According to the original paper (Alfven, 1942), I would like to prove his final equation


where $\partial$ denotes the mass density of the liquid, and other notations please see the attached pictures. enter image description here

My calculation is $$\frac{1}{\sigma}\nabla \times i=\nabla\times E+\nabla\times(\frac{v}{c}\times H)$$

$$=-\frac{1}{c}\frac{dH}{dt}+(H\cdot\nabla)\frac{v}{c}-(\frac{v}{c}\cdot\nabla)H =\frac{c}{4\pi\sigma}\nabla\times(\nabla\times H)$$

$$=\frac{c}{4\pi\sigma}(-\nabla^2 H+\nabla(\nabla\cdot H))=-\frac{c}{4\pi\sigma}\nabla^2 H $$ B=H in this case, it should be divergence-less too, and I assume incompressible fluid.

Take the time derivative again $$-\frac{1}{c}\frac{d^2H}{dt^2}+\frac{d}{dt}[(H\cdot\nabla)\frac{v}{c}]-\frac{d}{dt}[(\frac{v}{c}\cdot\nabla)H]=-\frac{d}{dt}\frac{c}{4\pi\sigma}\nabla^2 H $$

$$ {\partial}\frac{dv}{dt}=\frac{1}{4\pi}[(\nabla\times B)\times B]-\nabla P =\frac{1}{4\pi}[(B\cdot\nabla B-\frac{1}{2}(B^2)]-\nabla P $$

However, I'm stuck in this step. I know the $\frac{dH_0}{dt}$ and $\nabla\times H_0$ will be zero, but I have no idea to simply this equation, and especially how to eliminate the pressure term?

Thanks a lot!

  • $\begingroup$ there is a duplicate on math.stackexchange not sure if that user could solve this question but maybe you could try inquiring $\endgroup$
    – Dabed
    Commented Feb 10, 2020 at 3:16
  • $\begingroup$ I dont get why you have $\frac{1}{c\partial}(i\times H)-\frac{1}{\partial}\nabla P=\frac{1}{c\partial}((H\cdot\nabla) H-\nabla(H^2))-\frac{1}{\partial}\nabla P$, also I think $\frac{1}{\sigma}\nabla \times i=\nabla\times E+\frac{v}{c}\nabla\times H$ should actually be $\frac{1}{\sigma}\nabla \times i=\nabla\times E+\nabla\times(\frac{v}{c}\times H) $ and is not like you can't but $i=\sigma(\vec E + \frac{1}{c}\vec v \times \vec B)$ is like the Lorentz Force so taking the derivative feels a little weird and also because $\sigma=\infty$ , not sure what the correct way to proceed should be. $\endgroup$
    – Dabed
    Commented Feb 10, 2020 at 3:16
  • $\begingroup$ @dandide Thanks for the reminding, I updated the equations. Also thanks for the sharing of other post. $\endgroup$
    – Yui
    Commented Feb 10, 2020 at 12:27
  • $\begingroup$ reading wikipedia it seems the first four equation gives what is call the induction equation$ {\partial \vec{H} \over \partial t}=\frac{1}{\sigma}\nabla^2 \vec{H}+\vec{\nabla}\times(\frac{\vec{v}}{c}\times\vec{H})$ that is more or less inside your first lines so actually you only have two equation this one and the hydrodynamic one but as $\sigma=\infty$ then $\nabla^2 \vec{H}$ will dissapear and I don't see how combining this two will give the result. $\endgroup$
    – Dabed
    Commented Feb 10, 2020 at 23:05

1 Answer 1


Thanks to dandide for letting me know about this question.

Let's first have clear Alfven's assumptions (I use bold only on vector quantities):

  • $\sigma \rightarrow \infty$

  • $\mu = 1$. Thus $B = H$

  • Pressure is neglected.

  • Density is constant.

  • The magnetic field is uniform and equal to $\mathbf H = H_0' \hat z$ (here I have used the previous assumption $\mathbf B = \mathbf H$ of course).

Thus, Maxwell equations simplify to:

$$\nabla \times \mathbf H = \frac{4\pi}{c} \mathbf j \ \ \ \ \ \ \ \ \ (1)$$

$$\nabla \times \mathbf E = -\frac{1}{c} \frac{\partial \mathbf H}{\partial t} \ \ \ (2)$$

$$\mathbf E = -\frac{\mathbf v}{c} \times \mathbf H \ \ \ \ \ \ \ \ \ (3)$$

$$\rho \frac{\partial \mathbf v}{\partial t} = \frac{1}{c}(\mathbf j \times \mathbf H) \ \ (4)$$

The issue with this exercise is that Alfven did not explicitly show the applied perturbations.

These are (note that the primes do not mean derivatives here; they mean small perturbations):

$$\mathbf H = H_0 \hat z + H' \hat x \ \ (5)$$

$$\mathbf E = E' \hat y \ \ (6)$$

$$\mathbf j = j' \hat y \ \ (7)$$

$$\mathbf v = v' \hat x \ \ (8)$$

Once at this point it is just a matter of combining and manipulating the two sets of four equations to derive the wave equation $H'_{tt} = v^2 H'_{zz}$

I will tell you what algebra I applied to get it in several steps:

1) Combine $(1)$ and $(4)$ to get

$$4 \pi \rho \frac{\partial \mathbf v}{\partial t} = (\nabla \times \mathbf H) \times \mathbf H \ \ (9)$$

Plug equations $(5)$ and $(8)$ into $(9)$ and develop the cross product on the right hand side. After doing so, equate vector components. Note that you get three equations; two of them are zero (the derivative of a constant is of course zero) and the third equation you should get is:

$$4 \pi \rho \frac{\partial v'}{\partial t} = H_0 \frac{\partial H'}{\partial z} \ \ (10)$$

2) Plug equations $(5)$ and $(6)$ into $(2)$ to get

$$\frac{\partial E'}{\partial x} \hat z - \frac{\partial E'}{\partial z} \hat x = -\frac{\partial H'}{c\partial t} \hat x \ \ (11)$$

By equating vector components you of course get two equations out of $(11)$; one is simply zero and the other is

$$\frac{\partial E'}{\partial z} = \frac{\partial H'}{c\partial t} \ \ (12)$$

3) Plug equations $(5)$ and $(8)$ into $(3)$ to get

$$E' = \frac{1}{c} v' H_0 \ \ (13)$$

4) Take the partial derivative with respect to time on both sides of equation $(12)$ and plug it into $(13)$. Such a calculation yields an equation for $H'_{tt}$ in terms of $v'_{tz}$ (recall that second partial derivatives are commutative here). This equation is:

$$v'_{tz} = \frac{1}{H_0} H'_{tt} \ \ (14)$$

5) We just need to get another equation in terms of $v'_{tz}$. This can be achieved by taking the partial derivative with respect to $z$ on both sides of equation $(10)$. By doing so you get

$$v'_{tz} = \frac{H_0}{4 \pi \rho} H'_{zz} \ \ (15)$$

6) Finally combine $(14)$ and $(15)$ to get the desired result

$$H'_{tt} = v^2 H'_{zz}$$


$$v = \frac{H_0}{\sqrt{4 \pi \rho}}$$

  • $\begingroup$ Amazing! I got it, thanks a lot!! One more question: why could ignore the pressure first? $\endgroup$
    – Yui
    Commented Feb 18, 2020 at 14:04
  • $\begingroup$ great answer!,+1 $\endgroup$
    – Dabed
    Commented Feb 28, 2020 at 0:25

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