# Determining from its graph whether a vector field is conservative

Given the graph of a vector field, how can I tell whether it is conservative or non-conservative?

• Perhaps this link may help, particularly the first paragraph: citadel.sjfc.edu/faculty/kgreen/vector/Block4/vec_cons/… – imranfat Oct 18 '15 at 20:15
• I have seen that resource. However, I don't feel like it applies to all situations. For instance, take the a look at this question: imgur.com/D4Wm5OU. The second graph doesn't have a tendency to swirl around, however it is the answer and is considered non conservative. – John Fda Oct 18 '15 at 20:23
• +1) I looked at your link. Indeed the second graph does not "swirl" and thus according to the source it is not conservative, which is in line (Ha, what a word choice!) with what my link says. That said, I tend to agree with Travis, that solely a graph is in many cases just not sufficient to determine "conservativeness" (I know, it's bad English) and thus one should proceed with Calculus tools to confirm. That begs the question: Are there funky graphs that DO have some kind of swirlyness (another bad word...) but the Calculus would prove the conclusion otherwise? – imranfat Oct 18 '15 at 20:28
• @NishilShah In that case, one can use the same sort of argument indicated in the first item in my answer: Draw a rectangular loop $\gamma$ with sides parallel to the axes (say, oriented clockwise), and consider the path integral $\int_{\gamma} {\bf F} \cdot d{\bf s}$. The vertical segments contribute nothing to the integral, as ${\bf F}$ and $d{\bf s}$ are orthogonal there. Along the top segment ${\bf F}$ is large and pointing in the same direction as $d{\bf s}$, so this segment makes a large positive contribution... – Travis Willse Oct 18 '15 at 20:29
• ...On the bottom segment ${\bf F}$ is small and pointing in the opposite direction as $d{\bf s}$, so this segment makes a negative contribution, but one too small to cancel out the contribution from the top segment. So, the path integral over the loop $\gamma$ positive, and consequently ${\bf F}$ is not conservative. – Travis Willse Oct 18 '15 at 20:31

One cannot always make this determination visually, but one can apply some ad hoc tests that guarantee one or the other. For example (here we assume the given vector field $\bf F$ is $C^1$):

• If one can find, for example, an smooth, oriented loop $\gamma$ such that at every point of $\gamma$ the unit tangent vector makes a zero, acute, or right angle, and at least at one point where the vector field is nonzero makes a zero or acute angle with the vector field, then the vector field cannot be conservative: This condition guarantees that $\int_{\gamma} {\bf F} \cdot d{\bf s} > 0$, but for a conservative vector field that integral is zero for all loops $\gamma$.
• If the vector field is invariant under rotation about some point, then it is conservative: By translating we may take the distinguished point to be the origin, and by construction $\bf F$ has potential $f\left(\sqrt{x^2 + y^2}\right)$, where $f(r) := \int_a^r {\bf F}(x, 0) \cdot d{\bf x}$, where $d{\bf x}$ is the infinitesimal vector pointing in the positive $x$-direction and $(a, 0)$ is some point in the domain of $\bf F$ on the positive $x$--half-axis. (Strictly speaking, this construction assumes that the domain of $\bf F$ is connected.)
• I think this makes sense... – imranfat Oct 18 '15 at 20:30
• @imranfat If you'd like some clarification about something in particular, I'd be happy to give some. – Travis Willse Oct 18 '15 at 20:32
• I am somewhat "rusty" on Calc 3 material, Nihil's question I think is a legitimate one, but I couldn't word it better than what I did above. – imranfat Oct 18 '15 at 20:35
• @imranfat Yes, it's a natural question, and a very good one, too. For simple vector fields, finding a contour over which the line integral of the vector field is evidently nonzero is often a useful technique, but this is not always practical, as some vector fields are in some sense very close to conservative but are nonconservative. – Travis Willse Oct 18 '15 at 20:38

If it is conservative then $\vec F = \nabla \phi$ for some potential $\phi$, using a very useful identity, $\nabla \times \vec F = \nabla \times \nabla\phi = 0$, this mean that if the field is conservative, it won't curl around any point it will be straight lines, something that looks like electric or gravitationnal field !

• For instance look at this this question. The second graph doesn't have a tendency to swirl/curl around, however it is the answer and is considered non conservative. – John Fda Oct 18 '15 at 20:24
• $\vec F = f(x,y)\hat i$, so $\nabla\times\vec F = <0,0,\frac{\partial f(x,y)}{\partial y}>$, it infact does curls as it is not invariant under rotation ! – Noctisdark Oct 18 '15 at 21:17