Landau Damping
\[\]

Landau Damping #

The electrostatic dispersion relation for a plasma with a non-zero temperature leads to additional terms beyond the \( \omega^2 - \omega_p ^2 = 0 \) relation we got before.

Waves in Thermal, Unmagnetized Plasma using Vlasov Equations #

In a uniform plasma with \( T > 0 \), \( \vec E_0 = \vec B_0 = 0 \) and equilibrium distribution function \( f_0 (\vec v) \), a perturbation in \( f(\vec x, \vec v, t) \) can be written as

\[f(\vec x, \vec v, t) = f_0( \vec v) + f_1 (\vec x, \vec v , t) \]

Linearized Vlasov equation is

\[\pdv{f_1}{t} + \vec v \cdot \grad f_1 - \frac{e}{m} \vec E_1 \cdot \pdv{f_0}{ \vec v} = 0\]

Assuming high ion inertia and that the waves are plane waves in the \( \vu{x} \) direction

\[f_1 = \hat f_1 e^{i (kx - \omega t)}\]

Fourier transform to get

\[-i \omega f_1 + i k v_x f_1 - \frac{e}{m} E_x \pdv{f_0}{v_x} = 0\] \[f_1 = \frac{i e E_x}{m} \frac{\pdv{f_0}{v_x}}{\omega - k v_x}\]

What’s the electric field? We go ask Maxwell

\[\epsilon_0 \div \vec E_1 = - e n_1 = - e \int f_1 \dd ^3 v\]

Applying Fourier transform, we also assume an electrostatic form of the solution. That is, the wave is propagating along the x-direction, and the electric field also only varies in the x-direction.

\[i k \epsilon_0 E_x = -e \int f_1 \dd ^3 v\]

and substituting our expression for \( f_1 \), we arrive at our dispersion relation:

\[D(\omega, k) = 1 + \frac{e^2}{k m \epsilon_0} \int \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd ^3 v = 0\]

We can back out the background density from the distribution function, leaving us with the first moment of the normalized distribution function \( \hat f_0(v) \)

\[n_0 = \int f_0 \dd \vec v = n_0 \int \hat f_0 \dd v\] \[D(\omega, k) = 1 - \frac{\omega_p ^2}{k^2} \int \frac{\partial \hat f_0 / \partial v_x}{v_x - (\omega / k)}\]

Electrostatic Dispersion Relation - Vlasov #

Electrostatic dispersion relation is not easily solved. Note that if \( v_x = v_\phi = \omega / k \), the integral has a singularity. But, the integral only carries a singularity if \( \omega / k \) is not complex. If \( \omega = \omega_r + i \omega _i \) with \( \omega_i \neq 0 \), the singularity is no longer a problem. In practice, \( \omega \) is never purely real, so this is a fair assumption.

Vlasov’s approach is to ignore \( \omega_i \) and the singularity at \( v_\phi = \omega /k \) by considering only the Cauchy principal value of the integral.

It’s easier to examine the case when the phase velocity is much greater than the thermal velocity. If we take a small gap of size \( 2 \epsilon \) out of \( v_x \) near \( v_\phi \), the contribution from \( \partial \hat f_0 / \partial v_x \) is vanishingly small. Expanding the integral by parts,

\[\int _{-\infty} ^{\infty} \frac{\partial \hat f_0 / \partial v_x}{v_x - v_\phi} \dd v = \\ \lim _{\epsilon \rightarrow 0} \left[ \left. \frac{\hat f_0}{v_x - v_\phi} \right|_{- \infty} ^{v_\phi - \epsilon} - \int _{- \infty} ^{v_\phi - \epsilon} - \frac{\hat f_0 \dd v_x}{(v_x - v_\phi)^2} + \left. \frac{\hat f_0}{v _x - v_\phi} \right|_{v_\phi + \epsilon} ^\infty - \int _{v_\phi + \epsilon} ^\infty \frac{\hat f_0 \dd v_x}{(v_x - v_\phi)^2} \right] \\ = \lim _{\epsilon \rightarrow 0} \left[ - \int _{- \infty} ^{v_\phi - \epsilon} - \frac{\hat f_0 \dd v_x}{(v_x - v_\phi)^2} - \int _{v_\phi + \epsilon} ^\infty \frac{\hat f_0 \dd v_x}{(v_x - v_\phi)^2} \right] \\ \equiv \sout{\int} _{-\infty} ^\infty \frac{\hat f_0 \dd v_x }{(v_x - v_\phi)^2}\]

where we define the integral excluding \( v_\phi \pm \epsilon \) to be \( \sout{\int}_{-\infty} ^{\infty} \).

\[D(\omega, k) = 1 - \frac{\omega_p ^2}{k ^2} \sout{\int} _{-\infty} ^{\infty} \frac{\partial \hat f_0 / \partial v_x}{v_x - (\omega / k)} \dd v_x \\ = 1 - \frac{\omega_p ^2}{k ^2} \sout{\int} _{-\infty} ^{\infty} \frac{\hat f_0 \dd v_x}{(v_x - v_\phi)^2}\]

Assuming \( v_\phi \gg v_x \), then

\[\frac{1}{(v_x - v_\phi)^2} = \frac{1}{\left[- v_\phi(1 - v_x / v_\phi) \right]^2} \\ = \frac{1}{v_\phi ^2} (1 - \frac{v_x}{v_\phi})^{-2} \\ = \frac{1}{v_\phi ^2} \left( 1 + 2 \frac{v_x}{v_\phi} + 3 \frac{v_x ^2}{v_\phi ^2} + \ldots \right)\]

Then

\[D(\omega, k) = 1 - \frac{\omega_p ^2}{\frac{\omega ^2}{k^2} k^2} \left[ \sout{\int} _{-\infty} ^{\infty} \hat f_0 \dd v_x + \frac{2}{v_\phi} \sout{\int} _{-\infty} ^{\infty} v_x \hat f_0 \dd v_x + \frac{3}{v_\phi ^2} \sout{\int} _{-\infty} ^{\infty} v_x ^2 \hat f_0 \dd v_x + \ldots \right]\]

Because \( v_x \hat f_0 \) is an odd function,

\[\sout{\int} _{-\infty} ^{\infty} v_x \hat f_0 \dd x \approx \int _{-\infty} ^{\infty} v_x \hat f_0 \dd x = 0\]

\[\sout{\int} _{-\infty} ^{\infty} \hat f_0 \dd v_x \approx \int _{-\infty} ^{\infty} \hat f_0 \dd v_x = 1\] \[\int _{-\infty} ^{\infty} v_x ^2 \hat f_0 \dd v_x = \langle v_x ^2 \rangle\]

\[\rightarrow D(\omega, k) = 1 - \frac{\omega_p ^2}{\omega ^2} \left[ 1 + \frac{3 \langle v_x ^2 \rangle}{v_\phi ^2} \right] = 0\]

If \( T = 0 \) then \( \langle v_x ^2 \rangle = 0 \) then \( D(\omega, k) = 0 \) and \( \omega ^2 = \omega _p ^2 \) which is the cold plasma electrostatic dispersion relation. So we can think of the second term as a thermal correction to the electrostatic cold plasma dispersion relation.

If the thermal correction is small, that is \( \frac{3 \langle v_x ^2 \rangle}{v_\phi ^2} \ll 1 \), then \( \omega \approx \omega_p \) and noting that \( \langle v_x ^2 \rangle = v_t ^2 \) we have

\[\omega ^2 = \omega _p ^2 + 3 k ^2 v_t^2\]

which is the Bohm-Gross dispersion relation.

Electrostatic Dispersion Relation - Landau #

The Vlasov approach has some conceptual problems, because the assumption of an imaginary \( \omega \) is not sound in some situations. Landau argued that the electrostatic problem should be treated as an initial value problem in the time domain while using Fourier transformation for the spatial domain:

VlasovLandau
\( f_1 (\vec x, \vec v, t) = \tilde{f_1} e^{i (\vec k \cdot \vec x - \omega t)} \)\( f_1(\vec x, \vec v, t) = \tilde{f_1}(\vec v, t) e^{i\vec k \cdot \vec x} \)

Here, we Fourier transform only the spatial coordinates. Then, for an initial value problem in time, it is convenient to use the Laplace transform of \( f_1 \) and \( \vec E = \tilde{\vec E} (t) e^{i \vec k \cdot \vec x} \)

\[\mathcal{L}[f_1(\vec v, t)] \equiv F_1(\vec v, s) = \int _0 ^{\infty} e^{-s t} f_1(\vec v, t) \dd t\]

and its inverse

\[\mathcal{L}^{-1}[F_1(\vec v, s)] \equiv \frac{1}{2 \pi i} \int _{\sigma - i \infty} ^{\sigma + i \infty} e^{st} F(\vec v, s) \dd s\]

where \( \sigma > 0 \) defines our contour of integration in the complex plane, and by convention lies to the right of all of the singularities of \( F_1(\vec v, s) \). Note that this convention is equivalent to the one-sided Fourier transform, with \( s = - i \omega \).

For simplicity, let’s consider the unmagnetized, electrostatic case \( \hat{k} = \hat{x} \) and \( \vec E = E \hat{x} \). We then have a “1-dimensional” system in which \( \vec k \parallel \vu x \), \( \vec E \parallel \vu{x} \), so we can consider only \( (x, v_x) \) instead of \( \vec x, \vec v \). The linearized Vlasov equation in O(1) is

\[\pdv{f_1}{t} + \vec v \cdot \grad f_1 - \frac{e}{m} E_1 \cdot \pdv{f_0}{\vec v} = 0\]

Fourier transforming:

\[\pdv{\tilde{f_1}}{t} e^{i k x} + i k v_x \tilde{f_1} e^{i k x} - \frac{e}{m} \tilde{E} \pdv{\tilde{f_0}}{v_x} e^{i k x} = 0\]

We can divide out the \( e^{ikx} \):

\[\pdv{\tilde{f_1}}{t} + i k v_x \tilde{f_1} - \frac{e}{m} \tilde{E} \pdv{\tilde{f_0}}{v_x} = 0\]

Then, Laplace transform in time, where \( \mathcal{L}[\pdv{f}{t}] = s F(s) - f(0) \)

\[s \tilde{F_1} (\vec v, s) - \tilde{f_1}(\vec v, t = 0) + i k v_x \tilde{F_1} (\vec v, s) - \frac{e}{m} \tilde{E}(s) \pdv{f_0}{v_x} = 0\] \[(s + i k v_x) \tilde{F_1}(v, s) - \frac{e}{m} \tilde{E}(s) \pdv{f_0}{v_x} = \tilde{f_1}(\vec v, t = 0)\]

We can re-arrange the transformed Vlasov equation to solve for \( \tilde{F_1}(v_x, s) \):

\[\tilde{F_1}(v_x, s) = \frac{\tilde{f_1}(v, t = 0)}{s + i k v_x} + \frac{e}{m} \frac{\tilde{E}(s) \partial f_0 / \partial v_x}{s + i k v_x}\]

From Maxwell’s Poisson equation, the net charge density is just the perturbed electron density. The background charge density sums to zero and here we make the assumption that the ions remain static.

\[\epsilon_0 \div E_1(t) = -e n_1 = -e n_0 \int _{-\infty} ^{\infty} f_1 \dd \vec v\]

Applying the Fourier transform to Poisson:

\[i k \epsilon_0 \tilde{E}(t) = - e n_0 \int _{-\infty} ^{\infty} \tilde{f_1} \dd v_x\]

And applying the Laplace transform:

\[i k \epsilon_0 \tilde{E}(s) = - e n_0 \int _{-\infty} ^{\infty} \tilde{F_1} (v_x, s) \dd v_x\]

Substituting \( \tilde{F_1} \) back into the transformed Poisson equation:

\[i k \epsilon_0 \tilde{E}(s) = - e n_0 \int _{-\infty} ^{\infty} \left( \frac{\tilde{f_1}(v_1 t = 0)}{s + i k v_x} + \frac{e}{m} \frac{\tilde{E}(s) \partial f_0 / \partial v_x}{s + i k v_x} \right) \dd v_x\] \[\left[ i k \epsilon_0 + \frac{e^2 n_0}{m} \int _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{s + i k v_x} \dd v_x \right] \tilde{E} (s) = - e n_0 \int _{-\infty} ^{\infty} \frac{\tilde{f_1}(v_x, t=0)}{s + i k v_x} \dd v_x\]

Dividing by \( i k \epsilon_0 \),

\[\overbrace{\left[ 1 - i \frac{e^2 n_0}{\epsilon_0 m} \frac{1}{k} \int _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{s + i k v_x} \dd v_x \right]}^{D(k, s) = \epsilon(k, s)} \tilde{E} (s) = \overbrace{\frac{i e n_0}{k \epsilon_0} \int _{-\infty} ^{\infty} \frac{\tilde{f_1}(v_x, 0)}{s + i k v_x} \dd v_x}^{N(k, s)}\] \[\rightarrow \tilde{E}(s) = \frac{N(k, s)}{D(k, s)}\]

The inverse Laplace transform gives \( E(t) \).

\( \epsilon(k, s) \) is the plasma dielectric function. Here, \( s = - i \omega \) gets us back to the Vlasov solution.

Substituting \( \tilde E(s) \) into \( \tilde{F_1}(\vec v, s) \):

\[\tilde{F_1}(v_x, s) = \frac{\tilde{f_1}(\vec v, t=0)}{s + i k v_x} + \frac{e}{m} \frac{\partial f_0 / \partial v_x}{s + i k v_x} \frac{N(k, s)}{\epsilon(k, s)}\]

To invert the Laplace transform and solve for \( \tilde{f_1}(v_x, t) \), we need to integrate \( \tilde{F_1} e^{s t} \) over the complex plane:

\[\tilde{f_1}(v_x, t) = \frac{1}{2 \pi i} \int _C \tilde{F_1}(v_x, s) e^{st} \dd s\]

We compute the integral using the Cauchy residue theorem by locating all of the poles of \( \tilde{F_1}(v_x, s) \). So long as the initial distribution \( f_0(v_x) \) is both continuous and differentiable (smooth), the only singularities are the zeros of \( \epsilon(k, s) \), and the phase velocity \( s_0 = - i k v_x \). Let’s call the poles of \( \tilde{F_1} \) \( (s_1, s_2, \ldots) \), with \( s_1 \) having the most positive real component. Following Landau, we take the contour of integration to the left of all of the poles (that is, \( \sigma < \min(s_j) \)). By the Cauchy residue theorem, the contour integral will be the sum of the residues of \( s_j \). Since \( s_1 \) is the largest, it will be the dominant term. There are three cases:

  • If \( \text{Re}(s_1) > 0 \), then the residual of \( s_1 \) is the dominant, fastest-growing term, and \[\tilde{f_1}(v_x, t) = \text{Res}(s_1) e^{s_1 t} + \frac{1}{2 \pi i} \int_{C_1} \tilde{F_1} e^{st} \dd s\] where \( C_1 \) is a contour to the left of the singularities.
  • If \( \text{Re}(s_1) < 0 \), then the dominant term \( \text{Res}(s_1) e^{s_1 t} \) decays in time.
  • If the dominant zero of \( \epsilon(k, s) \) has \( \text{Re}(s_1) \) which is negative but very small, then the system is weakly damped and \[\epsilon(\omega, k) = 1 + \frac{n_0 e^2}{\epsilon_0 m} \frac{1}{k} \int _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x \\ = 1 + \frac{\omega_p ^2}{k} \sout{\int} _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x + \frac{\omega_p ^2}{k} \int_{C2} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x \] where \( C_2 \) is a contour in the lower half-plane that excludes \( s_1 \). Then, from the residue theorem:
\[\int _{C2} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x = \frac{1}{2} 2 \pi i \text{Res} \left[ \frac{\partial f_0 / \partial v_x}{\omega - k v_x}, v_x=\frac{\omega}{k} \right] \\ = \pi i \lim_{v_x \rightarrow \omega / k} (v_x - \frac{\omega}{k}) \frac{\left.\partial f_0 / \partial v_x\right|_{v_x = \omega / k}}{\omega - k v_x} \\ = \frac{- \pi i}{k} \lim_{v_x \rightarrow \omega / k} \frac{v_x - \omega / k}{v_x - \omega / k} \left.\pdv{f_0}{v_x}\right|_{v_x = \omega / k} \\ = - \frac{\pi i}{k} \left.\pdv{f_0}{v_x}\right|_{v_x = \omega / k}\]

Warm Plasma Dispersion Relation

\[\epsilon(\omega, k) = \underbrace{1 + \frac{\omega_p ^2}{k} \left[ \sout{\int} _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x \right.}_{\text{Vlasov's solution}} \underbrace{\left. - \frac{i \pi}{k} \left. \pdv{f_0}{v_x} \right|_{v_x = \omega / k} \right]}_{\text{Landau damping}}\]

We can evaluate the integral by parts:

\[\int _{-\infty} ^{\infty} \pdv{f_0}{v_x} \frac{\dd v_x}{\omega - k v_x} = \left. \frac{f_0}{\omega - k v_x} \right|_{- \infty} ^{\infty} - \frac{1}{k} \int _{-\infty} ^{\infty} \frac{- f_0 \dd v_x }{(v - v_\phi)^2} = \frac{1}{k}\int _{-\infty} ^{\infty} \frac{- f_0 \dd v_x }{(v_x - v_\phi)^2}\]

If \( v_\phi \gg v_x \) we can expand \( (v_x - v_\phi)^{-2} \)

\[(v_x - v_\phi)^{-2} = v_\phi ^{-2} \left( 1 - \frac{v_x}{v_\phi} \right)^{-2} \\ = v_\phi ^{-2} \left(1 + \frac{2 v_x}{v_\phi} + \frac{3 v_x ^2}{v_\phi ^2} + \frac{4 v_x ^3}{v_\phi ^3} + \ldots \right)\]

When taking the average over the distribution, the odd terms vanish. The average of \( f_0 v_x ^2 \) is just the thermal velocity \( v_t ^2 \):

\[\int _{-\infty} ^{\infty} f_0 v_x ^2 \dd v_x \equiv v_t ^2\] \[\rightarrow \frac{\omega_p ^2}{k} \sout{\int} _{-\infty} ^{\infty} \frac{\partial f_0 / \partial v_x}{\omega - k v_x} \dd v_x \approx - \frac{\omega_p ^2}{\omega^2} \left( 1 + \frac{3 v_t ^2}{v_\phi ^2} \right)\] \[D(k, \omega) = \underbrace{\left. 1 - \frac{\omega_{pe}^2}{\omega ^2} \right( 1}_{\text{CDPR}} \underbrace{\left. + \frac{3 v_t ^2}{v_\phi ^2} \right)}_{\text{Fluid Bohm-Gross}} \underbrace{\left. - \frac{i \pi \omega_{pe}^2}{k^2} \left.\pdv{f_0}{v_x} \right|_{v_x = \omega / k} \right.}_{\text{Kinetic description}} = 0\]

Landau Damping for Maxwellian Distribution #

Let \( f_0 \) be a Maxwellian distribution with thermal speed \( v_t = \sqrt{k T / m}\)

\[f_0 = \frac{1}{\sqrt{2 \pi} v_t} e^{- v_x ^2 / 2 v_t ^2}\] \[\pdv{f_0}{v_x} = \frac{1}{\sqrt{2 \pi} v_t} \left( - \frac{2 v_x}{2 v_t ^2} \right) e^{- v_x ^2 / 2 v_t ^2} = - \frac{v_x}{\sqrt{2 \pi} v_t ^3} e^{- v_x ^2 / 2 v_t ^2}\] \[\left. \pdv{f_0}{v_x} \right|_{v_x = \omega / k} = - \frac{\omega}{\sqrt{2\pi} k v_t ^3} e^{- \omega ^2 / 2 v_t ^2 k^2}\]

\[\rightarrow D(k, \omega) = 0 = 1 - \frac{\omega_p ^2}{\omega ^2} \left( 1 + \frac{3 k^2 v_t ^2}{\omega ^2} \right) + i \sqrt{\frac{\pi}{2}} \frac{\omega_{pe}^2 \omega}{k^3 v_t ^3} e^{-\omega ^2 / 2 k^2 v_t ^2}\]

For weakly damped electrostatic waves with \( \omega = \omega_r + i \gamma \) and \( \omega_r \approx \omega_p \),

\[\omega ^2 = \omega _r ^2 - \gamma ^2 + 2 i \gamma \omega _r\] \[\rightarrow 1 = \frac{\omega_p ^2}{\omega ^2} - \frac{\gamma ^2}{\omega ^2} + i \frac{2\gamma}{\omega_p} \\ \approx \frac{\omega_p ^2}{\omega ^2} + i \frac{2\gamma}{\omega_p} \]

If we compare this expression to the dispersion relation, we can pick out the Landau damping rate \( \gamma \):

\[\gamma = - \frac{1}{2} \sqrt{\frac{\pi}{2}} \frac{\omega_{pe}^4}{k^3 v_t ^3} e^{- \omega ^2 / 2 k^2 v_t ^2}\]

Since \( v_t / \omega_p = \lambda_D \) (the Debye length), then

\[\frac{\gamma}{\omega_p} = - \frac{1}{2} \sqrt{\frac{\pi}{2}} \frac{1}{(k \lambda_D)^3} e^{- \frac{1}{2 (k \lambda_D)^2}}\]

For small values of \( k \lambda_D \), we get small \( \gamma \) (weak damping). When \( \lambda \sim \lambda_D \), we get strong damping. However, recall that we made the assumption that \( v_\phi = \omega / k \gg v_t \), so large values of \( k \) break that assumption. Still, it suggests strong damping for large values of \( k \).

Ion Acoustic Waves and Ion Landau Damping #

When the electron and ion temperatures are comparable, there is generally no oscillation because when one moves some of the ions to form a slight imbalance which would induce the ions to oscillate, then the period is so slow that the electrons rush to fill the charge imbalance before the ions can respond. But, it is possible to consider the case where the ions are cold (\( v_i \ll v_p \)), but the electrons are hot (\( v_e \gg v_p \)). When the electron temperature is much higher than the ion temperature, the electrons have too much momentum to slow down and fill in the oscillations, so some ion oscillations may occur.

If we add the ion contribution to the governing equations:

\[\epsilon(\omega, k) = 1 + \sum_j \frac{\omega_{p,j} ^2}{k} \left[ \sout{\int} _{-\infty} ^{\infty} \frac{\partial f_{0, j} / \partial v_x}{\omega - k v_x} \dd v_x - \frac{i \pi}{k} \left. \pdv{f_{0,j}}{v_x} \right|_{v_x = \omega / k} \right]\]

Let’s consider the situation where:

  • \( v_\phi \ll v_{t, e} \)
  • \( v_\phi \gg v_{t, i} \)
  • \( k \lambda _D \ll 1\)
  • The initial distributions of the electrons and ions are Maxwellian

Under these assumptions, we can make some simplifications. Since \( v_\phi \ll v_{te} \), then for the electrons

\[\frac{1}{v_x - \omega / k} \approx \frac{1}{v_x}\]

across a large part of the distribution function, and

\[\pdv{f_{0e}}{v_x} \approx \left( \frac{1}{\sqrt{2 \pi} v_{te}} \exp \left[ - \frac{v_x ^2}{2 v_{te}^2} \right] \right) \left( - \frac{2 v_x}{2 v_{te}^2} \right)\] \[\sout{\int} _{-\infty} ^{\infty} \frac{\partial f_{0 e} / \partial v_x}{v_x - \omega / k} \dd v_x \approx - \frac{1}{\sqrt{2 \pi} v_{te}^3} \int _{-\infty} ^{\infty} e^{- v_x ^2 / 2 v_{te}^2} \dd v_x \]

Now, let \( X = v_x / \sqrt{2} v_t \), so that \( \dd X = \dd v_x / \sqrt{2} v_{te} \)

\[\sout{\int} _{-\infty} ^{\infty} \frac{\partial f_{0 e} / \partial v_x}{v_x - \omega / k} \dd v_x \approx - \frac{1}{\sqrt{\pi}} \frac{1}{v_{te} ^2} \left[ \int _{-\infty} ^{\infty} e^{-X ^2} \dd X \right] \\ = - \frac{1}{v_{te}^2}\]

Since \( v_\phi \gg v_{ti} \), then for the ions:

\[\frac{1}{v_x - \omega / k} = - \frac{1}{\frac{\omega}{k} - v_x} \\ = - \frac{k / \omega}{1 - \frac{v_x k}{\omega}}\\ = - \frac{k}{\omega} \left( 1 - \frac{v_x}{v_\phi} \right) ^{-1} \\ = - \frac{k}{\omega} \left( 1 + \frac{v_x}{v_\phi} + \ldots \right) \\ \approx - \frac{k}{\omega} - v_x \frac{k^2}{\omega ^2}\]

Them

\[\sout{\int} _{-\infty} ^{\infty} \frac{\partial f_{0i} / \partial v_x}{v_x - \omega / k} \approx \frac{k^2}{\omega ^2}\]

For a Maxwellian electron distribution with \( v_\phi / v_{te} \ll 1 \):

\[i \pi \frac{\omega_{pe}^2}{k^2} \left. \pdv{f_{0e}}{v_x} \right|_{v_x = \omega / k} = i \sqrt{\frac{\pi}{2}} \frac{\omega_{pe}^2 \omega}{k ^3 v_{te}^3} \exp \left[ - \frac{\omega ^2}{2 k^2 v_{te ^2}} \right] \\ \approx i \sqrt{\frac{\pi}{2}} \frac{\omega_{pe}^2 \omega}{k ^3 v_{te}^3}\]

And for Maxwellian ions:

\[i \pi \frac{\omega_{pi}^2}{k^2} \left. \pdv{f_{0i}}{v_x} \right|_{v_x = \omega / k} = i \sqrt{\frac{\pi}{2}} \frac{\omega_{pi}^2 \omega}{k^3 v_{ti}^3} \exp \left[ -\frac{\omega ^2}{2 k^2 v_{ti}^2} \right]\]

Putting it all together:

\[\epsilon(k, \omega) = 1 + \frac{\omega_{pe}^2}{k^2 v_{te}^2} - \frac{\omega_{pi}^2}{\omega ^2} + i \sqrt{\frac{\pi}{2}} \left[ \frac{\omega_{pe}^2 \omega}{k ^3 v_{te} ^3} + \frac{\omega_{pi}^2 \omega}{k^3 v_{ti}^3} \exp \left( - \frac{\omega^2}{2 k^2 v_{ti}^2} \right) \right] = 0\]

Assuming weak damping \( | \omega_i / \omega_r | \ll 1 \),

\[\text{Re}\left[ \epsilon (\omega, k) \right] = 1 + \frac{\omega_{pe}^2}{k^2 v_{te}^2} - \frac{\omega_{pi}^2}{\omega_r ^2} = 0\]

Since \( k \lambda_D \ll 1 \), \( k v_{te} / \omega_p \ll 1 \) so \( \frac{\omega_{pe}^2}{k^2 v_{te}^2} \gg 1 \)

\[\omega_r ^2 \approx \frac{\omega_{pi}^2}{\omega_{pe}^2} k^2 v_{te}^2 = k^2 \left( \frac{m_e}{m_i} \right) \left( \frac{k T_e}{m_e} \right) = k^2 c_s ^2\]

So \( \omega_r ^2 = k^2 c_s ^2 \), which is a linear dispersion relation with acoustic wave-like characteristics. Here, \( c_s \) is the ion acoustic speed for \( T_e \gg T_i \)

\[c_s ^2 \approx \frac{k T_e}{m_i}\]

The Landau damping rate is

\[\gamma_{Li} = - \text{Im}(\omega) = \frac{1}{2} \sqrt{\frac{\pi}{2}} k c_s \left[ \underbrace{\sqrt{\frac{m_e}{m_i}}}_{\text{Electron contribution}} + \underbrace{\left(\frac{T_e}{T_i} \right)^{3/2} e^{- T_e / 2 T_i}}_{\text{Ion contribution}} \right]\]

Ion Acoustic Instability #

If we suppose that the electrons have a drift speed \( v_e \) (this could correspond to a net electron current), then

\[f_{0e} = \frac{1}{\sqrt{2 \pi} v_{te}} \exp \left[ - \frac{(v_x - v_e)^2}{2 v_{te} ^2} \right]\] \[i \pi \frac{\omega_{pe}^2}{k^2} \left. \pdv{f_{0e}}{v_x} \right|_{v_x = \omega / k} \approx i \sqrt{\frac{\pi}{2}} \frac{\omega_{pe}^2(\omega - k v_e)}{k^3 v_{te}^3}\]

The electron contribution to \( \text{Im}[\omega] \) can switch sign from negative to positive, depending on the difference between \( v_\phi \) and \( v_e \). Setting the imaginary term in \( \epsilon(k, \omega) = 0 \) gives a threshold for the onset of ion acoustic turbulence (current-driven instability).