# Is the curl of every non-conservative vector field nonzero at some point?

Counterexamples? Intuitively, why?

Thanks for any answers.

As a side note, in what math class are gradient, divergence and curl taught typically?

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Here's a nice counterexample from wikipedia. en.wikipedia.org/wiki/Conservative_vector_field. See the section on "Irrotational vector fields." These concepts are typically introduced in a standard multivariable vector calculus class. –  AWertheim Jun 8 '13 at 3:21
Only if the domain is simply connected. –  Rahul Jun 8 '13 at 3:21
I learned gradient in multivariable calculus class, but didn't learn divergence and curl until vector calculus. –  Ataraxia Jun 8 '13 at 3:48
Suggestion: question body should not be treated as mere continuation of the title, they should be readable on their own. –  ˈjuː.zɚ79365 Jun 8 '13 at 4:48

The question to ask is:

If there is a smooth vector field $\mathbf{v}$ such that $$\nabla \times \mathbf{v} = 0 \quad \text{ in } \Omega,$$ is there a smooth function $\phi$ such that $$\mathbf{v} = \nabla\phi?$$

• If the domain is simply-connected (for example, a sphere, a box, but not cup with a handle), then this field must be a conservative field, i.e., $\mathbf{v} = \nabla \phi$.

• In general, we can have non-conservative field having zero curl. Let $$Z = \mathrm{ker}(\nabla\times) := \{\mathbf{w} \text{ is smooth}: \nabla \times \mathbf{w} = 0\},$$ which is the space of the fields having zero curl and $$B = \mathrm{im}(\nabla) = \{\nabla \psi: \psi \text{ is smooth}\},$$ which is the range of the gradient operator (i.e., all gradient fields). Then $$\mathrm{dim}\big(Z/B\big) = \beta_1\tag{1}$$ where $\beta_1$ is the first Betti number of the domain of interest $\Omega$, and $$\beta_1 = (\# \text{ of holes in the domain})$$ roughly speaking. (1) essentially means: $$\text{Field whose curl is zero} = \text{Conservative vector field } + \text{"Something"}. \tag{2}$$ This "something" here is a $\beta_1$-dimensional space.

• (1) is the by the coincidence of the dimension of the de Rham cohomology group and homology group. For simply connected domain $\beta_1 =0$. If the domain has $k$ holes, the difference is a $k$-dimensional space.

• Intuition: To be honest I don't know the intuition here either, if someone knows please enlighten me as well.

• How to find this "something": Let's use that example in the wiki's entry of conservative fields AWertheim mentioned in the first comment, but modify it a little bit. For an infinite cylinder $C_{x=y=0,z\in (-\infty,\infty)}(1)$ with radius 1 in $\mathbb{R}^3$ that contains the origin, we dig a hole of radius $\epsilon$ along $z$-axis: $$\Omega = C_{x=y=0,z\in (-\infty,\infty)}(1)\backslash C_{x=y=0,z\in (-\infty,\infty)}(\epsilon)$$ Then by above, that "something" is a dimension 1 space. Let $\mathbf{A}$ be a member of this something, we can pose a boundary value problem which is the technique used in Helmholtz decomposition of vector fields: \left\{ \begin{aligned} \nabla \times \mathbf{A} &= 0\quad \text{ in }\Omega \\ \nabla \cdot \mathbf{A} &= 0 \quad \text{ in }\Omega \\ \frac{1}{2\pi}\oint_{\gamma_C} \mathbf{A} \cdot d\mathbf{s} &= 1 \\ \mathbf{A} \cdot \mathbf{n} &= 0 \quad \text{ on }\Gamma, \end{aligned} \right. where $\gamma_C$ is a counterclockwise closed curve lived on that interior cylinder surface with winding number 1, and $\Gamma$ is the boundary of the exterior cylinder. We can find that this $\mathbf{A}$ is what wikipedia's entry has: $$\mathbf{A} = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right).$$ That "something" in (2) is just a multiple of $\mathbf{A}$. And now we can say that if $\nabla \times \mathbf{v} = 0$ in $\Omega$ above, then $$\mathbf{v} = \nabla \phi + c\mathbf{A},$$ where $c$ is a constant.

• Relevant question: What is the solution to Nash's problem presented in "A Beautiful Mind"?

• Your last question: You can learn div, curl, and grad in Calculus III I believe. Like Green theorem, Divergence theorem, and Stokes theorem.

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I don't know where my statement falls short, anyone please enlighten me? –  Shuhao Cao Jun 8 '13 at 4:36
Although I'm not the one who gave the downvote, I think it may have to do with presentation and also it would be nice if you expanded upon what you've said a little. Perhaps link to a page containing definitions or provide derivations of your own. –  ThisIsNotAnId Jun 8 '13 at 4:45
@user79365 Thanks, this makes sense, my bad. –  Shuhao Cao Jun 8 '13 at 4:52
@ThisIsNotAnId Thanks for the kind suggestion. –  Shuhao Cao Jun 8 '13 at 4:52