Generals 2005 I 4

From PlasmaWiki

Jump to: navigation, search

[edit] Part a

The canonical angular momentum is:

P_{\zeta}=Rmv_{\zeta}-\frac{e}{c}RA_{\zeta}


The flux may be defined as:

ψ = − RAζ


So:

P_{\zeta}=Rmv_{\zeta}-\frac{e}{c}\psi


Taking the lagrangian derivative, the left side is zero since it is conserved:

0=Rm\frac{dv_{\zeta}}{dt}-\frac{e}{c}\frac{d\psi}{dt}


Bounce averaging:

0=m\left\langle R\frac{dv_{\zeta}}{dt}\right\rangle -\frac{e}{c}\left\langle \frac{d\psi}{dt}\right\rangle


On average over a bounce, Rdvζ / dt is zero, since it returns to the same position with the same velocity after one complete bounce. Therefore:

\left\langle \frac{d\psi}{dt}\right\rangle =\left\langle \frac{\partial\psi}{\partial t}+\mathbf{v}\cdot\nabla\psi\right\rangle =0


By Faraday's law:

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


In terms of \mathbf{A}:

\nabla\times\mathbf{E}=-\frac{1}{c}\frac{\partial\left(\nabla\times\mathbf{A}\right)}{\partial t}


Interchanging derivatives:

\mathbf{E}=-\frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}


So:

\frac{\partial\psi}{\partial t}=-R\frac{\partial A_{\zeta}}{\partial t}=cRE_{\zeta}


Noting RB_{\theta}=-\partial\psi/\partial R, and over a bounce average assuming \mathbf{V}=\hat{r}v_{r}:

cREζ − vrRBθ = 0


So:

v_{r}=\frac{cE_{\zeta}}{B_{\theta}}


There is also diffusion, since there is a pressure gradient. We can estimate the banana diffusion by using a step size of Λi, and the ion-ion collision frequency, multiplying by fT = ε1 / 2, the fraction of trapped particles:

D=\Lambda_{i}^{2}\nu_{ii}=\frac{\rho^{2}q^{2}}{\epsilon^{3/2}}\nu_{ii}


The radial flux is then:

\Gamma_{r}=\frac{\rho^{2}q^{2}}{\epsilon^{3}}\nu_{ii}\frac{dn_{0}}{dr}-\sqrt{\epsilon}n_{0}\frac{cE_{\zeta}}{B_{\theta}}

[edit] Part b

The spitzer conductivity is modified to:

\sigma_{\|}=\sigma_{spitz}\left(1-\sqrt{\epsilon}\right)


because the trapped particles do not carry current. But the trapped particles do produce bootstrap current, due to the pressure gradient. The current is:

j_{BS}=e\frac{d\left(nv\right)}{dx}\Delta x


v_{T}=\sqrt{T/m_{i}}:

j_{BS}=\frac{e}{m_{i}v_{T}}\frac{d\left(nT\right)}{dx}\Delta x


Using p = nT and Δx = Λi = ρiq / ε1 / 2, the banana width:

j_{BS}=\frac{e}{mv_{T}}\frac{dp}{dx}\frac{\rho_{i}q}{\epsilon^{1/2}}=\frac{e}{m\Omega}\frac{q}{\epsilon^{1/2}}\frac{dp}{dx}=\frac{\epsilon^{1/2}c}{B_{\theta}}\frac{dp}{dx}


The current density is then:

\left\langle j_{\|}-j_{S}\right\rangle =\sigma_{\|}^{spitz}\left(1-\sqrt{\epsilon}\right)+\frac{\epsilon^{1/2}c}{B_{\theta}}\frac{dp}{dx}

[edit] Part c

Onzager symmetry states that the cross terms in the flux-driving force matrix will be equal (symmetric about the diagonal). In this problem, it states that the coefficient of the -n_{0}E_{\|} term in part a will be equal to the coefficient of the − Tedn / dx term in part b.

Retrieved from "http://localhost/wiki/Generals_2005_I_4"
Personal tools