Linearization of momentum equation of MHD equations

Question

The momentum equation of the MHD equations is as follows $$\rho \left(\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v} \cdot \nabla) \mathbf{v}\right) = -\nabla p + \frac{1}{\mu_0} (\nabla \times \mathbf{B}) \times \mathbf{B}$$

where $\rho$ is density, $p$ is pressure, $v$ is velocity vector, $B$ is magnetic field vector. Assuming $\rho_0$, $p_0$, $v_0$ and $B_0$ are the equilibrium parts of density, pressure, velocity vector, magnetic field vector, respectively, and $\delta\rho$, $\delta p$, $\delta v$ and $\delta B$ are the first order perturbed part of the corresponding parameters. The linearized equation should be

$$\rho_0 \frac{\partial \delta \mathbf{v}}{\partial t} + \rho_0 (\mathbf{v}_0 \cdot \nabla) \delta \mathbf{v} = -\nabla \delta p + \frac{1}{\mu_0} (\nabla \times \delta \mathbf{B}) \times \mathbf{B}_0$$

I tried as follows

"To linearize the given momentum equation, we assume that the perturbed variables ($\delta \rho$), ($\delta p$), ($\delta \mathbf{v}$), and ($\delta \mathbf{B}$) are small compared to their equilibrium values ($\rho_0$), ($p_0$), ($\mathbf{v}_0$), and ($\mathbf{B}_0$). We can express the perturbed quantities as follows:

$$\rho = \rho_0 + \delta \rho, \quad p = p_0 + \delta p, \quad \mathbf{v} = \mathbf{v}_0 + \delta \mathbf{v}, \quad \mathbf{B} = \mathbf{B}_0 + \delta \mathbf{B} $$

Now, let's substitute these perturbed quantities into the momentum equation and linearize the terms. Since we are interested in the first-order linear terms, we can ignore products of perturbed quantities:

$$ \begin{align*} \rho_0 \left(\frac{\partial (\mathbf{v}_0 + \delta \mathbf{v})}{\partial t} + ((\mathbf{v}_0 + \delta \mathbf{v}) \cdot \nabla) (\mathbf{v}_0 + \delta \mathbf{v})\right) =& -\nabla (p_0 + \delta p) \\&+ \frac{1}{\mu_0} \left(\nabla \times (\mathbf{B}_0 + \delta \mathbf{B})\right) \times (\mathbf{B}_0 + \delta \mathbf{B}) \end{align*}$$

the last term: $$\frac{1}{\mu_0} \left(\nabla \times (\mathbf{B}_0 + \delta \mathbf{B})\right) \times (\mathbf{B}_0 + \delta \mathbf{B}) = \frac{1}{\mu_0} [\left(\nabla \times (\mathbf{B}_0 + \delta \mathbf{B})\right) \times \mathbf{B}_0 + \left(\nabla \times (\mathbf{B}_0 + \delta \mathbf{B})\right) \times \delta \mathbf{B}]$$ $$ = \frac{1}{\mu_0} [\left(\nabla \times \mathbf{B}_0 \right) \times \mathbf{B}_0 + \left(\nabla \times \delta \mathbf{B} \right) \times \mathbf{B}_0 + \left(\nabla \times \mathbf{B}_0 \right) \times \delta \mathbf{B} + \left(\nabla \times \delta \mathbf{B} \right) \times \delta \mathbf{B}]$$

Expanding the terms and neglecting higher-order perturbations, we get:

$$ \begin{align*} \rho_0 \left(\frac{\partial \mathbf{v}_0}{\partial t} + (\mathbf{v}_0 \cdot \nabla) \mathbf{v}_0\right) + \rho_0 \frac{\partial \delta \mathbf{v}}{\partial t} + \rho_0 ((\delta \mathbf{v} \cdot \nabla) \mathbf{v}_0 + (\mathbf{v}_0 \cdot \nabla) \delta \mathbf{v}) =& -\nabla p_0 - \nabla \delta p + \frac{1}{\mu_0} [\left(\nabla \times \mathbf{B}_0 \right) \times \mathbf{B}_0 + \left(\nabla \times \delta \mathbf{B} \right) \times \mathbf{B}_0 + \left(\nabla \times \mathbf{B}_0 \right) \times \delta \mathbf{B} + \left(\nabla \times \delta \mathbf{B} \right) \times \delta \mathbf{B}] \end{align*}$$

Linearizing (taking first order perturbations) the above equations,

$$\begin{align*} \rho_0 \frac{\partial \delta \mathbf{v}}{\partial t} &= -\nabla \delta p + \frac{1}{\mu_0} \left(\nabla \times \delta \mathbf{B}\right) \times \mathbf{B}_0 + \frac{1}{\mu_0} \left(\nabla \times \mathbf{B}_0\right) \times \delta \mathbf{B} \end{align*}$$

Then how to get (the last term of the equation including perturbed part of magnetic field) the linearized MHD equation as the 2nd equation as:

$$\begin{align*} \rho_0 \frac{\partial \delta \mathbf{v}}{\partial t} &= -\nabla \delta p + \frac{1}{\mu_0} \left(\nabla \times \delta \mathbf{B}\right) \times \mathbf{B}_0 \end{align*}$$

Answer 1

The method of linearization is that each component is broken down into a static, constant value and a (small, relative to the constant value) perturbation: $$\mathbf{a}(\mathbf{x},\,t)\to\mathbf{a}_0+\delta\mathbf{a}(\mathrm{x},\,t).$$ For instance $\mathbf{a}_0=5\hat{\mathrm{e}}_z$ and $\delta\mathbf{a}=10^{-3}\sin(\omega t)\left(\alpha\hat{\mathrm{e}}_x+\beta\hat{\mathrm{e}}_y+\gamma\hat{\mathrm{e}}_z\right)$ (ignoring units for simplicity). So then when you apply the differential operators (for $\square\in\{\cdot,\,\times,\}$), $$\nabla\square\mathbf{a}=\underset{0}{\underbrace{\nabla\square\mathbf{a}_0}}+\nabla\square\delta\mathbf{a}\equiv\nabla\square\delta\mathbf{a}$$ because any differential operator acting on a constant value is always 0.

Knowing the above, you can replace all instances of the variables being acted on by the derivative operators with the perturbations in the original equation (i.e., $\partial_t\mathbf{v}\to\partial_t\delta\mathbf{v}$) and all other instances with both terms: $$\left(\rho_0+\delta\rho\right) \left(\frac{\partial \delta\mathbf{v}}{\partial t} + (\left(\mathbf{v}_0+\delta\mathbf{v}\right) \cdot \nabla) \delta\mathbf{v}\right) = -\nabla \delta p + \frac{1}{\mu_0} (\nabla \times \delta\mathbf{B}) \times \left(\mathbf{B}+\delta\mathbf{B}\right)$$ Since $\delta\rho$ is only going to multiply to two $\delta\mathbf{v}$ terms, this is ignorable and we have, $$\rho_0\left(\frac{\partial \delta\mathbf{v}}{\partial t} + \left(\mathbf{v}_0\cdot\nabla\right)\delta\mathbf{v}+\color{red}{\left(\delta\mathbf{v}\cdot\nabla\right)\delta\mathbf{v}}\right) = -\nabla \delta p + \frac{1}{\mu_0} \left(\nabla \times \delta\mathbf{B}\right) \times \mathbf{B}_0+\frac{1}{\mu_0}\color{red}{\left(\nabla \times \delta\mathbf{B}\right)\times\delta\mathbf{B}}$$ The two red terms are zero as they also involve the product of two perturbations. This leaves the expected equation, $$\rho_0\left(\frac{\partial \delta\mathbf{v}}{\partial t} + \left(\mathbf{v}_0\cdot\nabla\right)\delta\mathbf{v}\right) = -\nabla \delta p + \frac{1}{\mu_0} \left(\nabla \times \delta\mathbf{B}\right) \times \mathbf{B}_0$$