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Is $(\nabla \times \nabla)$ an operator?

I am wondering if its possible to compute \begin{align} \vec{f} \cdot (\nabla \times \nabla) \end{align}

Where $f$ is a vector. I would be interested in both cartesian and polar expressions.

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  • $\begingroup$ I guess you mean $(\nabla \times \nabla)\cdot \vec{f}$, don't you ? This is not the same. $\endgroup$
    – user65203
    Dec 15, 2015 at 10:40
  • $\begingroup$ What is exactly the difference? $\endgroup$ Jun 1, 2016 at 12:22

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By equality of mixed partials, that is the fact that: $$ \begin{align} \frac{\partial^2g}{\partial y \partial z}&=\frac{\partial^2g}{\partial z \partial y}\\ \frac{\partial^2g}{\partial z \partial x}&=\frac{\partial^2g}{\partial x \partial z}\\ \frac{\partial^2g}{\partial x \partial y}&=\frac{\partial^2g}{\partial y \partial x}, \end{align} $$ where $g$ is some function of $x$, $y$, and $z$, one has the following result for $C^2$ functions: $$ \begin{align} \vec{f}\cdot\left(\nabla \times \nabla\right) g &=\vec{f}\cdot\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \end{vmatrix} g\\ &= \vec{f}\cdot\left(\hat{x} \left(\frac{\partial^2g}{\partial y \partial z}-\frac{\partial^2g}{\partial z \partial y}\right) +\hat{y} \left(\frac{\partial^2g}{\partial z \partial x}-\frac{\partial^2g}{\partial x \partial z}\right) +\hat{z} \left(\frac{\partial^2g}{\partial x \partial y}-\frac{\partial^2g}{\partial y \partial x}\right)\right)\\ &=\vec{f}\cdot\vec{0}\\ &=0 \end{align} $$ So, this is, as user297767 said, merely a fancy way of writing the zero operator.


If this is too similar to user297767's answer I apologize, but being as my edit of their post was rejected for "deviating from the intent of the post", I feel as though it is reasonable for me to post this as an answer.

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Due to the equality of mixed partials, this is just the zero operator.

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    $\begingroup$ @Leucippus How does this not answer the question? Apply $\nabla \times \nabla$ or $f\cdot (\nabla \times \nabla)$ to any $C^2$ function on $\Bbb R^3$ and you get zero. So, assuming you're limiting yourself to twice continuously differentiable functions (as opposed to just twice differentiable functions) this operator is just a fancy way of writing the zero operator. $\endgroup$
    – user297767
    Dec 11, 2015 at 3:31
  • $\begingroup$ Edit your answer to include how the operators you have expressed lead to a zero result. Words are fine, but a demonstration is even better. $\endgroup$
    – Leucippus
    Dec 11, 2015 at 4:23

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