# Manipulation of the diffusive term in MHD induction equation

I am trying to solve the magnetohydrodynamic (MHD) equations with a spatially varying resistivity, $$\eta$$. To remove some of the numerical stiffness from my finite volume approach, I am trying to get rid of these curl expressions with some vector calculus identities. The expression that is causing me issue is: $$\nabla\times\left(\eta\nabla\times B\right)$$

I have also seen this written as: $$\nabla\cdot\left(\eta\left(\nabla B-\nabla B^{T}\right)\right)$$ such as in the paper: Space–time adaptive ADER-DG schemes for dissipative flows: Compressible Navier–Stokes and resistive MHD equations, Computer Physics Communications.

My question is: are these two expressions equal? I can kind of see how they might be using the cross product rule: https://en.wikipedia.org/wiki/Vector_calculus_identities but I'm a little uneasy of using this identity with the vector operator $$\nabla$$.

Would anybody kindly be able to shed any light on this for me, and possibly take me through the steps to cast the first expression as the second form?

Thank you in advance.

P.S. This is my first post, I hope it's OK.

There are several vector/tensor calculus rules that will come in handy, so I will defined them here first (in no particular order): \begin{align} \nabla \cdot \left[ \nabla \mathbf{A} - \left( \nabla \mathbf{A} \right)^{T} \right] & = \nabla \times \left( \nabla \times \mathbf{A} \right) \tag{0a} \\ \nabla \cdot \left( f \ \mathbf{A} \right) & = f \nabla \cdot \mathbf{A} + \nabla f \cdot \mathbf{A} \tag{0b} \\ \nabla \times \left( f \ \mathbf{A} \right) & = f \nabla \times \mathbf{A} + \nabla f \times \mathbf{A} \tag{0c} \\ \nabla \left( \mathbf{A} \cdot \mathbf{B} \right) & = \mathbf{A} \times \left( \nabla \times \mathbf{B} \right) + \mathbf{B} \times \left( \nabla \times \mathbf{A} \right) + \left( \mathbf{A} \cdot \nabla \right) \mathbf{B} + \left( \mathbf{B} \cdot \nabla \right) \mathbf{A} \tag{0d} \\ \mathbf{A} \cdot \left( \nabla \mathbf{B} \right)^{T} & = \left( \mathbf{A} \times \nabla \right) \times \mathbf{B} + \mathbf{A} \left( \nabla \cdot \mathbf{B} \right) \tag{0e} \\ \left( \nabla \mathbf{B} \right) \cdot \mathbf{A} & = \mathbf{A} \times \left( \nabla \times \mathbf{B} \right) + \left( \mathbf{A} \cdot \nabla \right) \mathbf{B} \tag{0f} \\ \nabla \cdot \left( \mathbf{A} \times \mathbf{B} \right) & = \mathbf{B} \cdot \left( \nabla \times \mathbf{A} \right) - \mathbf{A} \cdot \left( \nabla \times \mathbf{B} \right) \tag{0g} \\ \nabla \times \left( \nabla \times \mathbf{A} \right) & = \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla^{2} \mathbf{A} \tag{0h} \\ \nabla \times \left( \mathbf{A} \times \mathbf{B} \right) & = \nabla \cdot \left( \mathbf{A} \mathbf{B} \right)^{T} - \left( \mathbf{A} \mathbf{B} \right) \tag{0i} \end{align}

From these relations, one can show the following: $$\nabla \cdot \left\{ \eta \left[ \nabla \mathbf{B} - \left( \nabla \mathbf{B} \right)^{T} \right] \right\} = \nabla \times \left( \eta \nabla \times \mathbf{B} \right) + \left( \nabla \eta \cdot \nabla \right) \mathbf{B} - \left( \nabla \eta \times \nabla \right) \times \mathbf{B} \tag{1}$$ where we have taken advantage of Maxwell's equations to eliminate the divergence of the magnetic field term.

The second term on the right-hand side can be expanded to the following form: $$\left( \nabla \eta \cdot \nabla \right) \mathbf{B} = \left( \mathbf{B} \cdot \nabla \right) \nabla \eta - \nabla^{2} \eta \mathbf{B} - \nabla \times \left( \nabla \eta \times \mathbf{B} \right) \tag{2}$$ where all the terms on the right-hand side involve second order derivatives of $$\eta$$.

Generally, to simplify this down to make the two expressions of interest in your question equal one needs to make assumptions about the properties of the system. For instance, the $$\nabla \times \left( \eta \nabla \times \mathbf{B} \right)$$ comes from an approximation of Ohm's law and Ampere's law, i.e., $$\mathbf{E} \approx \eta \mathbf{j}$$ and $$\mathbf{j} \propto \nabla \times \mathbf{B}$$. If there are no local electric field sources (i.e., no excess charges), then $$\nabla \cdot \mathbf{E} = 0$$, which implies: $$\nabla \cdot \left( \eta \mathbf{j} \right) = \eta \nabla \cdot \mathbf{j} + \nabla \eta \cdot \mathbf{j} = 0 \tag{3}$$ If $$\mathbf{j} \propto \nabla \times \mathbf{B}$$ is true, then the first term is zero as the divergence of the curl of a vector is always zero so we are left with: $$\nabla \cdot \left( \eta \mathbf{j} \right) \approx \nabla \eta \cdot \mathbf{j} = 0 \tag{4}$$ The right-hand side can be rewritten as $$\nabla \eta \cdot \left( \nabla \times \mathbf{B} \right) = 0$$. We can then use Equation 0g above to show that the following is also true: $$\nabla \cdot \left( \nabla \eta \times \mathbf{B} \right) = 0 \tag{5}$$ where we have used the fact that the curl of the gradient of a scalar is always zero.

We also know another relationship from Faraday's law where: \begin{align} \nabla \times \mathbf{E} & = - \frac{ \partial \mathbf{B} }{ \partial t } \tag{6a} \\ & = \nabla \times \left( \eta \ \mathbf{j} \right) \tag{6b} \\ & = \eta \nabla \times \mathbf{j} + \nabla \eta \times \mathbf{j} \tag{6c} \\ & = \frac{ 1 }{ \mu_{o} } \left[ \eta \nabla \times \left( \nabla \times \mathbf{B} \right) + \nabla \eta \times \left( \nabla \times \mathbf{B} \right) \right] \tag{6d} \\ & = \frac{ 1 }{ \mu_{o} } \nabla \times \left( \eta \nabla \times \mathbf{B} \right) \tag{6e} \end{align} where $$\mu_{o}$$ is the permeability of free space.

My question is: are these two expressions equal?

In general, no. Under the right approximations, yes.