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Categories: Perturbation, Physics, Plasma physics, Plasma waves.

Ion sound wave

In a plasma, electromagnetic interactions allow compressional longitudinal waves to propagate at lower temperatures and pressures than would be possible in a neutral gas.

We start from the two-fluid model’s momentum equations, rewriting the electric field $\vb{E} = - \nabla \phi$ and the pressure gradient $\nabla p = \gamma k_B T \nabla n$, and arguing that $m_e \approx 0$ because $m_e \ll m_i$:

$$\begin{aligned} m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t} &= - q_i n_i \nabla \phi - \gamma_i k_B T_i \nabla n_i \\ 0 &= - q_e n_e \nabla \phi - \gamma_e k_B T_e \nabla n_e \end{aligned}$$

Note that we neglect ion-electron collisions, and allow for separate values of $\gamma$. We split $n_i$, $n_e$, $\vb{u}_i$ and $\phi$ into an equilibrium (subscript $0$) and a perturbation (subscript $1$):

$$\begin{aligned} n_i = n_{i0} + n_{i1} \qquad n_e = n_{e0} + n_{e1} \qquad \vb{u}_i = \vb{u}_{i0} + \vb{u}_{i1} \qquad \phi = \phi_0 + \phi_1 \end{aligned}$$

Where the perturbations $n_{i1}$, $n_{e1}$, $\vb{u}_{i1}$ and $\phi_1$ are tiny, and the equilibrium components $n_{i0}$, $n_{e0}$, $\vb{u}_{i0}$ and $\phi_0$ are assumed to satisfy:

$$\begin{aligned} \pdv{n_{i0}}{t} = 0 \qquad \frac{\mathrm{D} \vb{u}_{i0}}{\mathrm{D} t} = 0 \qquad \nabla n_{i0} = \nabla n_{e0} = 0 \qquad \vb{u}_{i0} = 0 \qquad \phi_0 = 0 \end{aligned}$$

Inserting this decomposition into the momentum equations yields new equations:

$$\begin{aligned} m_i (n_{i0} \!+\! n_{i1}) \frac{\mathrm{D} (\vb{u}_{i0} \!+\! \vb{u}_{i1})}{\mathrm{D} t} &= - q_i (n_{i0} \!+\! n_{i1}) \nabla (\phi_0 \!+\! \phi_1) - \gamma_i k_B T_i \nabla (n_{i0} \!+\! n_{i1}) \\ 0 &= - q_e (n_{e0} \!+\! n_{e1}) \nabla (\phi_0 \!+\! \phi_1) - \gamma_e k_B T_e \nabla (n_{e0} \!+\! n_{e1}) \end{aligned}$$

Using the assumed properties of $n_{i0}$, $n_{e0}$, $\vb{u}_{i0}$ and $\phi_0$, and discarding products of perturbations for being too small, we arrive at the below equations. Our choice $\vb{u}_{i0} = 0$ lets us linearize the material derivative $\mathrm{D}/\mathrm{D} t = \ipdv{}{t}$ for the ions:

$$\begin{aligned} m_i n_{i0} \pdv{\vb{u}_{i1}}{t} &\approx - q_i n_{i0} \nabla \phi_1 - \gamma_i k_B T_i \nabla n_{i1} \\ 0 &\approx - q_e n_{e0} \nabla \phi_1 - \gamma_e k_B T_e \nabla n_{e1} \end{aligned}$$

Because we are interested in linear waves, we make the following plane-wave ansatz:

$$\begin{aligned} n_{i1}(\vb{r}, t) &= n_{i1} \exp(i \vb{k} \cdot \vb{r} - i \omega t) \\ n_{e1}(\vb{r}, t) &= n_{e1} \exp(i \vb{k} \cdot \vb{r} - i \omega t) \\ \vb{u}_{i1}(\vb{r}, t) &= \vb{u}_{i1} \exp(i \vb{k} \cdot \vb{r} - i \omega t) \\ \phi_1(\vb{r}, t) &= \phi_1 \,\,\exp(i \vb{k} \cdot \vb{r} - i \omega t) \end{aligned}$$

Which we then insert into the momentum equations for the ions and electrons:

$$\begin{aligned} - i \omega m_i n_{i0} \vb{u}_{i1} &= - i \vb{k} q_i n_{i0} \phi_1 - i \vb{k} \gamma_i k_B T_i n_{i1} \\ 0 &= - i \vb{k} q_e n_{e0} \phi_1 - i \vb{k} \gamma_e k_B T_e n_{e1} \end{aligned}$$

The electron equation can easily be rearranged to get a relation between $n_{e1}$ and $n_{e0}$:

$$\begin{aligned} i \vb{k} \gamma_e k_B T_e n_{e1} = - i \vb{k} q_e n_{e0} \phi_1 \qquad \implies \qquad n_{e1} = - \frac{q_e \phi_1}{\gamma_e k_B T_e} n_{e0} \end{aligned}$$

Due to their low mass, the electrons’ heat conductivity can be regarded as infinite compared to the ions’. In that case, all electron gas compression is isothermal, meaning it obeys the ideal gas law $p_e = n_e k_B T_e$, so that $\gamma_e = 1$. Note that this yields the first-order term of a Taylor expansion of the Boltzmann relation.

At equilibrium, quasi-neutrality demands that $n_{i0} = n_{e0} = n_0$, so we can rearrange the above relation to $n_0 = - k_B T_e n_{e1} / (q_e \phi_1)$, which we insert into the ion equation to get:

$$\begin{gathered} i \omega m_i \frac{k_B T_e n_{e1}}{q_e \phi_1} \vb{u}_{i1} = - i q_i \frac{k_B T_e n_{e1}}{q_e \phi_1} \phi_1 \vb{k} - i \gamma_i k_B T_i n_{i1} \vb{k} \\ \implies \qquad \omega m_i \frac{T_e n_{e1}}{q_e \phi_1} \vb{k} \cdot \vb{u}_{i1} = T_e n_{e1} |\vb{k}|^2 - \gamma_i T_i n_{i1} |\vb{k}|^2 \end{gathered}$$

Where we have taken the dot product with $\vb{k}$, and used that $q_i / q_e = -1$. In order to simplify this equation, we turn to the two-fluid ion continuity relation:

$$\begin{aligned} 0 &= \pdv{(n_{i0} \!+\! n_{i1})}{t} + \nabla \cdot \Big( (n_{i0} \!+\! n_{i1}) (\vb{u}_{i0} \!+\! \vb{u}_{i1}) \Big) \approx \pdv{n_{i1}}{t} + n_{i0} \nabla \cdot \vb{u}_{i1} \end{aligned}$$

Into which we insert our plane-wave ansatz, and substitute $n_{i0} = n_0$ as before, yielding:

$$\begin{aligned} 0 = - i \omega n_{i1} + i n_{i0} \vb{k} \cdot \vb{u}_{i1} \qquad \implies \qquad \vb{k} \cdot \vb{u}_{i1} = \omega \frac{n_{i1}}{n_{i0}} = \omega \frac{q_e n_{i1} \phi_1}{k_B T_e n_{e1}} \end{aligned}$$

Substituting this in the ion momentum equation leads us to a dispersion relation $\omega(\vb{k})$:

$$\begin{gathered} \omega^2 m_i \frac{T_e n_{e1}}{q_e \phi_1} \frac{q_e n_{i1} \phi_1}{k_B T_e n_{e1}} = \omega^2 m_i \frac{n_{i1}}{k_B} = |\vb{k}|^2 \big( T_e n_{e1} - \gamma_i T_i n_{i1} \big) \\ \implies \qquad \omega^2 = \frac{|\vb{k}|^2}{m_i} \Big( k_B T_e \frac{n_{e1}}{n_{i1}} - \gamma_i k_B T_i \Big) \end{gathered}$$

Finally, we would like to find an expression for $n_{e1} / n_{i1}$. It cannot be $1$, because then $\phi_1$ could not be nonzero, according to Gauss’ law. Nevertheless, some authors tend to ignore this fact, thereby making the so-called plasma approximation. We will not, and thus turn to Gauss’ law:

$$\begin{aligned} \varepsilon_0 \nabla \cdot \vb{E} = - \varepsilon_0 \nabla^2 \phi_1 = q_i n_i - q_e n_e = - q_e (n_{i1} - n_{e1}) \end{aligned}$$

One final time, we insert our plane-wave ansatz, and use our Boltzmann-like relation between $n_{e1}$ and $n_{e0}$ to substitute $\phi_1 = - k_B T_e n_{e1} / (q_e n_{e0})$:

$$\begin{gathered} q_e (n_{e1} - n_{i1}) = |\vb{k}|^2 \varepsilon_0 \phi_1 = - |\vb{k}|^2 \varepsilon_0 \frac{k_B T_e n_{e1}}{q_e n_{e0}} \\ \implies \qquad n_{i1} = n_{e1} + |\vb{k}|^2 \varepsilon_0 \frac{k_B T_e n_{e1}}{q_e^2 n_{e0}} = n_{e1} \big( 1 + |\vb{k}|^2 \lambda_{De}^2 \big) \end{gathered}$$

Where $\lambda_{De}$ is the electron Debye length. We thus reach the following dispersion relation, which governs ion sound waves or ion acoustic waves:

$$\begin{aligned} \boxed{ \omega^2 = \frac{|\vb{k}|^2}{m_i} \bigg( \frac{k_B T_e}{1 + |\vb{k}|^2 \lambda_{De}^2} + \gamma_i k_B T_i \bigg) } \end{aligned}$$

The aforementioned plasma approximation is valid if $|\vb{k}| \lambda_{De} \ll 1$, which is often reasonable, in which case this dispersion relation reduces to:

$$\begin{aligned} \omega^2 = \frac{|\vb{k}|^2}{m_i} \bigg( k_B T_e + \gamma_i k_B T_i \bigg) \end{aligned}$$

The phase velocity $v_s$ of these waves, i.e. the speed of sound, is then given by:

$$\begin{aligned} \boxed{ v_s = \frac{\omega}{k} = \sqrt{\frac{k_B T_e}{m_i} + \frac{\gamma_i k_B T_i}{m_i}} } \end{aligned}$$

Curiously, unlike in a neutral gas, this velocity is nonzero even if $T_i = 0$, meaning that the waves still exist then. In fact, usually the electron temperature $T_e$ dominates $T_e \gg T_i$, even though the main feature of these waves is that they involve ion density fluctuations $n_{i1}$.

References

1. F.F. Chen, Introduction to plasma physics and controlled fusion, 3rd edition, Springer.
2. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

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