Generals 1998 9

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1 Basic Plasma Physics

[edit] Basic Plasma Physics

[edit] Part a

The fluid equations are:

$\frac{\partial n}{\partial t}+\frac{\partial}{\partial x}\left(nv\right)=0$

$mn\frac{dv}{dt}=-neE-\frac{\partial p}{\partial x}$

$\frac{\partial E}{\partial x}=4\pi e\left(n_{i}-n_{e}\right)$

We assume the equation of state:

$p=p_{0}\left(n/n_{0}\right)^{\Gamma}$

[edit] Part b

If $n = n 0 + n 1$ , $p = p 0 + p 1$ , $v = v 1$ , $E = E 1$ :

$\frac{\partial n_{1}}{\partial t}+n_{0}\frac{\partial v_{1}}{\partial x}=0$

$mn_{0}\frac{\partial v_{1}}{\partial t}=-n_{0}eE_{1}-\frac{\partial p_{1}}{\partial x}$

$\frac{\partial E_{1}}{\partial x}=-4\pi en_{1}$

Assuming waves like $\exp\left[i\left(kx-\omega t\right)\right]$ :

- i ω n 1 + i k n 0 v 1 = 0

$-i\omega mn_{0}v_{1}=-n_{0}eE_{1}-\Gamma p_{0}\frac{ikn_{1}}{n_{0}}$

i k E 1 = - 4π e n 1

Combining the last two:

$\omega mn_{0}v_{1}=\left(\frac{4\pi e^{2}n_{0}}{k}\right)n_{1}+\Gamma p_{0}\frac{kn_{1}}{n_{0}}$

Plugging into the first equation and using $\omega_{pe}^{2}$ :

$-i\omega n_{1}+\frac{ik}{\omega}\left[\omega_{pe}^{2}\frac{n_{1}}{k}+\frac{\Gamma p_{0}}{m}\frac{kn_{1}}{n_{0}}\right]=0$

Simplifying:

$-\omega^{2}+\omega_{pe}^{2}+\frac{\Gamma p_{0}}{mn_{0}}k^{2}=0$

[edit] Part c

The group velocity is:

$v_{g}=\frac{\partial\omega}{\partial k}=\frac{\Gamma p_{0}}{mn_{0}}\frac{k}{\omega}=\frac{\Gamma p_{0}}{mn_{0}}k\left(\omega_{pe}^{2}+\frac{\Gamma p_{0}}{mn_{0}}k^{2}\right)^{-1/2}$

So the time it will take for the antenna to detect it is:

$\tau=\frac{L}{v_{g}}=\frac{mn_{0}}{\Gamma p_{0}k}\left(\omega_{pe}^{2}+\frac{\Gamma p_{0}}{mn_{0}}k^{2}\right)^{1/2}$

If $p 0 = n 0 T 0$ :

$\tau=\frac{L}{v_{g}}=\frac{m}{\Gamma T_{0}k}\left(\omega_{pe}^{2}+\frac{\Gamma T_{0}}{m}k^{2}\right)^{1/2}$

If $T_{0}\rightarrow0$ , this $\tau\rightarrow\infty$ .

[edit] Part d

The phase velocity is:

$v_{p}=\frac{\omega}{k}=\left(\frac{\omega_{pe}^{2}}{k^{2}}+\frac{\Gamma T_{0}}{m}\right)^{1/2}$

The debye length is $\lambda_{De}^{2}=T_{0}/4\pi n_{0}e^{2}$ . Using this:

$v_{p}=\left(\frac{T_{0}}{k^{2}\lambda_{De}^{2}m}+\frac{\Gamma T_{0}}{m}\right)^{1/2}=\sqrt{\frac{T_{0}}{m}}\left(\frac{1}{k^{2}\lambda_{De}^{2}}+\Gamma\right)^{1/2}$

If $k\lambda_{De}\ll1$ , since $Γ˜1$ , this is approximately: