---
title: "Langmuir waves"
sort_title: "Langmuir waves"
date: 2021-10-30
categories:
- Physics
- Plasma physics
- Plasma waves
- Perturbation
layout: "concept"
---

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](/know/concept/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](/know/concept/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](/know/concept/maxwells-equations/):

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

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](/know/concept/maxwell-boltzmann-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.