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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}) $ ?

EDIT: I could derive an expression that I think it is equivalent to Carlo's comment.

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A-\nabla \mathbf A^T) \cdot \nabla \phi $$ 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 $\nabla^2 \mathbf A$.

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}) $ ?

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 $\nabla^2 \mathbf A$.

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}) $ ?

EDIT: I could derive an expression that I think it is equivalent to Carlo's comment.

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi $$ Now, from the identity $\nabla \cdot (\phi \nabla \mathbf A)=\phi\nabla\cdot (\nabla \mathbf A)+(\nabla \mathbf A)^T \nabla \phi$ 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 $\nabla \cdot(\nabla \mathbf A)=\nabla (\nabla \cdot \mathbf A)$ and $\nabla \cdot(\nabla \mathbf A^T)=\nabla^2 \mathbf A$:

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

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}) $ ?

EDIT: I could derive an expression that I think it is equivalent to Carlo's comment.

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A-\nabla \mathbf A^T) \cdot \nabla \phi $$ 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 $\nabla^2 \mathbf A$.

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}) $ ?

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}) $ ?

EDIT: I could derive an expression that I think it is equivalent to Carlo's comment.

$$ \nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi $$ Now, from the identity $\nabla \cdot (\phi \nabla \mathbf A)=\phi\nabla\cdot (\nabla \mathbf A)+(\nabla \mathbf A)^T \nabla \phi$ 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 $\nabla \cdot(\nabla \mathbf A)=\nabla (\nabla \cdot \mathbf A)$ and $\nabla \cdot(\nabla \mathbf A^T)=\nabla^2 \mathbf A$:

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