I am trying to derive the wave equation presented by Alfven in his 1942 paper
Based on the electrodynamic equations:
$$\nabla \times H = \frac{4\pi}{c}J$$
$$\nabla \times E = -\frac{1}{c} \frac{dB}{dt}$$
$$B = \mu H$$
$$J = \sigma(E + \frac{v}{c} \times B)$$
Together with the hydrodynamic equation:
$$\rho \frac{dv}{dt} = \frac{1}{c}(J \times B) - \nabla P$$
Derive the following partial differential equation (which is a one dimensional wave equation):
$$\frac{d^2 H'}{dz^2} = \frac{4\pi\rho}{H_0^2}\frac{d^2 H'}{dt^2}$$
Where:
1) $\sigma$ is the electric conductivity.
2) $\mu$ permeability.
3) $\rho$ is the mass density of the liquid.
4) $J$ is the current density.
5) $v$ is the velocity of the liquid.
6) $P$ is the pressure.
7) $H$ is a quantity used in Physics to express Ampere's Law in terms of the current alone.
8) $c$ is the speed of light.
9) $B$ is the magnetic field.
10) $E$ is the electric field.
11) $H'$ is the variable magnetic field.
I am working assuming perfect conductivity (i.e. $\sigma = \infty$) and no resistance by the medium to allow fluids to pass through it (i.e. $\mu = 1$).
I know how to proceed if we are given Maxwell's equations (which is a set of coupled, first-order, partial differential equations) by applying the curl. However, here we have an additional equation: the hydrodynamic.
How to derive it then? Could you give me hints? The paper doesn't provide more details on the calculation.
Thanks.
EDIT
Here's a piece of the original paper:
Note that what I am trying to do is what Alfven said to be 'elementary calculation'; deriving the following wave equation:
$$\frac{d^2 H'}{dz^2} = \frac{4\pi\rho}{H_0^2}\frac{d^2 H'}{dt^2}$$
I know how to get the wave equation out of Maxwell's equations in free space, which are as follows:
$$\nabla \cdot E = 0$$
$$\nabla \cdot B = 0$$
$$\nabla \times E = -\frac{1}{c} \frac{dB}{dt}$$
$$\nabla \times B = \mu \epsilon \frac{dE}{dt}$$
How? Decoupling $E$ and $B$ by applying the curl to both $\nabla \times E$ and $\nabla \times B$. By doing so you get two wave equations; one for $E$ and another for $B$.
Let me show how to do so explicitly for $E$ (more details in Introduction to Electrodynamics; Electromagnetic Waves' chapter; Griffiths):
$$\nabla \times (\nabla \cdot E) = \nabla (\nabla \cdot E) - \nabla^2 E = \nabla \times (-\frac{\partial B}{\partial t}) = -\frac{\partial}{\partial t}(\nabla \times B) = -\mu \epsilon \frac{\partial ^2 E}{\partial t^2}$$
As $\nabla \cdot E = 0$, we end up getting:
$$\nabla^2 E = \mu \epsilon \frac{\partial ^2 E}{\partial t^2}$$
As expected, we end up with the wave equation for $E$.
So in this problem I think it is just about doing the same but this time with H:
$$\nabla \times H = \frac{4\pi}{c}J$$
What I have done is the following (with all steps):
$$\nabla \times (\nabla \cdot H) = \nabla (\nabla \cdot H) - \nabla^2 H = \frac{4\pi}{c} \nabla \times J$$
We're given $J$, so:
$$\vec \nabla \times \vec J = \sigma \vec \nabla \times ( \vec E + \frac{1}{c}(\vec v \times \vec B))$$
I am stuck in this product:
$$\nabla \times (\vec v \times \vec B)$$
At first I thought $\nabla \times (\vec v \times \vec B)$ could be zero, but this is not the case, as we're talking about transverse waves; thus $v$ and $B$ are perpendicular and not parallel.
Any help is appreciated.