# Mystery (?) operator

Has anyone come across this operator ?

$$\frac{\partial^2}{\partial_x\partial_y} + \frac{\partial^2}{\partial_x\partial_z} + \frac{\partial^2}{\partial_z\partial_y}$$

I've never seen it until it came up in a field theory context. What can it mean ?

Has anyone come across this operator ?

$$\frac{\partial^2}{\partial_x\partial_y} + \frac{\partial^2}{\partial_x\partial_z} + \frac{\partial^2}{\partial_z\partial_y}$$

I've never seen it until it came up in a field theory context. What can it mean ?

Hm, I've never seen it used in any context before. Where did you see it, maybe it's interesting to me, as well. I was just fiddling around a bit and I noticed that the operator above equals

$$\frac{1}{2} \left ( \begin{bmatrix} 0 &1 &1\\ 1 &0 &1\\ 1 &1 &0 \end{bmatrix} \nabla \right ) \cdot \nabla$$

which kind of reminds me of the scalar triple product

$$\frac{1}{2} \nabla \cdot \left ( \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} \times \nabla \right )$$

The operator is half the divergence of a 3D symmetric curl,

$$\nabla^{i} \cdot ( s_{ijk}\partial^{j}A^{k})$$.

where s is a symmetric permutation operator, which is like the Levi-Civita symbol, but with positive value where the L-C has a negative.( I can't write this in vector notation just now)

I'm not sure it has any physical interpretation. Given that the divergence of the usual curl is identically zero, I thought this thing might mean something.

foxjwill, you are right, I noticed also what you say. I haven't come across that weird matrix in other context.

M

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foxjwill,

I've realized that this

$$\frac{1}{2} \left ( \begin{bmatrix}0 &1 &1\\1 &0 &1\\1 &1 &0\end{bmatrix} \nabla \right )$$

is another way to write the symmetric curl.