Generals 2004 I 6

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$δ W$ is given as:

$\delta W=\frac{1}{2}\int_{V}\left[\gamma p\left|\nabla\cdot\xi\right|^{2}-\xi^{\star}\cdot\nabla\left(\xi\cdot\nabla p\right)+\left|Q\right|^{2}/\mu-\xi^{\star}\cdot\mathbf{j}\times\mathbf{Q}\right]d^{3}x$

The MHD equations are:

$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf{v}\right)=0$

$\rho\frac{d\mathbf{v}}{dt}=\frac{1}{c}\mathbf{j}\times\mathbf{B}-\nabla p+\rho\mathbf{g}$

$\frac{d}{dt}\left(\frac{p}{\rho^{\gamma}}\right)=0$

$\mathbf{E}+\frac{1}{c}\mathbf{v}\times\mathbf{B}=0$

$\nabla\times\mathbf{E}=-\frac{1}{c}\frac{\partial\mathbf{B}}{\partial t}$

The linearized equations are:

$\frac{\partial\rho_{1}}{\partial t}+\rho_{0}\nabla\cdot\mathbf{v}_{1}=0$

$\rho_{0}\frac{\partial\mathbf{v}_{1}}{\partial t}=\frac{1}{4\pi}\left(\nabla\times\mathbf{B}_{1}\right)\times\mathbf{B}_{0}+\frac{1}{4\pi}\left(\nabla\times\mathbf{B}_{0}\right)\times\mathbf{B}_{1}-\nabla p_{1}+\rho_{1}\mathbf{g}$

$\frac{\partial\mathbf{B}_{1}}{\partial t}=\nabla\times\left(\mathbf{v}_{1}\times\mathbf{B}_{0}\right)$

$\frac{\partial p_{1}}{\partial t}+\mathbf{v}_{1}\cdot\nabla p_{0}+ \frac{\gamma p_{0}}{\rho_{0}}\frac{\partial\rho_{1}}{\partial t}+ \frac{\gamma p_{0}}{\rho_{0}}\mathbf{v}_1\cdot \nabla \rho_0 =0$

We can combine the first and fourth equations:

$\frac{\partial p_{1}}{\partial t}=-\mathbf{v}_{1}\cdot\nabla p_{0}+\gamma p_{0}\nabla\cdot\mathbf{v}_{1}$

For displacement $ξ$ , the equations become:

$\frac{\partial\rho_{1}}{\partial t}+\nabla\cdot\left(\rho_{0}\frac{d\xi}{dt}\right)=0$

$\rho_{0}\frac{d^{2}\xi}{dt^{2}}=\frac{1}{c}\left(\nabla\times\mathbf{B}_{1}\right)\times\mathbf{B}_{0}+\frac{1}{c}\left(\nabla\times\mathbf{B}_{0}\right)\times\mathbf{B}_{1}-\nabla p_{1}+\rho_{1}\mathbf{g}$

$\frac{\partial\mathbf{B}_{1}}{\partial t}=\nabla\times\left(\frac{d\xi}{dt}\times\mathbf{B}_{0}\right)$

$\frac{\partial p_{1}}{\partial t}=-\frac{d\xi}{dt}\cdot\nabla p_{0}+\gamma p_{0}\nabla\cdot\frac{d\xi}{dt}$

Integrating in time:

$\rho_{1}=-\nabla\cdot\left(\xi\rho_{0}\right)$

$p_{1}=-\xi\cdot\nabla p_{0}+\gamma p_{0}\nabla\cdot\xi$

$\mathbf{B}_{1}=\nabla\times\left(\xi\times\mathbf{B}_{0}\right)=\mathbf{Q}$

So the momentum equation becomes:

$\mathbf{F}=\rho\frac{d^{2}\xi}{dt^{2}}=\ldots-\nabla\cdot\left(\xi\rho_{0}\right)\mathbf{g}$

And the potential energy becomes:

$\delta W=-\frac{1}{2}\int\xi^{\star}\cdot\mathbf{F}d\mathbf{r}=\frac{1}{2}\int \left[\ldots +\rho_{0}\left(\xi^{\star}\cdot\mathbf{g}\right)\left(\nabla\cdot\xi\right) +\left(\xi^{\star}\cdot\mathbf{g}\right)\left(\xi\cdot \nabla \rho_0\right) \right]d\mathbf{r}$

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Generals 2004 I 6

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