Categories: Physics, Plasma physics.

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 $$\phi_0$$ by definition 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. Note that we will implicitly use $$\vb{u}_{i0} = 0$$ to pretend that the material derivative $$\mathrm{D}/\mathrm{D} t$$ is linear:

\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 defined properties of the equilibrium components $$n_{i0}$$, $$n_{e0}$$, $$\vb{u}_{i0}$$ and $$\phi_0$$, and neglecting all products of perturbations for being small, this reduces to:

\begin{aligned} m_i n_{i0} \pdv{\vb{u}_{i1}}{t} &= - q_i n_{i0} \nabla \phi_1 - \gamma_i k_B T_i \nabla n_{i1} \\ 0 &= - 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 \quad \implies \quad 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}

Then 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} \quad \implies \quad \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, authors often ignore this fact, thereby making the so-called plasma approximation. We will not, and therefore 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 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 $$n_{i1}$$.

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.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.