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 modelâ€™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 electronsâ€™ heat conductivity can be regarded as infinite compared to the ionsâ€™. 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 Gaussâ€™ law. Nevertheless, some authors tend to ignore this fact, thereby making the so-called plasma approximation. We will not, and thus turn to Gaussâ€™ 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.