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Chapter 5 • Theory & Derivations

Electrostatic Waves in Unmagnetized and Magnetized Plasmas

Comprehensive theory of electrostatic plasma oscillations and waves: harmonic representations, phase and group velocity, perturbation linearization of the multi-fluid equations, cold electron plasma oscillations, thermal electron pressure and the complete Bohm-Gross dispersion relation derivation, ion acoustic waves (plasma sound waves), ion sound speed, short-wavelength electron screening breakdown at k lambda_D ~ 1, comprehensive comparison between electron and ion modes, and electrostatic waves in magnetized plasmas including Upper Hybrid, Lower Hybrid, and electrostatic ion cyclotron (EIC) waves.

§5.1 Harmonic Wave Representations, Phase Velocity & Group Velocity

1. Complex Harmonic Wave Representations

Small-amplitude perturbations in plasma fluid quantities (density, velocity, electric and magnetic fields) are expanded as superpositions of plane waves:

$$\psi(\vec{r}, t) = \psi_0 + \psi_1 \exp\left[ i(\vec{k}\cdot\vec{r} - \omega t) \right]$$

where $\psi_0$ is the unperturbed background equilibrium state, $\psi_1$ is the first-order perturbation amplitude ($|\psi_1| \ll |\psi_0|$), $\vec{k}$ is the wavevector, and $\omega$ is the angular frequency. Under this harmonic convention, differential operators transform into algebraic multipliers:

$$\nabla \to i\vec{k}, \quad \frac{\partial}{\partial t} \to -i\omega$$

2. Phase Velocity vs Group Velocity

The speed at which surfaces of constant wave phase propagate is the phase velocity $\vec{v}_{ph}$:

$$\vec{v}_{ph} = \frac{\omega}{k} \hat{k}$$

The speed and direction at which a modulated wave packet—and consequently physical energy and information—transmits through the plasma is the group velocity $\vec{v}_g$:

$$\vec{v}_g = \nabla_k \omega = \frac{\partial \omega}{\partial k} \hat{k}$$

In dispersive plasma media where $\omega(k)$ is non-linear, $v_{ph} \ne v_g$.

§5.2 Linearization of Multi-Fluid Equations & Electrostatic Perturbations

1. The Linearization Procedure

For electrostatic waves, the magnetic field perturbation vanishes ($\vec{B}_1 = 0$), so the electric field is purely curl-free and derived from an electrostatic potential: $\vec{E}_1 = -\nabla \phi_1 = -i\vec{k}\phi_1$.

Each fluid variable is decomposed into equilibrium plus first-order perturbation:

$$n_\alpha = n_0 + n_{\alpha 1}, \quad \vec{u}_\alpha = 0 + \vec{u}_{\alpha 1}, \quad \vec{E} = 0 + \vec{E}_1$$

Neglecting second-order nonlinear terms ($n_1 \vec{u}_1 \approx 0$ and $(\vec{u}_1\cdot\nabla)\vec{u}_1 \approx 0$):

  • Linearized Continuity: $$-i\omega n_{\alpha 1} + i n_0 \vec{k}\cdot\vec{u}_{\alpha 1} = 0 \implies n_{\alpha 1} = n_0 \frac{\vec{k}\cdot\vec{u}_{\alpha 1}}{\omega}$$
  • Linearized Momentum: $$-i\omega m_\alpha n_0 \vec{u}_{\alpha 1} = q_\alpha n_0 \vec{E}_1 - \gamma_\alpha k_B T_\alpha (i\vec{k} n_{\alpha 1})$$
  • Poisson's Equation: $$i\vec{k}\cdot\vec{E}_1 = \frac{e(n_{i1} - n_{e1})}{\varepsilon_0}$$

§5.3 Cold Electron Plasma Oscillations (Langmuir Waves) & Plasma Cutoff

1. The Cold Plasma Limit

In the cold plasma limit ($T_e = 0$), electron pressure vanishes. Because the oscillation frequency is high, massive ions cannot respond ($n_{i1} = 0, \vec{u}_{i1} = 0$). The linearized electron momentum equation reduces to:

$$-i\omega m_e \vec{u}_{e1} = -e \vec{E}_1 \implies \vec{u}_{e1} = \frac{e}{i\omega m_e}\vec{E}_1$$

Substitute $\vec{u}_{e1}$ into continuity:

$$n_{e1} = n_0 \frac{\vec{k}\cdot\vec{u}_{e1}}{\omega} = \frac{n_0 e}{i\omega^2 m_e}\vec{k}\cdot\vec{E}_1$$

Substitute $n_{e1}$ into Poisson's equation $i\vec{k}\cdot\vec{E}_1 = -\frac{e n_{e1}}{\varepsilon_0}$:

$$i\vec{k}\cdot\vec{E}_1 = -\frac{e}{\varepsilon_0}\left( \frac{n_0 e}{i\omega^2 m_e}\vec{k}\cdot\vec{E}_1 \right) = \frac{n_0 e^2}{\varepsilon_0 m_e \omega^2} (i\vec{k}\cdot\vec{E}_1)$$

For a non-trivial wave solution ($\vec{k}\cdot\vec{E}_1 \ne 0$):

$$1 - \frac{n_0 e^2}{\varepsilon_0 m_e \omega^2} = 0 \implies \omega^2 = \frac{n_0 e^2}{\varepsilon_0 m_e} \equiv \omega_{pe}^2$$

In a cold plasma, electron oscillations occur at the single constant frequency $\omega = \omega_{pe}$, independent of wavevector $k$. Consequently:

$$v_g = \frac{d\omega}{dk} = 0$$

Cold electron plasma oscillations do not propagate; they are purely local stationary oscillations!

§5.4 Thermal Electron Pressure & Derivation of the Bohm-Gross Dispersion Relation

1. Thermal Pressure Correction

When electron temperature is non-zero ($T_e > 0$), thermal pressure $\nabla P_{e1} = \gamma_e k_B T_e \nabla n_{e1}$ provides an additional restoring force. For one-dimensional high-frequency compressions along $\vec{k}$, there is only one translational degree of freedom ($d=1$), so the adiabatic index is $\gamma_e = (d+2)/d = 3$.

The linearized electron momentum equation becomes:

$$-i\omega m_e n_0 u_{e1} = -e n_0 E_1 - 3 k_B T_e (i k n_{e1})$$

Using continuity $n_{e1} = n_0 \frac{k u_{e1}}{\omega}$:

$$-i\omega m_e u_{e1} = -e E_1 - 3 k_B T_e i k \left( \frac{k u_{e1}}{\omega} \right) \implies u_{e1}\left( -i\omega + \frac{3 k_B T_e i k^2}{\omega m_e} \right) = -\frac{e}{m_e} E_1$$ $$u_{e1} = \frac{e E_1}{i m_e \omega} \left( 1 - \frac{3 k^2 v_{\text{th},e}^2}{\omega^2} \right)^{-1}$$

where $v_{\text{th},e} = \sqrt{k_B T_e / m_e}$ is the electron thermal speed.

2. The Bohm-Gross Dispersion Relation

Substituting $n_{e1}$ into Poisson's equation $i k E_1 = -\frac{e n_{e1}}{\varepsilon_0} = -\frac{e n_0 k u_{e1}}{\varepsilon_0 \omega}$:

$$1 - \frac{\omega_{pe}^2}{\omega^2 - 3 k^2 v_{\text{th},e}^2} = 0 \implies \omega^2 = \omega_{pe}^2 + 3 k^2 v_{\text{th},e}^2$$

This is the famous Bohm-Gross dispersion relation for warm electron plasma waves (Langmuir waves).

Differentiating with respect to $k$ yields a non-zero group velocity:

$$2\omega \frac{d\omega}{dk} = 6 k v_{\text{th},e}^2 \implies v_g = \frac{3 k v_{\text{th},e}^2}{\omega} = \frac{3 v_{\text{th},e}^2}{v_{ph}}$$

Thermal pressure allows electron plasma waves to propagate and carry energy across the plasma!

§5.5 Ion Acoustic Waves (Plasma Sound Waves): Derivation & Sound Speed c_s

1. Low-Frequency Dynamics

At low frequencies ($\omega \ll \omega_{pe}$), electrons move so rapidly compared to the wave that they remain in continuous thermodynamic equilibrium, establishing a Boltzmann distribution in the wave potential $\phi_1$:

$$n_{e1} = n_0 \frac{e \phi_1}{k_B T_e}$$

The heavy ions, however, are accelerated dynamically by the wave electric field $E_1 = -ik\phi_1$. Linearized ion continuity and momentum equations with ion temperature $T_i$:

$$-i\omega n_{i1} + i n_0 k u_{i1} = 0 \implies n_{i1} = n_0 \frac{k u_{i1}}{\omega}$$ $$-i\omega M_i n_0 u_{i1} = e n_0 (-i k \phi_1) - \gamma_i k_B T_i (i k n_{i1})$$

Solving for ion density perturbation $n_{i1}$:

$$n_{i1} = \frac{n_0 e k^2}{M_i (\omega^2 - \gamma_i k^2 v_{\text{th},i}^2)} \phi_1$$

2. The Ion Acoustic Dispersion Relation

Substituting $n_{e1}$ and $n_{i1}$ into Poisson's equation $k^2 \phi_1 = \frac{e(n_{i1} - n_{e1})}{\varepsilon_0}$:

$$k^2 \phi_1 = \frac{e}{\varepsilon_0}\left[ \frac{n_0 e k^2}{M_i (\omega^2 - \gamma_i k^2 v_{\text{th},i}^2)}\phi_1 - \frac{n_0 e \phi_1}{k_B T_e} \right]$$

Dividing by $\phi_1$ and using $\lambda_{De}^2 = \frac{\varepsilon_0 k_B T_e}{n_0 e^2}$ and $\omega_{pi}^2 = \frac{n_0 e^2}{\varepsilon_0 M_i}$:

$$k^2 + \frac{1}{\lambda_{De}^2} = \frac{\omega_{pi}^2 k^2}{\omega^2 - \gamma_i k^2 v_{\text{th},i}^2}$$

For cold ions ($T_i \ll T_e$):

$$\omega^2 = \frac{k^2 \omega_{pi}^2 \lambda_{De}^2}{1 + k^2 \lambda_{De}^2}$$

Noting that $\omega_{pi} \lambda_{De} = \sqrt{\frac{n_0 e^2}{\varepsilon_0 M_i}}\sqrt{\frac{\varepsilon_0 k_B T_e}{n_0 e^2}} = \sqrt{\frac{k_B T_e}{M_i}} \equiv c_s$, where $c_s$ is the ion sound speed:

$$\omega^2 = \frac{k^2 c_s^2}{1 + k^2 \lambda_{De}^2}$$

Including finite ion temperature ($T_i > 0$ with 1D adiabatic compression $\gamma_i = 3$):

$$c_s = \sqrt{\frac{k_B T_e + 3 k_B T_i}{M_i}}$$

§5.6 Acoustic-to-Ion-Plasma Transition (k lambda_D ~ 1) & Electron vs Ion Waves

1. Limiting Regimes of Ion Acoustic Waves

The ion acoustic dispersion relation exhibits two fundamentally distinct physical behaviors depending on wavelength relative to the Debye length:

  1. Long-Wavelength Acoustic Limit ($k \lambda_{De} \ll 1$, $\lambda \gg \lambda_{De}$): $$\omega \approx k c_s, \quad v_{ph} = v_g = c_s = \text{const}$$ The wave is non-dispersive and behaves identically to an ordinary acoustic sound wave in neutral gas. In this limit, electrons perfectly shield ion charge fluctuations, maintaining quasi-neutrality ($n_{e1} \approx n_{i1}$).
  2. Short-Wavelength Ion Plasma Limit ($k \lambda_{De} \gg 1$, $\lambda \ll \lambda_{De}$): $$\omega \approx \frac{k c_s}{k \lambda_{De}} = \frac{c_s}{\lambda_{De}} = \omega_{pi}$$ At wavelengths shorter than the Debye length, electron Debye shielding breaks down completely. The electrons can no longer cluster to screen the ions, and the wave degenerates into constant-frequency ion plasma oscillations at $\omega = \omega_{pi}$, with zero group velocity ($v_g \to 0$).

2. Fundamental Comparison: Electron Waves vs Ion Waves

Feature Electron Plasma Wave (Langmuir) Ion Acoustic Wave (Sound)
Restoring Force Electric field + Electron thermal pressure Electron thermal pressure ($T_e$)
Inertia Electron mass $m_e$ Ion mass $M_i$
Characteristic Frequency High ($\omega \ge \omega_{pe} \sim 10^{11}\text{ rad/s}$) Low ($\omega \le \omega_{pi} \sim 10^9\text{ rad/s}$)
Ion Motion Stationary neutralizing background Oscillating fluid elements

§5.7 Electrostatic Waves in Magnetized Plasmas: Upper Hybrid, Lower Hybrid & EIC Waves

1. Waves Propagating Perpendicular to B0: Upper Hybrid Oscillations

When a static magnetic field $\vec{B}_0 = B_0 \hat{z}$ is present, consider electrostatic electron waves propagating perpendicular to the field ($\vec{k} = k \hat{x} \perp \vec{B}_0$).

The electrons experience two restoring forces simultaneously:

  • The electrostatic space-charge restoring force ($-\omega_{pe}^2$).
  • The magnetic Lorentz force restoring gyration ($-\omega_{ce}^2$).

Solving the linearized electron equations of motion yields the Upper Hybrid frequency $\omega_{UH}$:

$$\omega^2 = \omega_{UH}^2 = \omega_{pe}^2 + \omega_{ce}^2$$

Including electron thermal pressure: $\omega^2 = \omega_{UH}^2 + 3 k^2 v_{\text{th},e}^2$.

2. Lower Hybrid Oscillations

When both electron and ion motions are included for perpendicular propagation, an intermediate resonance occurs between the electron and ion cyclotron frequencies, termed the Lower Hybrid frequency $\omega_{LH}$:

$$\frac{1}{\omega_{LH}^2} = \frac{1}{\omega_{pi}^2 + \omega_{ci}^2} + \frac{1}{\omega_{ce}\omega_{ci}} \implies \omega_{LH} \approx \sqrt{\omega_{ce}\omega_{ci}}$$

3. Electrostatic Ion Cyclotron (EIC) Waves

For waves propagating at an oblique angle nearly perpendicular to $\vec{B}_0$ ($k_\perp \gg k_\parallel$) with frequencies near the ion cyclotron frequency $\omega \sim \omega_{ci}$:

$$\omega^2 = \omega_{ci}^2 + k_\perp^2 c_s^2$$

These electrostatic ion cyclotron (EIC) waves are driven by parallel electron currents and play a major role in auroral ion heating and tokamak scrape-off layers.

Honors Examination Worked Problems & Solutions

Rigorous step-by-step mathematical proofs and solutions to university degree examination problems.

Advanced Honors Exam Problem Example 5.1: Complete Linearized Fluid Derivation of the Bohm-Gross Dispersion Relation with 1D Adiabatic Pressure

Derive the Bohm-Gross dispersion relation for high-frequency electron plasma waves from the linearized 1D multi-fluid equations. (a) Write down the linearized continuity, momentum, and Poisson equations for 1D perturbations along $\hat{x}$ with isothermal/adiabatic index $\gamma_e = 3$. (b) Eliminate $u_{e1}$ and $n_{e1}$ to derive the algebraic wave equation in terms of $E_1$. (c) Deduce the Bohm-Gross dispersion relation $\omega^2 = \omega_{pe}^2 + 3 k^2 v_{\text{th},e}^2$, calculate the phase velocity $v_{ph}$ and group velocity $v_g$, and show that $v_{ph} v_g = 3 v_{\text{th},e}^2$ in the limit $k\lambda_D \ll 1$.

Full Rigorous Analytical Solution

(a) Linearized 1D Fluid Equations: Consider 1D perturbations $\propto e^{i(kx - \omega t)}$ in unmagnetized plasma with stationary ions:

  1. Linearized electron continuity:
$$\frac{\partial n_{e1}}{\partial t} + n_0 \frac{\partial u_{e1}}{\partial x} = 0 \implies -i\omega n_{e1} + i k n_0 u_{e1} = 0 \implies n_{e1} = n_0 \frac{k u_{e1}}{\omega}$$
  1. Linearized electron momentum with 1D adiabatic pressure $\nabla P_{e1} = 3 k_B T_e \nabla n_{e1}$:
$$m_e n_0 \frac{\partial u_{e1}}{\partial t} = -e n_0 E_1 - \frac{\partial P_{e1}}{\partial x} = -e n_0 E_1 - 3 k_B T_e \frac{\partial n_{e1}}{\partial x}$$
$$-i\omega m_e n_0 u_{e1} = -e n_0 E_1 - 3 k_B T_e (i k n_{e1})$$
  1. Poisson's equation:
$$\varepsilon_0 \frac{\partial E_1}{\partial x} = -e n_{e1} \implies i k \varepsilon_0 E_1 = -e n_{e1}$$

(b) Elimination & Wave Equation: Substitute $n_{e1}$ from continuity into momentum:

$$-i\omega m_e n_0 u_{e1} = -e n_0 E_1 - 3 k_B T_e i k \left( n_0 \frac{k u_{e1}}{\omega} \right) = -e n_0 E_1 - \frac{3 k_B T_e n_0 k^2}{\omega} i u_{e1}$$

Rearrange to group $u_{e1}$ terms:

$$-i u_{e1}\left( m_e n_0 \omega - \frac{3 k_B T_e n_0 k^2}{\omega} \right) = -e n_0 E_1$$

Multiply by $\omega / n_0$:

$$-i u_{e1}\left( m_e \omega^2 - 3 k_B T_e k^2 \right) = -e \omega E_1 \implies u_{e1} = \frac{-i e \omega E_1}{m_e \omega^2 - 3 k_B T_e k^2}$$

Now express $n_{e1}$ in terms of $E_1$:

$$n_{e1} = n_0 \frac{k u_{e1}}{\omega} = -\frac{i n_0 e k E_1}{m_e \omega^2 - 3 k_B T_e k^2}$$

Substitute $n_{e1}$ into Poisson's equation $i k \varepsilon_0 E_1 = -e n_{e1}$:

$$i k \varepsilon_0 E_1 = -e \left( -\frac{i n_0 e k E_1}{m_e \omega^2 - 3 k_B T_e k^2} \right) = \frac{i n_0 e^2 k E_1}{m_e \omega^2 - 3 k_B T_e k^2}$$

For non-trivial wave amplitude ($E_1 \ne 0$):

$$\varepsilon_0 = \frac{n_0 e^2}{m_e \omega^2 - 3 k_B T_e k^2} \implies 1 = \frac{n_0 e^2 / (\varepsilon_0 m_e)}{\omega^2 - \frac{3 k_B T_e}{m_e} k^2}$$

(c) Bohm-Gross Relation & Velocity Analysis: Recognizing $\omega_{pe}^2 = \frac{n_0 e^2}{\varepsilon_0 m_e}$ and $v_{\text{th},e}^2 = \frac{k_B T_e}{m_e}$:

$$1 = \frac{\omega_{pe}^2}{\omega^2 - 3 k^2 v_{\text{th},e}^2} \implies \omega^2 = \omega_{pe}^2 + 3 k^2 v_{\text{th},e}^2$$

This is the Bohm-Gross dispersion relation.

1. Phase Velocity:

$$v_{ph} = \frac{\omega}{k} = \frac{\sqrt{\omega_{pe}^2 + 3 k^2 v_{\text{th},e}^2}}{k} = \sqrt{\frac{\omega_{pe}^2}{k^2} + 3 v_{\text{th},e}^2}$$

In the long-wavelength limit ($k \lambda_D \ll 1$, where $\lambda_D = v_{\text{th},e}/\omega_{pe}$):

$$v_{ph} \approx \frac{\omega_{pe}}{k} \gg v_{\text{th},e}$$

2. Group Velocity:

Differentiating $\omega^2 = \omega_{pe}^2 + 3 k^2 v_{\text{th},e}^2$ with respect to $k$:

$$2\omega \frac{d\omega}{dk} = 6 k v_{\text{th},e}^2 \implies v_g = \frac{d\omega}{dk} = \frac{3 k v_{\text{th},e}^2}{\omega}$$

3. Product $v_{ph} v_g$:

$$v_{ph} v_g = \left( \frac{\omega}{k} \right)\left( \frac{3 k v_{\text{th},e}^2}{\omega} \right) = 3 v_{\text{th},e}^2$$

This product is strictly independent of frequency and wavevector! In the long-wavelength limit where $v_{ph} \to \infty$, the group velocity $v_g \to 0$, ensuring causality is preserved.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.

Advanced Honors Exam Problem Example 5.2: Derivation of the Ion Acoustic Dispersion Relation with Finite Ion Temperature and Electron Screening

Consider low-frequency electrostatic waves in a two-component plasma with electron temperature $T_e$ and ion temperature $T_i$. (a) From the linearized fluid equations, derive the general ion acoustic dispersion relation:

$$\omega^2 = \frac{k^2 c_s^2}{1 + k^2 \lambda_{De}^2} + \gamma_i k^2 v_{\text{th},i}^2$$

(b) For an argon plasma ($M_i = 40\text{ amu} = 6.64 \times 10^{-26}\text{ kg}$) with $k_B T_e = 3.0\text{ eV}$ and $k_B T_i = 0.10\text{ eV}$ at density $n_0 = 10^{16}\text{ m}^{-3}$, calculate the ion sound speed $c_s$ and the Debye length $\lambda_{De}$. (c) Evaluate the wave frequency $f = \omega / (2\pi)$ and phase velocity $v_{ph}$ at two distinct wavelengths: $\lambda_1 = 10.0\text{ cm}$ and $\lambda_2 = 0.50\text{ mm}$.

Full Rigorous Analytical Solution

(a) Derivation of General Ion Acoustic Dispersion: For low frequencies ($\omega \ll \omega_{pe}$), electrons obey the Boltzmann distribution:

$$n_{e1} = n_0 \frac{e\phi_1}{k_B T_e}$$

For ions, linearized continuity and momentum equations with 1D adiabatic compression $\gamma_i = 3$:

$$-i\omega n_{i1} + i n_0 k u_{i1} = 0 \implies u_{i1} = \frac{\omega}{k}\frac{n_{i1}}{n_0}$$
$$-i\omega M_i n_0 u_{i1} = -e n_0 (i k \phi_1) - 3 k_B T_i (i k n_{i1})$$

Substitute $u_{i1}$:

$$-i\omega M_i n_0 \left( \frac{\omega}{k}\frac{n_{i1}}{n_0} \right) = -i e n_0 k \phi_1 - 3 i k_B T_i k n_{i1}$$
$$-i M_i \frac{\omega^2}{k} n_{i1} + 3 i k_B T_i k n_{i1} = -i e n_0 k \phi_1$$

Multiply by $i k / M_i$:

$$(\omega^2 - 3 k^2 v_{\text{th},i}^2) n_{i1} = \frac{n_0 e k^2}{M_i}\phi_1 \implies n_{i1} = \frac{n_0 e k^2 \phi_1}{M_i (\omega^2 - 3 k^2 v_{\text{th},i}^2)}$$

Substitute $n_{e1}$ and $n_{i1}$ into Poisson's equation $k^2 \phi_1 = \frac{e(n_{i1} - n_{e1})}{\varepsilon_0}$:

$$k^2 \phi_1 = \frac{e^2 n_0}{\varepsilon_0}\left[ \frac{k^2 \phi_1}{M_i (\omega^2 - 3 k^2 v_{\text{th},i}^2)} - \frac{\phi_1}{k_B T_e} \right]$$

Divide by $\phi_1$ and use $\omega_{pi}^2 = \frac{n_0 e^2}{\varepsilon_0 M_i}$ and $\frac{1}{\lambda_{De}^2} = \frac{n_0 e^2}{\varepsilon_0 k_B T_e}$:

$$k^2 + \frac{1}{\lambda_{De}^2} = \frac{\omega_{pi}^2 k^2}{\omega^2 - 3 k^2 v_{\text{th},i}^2}$$

Rearranging:

$$\omega^2 - 3 k^2 v_{\text{th},i}^2 = \frac{\omega_{pi}^2 k^2}{k^2 + 1/\lambda_{De}^2} = \frac{\omega_{pi}^2 \lambda_{De}^2 k^2}{1 + k^2 \lambda_{De}^2} = \frac{k^2 c_s^2}{1 + k^2 \lambda_{De}^2}$$

where $c_s = \omega_{pi} \lambda_{De} = \sqrt{\frac{k_B T_e}{M_i}}$.

$$\omega^2 = \frac{k^2 c_s^2}{1 + k^2 \lambda_{De}^2} + 3 k^2 v_{\text{th},i}^2$$

(b) Numerical Calculation of $c_s$ and $\lambda_{De}$: Given: $k_B T_e = 3.0\text{ eV} = 3.0 \times 1.6022\times 10^{-19}\text{ J} = 4.807\times 10^{-19}\text{ J}$ $M_i = 40 \times 1.6605\times 10^{-27}\text{ kg} = 6.642 \times 10^{-26}\text{ kg}$ Ion sound speed:

$$c_s = \sqrt{\frac{k_B T_e}{M_i}} = \sqrt{\frac{4.807\times 10^{-19}}{6.642\times 10^{-26}}} = \sqrt{7.237\times 10^6} \approx 2.690 \times 10^3\text{ m/s} = 2.69\text{ km/s}$$

Debye length:

$$\lambda_{De} = \sqrt{\frac{\varepsilon_0 k_B T_e}{n_0 e^2}} = 7434 \sqrt{\frac{3.0}{10^{16}}} = 7434 \times (1.732 \times 10^{-8}) = 1.288 \times 10^{-4}\text{ m} = 0.129\text{ mm}$$

(c) Evaluation at Two Wavelengths:

1. For $\lambda_1 = 10.0\text{ cm} = 0.10\text{ m}$:

$$k_1 = \frac{2\pi}{\lambda_1} = \frac{2\pi}{0.10} = 62.83\text{ m}^{-1}$$
$$k_1 \lambda_{De} = 62.83 \times 1.288\times 10^{-4} = 8.09 \times 10^{-3} \ll 1$$

Here $k_1 \lambda_{De} \ll 1$, so the wave is in the pure acoustic regime:

$$v_{ph,1} \approx c_s = 2.690\times 10^3\text{ m/s}$$
$$f_1 = \frac{v_{ph,1}}{\lambda_1} = \frac{2690\text{ m/s}}{0.10\text{ m}} = 2.69 \times 10^4\text{ Hz} = 26.9\text{ kHz}$$

2. For $\lambda_2 = 0.50\text{ mm} = 5.0\times 10^{-4}\text{ m}$:

$$k_2 = \frac{2\pi}{5.0\times 10^{-4}} = 12,566\text{ m}^{-1}$$
$$k_2 \lambda_{De} = 12566 \times 1.288\times 10^{-4} = 1.619$$

Since $k_2 \lambda_{De} \sim 1$, dispersion is significant:

$$\omega_2 = \frac{k_2 c_s}{\sqrt{1 + (k_2 \lambda_{De})^2}} = \frac{12566 \times 2690}{\sqrt{1 + (1.619)^2}} = \frac{3.380\times 10^7}{\sqrt{1 + 2.621}} = \frac{3.380\times 10^7}{1.903} = 1.776 \times 10^7\text{ rad/s}$$
$$f_2 = \frac{\omega_2}{2\pi} = \frac{1.776\times 10^7}{2\pi} \approx 2.827 \times 10^6\text{ Hz} = 2.83\text{ MHz}$$
$$v_{ph,2} = \frac{\omega_2}{k_2} = \frac{1.776\times 10^7}{12566} = 1.413 \times 10^3\text{ m/s} = 1.41\text{ km/s}$$

Notice that $v_{ph,2}$ is substantially lower than $c_s$ (1.41 km/s vs 2.69 km/s), demonstrating the dispersive roll-off toward the ion plasma frequency.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.

Advanced Honors Exam Problem Example 5.3: Electrostatic Ion Cyclotron (EIC) Wave Dispersion and Resonance in Magnetized Geometry

Consider electrostatic waves in a magnetized plasma with $\vec{B}_0 = B_0 \hat{z}$ propagating at an oblique angle nearly perpendicular to the magnetic field ($k_\perp \gg k_\parallel$) with wavevector $\vec{k} = k_\perp \hat{x} + k_\parallel \hat{z}$. (a) Assuming electrons respond isothermally along the magnetic field to satisfy the parallel Boltzmann relation $n_{e1} = n_0 \frac{e\phi_1}{k_B T_e}$, and treating ions via magnetized fluid momentum equations with cyclotron frequency $\omega_{ci}$, derive the Electrostatic Ion Cyclotron (EIC) wave dispersion relation:

$$\omega^2 = \omega_{ci}^2 + k_\perp^2 c_s^2$$

(b) In a fusion tokamak edge plasma where $B_0 = 3.0\text{ Tesla}$, deuterium ions ($M_i = 3.34\times 10^{-27}\text{ kg}$), and $k_B T_e = 50\text{ eV}$, compute the ion cyclotron frequency $f_{ci}$ and the EIC wave frequency for perpendicular wavelength $\lambda_\perp = 2.0\text{ cm}$.

Full Rigorous Analytical Solution

(a) Derivation of EIC Dispersion Relation: Let the electrostatic potential perturbation be $\phi_1(x, z, t) = \phi_1 e^{i(k_\perp x + k_\parallel z - \omega t)}$.

1. Electron Response:

Because electrons move rapidly along field lines ($v_{\text{th},e} \gg \omega / k_\parallel$), they shield the parallel electric field $E_z = -i k_\parallel \phi_1$ adiabatically:

$$n_{e1} = n_0 \frac{e\phi_1}{k_B T_e}$$

2. Ion Dynamics:

For cold ions ($T_i \approx 0$), the linearized ion momentum equation in $\vec{B}_0 = B_0 \hat{z}$ is:

$$-i\omega M_i \vec{u}_{i1} = e(-\nabla\phi_1 + \vec{u}_{i1}\times\vec{B}_0)$$

Resolving into Cartesian components:

$$-i\omega M_i u_{ix} = -e (i k_\perp \phi_1) + e B_0 u_{iy}$$
$$-i\omega M_i u_{iy} = -e B_0 u_{ix}$$
$$-i\omega M_i u_{iz} = -e (i k_\parallel \phi_1)$$

From the $y$-equation: $u_{iy} = \frac{e B_0}{i\omega M_i} u_{ix} = \frac{\omega_{ci}}{i\omega} u_{ix}$, where $\omega_{ci} = \frac{e B_0}{M_i}$. Substitute $u_{iy}$ into the $x$-equation:

$$-i\omega M_i u_{ix} = -i e k_\perp \phi_1 + e B_0 \left( \frac{\omega_{ci}}{i\omega} u_{ix} \right) = -i e k_\perp \phi_1 - i \frac{M_i \omega_{ci}^2}{\omega} u_{ix}$$

Multiply by $i\omega / M_i$:

$$\omega^2 u_{ix} = \frac{e k_\perp \omega}{M_i} \phi_1 + \omega_{ci}^2 u_{ix} \implies (\omega^2 - \omega_{ci}^2) u_{ix} = \frac{e k_\perp \omega}{M_i}\phi_1$$
$$u_{ix} = \frac{e k_\perp \omega}{M_i (\omega^2 - \omega_{ci}^2)}\phi_1$$

From the ion continuity equation:

$$-i\omega n_{i1} + i n_0 (k_\perp u_{ix} + k_\parallel u_{iz}) = 0$$

Since $k_\perp \gg k_\parallel$, the perpendicular divergence dominates ($k_\perp u_{ix} \gg k_\parallel u_{iz}$):

$$n_{i1} \approx n_0 \frac{k_\perp u_{ix}}{\omega} = \frac{n_0 e k_\perp^2}{M_i (\omega^2 - \omega_{ci}^2)}\phi_1$$

In a dense plasma with $k_\perp \lambda_D \ll 1$, quasi-neutrality requires $n_{e1} \approx n_{i1}$:

$$n_0 \frac{e\phi_1}{k_B T_e} = \frac{n_0 e k_\perp^2}{M_i (\omega^2 - \omega_{ci}^2)}\phi_1$$

Canceling $n_0 e \phi_1$:

$$\frac{1}{k_B T_e} = \frac{k_\perp^2}{M_i (\omega^2 - \omega_{ci}^2)} \implies \omega^2 - \omega_{ci}^2 = \frac{k_B T_e}{M_i} k_\perp^2 = c_s^2 k_\perp^2$$
$$\omega^2 = \omega_{ci}^2 + k_\perp^2 c_s^2$$

This is the Electrostatic Ion Cyclotron (EIC) dispersion relation.

(b) Numerical Evaluation in Tokamak Edge: Given: $B_0 = 3.0\text{ T}$, $M_i = 3.344\times 10^{-27}\text{ kg}$ (deuteron) $k_B T_e = 50\text{ eV} = 50 \times 1.6022\times 10^{-19}\text{ J} = 8.011\times 10^{-18}\text{ J}$ $\lambda_\perp = 0.02\text{ m} \implies k_\perp = \frac{2\pi}{0.02} = 314.16\text{ m}^{-1}$

1. Ion Cyclotron Frequency:

$$\omega_{ci} = \frac{e B_0}{M_i} = \frac{(1.6022\times 10^{-19})(3.0)}{3.344\times 10^{-27}} = 1.437 \times 10^8\text{ rad/s}$$
$$f_{ci} = \frac{\omega_{ci}}{2\pi} = \frac{1.437\times 10^8}{2\pi} \approx 2.288 \times 10^7\text{ Hz} = 22.88\text{ MHz}$$

2. Ion Sound Speed:

$$c_s = \sqrt{\frac{k_B T_e}{M_i}} = \sqrt{\frac{8.011\times 10^{-18}}{3.344\times 10^{-27}}} = \sqrt{2.396\times 10^9} \approx 4.895 \times 10^4\text{ m/s} = 48.95\text{ km/s}$$

3. EIC Wave Frequency:

$$k_\perp c_s = (314.16\text{ m}^{-1})(4.895\times 10^4\text{ m/s}) = 1.538 \times 10^7\text{ rad/s}$$
$$\omega = \sqrt{\omega_{ci}^2 + (k_\perp c_s)^2} = \sqrt{(1.437\times 10^8)^2 + (1.538\times 10^7)^2} = \sqrt{2.065\times 10^{16} + 2.365\times 10^{14}} = \sqrt{2.089\times 10^{16}}$$
$$\omega = 1.445 \times 10^8\text{ rad/s}$$
$$f = \frac{\omega}{2\pi} = 2.300 \times 10^7\text{ Hz} = 23.00\text{ MHz}$$

The thermal pressure shifts the oscillation frequency slightly above the fundamental cyclotron resonance by $\approx 0.12\text{ MHz}$.

Final Answer & Physical Verification

Complete rigorous derivation and proof detailed above.

Final Answer & Physical Insight

Complete rigorous derivation and proof detailed above.