The curl operator $\vec\nabla\times\mathbb{1}$ can be written as a skew-symmetric 3x3 matrix

$$\mathrm{curl} = \begin{pmatrix}0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0\end{pmatrix}$$

(in Cartesian coordinates). Since $\mathrm{curl}\,\mathrm{grad}=0$ and $\mathrm{div}\,\mathrm{curl}=0$, the Eigenvector1 to the Eigenvalue 0 is $\vec\nabla$. And by calculating $\det(\mathrm{curl}-\lambda)\stackrel!=0=-\lambda(\lambda^2+\Delta)$ one obtains that the other two Eigenvalues (with opposite signs) both satisfy $\lambda_\pm^2=-\Delta$, i.e. they are the two roots of the negative Laplacian.

Since $\mathrm{curl}^2=\mathrm{grad}\,\mathrm{div}-\Delta$, the Eigenvectors $\vec f_\pm$, satisfying $\mathrm{curl}^2\vec f_\pm = -\Delta\vec f_\pm$, they have a constant divergence. More precisely, since they must be orthogonal to $\vec\nabla$ due to the different Eigenvalues, they can be written as $\vec f_\pm = \vec g_\pm\times\vec\nabla$. Since the $\vec f_\pm$ are also orthogonal to each other, one can obtain

$$\vec f_\pm = \pm\lambda_\pm^{-1}\,\vec f_\mp\times\vec\nabla$$

where the factor $\pm\lambda_\pm^ {-1}$ was chosen for symmetry and consistency. But what is an analytical (non-recursive) expression for them?

You may already have noticed that the $\lambda_\pm$ are Dirac operators, so I wouldn't be too surprised if an answer included spinors and $\partial\!\!/$, even though that is in 4D... In fact, it is most likely that this is the case, since apart from $\vec\nabla$, the vector of $\gamma$ (or Pauli) matrices would not set any preferred direction.

1 In the sense that for any scalar function $s(\vec x)$, $\vec\nabla s$ is an Eigenvector of $\mathrm{curl}$ with Eigenvalue 0, i.e. $\forall s:\mathrm{curl}\vec\nabla s = 0\cdot\vec\nabla s$. It's just like stating $[x_i,\partial_j] = \delta_{ij}$ while this only makes sense when acting on something.

  • $\begingroup$ "The Eigenvector is $\nabla$"?? $\endgroup$ – Cocopuffs Aug 24 '12 at 7:25
  • $\begingroup$ @Cocopuffs I mean if you accept calling operators vectors as well. It has to act on a scalar field of course. Disclaimer: I'm a Physicist and probably a bit lax on definitions here... $\endgroup$ – Tobias Kienzler Aug 24 '12 at 7:27
  • $\begingroup$ I'm not convinced by the whole setup. If you want to solve $\nabla \times u = \lambda u$ for $\lambda \in \mathbb{R}$ you're dealing with an infinite-dimensional space and you can't use linear algebra and determinants like you do above $\endgroup$ – Cocopuffs Aug 24 '12 at 7:36
  • $\begingroup$ This paper "Eigenfunctions of the Curl Operator..." (Moses, 1971) turned up in a quick Google search, but given that you're considering $\nabla$ to be an eigenvector I'm not at all sure it's what you're looking for... $\endgroup$ – Rahul Aug 24 '12 at 7:36
  • $\begingroup$ @Cocopuffs apart from $\lambda=0$ I'm pretty sure $\lambda$ will not be a mere number but an operator. The question is whether there exist two scalar operators that span the Eigenspace of $\mathrm{curl}$ $\endgroup$ – Tobias Kienzler Aug 24 '12 at 7:38

Be $f(x,y,z)=\mathrm e^{\alpha x+\beta y+\gamma z}$ and $v$ an eigenvector of $\begin{pmatrix} 0 & -\gamma & \beta\\ \gamma & 0 & -\alpha\\ -\beta & \alpha & 0\end{pmatrix}$. Then $f(x,y,z)v$ is an eigenvector of $\rm curl$.

Proof: $\partial_x f(x,y,z) = \alpha f(x,y,z)$, $\partial y f(x,y,z) = \beta f(x,y,z)$, $\partial z f(x,y,z) = \gamma f(x,y,z)$. Therefore the "curl matrix" acts on $f(x,y,z)v$ as if all partial derivatives were replaced by the corresponding factor, which gives the matrix above. Thus, we arrive at an ordinary eigenvector equation, and $v$ (and thus each multiple of $v$) is an eigenvector by assumption.

I'm not sure if those are all eigenvectors, though.

  • $\begingroup$ Ah of course, going to Fourier space makes this a lot less operator-y :-) For a shorter notation, take $\vec k = -i(\alpha,\beta,\gamma)$ such that $f=e^{i\vec k\vec x}$, then $\vec\nabla\times f\vec\nu = i\vec k\times f\vec\nu = \lambda(\vec k) e^ {i\vec k\vec x}\vec\nu(\vec k)$. I guess there should exist analytical formulae for $\lambda$ and $\vec\nu$. $\endgroup$ – Tobias Kienzler Aug 24 '12 at 8:11
  • $\begingroup$ Ok, it should be $\vec\nu(\lambda=0)=\vec k$ and $\vec\nu(\lambda=\pm|\vec k|\equiv\pm k) = (-(\beta^2+\gamma^2)k,\, (\alpha\beta\mp i\gamma k)k,\, \mp(\beta^3+\beta\gamma^2+\alpha^2\beta\pm\alpha\gamma k))$. I guess with some cyclic permutations that could be put in a nicer form... $\endgroup$ – Tobias Kienzler Aug 24 '12 at 8:34
  • $\begingroup$ (I am accepting this as answer for now, although for completeness the Fourier transform needs to be reversed, i.e. $\vec f_\pm \propto \iiint_{\mathbb{R}^3}d^3k\,e^{i\vec k\vec x}\vec\nu_\pm(\vec k)$ to obtain an operator again) $\endgroup$ – Tobias Kienzler Aug 24 '12 at 9:08
  • $\begingroup$ I don't understand that last comment. An eigenvector is not an operator. An eigenvector is a vector for which applying the operator gives a multiple of the vector. You don't expect the eigenvector of a matrix to be a matrix, do you? $\endgroup$ – celtschk Aug 24 '12 at 9:23
  • $\begingroup$ No, I mean that I think that the (infinitely many) Eigen-functions can be decomposed into (likely three) orthogonal (in a Cartesian meaning) vectors with operators as components (which is what I mean by Eigen-vector), acting on arbitrary scalar (or maybe matrix valued) functions. I.e. I'd like to call "$\vec\nabla s(\vec x)$" an Eigenfunction, and "$\vec\nabla$" an Eigenvector(-operator) $\endgroup$ – Tobias Kienzler Aug 24 '12 at 9:34

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