begingroup$ @CAF Thanks I got the curl as 0. But pppqqq says that this theorem is not strictly valid. Is pppqqq correct ? $\endgroup$
    – user135951
    Dec 7, 2016 at 10:21