Plasma Physics: Brief Overview

A plasma screens electric fields over the Debye length:

$\lambda_D = \sqrt{\frac{\varepsilon_0 k_B T}{n_e e^2}}$

For $n_e = 10^{18}$ m$^{-3}$, $T = 10^4$ K: $\lambda_D = 7.4 \times 10^{-5}$ m $= 74\,\mu$M.

The plasma frequency:

$\omega_p = \sqrt{\frac{n_e e^2}{m_e \varepsilon_0}}$

For $n_e = 10^{18}$ m$^{-3}$: $\omega_p = 5.64 \times 10^{10}$ rad/s, $f_p = 8.98$ GHz. EM waves with $\omega < \omega_p$ cannot propagate (evanescent).

Small displacements of the electron cloud create restoring forces, leading to Langmuir waves:

$\omega_{\text{Langmuir} = \omega_p\left(1 + \frac{3k_BT}{2m_e}\frac{k^2}{\omega_p^2}\right)^{-1/2}}$

At long wavelengths ($k \to 0$): $\omega \to \omega_p$ (undamped). With ion motion: the ion-acoustic wave has $\omega^2 = k^2 c_s^2/(1 + k^2\lambda_D^2)$ where $c_s = \sqrt{k_BT/m_i}$.

Problem. A uniformly charged sphere of radius $R$ has total charge $Q$. Find $E$ inside and outside.

Solution. Outside ($r > R$): $\oint E \cdot dA = Q/\varepsilon_0 \implies E \cdot 4\pi r^2 = Q/\varepsilon_0 \implies E = \frac{Q}{4\pi\varepsilon_0 r^2}$. Inside ($r < R$): enclosed charge $= Q(r/R)^3$. $E = \frac{Qr}{4\pi\varepsilon_0 R^3}$.

$\blacksquare$

Problem. An EM wave has $E_0 = 100 \mathrm{ V/m}$ in vacuum. Find the average Poynting vector magnitude.

Solution. ${\langle S \rangle = \frac{E_0^2}{2\mu_0 c} = \frac{100^2}{2 \times 4\pi \times 10^{-7} \times 3 \times 10^8} = \frac{10000}{754} \approx 13.3 \mathrm{ W/m}^2}$.

$\blacksquare$

• Confusing Gauss’s law applications. Gauss’s law is most useful for systems with high symmetry (spherical, cylindrical, planar). Fix: Choose a Gaussian surface matching the symmetry; the flux through the surface equals the enclosed charge divided by $\varepsilon_0$.
• Wrong Maxwell equation sign. Faraday’s law has a negative sign: $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Fix: The minus sign reflects Lenz’s law — the induced EMF opposes the change in flux.
• Confusing $\vec{D}$and $\vec{E}$, $\vec{H}$and $\vec{B}$. $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$; $\vec{H} = \vec{B}/\mu_0 - \vec{M}$. Fix: In vacuum: $\vec{D} = \varepsilon_0\vec{E}$, $\vec{H} = \vec{B}/\mu_0$.

• Maxwell’s equations: Gauss’s law, Gauss’s law for magnetism, Faraday’s law, Ampère-Maxwell law.
• Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$.
• EM waves: $E_0 = cB_0$; $c = 1/\sqrt{\mu_0\varepsilon_0}$; Poynting vector $\vec{S} = \vec{E} \times \vec{H}/\mu_0$.
• Boundary conditions: tangential $E$ and normal $B$ are continuous across interfaces.

• Confusing Debye shielding with perfect neutrality. A plasma is quasineutral ($n_i \approx n_e$) on scales large compared to $\lambda_D$, but charge separation exists on Debye-length scales. Fix: Use $\lambda_D$ as the scale below which individual charges matter.
• Assuming all EM waves propagate in a plasma. Waves with $\omega < \omega_p$ are evanescent — they decay exponentially. Fix: The cut-off condition is $\omega > \omega_p$ for propagation; below $\omega_p$ the refractive index becomes imaginary.
• Forgetting ion motion in low-frequency waves. The ion-acoustic wave requires mobile ions; at frequencies above $\omega_{pi}$ (ion plasma frequency), ions cannot respond. Fix: Check whether $\omega \ll \omega_{pi}$ before using the ion-acoustic dispersion.
• Treating Coulomb collisions as rare. While high-temperature plasmas are often collisionless, the collision frequency scales as $T^{-3/2}$; cold, dense plasmas can be collisional. Fix: Compare the mean free path to system size using $\nu_{ei} \propto n_e T_e^{-3/2}$.

• Fusion energy (tokamaks): Magnetic confinement of deuterium-tritium plasmas at $T \sim 10^8$ K requires understanding of MHD stability, transport, and wave heating.
• Space physics: The solar wind ($n_e \sim 10^7$ m$^{-3}$, $T \sim 10^5$ K) is a plasma that interacts with Earth’s magnetosphere, causing aurorae and geomagnetic storms.
• Semiconductor processing: Low-temperature plasmas ($T_e \sim 10^4$ K, $n_e \sim 10^{16}$ m$^{-3}$) are used for etching and deposition in microchip fabrication.
• Radio astronomy: Pulsar signals propagate through the interstellar medium (ISM) plasma; dispersion measurements give the column density ${\rm DM} = \int n_e\, dl$.

Problem. The solar corona has $n_e \approx 10^{14}$ m$^{-3}$ and $T \approx 10^6$ K. Compute the Debye length. How many electrons are in a Debye sphere?

Solution.

$\lambda_D = \sqrt{\frac{\varepsilon_0 k_B T}{n_e e^2}} = \sqrt{\frac{(8.85\times10^{-12})(1.38\times10^{-23})(10^6)}{(10^{14})(1.6\times10^{-19})^2}}$

$\lambda_D \approx \sqrt{\frac{1.22\times10^{-28}}{2.56\times10^{-24}}} = \sqrt{4.77\times10^{-5}} \approx 6.9\times10^{-3}\;\mathrm{m} = 6.9\;\mathrm{mm}$

The Debye sphere volume is $\frac{4}{3}\pi\lambda_D^3 \approx 1.38\times10^{-6}$ m$^3$, containing $N_D = n_e \cdot \frac{4}{3}\pi\lambda_D^3 \approx 10^{14} \times 1.38\times10^{-6} \approx 1.4\times10^8$ electrons. Since $N_D \gg 1$, the corona satisfies the plasma criterion for collective behaviour.

$\blacksquare$