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2005 II 1

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The energy density in the plasma is:

$w_{p}=\frac{1}{2}\rho v^{2}+\frac{B^{2}}{8\pi}$

So the total energy in the plasma is:

$E_{p}=\int_{V}d\mathbf{x}\left(\frac{1}{2}\rho v^{2}+\frac{B^{2}}{8\pi}\right)$

The ideal MHD equations with $p = 0$ give:

$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf{v}\right)=0$ $\rho\frac{d\mathbf{v}}{dt}=\frac{1}{c}\mathbf{J}\times\mathbf{B}$ $\nabla\times\mathbf{E}=-\frac{1}{c}\frac{\partial\mathbf{B}}{\partial t}$ $\mathbf{E}+\frac{\mathbf{v}}{c}\times\mathbf{B}=0$ $\mathbf{J}=\frac{c}{4\pi}\nabla\times\mathbf{B}$

Combining:

$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf{v}\right)=0$ $\rho\frac{\partial\mathbf{v}}{\partial t}+\rho\mathbf{v}\cdot\nabla\mathbf{v}=\frac{1}{4\pi}\left(\nabla\times\mathbf{B}\right)\times\mathbf{B}$ $\frac{\partial\mathbf{B}}{\partial t}=\nabla\times\left(\mathbf{v}\times\mathbf{B}\right)$

So:

$\frac{dE_{p}}{dt}=\int_{V}d\mathbf{x}\left(\frac{1}{2}\frac{\partial\rho}{\partial t}v^{2}+\rho\mathbf{v}\cdot\frac{\partial\mathbf{v}}{\partial t}+\frac{\mathbf{B}}{4\pi}\cdot\frac{\partial\mathbf{B}}{\partial t}\right)+\int_{S}d\mathbf{x}_{\partial}w_{F}\hat{n}\cdot\mathbf{v}$

Plugging in:

$\begin{array}{rcl} \frac{dE_{p}}{dt} & = & \int_{V}d\mathbf{x}\left(-\frac{1}{2}v^{2}\nabla\cdot\left(\rho\mathbf{v}\right)+\mathbf{v}\cdot\left(\frac{1}{4\pi}\left(\nabla\times\mathbf{B}\right)\times\mathbf{B}-\rho\mathbf{v}\cdot\nabla\mathbf{v}\right)\right)\\ & & \int_{V}d\mathbf{x}\frac{\mathbf{B}}{4\pi}\cdot\nabla\times\left(\mathbf{v}\times\mathbf{B}\right)+\int_{S}d\mathbf{x}_{\partial}\left(\right)\hat{n}\cdot\mathbf{v}\end{array}$

Rearranging:

$\begin{array}{rcl} \frac{dE_{p}}{dt} & = & \frac{1}{2}\int_{V}d\mathbf{x}\left(-v^{2}\nabla\cdot\left(\rho\mathbf{v}\right)-\rho\left(\mathbf{v}\cdot\nabla\right)v^{2}\right)\\ & & +\frac{1}{4\pi}\int_{V}d\mathbf{x}\left[-\left(\nabla\times\mathbf{B}\right)\cdot\left(\mathbf{v}\times\mathbf{B}\right)+\mathbf{B}\cdot\nabla\times\left(\mathbf{v}\times\mathbf{B}\right)\right]\\ & & +\int_{S}d\mathbf{x}_{\partial}\left(\frac{1}{2}\rho v^{2}+\frac{B^{2}}{8\pi}\right)\hat{n}\cdot\mathbf{v}\end{array}$

Or, using vector identities 1, 7 and 9:

$\begin{array}{rcl} \frac{dE_{p}}{dt} & = & -\frac{1}{2}\int_{V}d\mathbf{x}\nabla\cdot\left(v^{2}\rho\mathbf{v}\right)\\ & & +\frac{1}{4\pi}\int_{V}d\mathbf{x}\nabla\cdot\left[\left(\mathbf{v}\times\mathbf{B}\right)\times\mathbf{B}\right]\\ & & +\int_{S}d\mathbf{x}_{\partial}\left(\frac{1}{2}\rho v^{2}+\frac{B^{2}}{8\pi}\right)\hat{n}\cdot\mathbf{v}\end{array}$

Then:

$\begin{array}{rcl} \frac{dE_{p}}{dt} & = & -\frac{1}{2}\int_{S}d\mathbf{x}_{\partial}v^{2}\rho\mathbf{v}\cdot\hat{n}\\ & & +\frac{1}{4\pi}\int_{S}d\mathbf{x}_{\partial}\left[\left(\mathbf{v}\times\mathbf{B}\right)\times\mathbf{B}\right]\cdot\hat{n}\\ & & +\int_{S}d\mathbf{x}_{\partial}\left(\frac{1}{2}\rho v^{2}+\frac{B^{2}}{8\pi}\right)\hat{n}\cdot\mathbf{v}\end{array}$

Or:

$\frac{dE_{p}}{dt}=\int_{S}d\mathbf{x}_{\partial}\left(\frac{1}{4\pi}\left(\mathbf{v}\times\mathbf{B}\right)\times\mathbf{B}+\frac{B^{2}}{8\pi}\mathbf{v}\right)\cdot\hat{n}$

Expanding the double cross product:

$\frac{dE_{p}}{dt}=\int_{S}d\mathbf{x}_{\partial}\left(\frac{1}{4\pi}\left(\mathbf{B}\cdot\mathbf{v}\right)\mathbf{B}-\frac{B^{2}}{8\pi}\mathbf{v}\right)\cdot\hat{n}$

$\mathbf{B}\cdot\hat{n}=0$ on the boundary:

$\frac{dE_{p}}{dt}=-\frac{1}{8\pi}\int_{S}d\mathbf{x}_{\partial}B^{2}\mathbf{v}\cdot\hat{n}$

The vacuum energy density is:

$w_{v}=\frac{B^{2}}{8\pi}$

So:

$\frac{dE_{v}}{dt}=\frac{1}{4\pi}\int_{V}\mathbf{B}\cdot\frac{\partial\mathbf{B}}{\partial t}d\mathbf{x}+\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Plugging in for $\partial B/\partial t$ :

$\frac{dE_{v}}{dt}=-\frac{c}{4\pi}\int_{V}\mathbf{B}\cdot\left(\nabla\times\mathbf{E}\right)d\mathbf{x}+\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Using the vector identity:

$\frac{dE_{v}}{dt}=-\frac{c}{4\pi}\int_{V}\left[\nabla\times\left(\mathbf{E}\times\mathbf{B}\right)+\mathbf{E}\cdot\left(\nabla\times\mathbf{B}\right)\right]d\mathbf{x}+\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Since $\nabla\times\mathbf{B}=\mathbf{j}=0$ in the vacuum, and plugging in $\mathbf{E}=-\mathbf{v}\times\mathbf{B}/c$ :

$\frac{dE_{v}}{dt}=\frac{1}{4\pi}\int_{V}\nabla\cdot\left(\left(\mathbf{v}\times\mathbf{B}\right)\times\mathbf{B}\right)d\mathbf{x}+\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Expanding the double cross product, and changing the integral to a surface integral:

$\frac{dE_{v}}{dt}=\frac{1}{4\pi}\int_{S}\hat{n}\cdot\left(B^{2}\mathbf{v}-\left(\mathbf{B}\cdot\mathbf{v}\right)\mathbf{B}\right)d\mathbf{x}_{\partial}+\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Since $\mathbf{B}\cdot\hat{n}$ is zero on the surface:

$\frac{dE_{v}}{dt}=-\frac{1}{8\pi}\int_{S}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

There will be no jump conditions at the plasma-vacuum interface, since there are no surface currents. The time derivative of the total energy will then be:

$\frac{dE}{dt}=\frac{dE_{p}}{dt}+\frac{dE_{v}}{dt}=-\frac{1}{8\pi}\int_{S_{p}}d\mathbf{x}_{\partial}B^{2}\mathbf{v}\cdot\hat{n}-\frac{1}{8\pi}\int_{S_{v}}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

The vacuum boundary contains the plasma-vacuum boundary and the vacuum-container boundary:

$\frac{dE}{dt}=-\frac{1}{8\pi}\int_{S_{p}}d\mathbf{x}_{\partial}B^{2}\mathbf{v}\cdot\hat{n}-\frac{1}{8\pi}\int_{S_{p}}B^{2}\mathbf{v}\cdot\left(-\hat{n}\right)d\mathbf{x}_{\partial}-\frac{1}{8\pi}\int_{S_{w}}B^{2}\mathbf{v}\cdot\hat{n}d\mathbf{x}_{\partial}$

Where we have used the negative sign since the normal faces out of the vacuum, and so into the plasma. The total energy time derivative is then: