Starting from the time evolution equation of the magnetic field for incompressible MHD (magnetohydrodynamics)

$$\frac{\partial \vec{B}}{\partial t} = \nabla \times (\vec{v} \times \vec{B}) + \frac{\eta}{\mu_{0}} \nabla^{2} \vec{B}$$

and the definition of the vector potential $\vec{A}$

$$ \nabla \times \vec{A} = \vec{B}$$

How is it that one can arrive at the time evolution equation of the vector potential? Which is

$$\frac{\partial \vec{A}}{\partial t} + (\vec{v} \cdot \nabla) \vec{A} = \frac{\eta}{\mu_{0}} \nabla^{2} \vec{A}$$

according to these lecture notes (NB: PDF) from Rony Keppens.

I have derived that

\begin{align} \nabla \times (\vec{v} \times \vec{B}) &= -(\nabla \cdot \vec{v})\vec{B} - (\vec{v} \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{v} + (\nabla \cdot \vec{B})\vec{v}\\ &= -(\vec{v} \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla)\vec{v} \end{align} where I have used the Maxwell equation that $\nabla \cdot \vec{B} = 0$ and the continuity equation for an incompressible fluid $\nabla \cdot \vec{v} = 0$. However, I don't think this helps me at all.

Chiefly, I think my difficulty is understanding how to recover $\frac{\partial \vec{A}}{\partial t}$ from setting $\frac{\partial \vec{B}}{\partial t} = \frac{\partial (\nabla \times \vec{A})}{\partial t}$

But in general my question is: how does one derive the time evolution equation for the vector potential in the form written above?


2 Answers 2


You are making the problem too difficult for yourself. You should be looking for vector calculus identities and space-time orthogonality. Specifically, $$ \frac{\partial}{\partial t}\nabla\times\mathbf A=\nabla\times\frac{\partial\mathbf A}{\partial t}\\ \nabla^2\left(\nabla\times\mathbf A\right)=\nabla\times\left(\nabla^2\mathbf A\right) $$ You'll then have three terms with $\nabla\times$ in front (ignoring constants): $$ \nabla\times\frac{\partial\mathbf A}{\partial t}=\nabla\times\mathbf u\times\nabla\times\mathbf A+\nabla\times\nabla^2\mathbf A\tag{1} $$ This is a vector relation of the form, $$ \nabla\times\mathbf f=\nabla\times\mathbf g $$ which implies $$ \mathbf f=\mathbf g+\nabla h $$ where $h$ is some scalar.

Thus, Equation (1) can be 'uncurled' by considering the equivalent relation $$ \frac{\partial\mathbf A}{\partial t}=\mathbf u\times\nabla\times\mathbf A+\nabla^2\mathbf A+\nabla\phi $$ (taking the curl of this returns (1)) where $\phi$ is your gauge. Then using the BAC-CAB rule, you get $$ \frac{\partial\mathbf A}{\partial t}=\nabla\left(\mathbf u\cdot\mathbf A\right)-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A+\nabla\phi\tag{2} $$ You can then fix the gauge, choosing $\phi=-\mathbf u\cdot\mathbf A$ such that (2) becomes \begin{align} \frac{\partial\mathbf A}{\partial t}&=\nabla\left(\mathbf u\cdot\mathbf A\right)-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A-\nabla\left(\mathbf u\cdot\mathbf A\right)\\ &=-\left(\mathbf u\cdot\nabla\right)\mathbf A + \nabla^2\mathbf A\tag{$\star$} \end{align} where ($\star$) is the result you are after.

  • $\begingroup$ First, can you give a reference for why curl(x) = curl(y) implies x = y (or explain it)? I think that is only true in limited situations since curl(x) applies some derivatives to x and integrating back to x introduces a constant. Also, I am still not sure how to get (v dot del)A from v x curl(A); I think v x curl(A) is equal to a sum of (del dot A)v and (v dot del)A but I have not yet successfully derived the relation. The vector calculus link you provided gives a relation for curl(A x B) but since curl is not associative, I don't know what the corresponding identity is in this case. $\endgroup$ Apr 28, 2015 at 0:33
  • $\begingroup$ I didn't say that $\nabla\times\mathbf a=\nabla\times\mathbf b$ means $\mathbf a=\mathbf b$, I said that the curl could be removed; you would obviously have to add a gauge, $\nabla\phi$, to the system. Use this gauge to eliminate the "extra" terms you can't get rid of. $\endgroup$
    – Kyle Kanos
    Apr 28, 2015 at 2:11
  • $\begingroup$ Still, how can I get from $\vec{v} \times (\nabla \times \vec{A})$ to $(\vec{v} \cdot \nabla)\vec{A}$? I tried using the Jacobi identity to re-write it in a form where I could use the vector cross product identity but it didn't help at all. $\endgroup$ Apr 28, 2015 at 22:31
  • $\begingroup$ (1) Recognize that $a\times b\times c=-b\times a\times c$, then (2) use the last identity on the Curl section and (3) find the gauge $\nabla\phi=\rm something$ to eliminate all terms that aren't what you want. $\endgroup$
    – Kyle Kanos
    Apr 29, 2015 at 1:58
  • $\begingroup$ I don't think I can use (1). I think the cross product is not associative (see link), so that while I understand (a x b) x c = (-b x a) x c by anticommutativity, I don't think that helps since I am dealing with expressions of the form a x (b x c). $\endgroup$ May 2, 2015 at 21:36

The answer posted by @KyleKanos and associated comments fully answered my question. Here, I am gathering that information together and hope to represent it in a way so that future readers with my limited physical/mathematical knowledge may more rapidly perceive the solution to my question. Note that I'll ignore the constants that I included in my question, since they pose no obstacle and are merely cumbersome to include.

First, substitute $\nabla \times \vec{A}$ for each $\vec{B}$ occurring in the time evolution equation for the magnetic field.

Second, use spacetime orthogonality to pull the partial time derivative outside the curl.

Third, learn the curl of the curl vector identity:

$$\nabla \times (\nabla \times \vec{f}) = \nabla(\nabla \cdot \vec{f}) - \nabla^{2}\vec{f}$$ which is true for any vector field $\vec{f}$. Note that in our case the first term on the right hand side (RHS) is equal to zero since the vector potential is divergence free. Use this identity to make the curl and the Laplacian switch places, arriving at

$$\nabla \times \frac{\partial \vec{A}}{\partial t} = \nabla \times ( \vec{v} \times (\nabla \times \vec{A})) + \nabla \times (\nabla^{2} \vec{A})$$

Fourth, do a similar manipulation as I provided in my question to find

$$\nabla \times \frac{\partial \vec{A}}{\partial t} = -\nabla \times ((\vec{v} \cdot \nabla)\vec{A}) + \nabla \times (\nabla^{2} \vec{A})$$

In the substitution, two terms dropped out since both the velocity and the vector potential are divergence free and a third term dropped out since it was the gradient of a scalar function (the curl of the gradient of a scalar function is identically zero).

Fifth, use the fact that there exists a distributive property for curl

$$\nabla \times \frac{\partial \vec{A}}{\partial t} = -\nabla \times \Big( (\vec{v} \cdot \nabla)\vec{A} + \nabla^{2} \vec{A} \Big)$$

Sixth, remove the curls (the removal happens [I think] by taking a path integral on both sides; as with ordinary integration introducing a constant C, the path integration introduces an arbitrary vector field whose curl is zero [which I'll write as the gradient of a scalar function to follow Kyle Kanos and tradition])

$$\frac{\partial \vec{A}}{\partial t} = \vec{u} \times (\nabla \times \vec{A}) + \nabla^{2} \vec{A} + \nabla \phi$$

The last step is to set $\nabla \phi$ equal to zero. This is permissible (in my case at least) because we don't really care what the value of the vector potential is; we're just evolving it through time so that when we take the curl of it, we get a divergence free magnetic field. (And recall that $\vec{B} = \nabla \times \vec{A}$ so that any $\nabla \phi$ will have no effect on the magnetic field. [Remember that the curl of the gradient of a scalar field is zero!])


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