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+---
+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.