Linearised MHD Equations

Before linearising we need to define the equilibrium. Assuming the basic state is static we need to satisfy,

$$\nabla p_{0} = {\bf j}_{0} \times {\bf B}_{0} + \rho_{0}{\bf g},$$ (4.1)

along with $\nabla \cdot {\bf B}_{0} = 0$, the gas law and an energy equation. Having obtained the equilibrium we then set

$${\bf B} = {\bf B}_{0} + {\bf B}_{1}({\bf r}, t),$$ (4.2)

$${\bf v} = {\bf0} + {\bf v}_{1}({\bf r}, t),$$ (4.3)

$$\rho = \rho_{0} + \rho_{1}({\bf r}, t),$$ (4.4)

$$p = p_{0} + p_{1}({\bf r}, t),$$ (4.5)

where the perturbed quantities, denoted by a subscript `1' are smaller than the equilibrium quantities, denoted by a subscript `0'. Note that the equilibrium quantities are independent of time, $t$. Next we substitute these expressions into the MHD equations and neglect any products of small terms. Thus, the mass continuity equation becomes

$$\begin{eqnarray*} {\partial \rho \over \partial t} + \nabla \cdot (\rho {\bf v}... ... t} + \nabla \cdot \left (\rho_{0}{\bf v}_{1}\right ) & = & 0. \end{eqnarray*}$$

Hence, the linearised mass continuuity equation is

$${\partial \rho_{1}\over \partial t} + \nabla \cdot \left (\rho_{0}{\bf v}_{1}\right ) = 0.$$

In a similar manner () - () in the ideal MHD limit (i.e. $R_{m} \rightarrow \infty$) reduce to

$${\partial \rho_{1}\over \partial t} + \nabla \cdot \left (\rho_{0}{\bf v}_{1}\right ) = 0,$$ (4.6)

$$\rho_{0}{\partial {\bf v}_{1}\over \partial t} = - \nabla p... ...es {\bf B}_{0}\over \mu} \times {\bf B}_{1} + \rho_{1}{\bf g},$$ (4.7)

$${\partial p_{1}\over \partial t} + {\bf v}_{1}\cdot \nabla p_{0} = - \gamma p_{0} \nabla \cdot {\bf v}_{1},$$ (4.8)

$${\partial {\bf B}_{1}\over \partial t} = \nabla \times \left ({\bf v}_{1} \times {\bf B}_{0}\right ),$$ (4.9)

and

$$\nabla \cdot {\bf B}_{1} = 0.$$ (4.10)

Example 4.1.1   Show that if $\nabla \cdot {\bf B}_{1} = 0$ at time $t=0$, (4.9) implies that (4.10) is true for all time. Take the divergence of (4.9) so that

$$\nabla \cdot {\partial {\bf B}_{1}\over \partial t} = {\par... ... ( \nabla \times ({\bf v}_{1} \times {\bf B}_{0})\right ) = 0.$$

Therefore, $\nabla \cdot {\bf B}_{1}$ is constant in time. However, if it is zero at $t=0$ then it is zero for all time.

The linearised equations may now be combined into one equation in the following manner. Take the time derivative of (4.7), the linearised equation of motion, to get

$$\rho_{0}{\partial^{2}{\bf v}_{1}\over \partial t^{2}} = - \... ...r \partial t} + {\partial \rho_{1}\over \partial t} {\bf g},$$

and use (4.6), (4.8) and (4.9) to eliminate the perturbed density, pressure and magnetic field and get

$$\rho_{0}{\partial^{2}{\bf v}_{1}\over \partial t^{2}} = + \nabla \left [{\bf v}_{1}\cdot \nabla p_{0} + \gamma p_{0} \nab... ...\times \nabla \times ({\bf v}_{1}\times {\bf B}_{0})\right ] \times {\bf B}_{0}$$

$$+ {1\over \mu}\left [\nabla \times {\bf B}_{0}\right ]\times \nab... ...s ({\bf v}_{1}\times {\bf B}_{0}) - \nabla \cdot (\rho_{0}{\bf v}_{1}) {\bf g}.$$ (4.11)

(4.11) is the linearised equation of motion and forms the basis of the study of MHD waves and MHD instabilities.