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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?

Ryan Farber
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2 Answers2

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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.

Kyle Kanos
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  • 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. – Ryan Farber Apr 28 '15 at 00:33
  • 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. – Kyle Kanos Apr 28 '15 at 02:11
  • 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. – Ryan Farber Apr 28 '15 at 22:31
  • (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. – Kyle Kanos Apr 29 '15 at 01:58
  • 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). – Ryan Farber May 02 '15 at 21:36
  • Additionally, I don't think I have learned gauge transforms before. I looked them up on wikipedia and here. I understand why they can be introduced, but I don't know how to use them to transform equations. Additionally, I talked to a friend today about this problem, and she was also unsure how a gauge transformation could be used to remove the curls. Can you point me to a resource that discusses how to apply them to do what you suggest or be more clear how I seek my gauge? – Ryan Farber May 02 '15 at 21:41
  • If you want, I can edit my question to being about moving curls outside of expressions, which you adequately explained and then ask a new question about gauge transformation. I haven't accepted your answer yet because I don't believe the formulation of my question as it stands has been answered adequately to my level of understanding. – Ryan Farber May 02 '15 at 21:43
  • @RyanFarber: Actually, it turns out it's a form of the $BAC-CAB$ rule (you'll need it as $BAC-ABC$), not rotating the cross products that I incorrectly thought of (as you well pointed out the non-associativity). It's actually really trivial once you toss that in there; if you still don't see it (after this new hint), I'll update my answer with a ever-so-slightly-more-complete derivation (we have a strong aversion to doing homework here, and this work is pretty close to HW, hence my reluctance to give a full working solution) – Kyle Kanos May 03 '15 at 01:30
  • Ah, and my BAC term drops out since curl(grad(f)) = 0. Thanks and to future readers: the BAC - CAB rule is also known as the vector triple product rule. – Ryan Farber May 03 '15 at 01:51
  • Well the BAC term drops out due to the gauge fixing, not due to the curl. – Kyle Kanos May 03 '15 at 01:51
  • Unless I've implicitly chosen the gauge to be zero (I'm not sure if that's phrased right; I mean $\nabla(\psi)$ = 0), I don't think I've fixed the gauge yet. I think BAC drops out independent of the gauge by the "Curl of the gradient" identity. – Ryan Farber May 03 '15 at 02:09
  • Do you have any resource (besides the wikipedia article you just linked) you can point me to for figuring out how to fix the gauge so I can remove the curl from both sides? That's the only piece I'm still missing in this derivation. – Ryan Farber May 03 '15 at 02:11
  • I do not at all see how you can say that I think BAC drops out independent of the gauge by the "Curl of the gradient" identity. There are no more curls here, you've taken them all out. You should have something like $\partial_tA=\nabla(u\cdot A)-(u\cdot\nabla)A+\nabla^2A+\nabla\phi$. There's nothing extra here. You need to define $\phi$ such that the $\nabla(u\cdot A)$ term is subtracted away. – Kyle Kanos May 03 '15 at 02:17
  • That definition of $\phi$ is what is meant by gauge fixing. You are choosing the definition of your gauge, $\phi$, such that what is left is easier to handle. Generally you want it to apply to a symmetry of some sort, but here we have something just jumping out at us. – Kyle Kanos May 03 '15 at 02:20
  • I agree that once the curls are removed, choosing a gauge is trivial. I said BAC drops out independent of the gauge because I had not removed the curls yet. Instead, I used the vector triple product identity with all curls present and so the curl of $\nabla(\vec{u} \cdot \vec{A})$ is present which is zero. – Ryan Farber May 03 '15 at 18:55
  • I had not removed the curls yet because I still don't understand why that is permissible; I think you said you have to introduce the gauge to do so. Also, I just noticed (or it just appeared) a message saying to avoid extended discussions in comments: should we move this to "chat" whatever that is? – Ryan Farber May 03 '15 at 18:59
  • If you haven't "uncurled" the equation, then you have no gauge (no $\phi$ term). You cannot simply add $\nabla\phi$ to the equation because then you're solving a different equation. You can remove them because it is an equivalent problem: $\partial_t(\nabla\times \mathbf A)=F(\nabla\times\mathbf A)$ is similar to $\partial_t\mathbf A=F'(\mathbf A)+\nabla\phi$ (the proof is obvious: take the curl of the latter!) – Kyle Kanos May 03 '15 at 19:03
  • Isn't it rather $\nabla \times \vec{f} = \nabla \times \vec{g} \Rightarrow \vec{f} = \vec{g} + \vec{h}$ where $\vec{f}$ and $\vec{g}$ are any vectors and $\vec{h}$ is any vector with zero curl (which can also be written as the gradient of a scalar function)? The direction of the implication arrow matters! It is trivial to prove the right hand side (RHS) implies the left hand side (LHS) [as you say, just take the curl of the RHS]. For the direction of implication I want, am I correct in thinking it arises by performing a path integration on each side of the equation of the LHS (of implication)? – Ryan Farber May 07 '15 at 18:31
  • Your first sentence is precisely what I said in the comment directly above it. We're using the implication as an expression of equivalent relations between the two equations. Since you need/want to solve for $f$ anyway, why bother taking the curl when $f=g+h$ works equally as well? – Kyle Kanos May 07 '15 at 18:44
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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!])

Ryan Farber
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