I'm trying to convert the next equations for a cold plasma (consider $P=0$ the pressure), for low frequencies and ignoring dissipative effects: $$\rho \left(\left(\frac{\partial \vec{v} }{\partial t}\right)+ (\vec{v}\cdot \nabla)\vec{v}\right)= \frac{1}{c}\vec{j}\times \vec{B}$$ $$\vec{E}=-\frac{1}{c}\vec{v}\times\vec{B}$$ $$\nabla \times \vec{E}=-\frac{1}{c}\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}=\frac{4\pi}{c}\vec{j}+\frac{1}{c}\frac{\partial \vec{E}}{\partial t}$$ Assuming $\vec{B}=\vec{B}_0+\vec{B}_1$ with $\vec{B}_0=B_0 \hat{e}_z$ constant and homogeneus. Using a linearization for plane waves and considering the equilibrium velocity as $v_0=0$ convert those equations onto: $$\vec{v}_1=\frac{i}{\rho_0 \omega c}\vec{j}_1 \times \vec{B}_0$$ $$\vec{E}_1=-\frac{1}{c}\vec{v}_1 \times \vec{B}_0$$ $$\vec{B}_1=\frac{c}{\omega}\vec{k}\times \vec{E}_1$$ $$\vec{k}\times(\vec{k}\times \vec{E}_1)=-\left(\frac{1+V_A^2/c^2}{V_A^2}\right)\omega^2 \vec{E}_1$$

Where $V_A=\left(\frac{B_0^2}{\rho_0 4\pi}\right)^{1/2}$ is the Alfven velocity.

I could transform the first three equations but the one involving Alfven velocity gave me a lot of problems. Could you help me?


As a matter of preference, I will use $\mathbf{Q}_{o}$ and $\delta \mathbf{Q}$ for the quasi-static and fluctuating terms.

Next, recall from vector calculus that the following holds: $$ \mathbf{A} \times \left( \mathbf{B} \times \mathbf{C} \right) = \left( \mathbf{A} \cdot \mathbf{C} \right) \mathbf{B} - \left( \mathbf{A} \cdot \mathbf{B} \right) \mathbf{C} \tag{1} $$

Next we use the expression for $\delta \mathbf{v}$ in the convective electric field term to get: $$ \begin{align} \delta \mathbf{E} & = - \frac{ 1 }{ c } \ \delta \mathbf{v} \times \mathbf{B}_{o} \tag{2a} \\ & = - \frac{ i }{ \rho_{o} \ \omega \ c } \left[ \left( \delta \mathbf{j} \times \mathbf{B}_{o} \right) \times \mathbf{B}_{o} \right] \tag{2b} \\ & = \frac{ i \ B_{o}^{2} }{ \rho_{o} \ \omega \ c^{2} } \left[ \frac{ \mathbf{B}_{o} \times \left( \delta \mathbf{j} \times \mathbf{B}_{o} \right) }{ B_{o}^{2} } \right] \tag{2c} \\ & = \frac{ 4 \ \pi \ i }{ \omega } \left( \frac{ V_{A} }{ c \ B_{o} } \right)^{2} \left[ B_{o}^{2} \ \delta \mathbf{j} - \left( \delta \mathbf{j} \cdot \mathbf{B}_{o} \right) \mathbf{B}_{o} \right] \tag{2d} \end{align} $$

The next thing to notice is that Equation 2a tells us that $\delta \mathbf{E}$ is orthogonal to $\mathbf{B}_{o}$, therefore the last term in Equation 2d must be zero. This is another way of saying there are no field-aligned current perturbations. Then we can rewrite Equation 2d in terms of the current to get: $$ \delta \mathbf{j} = \frac{ \omega }{ 4 \ \pi \ i } \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} \tag{3} $$

Finally, Ampere's law goes to: $$ \begin{align} i \mathbf{k} \times \delta \mathbf{B} & = \frac{ 4 \ \pi }{ c } \delta \mathbf{j} - \frac{ i \ \omega }{ c } \delta \mathbf{E} \tag{4a} \\ \frac{ i \ c }{ \omega } \mathbf{k} \times \left( \mathbf{k} \times \delta \mathbf{E} \right) & = \frac{ 4 \ \pi }{ c } \left[ \frac{ \omega }{ 4 \ \pi \ i } \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} \right] - \frac{ i \ \omega }{ c } \delta \mathbf{E} \tag{4b} \\ \mathbf{k} \times \left( \mathbf{k} \times \delta \mathbf{E} \right) & = - \left( \frac{ \omega }{ c } \right)^{2} \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} - \left( \frac{ \omega }{ c } \right)^{2} \delta \mathbf{E} \tag{4c} \\ & = - \left( \frac{ \omega }{ c } \right)^{2} \left[ 1 + \left( \frac{ c }{ V_{A} } \right)^{2} \right] \delta \mathbf{E} \tag{4d} \\ & = - \omega^{2} \left[ \frac{ 1 }{ c^{2} } + \frac{ 1 }{ V_{A}^{2} } \right] \delta \mathbf{E} \tag{4e} \\ & = - \left( \frac{ \omega }{ V_{A} } \right)^{2} \left[ 1 + \left( \frac{ V_{A} }{ c } \right)^{2} \right] \delta \mathbf{E} \tag{4f} \end{align} $$ where Equation 4f is the same as that which you seek.

The only part that may not be immediately obvious here is that $\delta \mathbf{j}$ is orthogonal to $\mathbf{B}_{o}$. Generally if the only contribution to the electric field is from the convective term, then there can be no field-aligned currents. If you allow for other terms in the generalized Ohm's law, then you can get field-aligned currents, which are equivalent to a specific form of Alfven wave called kinetic or shear Alfven waves, depending on the limits/boundary conditions of interest.

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