Categories: Physics, Plasma physics.

Langmuir waves

In plasma physics, Langmuir waves are oscillations in the electron density, which may or may not propagate, depending on the temperature.

Assuming no magnetic field $$\vb{B} = 0$$, no ion motion $$\vb{u}_i = 0$$ (since $$m_i \gg m_e$$), and therefore no ion-electron momentum transfer, the two-fluid equations tell us that:

\begin{aligned} m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t} = q_e n_e \vb{E} - \nabla p_e \qquad \quad \pdv{n_e}{t} + \nabla \cdot (n_e \vb{u}_e) = 0 \end{aligned}

These are the electron momentum and continuity equations. We also need Gauss’ law:

\begin{aligned} \varepsilon_0 \nabla \cdot \vb{E} = q_e (n_e - n_i) \end{aligned}

We split $$n_e$$, $$\vb{u}_e$$ and $$\vb{E}$$ into a base component (subscript $$0$$) and a perturbation (subscript $$1$$):

\begin{aligned} n_e = n_{e0} + n_{e1} \qquad \quad \vb{u}_e = \vb{u}_{e0} + \vb{u}_{e1} \qquad \quad \vb{E} = \vb{E}_0 + \vb{E}_1 \end{aligned}

Where the perturbations $$n_{e1}$$, $$\vb{u}_{e1}$$ and $$\vb{E}_1$$ are very small, and the equilibrium components $$n_{e0}$$, $$\vb{u}_{e0}$$ and $$\vb{E}_0$$ by definition satisfy:

\begin{aligned} \pdv{n_{e0}}{t} = 0 \qquad \pdv{\vb{u}_{e0}}{t} = 0 \qquad \nabla n_{e0} = 0 \qquad \vb{u}_{e0} = 0 \qquad \vb{E}_0 = 0 \end{aligned}

We insert this decomposistion into the electron continuity equation, arguing that $$n_{e1} \vb{u}_{e1}$$ is small enough to neglect, leading to:

\begin{aligned} 0 &= \pdv{(n_{e0} \!+\! n_{e1})}{t} + \nabla \cdot \Big( (n_{e0} \!+\! n_{e1}) \: (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \Big) \\ &= \pdv{n_{e1}}{t} + \nabla \cdot \Big( n_{e0} \vb{u}_{e1} + n_{e1} \vb{u}_{e1} \Big) \\ &\approx \pdv{n_{e1}}{t} + \nabla \cdot (n_{e0} \vb{u}_{e1}) = \pdv{n_{e1}}{t} + n_{e0} \nabla \cdot \vb{u}_{e1} \end{aligned}

Likewise, we insert it into Gauss’ law, and use the plasma’s quasi-neutrality $$n_i = n_{e0}$$ to get:

\begin{aligned} \varepsilon_0 \nabla \cdot \big( \vb{E}_0 \!+\! \vb{E}_1 \big) = q_e (n_{e0} + n_{e1} - n_i) \quad \implies \quad \varepsilon_0 \nabla \cdot \vb{E}_1 = q_e n_{e1} \end{aligned}

Since we are looking for linear waves, we make the following ansatz for the perturbations:

\begin{aligned} n_{e1}(\vb{r}, t) &= n_{e1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t) \\ \vb{u}_{e1}(\vb{r}, t) &= \vb{u}_{e1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t) \\ \vb{E}_1(\vb{r}, t) &= \vb{E}_1 \:\exp\!(i \vb{k} \cdot \vb{r} - i \omega t) \end{aligned}

Inserting this into the continuity equation and Gauss’ law yields, respectively:

\begin{aligned} - i \omega n_{e1} = - i n_{e0} \vb{k} \cdot \vb{u}_{e1} \qquad \quad -\! i \varepsilon_0 \vb{k} \cdot \vb{E}_1 = q_e n_{e1} \end{aligned}

However, there are three unknowns $$n_{e1}$$, $$\vb{u}_{e1}$$ and $$\vb{E}_1$$, so one more equation is needed.

Cold Langmuir waves

We therefore turn to the electron momentum equation. For now, let us assume that the electrons have no thermal motion, i.e. the electron temperature $$T_e = 0$$, so that $$p_e = 0$$, leaving:

\begin{aligned} m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t} = q_e n_e \vb{E} \end{aligned}

Inserting the decomposition then gives the following, where we neglect $$(\vb{u}_{e1} \cdot \nabla) \vb{u}_{e1}$$ because $$\vb{u}_{e1}$$ is so small by assumption:

$\begin{gathered} m_e (n_{e0} \!+\! n_{e1}) \Big( \pdv{(\vb{u}_{e0} \!+\! \vb{u}_{e1})}{t} + \big( (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \cdot \nabla \big) (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \Big) = q_e \big( n_{e0} \!+\! n_{e1} \big) \big( \vb{E}_0 \!+\! \vb{E}_1 \big) \\ \implies \qquad q_e \vb{E}_1 = m_e \Big( \pdv{\vb{u}_{e1}}{t} + \big(\vb{u}_{e1} \cdot \nabla \big) \vb{u}_{e1} \Big) \approx m_e \pdv{\vb{u}_{e1}}{t} \end{gathered}$

And then inserting our plane-wave ansatz yields the third equation we were looking for:

\begin{aligned} -i \omega m_e \vb{u}_{e1} = q_e \vb{E}_1 \end{aligned}

Solving this system of three equations for $$\omega^2$$ gives the following dispersion relation:

\begin{aligned} \omega^2 = \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1} = \frac{i \omega n_{e0} q_e}{\omega m_e n_{e1}} \vb{k} \cdot \vb{E}_1 = \frac{i n_{e0} n_{e1} q_e^2}{i \varepsilon_0 m_e n_{e1}} = \frac{n_{e0} q_e^2}{\varepsilon_0 m_e} \end{aligned}

This result is known as the plasma frequency $$\omega_p$$, and describes the frequency of cold Langmuir waves, otherwise known as plasma oscillations:

\begin{aligned} \boxed{ \omega_p = \sqrt{\frac{n_{0e} q_e^2}{\varepsilon_0 m_e}} } \end{aligned}

Note that this is a dispersion relation $$\omega(k) = \omega_p$$, but that $$\omega_p$$ does not contain $$k$$. This means that cold Langmuir waves do not propagate: the oscillation is “stationary”.

Warm Langmuir waves

Next, we generalize this result to nonzero $$T_e$$, in which case the pressure $$p_e$$ is involved:

\begin{aligned} m_e n_{e0} \pdv{\vb{u}_{e1}}{t} = q_e n_{e0} \vb{E}_1 - \nabla p_e \end{aligned}

From the two-fluid thermodynamic equation of state, we know that $$\nabla p_e$$ can be written as:

\begin{aligned} \nabla p_e = \gamma k_B T_e \nabla n_e = \gamma k_B T_e \nabla (n_{e0} + n_{e1}) = \gamma k_B T_e \nabla n_{e1} \end{aligned}

With this, insertion of our plane-wave ansatz into the electron equation results in:

\begin{aligned} -i \omega m_e n_{e0} \vb{u}_{e1} = q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k} \end{aligned}

Which once again closes the system of three equations. Solving for $$\omega^2$$ then gives:

\begin{aligned} \omega^2 = \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1} &= \frac{i \omega n_{e0}}{\omega n_{e0} m_e n_{e1}} \vb{k} \cdot \Big( q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k} \Big) \\ &= \frac{n_{e0} q_e^2}{\varepsilon_0 m_e} - \frac{i \omega}{\omega m_e n_{e1}} i \gamma k_B T_e n_{e1} \big(\vb{k} \cdot \vb{k}\big) \end{aligned}

Recognizing the first term as the plasma frequency $$\omega_p^2$$, we therefore arrive at the Bohm-Gross dispersion relation $$\omega(\vb{k})$$ for warm Langmuir waves:

\begin{aligned} \boxed{ \omega^2 = \omega_p^2 + \frac{\gamma k_B T_e}{m_e} |\vb{k}|^2 } \end{aligned}

This expression is typically quoted for 1D oscillations, in which case $$\gamma = 3$$ and $$k = |\vb{k}|$$:

\begin{aligned} \omega^2 = \omega_p^2 + \frac{3 k_B T_e}{m_e} k^2 \end{aligned}

Unlike for $$T_e = 0$$, these “warm” waves do propagate, carrying information at group velocity $$v_g$$, which, in the limit of large $$k$$, is given by:

\begin{aligned} v_g = \pdv{\omega}{k} \to \sqrt{\frac{3 k_B T_e}{m_e}} \end{aligned}

This is the root-mean-square velocity of the Maxwell-Boltzmann speed distribution, meaning that information travels at the thermal velocity for large $$k$$.

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.