Oscillations

Collision Frequencies

Gases have three degrees of freedom to oscillate, but plasmas have four: the electron fluid adds an additional parameter.

One way to see how the electron fluid oscillates, take a region of plasma and displace the electrons by a distance $x$ and release. Then, watch how the electrons oscillate.

Gauss's law gives the resulting electric field (same as a parallel-plate capacitor.

$E = \frac{n e x}{\epsilon_0}$

The restoring force is therefore

$F = \frac{x n e^2}{\epsilon_0}$

We know forces of that form. When released, it will undergo simple harmonic motion

$\omega = \sqrt{\frac{K}{m}} \qquad K = \frac{ ne^2}{\epsilon_0}$

We define the plasma frequency as

$\omega_{pe} = \sqrt{ \frac{ne^2}{m_e \epsilon_0}}$

If we take the displacement out to $\lambda_D$ then the peak velocity is the electron thermal speed $v_e$

$\omega_{pe} \lambda_D = v_e$

Taking a look back at the collision frequencies, the electron-electron collision frequency $\nu_{ee}$ (associated with the energy transfer) is close to the electron-ion collision frequency $\nu_{ei}$ associated with momentum transfer, and both are about

$\nu_{ee} \approx \nu_{ei} \approx \frac{v_e}{\lambda_D \Lambda} = \frac{\omega_{pe}}{\Lambda}$

So the electron-electron collision frequency will be much less than the plasma frequency if $\Lambda \gg 1$

$\Lambda \gg 1 \rightarrow \nu_{ee} \ll \omega_{pe}$

Radiation

Cyclotron radiation

As an electron moves through a magnetic field, it takes a helical path. The acceleration is

$\vec a = \frac{v^2}{r} = v_{th} \omega _c$

Only the component of $\vec a$ perpendicular to $r$ can cause radiation in the $r$ direction.

$\theta$-pinch

Called theta-pinch because current $I$ is in the $\theta$ direction. The magnetic field is constant within the plasma, and constant outside the plasma (inside the device).

$\frac{B_0 ^2}{2 \mu_0}$

Z-pinch

Talk about a Z-pinch that's stationary and compressing. Assume T is a constant (burning through). The radiated power is then

$P_{rad} \propto n^2$

The ohmic heating goes like $j^2$

$P_{ohmic} \propto j^2$

So the key parameter in the ratio is $j/n$.

There is no $B_{\parallel}$ so we need to balance the pressure as well.

Consider ohmically heating a pure Z-pinch like FRC or ZaP.

First, assume that the power radiated per unit volume goes like

$n^2 f (T) = P/V$

The dependence is true for a fixed impurity fraction for line radiation in steady state, and for Bremsstrahlung.

Ignore $\ln \Lambda$ dependency in ohmic heating so

$P/V = \eta(T) j^2$

The pressure balance demands

$\frac{B^2}{2 \mu_0} = p = 2 nk T_0$

For Bremsstrahlung temperature and density dependence are

$P_{rad} = c_1 n^2 T^{1/2}$

$P_{ohm} = c_2 j^2 T^{-3/2}$

Pressure balance

$n T \propto B^2 \propto j^2 r^2$

$nT = c_3 j^2 r^2$

$T = c_e \frac{j^2 r^2}{n}$

So if we're going to get any heating, we need the ohmic heating to be greater than the radiation

$c_2 j^2 T^{-3/2} \geq c_1 n^2 T^{1/2}$

using

$T = c_3 \frac{j^2 r^2}{n}$

$\rightarrow \frac{c_2}{c_1 c_3 ^2} > j^2 r^4 \sim I^2$

Density cancels out, and we're left with a current term ($I^2$) called the Pease current

$I \leq \approx 1 MA$

Z-pinch radiatively collapses above 1MA for Bremsstralung radiation. Counter-intuitively, the current must be less than the Pease current in order to continue heating. There is an even lower current for line radiation going up through the burn-through curve.

Skin Depths in Plasma

Two types of skin depths in plasma: collisional and collisionless

$\delta = \sqrt{ \frac{ 2}{\mu_0 \omega \sigma}} \quad \text{Collisional}$

$\delta = \frac{c}{\omega_{pe}} \quad \text{Electron collisionless skin depth}$

$\delta = \frac{c}{\omega_{pi}} \quad \text{Ion collisionless skin depth}$

Collisionless skin depth is due to electron inertia, which allows the field to penetrate even if it is very hot and a good conductor. What is magnitude? Look at E production and B shielding

$\curl E = - \pdv{B}{t}$

$\curl B = \mu_0 j + \mu_0 \epsilon_0 \pdv{E}{t}$

Electrons accelerate with E

$m \pdv{v}{t} = E e$

Doing a wave analysis assuming spatial solutions that decay off as $1 / \delta$ and oscillate with frequency $\omega$, we have

$\frac{1}{\delta} \sim \grad$

$\omega \sim \pdv{}{t}$

$\frac{E}{\delta} = - \omega B$

$m \omega v = - e E$

$j = - n e v$

$\frac{B}{\delta} = \mu_0 (- n e v) + \mu_0 \epsilon_0 \omega E = \frac{E}{\delta ^2 \omega}$

$\frac{E}{\delta ^2 \omega} = \frac{\mu_0 n e^2}{m} \frac{E}{\omega} + \mu_0 \epsilon_0 \omega E$

That's the dispersion relation for our wave solutions. Plugging in $c$

$- \frac{c^2}{\delta ^2} \frac{1}{\omega} = \frac{n e^2}{\epsilon_0 m }\frac{1}{\omega} + \omega$

That lets us identify the electron plasma frequency