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In multivariable calculus, a common mnemonic to remember $\text{div}(\text{curl}(\mathbf{F}))=0$ for vector fields $\mathbf{F}$ on $\mathbb{R}^3$ is treating the gradient operator $\nabla=\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\rangle$ as a vector, writing

$$\text{div}(\text{curl}(\mathbf{F}))=\nabla\cdot\nabla\times \mathbf{F},$$

and using the fact that for any vectors $\textbf{u},\textbf{v}\in\mathbb{R}^3$, there holds $\textbf{u}\cdot \textbf{u}\times\textbf{v}=0.$

Is there a simple classification of vector fields $\mathbf{F}$ on $\mathbb{R}^3$ that satisfy $\mathbf{F}\cdot \nabla\times\mathbf{F}=0$?

A trivial example of such a vector field is any $\mathbf{F}(x,y)=\langle P(x,y), Q(x,y), 0\rangle$. This leads to a less general question: are there vector fields on $\mathbb{R}^3$ not contained in a plane for which the above equation holds? Does this equation have any special physical meaning?

Thank you.

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  • $\begingroup$ Obviously it means that the curl of the field is orthogonal to the field itself. $\endgroup$
    – md2perpe
    Commented Nov 13, 2017 at 19:40
  • $\begingroup$ @symplectomorphic: I think you have a better handle on this subject that I do, and so I'm going to ask: If the physical interpretation of $\nabla \times F$ is the rotation of a vector field (i.e the curl) then isn't this user asking for vector fields which are perpendicular to its own rotation? $\endgroup$ Commented Nov 13, 2017 at 19:41
  • $\begingroup$ @md2perpe: Nice! Okay, we are thinking the same. Does this question then make sense? How can a vector field be orthogonal to its own rotation? The rotation is determined by the vector field. $\endgroup$ Commented Nov 13, 2017 at 19:42
  • $\begingroup$ The curl is a vector orthogonal to the plane in which the field rotates. $\endgroup$
    – md2perpe
    Commented Nov 13, 2017 at 19:54

1 Answer 1

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It's helpful to translate this question into the notation of differential forms. If the vector field $\mathbf F$ corresponds to the $1$-form $F_1\,dx + F_2\,dy + F_3\,dz$, then $\text{curl}\,\mathbf F$ corresponds to the $2$-form $d\omega$ and the dot product $\mathbf F\cdot \text{curl}\,\mathbf F$ is the coefficient of the $3$-form $\omega\wedge d\omega$. So we're asking when $\omega\wedge d\omega = 0$.

This is precisely the integrability condition for the two-planes given by $\omega=0$ (i.e., the two-planes orthogonal to $\mathbf F$) to have everywhere integrable submanifolds. That is, $\Bbb R^3$ is foliated by surfaces with normal vector $\mathbf F$. [Of course, I'm assuming $\mathbf F$ is nowhere-zero.] More analytically, you can rescale $\mathbf F$ by a smooth (nowhere-zero) function to make it conservative.

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  • $\begingroup$ Thank you, Ted. This is exactly the kind of answer I was hoping for. $\endgroup$ Commented Nov 13, 2017 at 20:03
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    $\begingroup$ It's a nice question, @MaximG. :) $\endgroup$ Commented Nov 13, 2017 at 20:04
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    $\begingroup$ So, in the plane, any system $\omega = P\,dx + Q\,dy=0$ is integrable (away from zeroes you can actually find an integrating factor). So, yes, you're finding curves orthogonal to $\langle P,Q\rangle$, and this gives the foliation of $\Bbb R^3$ by right cylinders built along those curves. $\endgroup$ Commented Nov 13, 2017 at 20:16

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