# Conservative vector fields

I was always told that to find whether or not a vector field is conservative, see if the curl is zero.

I have now been told that just because the curl is zero does not necessarily mean it is conservative. However, if it is non-zero it is definitely not-conservative.

IE: $\vec{F}=\frac{K}{r^3}\hat{r}$ (Spherical coordinates, $K$ is constant)

To show this is conservative I would go ahead and take the curl. It will be zero - but that is not definitive proof it is conservative? How would I show it is?

• mathinsight.org/path_dependent_zero_curl – sci-guy Sep 11 '14 at 5:46
• In short, a conservative field is path-independent - it depends only on the initial and final points. As such, the closed line integral vanishes, and by Green's theorem the curl must also vanish. So conservativeness implies zero curl, but the reverse ($\nabla \times f = 0 \implies \mathrm{cons.}$) does not hold. – JamalS Sep 11 '14 at 7:07
• @JamalS perhaps that should be an answer? – David Z Sep 11 '14 at 7:17
• Also I think perhaps this question should go to Mathematics...? (Andrew: we can migrate it if that's the case, so don't repost it there.) – David Z Sep 11 '14 at 7:18
• @JamalS: Your comment is absolutely correct, but more can be said about when and why the reverse does not hold! – DanielSank Sep 11 '14 at 7:20

I was always told that to find whether or not a field is conservative, see if the curl is zero.

This is almost always true, but not always true.

I have now been told that just because the curl is zero does not necessarily mean it is conservative.

Correct!

To illustrate what's going on, let's do an example. Conside the following vector field: $$\vec{v}(x,y)=\frac{-y\hat{x}+x\hat{y}}{x^2+y^2}.$$ Note that $\vec{v}$ is not defined at the origin. Is $\vec{v}$ conservative? Let's define "conservative" as follows

A vector field $\vec{v}$ is conservative if for any closed path $C$, the integral $\int_C \vec{v}\dot\,d\vec{l}=0.$

Consider the path parametrized as $x(t)=r\cos(2\pi t)$ and $y(t)=r\sin(2 \pi t)$ for $t$ going from 0 to 1. This path is just a circle of radius $r$ centered on the origin. The displacement on the path is

$$\frac{d\vec{l}}{dt} = 2\pi r \left( - \hat{x}\sin(2\pi t) + \hat{y}\cos(2\pi t) \right).$$

If we integrate our example $\vec{v}$ on this path we get

\begin{align} \int_C \vec{v}\cdot d\vec{l} &=\int_{t=0}^1 \left( \frac{-y\hat{x}+x\hat{y}}{x^2 + y^2} \right) \cdot (2\pi r) \left( -\hat{x}\sin(2\pi t) + \hat{y}\cos(2\pi t) \right)\,dt\\ &= 2\pi \end{align} which shows that $\vec{v}$ is definitely not conservative. Note that the integral does not depend on the radius $r$ of the path.

Now, we compute the curl of $\vec{v}$. For convenience, define $r\equiv x^2 + y^2$, i.e. $r$ is the radial polar coordinate.

\begin{align} \nabla \times \vec{v} &\equiv \frac{d\vec{v}_y}{dx} - \frac{d\vec{v}_x}{dy}\\ &=\frac{r^2 - 2 x^2}{r^4} - \frac{-r^2 + 2 y^2}{r^4} \\ &= \frac{2r^2 - 2r^2}{r^4}\\ &= 0. \end{align} We have now shown that $\vec{v}$ has zero curl. A consequence of this is that if we were to integrate $\vec{v}$ along any little path around a point where $\vec{v}$ is defined, we are guaranteed to get zero.

Thus, $\vec{v}$ has zero curl but is not conservative. What's going on?

If you picture $\vec{v}$ you'll see that it is a swirl of vector lines circling the origin. The magnitude of the lines decreases as you move away from the origin. This decrease is just right so that if you were to integrate around a little loop which does not encircle the origin (i.e. if you check the curl) you get zero. However, because of the global circling around the origin, if you integrate along a loop which does enclose the origin, you get something non-zero. Thus, you can think of the integral as either "feeling" the presence of the origin and picking up the $2\pi$ we calculated, or not feeling the origin and giving zero. It's like the origin is a special point worth $2\pi$.

This is really interesting! Our field $\vec{v}$ is conservative everywhere locally, but if you make a path around the origin you can get a nonzero integral, so $\vec{v}$ is not conservative globally.

Remember we pointed out that $\vec{v}$ is not defined at the origin? This is not an accident. Vector fields which are conservative locally but not globally must have "holes" at which they are not defined. In fact, these vector fields must be approaching infinity near their holes, which $\vec{v}$ most certainly is, as you can check . Those infinite points have "residues" which show up in integrals which go around them. For the experts in the audience, this is exactly the same residue you get from integrating around a simple pole in the complex plane.

Let's get back to your question

To show this is conservative I would go ahead and take the curl. It will be zero - but that is not definitive proof it is conservative? How would I show it is?

As you have said, and we have demonstrated, having zero curl does not guarantee that a field is conservative. What does guarantee that a field is conservative is that you can express it as the gradient of a scalar function. In more general mathematical terms, if there exists a function $f$ such that $\nabla f = \vec{v}$, then $\vec{v}$ is said to be exact. An exact vector field is absolutely 100% guaranteed to conservative.

So, one answer to your question is that to show a vector field is conservative, just show that it can be written as the gradient of a function. Another answer is, calculate the general closed path integral of the vector field and show that it's identically zero in all cases.

Let's go on though, because this is really super interesting stuff.

Vector fields with zero curl are guaranteed to be exact, meaning that zero curl guarantees conservativeness, unless the vector field has holes (i.e. points at which it's not defined). So, the mantra you learned that zero curl indicates conservativeness is almost always true, but it fails for vector fields which have holes, like our example $\vec{v}$ does at the origin.

Now here's the really amazing part. If I tell you that a vector field has exactly 1 hole, and that it has zero curl but is not exact, there is only one vector field it can possibly be (up to addition of other exact vector fields). In other words, if I tell you that a vector field $\vec{W}$ has zero curl, has one hole, and is not a gradient of any function, then you know for sure that $\vec{W}=\vec{v} + \vec{\lambda}$ where $\vec{\lambda} = \nabla f$ for some $f$. If repeat the same situation but with two holes, then you know that $\vec{W}$ is expressible as a linear combination of specific curl-less but not exact vector fields associated to the two holes. This whole business generalizes to high dimensional spaces. If you like it, read up on differential forms. You can try the book "Analysis on Manifolds" by Munkres, although that is a very "mathy" book.

One last thing. Instead of talking about having zero curl and being the gradient of a function, you can talk about having zero divergence and being the curl of another vector field. Normally, if a vector field has zero divergence, you can write it as the curl of something else. The electric field of a point charge is conservative and has zero divergence. However, it is not the curl of any vector field. In fact, it is the only $^{}$ vector field in three dimensions which has zero divergence and is not a curl of something else. And of course, the electric field of a point charge goes to infinity at the charge point, so this is one of those fields where having a "hole" allows it to break the usual rules. How did Nature know to do that?

 The numerator of $\vec{v}$ goes as $r$ while the denominator goes as $r^2$. Therefore the whole thing goes as $1/r$ which diverges near the origin.

 This is a slightly incorrect statement, but it's ok for now.

• what is incorrect about the "only" in ? – hyportnex Sep 11 '14 at 14:11
• @user31748: In 3D, having zero divergence, one hole, and not being a curl means you must be the point charge electric field up to addition of other vector fields which are curls. – DanielSank Sep 11 '14 at 15:09
• "What does guarantee that a field is conservative is that you can express it as the gradient of a scalar function. In more general mathematical terms, if there exists a function $f$ such that $\nabla f = \vec{v}$, then $\vec{v}$ is said to be exact. An exact vector field is absolutely 100% > guaranteed to [be] conservative." What about vector fields of the form $\nabla \Big(\frac{k}{ \;|\,\vec{r} \, |\,^2 } \Big)$? I find this very confusing: we the potential, but isn't the resulting field only locally conservative (& therefore this condition doesn't guarantee exactness always)? – Rax Adaam Feb 19 '16 at 15:41
• @RaxAdaam what makes you think that vector field is not globally conservative? – DanielSank Feb 19 '16 at 17:26
• @RaxAdaam Well, yeah, but that function is totally pathological. It has a branch cut. I guess I should have explained that. – DanielSank Feb 20 '16 at 17:27