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 $$k$$.

## 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.

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