Curl as a divergence... Is it possible? [closed]

Question

I want to know if it is possible to express the operation

$$ \nabla \phi \times (\nabla \times \mathbf A) $$

as the divergence of second order tensor field $T$. Here $ \phi$ is a scalar field and $\mathbf A$ is a solenoidal vector field ($\nabla \cdot \mathbf A=0$)

I have used all possible identities and finally I can only get

$$ \nabla \phi \times (\nabla \times \mathbf A) = \nabla \phi \cdot(\nabla \mathbf A - \nabla \mathbf A^T) $$

Is it possible to find something like $ \nabla \phi \times (\nabla \times \mathbf A) = \nabla \cdot(\text{tensor}) $ ?

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EDIT: I could derive an expression that may be equivalent to Carlo's answer.

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A-\nabla \mathbf A^T) \cdot \nabla \phi $$ Now, from the identity $\nabla \cdot (\phi \mathbf T)=\phi\nabla\cdot \mathbf T+ \mathbf T^T \nabla \phi$ we can write

$$ (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla \cdot(\phi(\nabla \mathbf A - \nabla \mathbf A^T))+\phi\nabla\cdot(\nabla \mathbf A - \nabla \mathbf A^T) $$ Considering that $\nabla \cdot(\nabla \mathbf A)=\nabla^2 \mathbf A$ and $\nabla \cdot(\nabla \mathbf A^T)=\nabla (\nabla \cdot \mathbf A)=0 $:

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla\cdot(\phi(\nabla \mathbf A-\nabla \mathbf A^T)+\phi\nabla^2 \mathbf A $$ This form is convenient since the LHS of the equation already contains $\nabla^2 \mathbf A$.

Answer 1

Let me expand my comment that I think this is only possible if the Laplacian of $\phi$ vanishes.
The object we are considering is $$\mathbf V\equiv\nabla\phi\times(\nabla\times \mathbf A)=(\nabla \mathbf A)·\nabla \phi - (\nabla\phi)·\nabla \mathbf A.$$ Writing this in components, $$V_i=\sum_j(\partial_i A_j) (\partial_j\phi)-(\partial_j\phi)(\partial_j A_i),$$ using the chain rule and $\nabla\cdot\mathbf A=0$, $$V_i=\sum_j\partial_j(\phi\partial_i A_j)-\partial_j(A_i\partial_j\phi)+A_i(\partial_j\partial_j\phi),$$ hence we arrive at $$\mathbf V=\nabla\cdot\mathbf M^T+\mathbf A(\Delta\phi),$$ with $\Delta$ the Laplacian and the tensor $$\mathbf M=\phi\nabla\mathbf A-\mathbf A\nabla\phi.$$ This has the form required in the OP if the Laplacian of $\phi$ vanishes.

Answer 2

For me $\mathrm{div}$ transforms a $3$-components vector field on $\mathbb{R}^3$ into a scalar function on $\mathbb{R}^3$. So, being you expression $\boldsymbol{\nabla}\phi\times (\boldsymbol{\nabla}\times \boldsymbol{A})$ a $3$-component vector field (for $\boldsymbol{A}$ a vector field of course), your question as stated doesn't make sense. [Edit: after the OP's edit, it now makes sense as $\mathrm{div}$ of a tensor field]

Let me try to interpret what you want to ask. To do this, I will translate everything in the language of differential forms on a $3$-manifold $M$ (which could be $\mathbb{R}^3$).

Your expression corresponds to $d\phi\wedge d\alpha$ where $\phi$ is a smooth (scalar) function and $\alpha$ is a $1$-form on $M$. I think that you may want to ask whether the resulting $3$-form is exact, i.e. whether

$$d\phi \wedge d\alpha=d\omega$$

for a $2$-form $\omega$ on $M$. The answer is clearly yes: just take $\omega:=\phi d \alpha$.