Solving the Mystery of the Operator: $\partial^2$

[Mentz114]

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 ?

 

[Pere Callahan]

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.

 

[foxjwill]

I was just fiddling around a bit and I noticed that the operator above equals

\frac{1}{2} \left ( \begin{bmatrix}<br /> 0 &amp;1 &amp;1\\<br /> 1 &amp;0 &amp;1\\<br /> 1 &amp;1 &amp;0<br /> \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 )

 

[Mentz114]

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

 

[Mentz114]

foxjwill,

I've realized that this

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

is another way to write the symmetric curl.