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Plasma Parameters of Six Real Systems

The problem

Apply the three ideal-plasma conditions to six systems spanning seventeen orders of magnitude in density, and decide which ones are actually plasmas.

The formulas

\[\lambda_D = \sqrt{\frac{\epsilon_0 k_BT_e}{n_ee^2}} = 7430\sqrt{\frac{T_e[\text{eV}]}{n_e[\text{m}^{-3}]}}\ \text{m}, \qquad N_D = \frac{4}{3}\pi n_e\lambda_D^3, \qquad \frac{\omega_{pe}}{2\pi} = 8980\sqrt{n_e[\text{cm}^{-3}]}\ \text{Hz}\]

The three conditions: \(\lambda_D \ll L\) (shielding fits inside the system), \(N_D \gg 1\) (enough particles to do the shielding), and \(\omega_{pe} \gg \nu_{coll}\) (collective response outruns collisions).

The survey

System \(n_e\) (m⁻³) \(T_e\) (eV) \(L\) \(\lambda_D\) \(N_D\) \(\lambda_D/L\) Verdict
Candle flame \(10^{14}\) 0.2 1 cm 0.33 mm \(1.5\times10^4\) 0.03 plasma (weakly ionised)
Solar wind at 1 AU \(10^{7}\) 10 \(10^{11}\) m 7.4 m \(1.7\times10^{10}\) \(7\times10^{-11}\) ideal plasma
Glow discharge \(10^{16}\) 3 10 cm 0.13 mm \(8.9\times10^4\) \(10^{-3}\) ideal plasma
Tokamak core \(10^{20}\) \(10^4\) 1 m 74 µm \(1.7\times10^8\) \(7\times10^{-5}\) ideal plasma
Copper conduction electrons \(8.5\times10^{28}\) 7 1 mm 0.067 Å 0.11 \(7\times10^{-8}\) not a plasma
ICF compressed core \(10^{31}\) \(10^4\) 0.1 mm 0.0024 Å \(5.4\times10^2\) \(2\times10^{-6}\) marginal

Working one row: the candle flame

\[\lambda_D = 7430\sqrt{\frac{0.2}{10^{14}}} = 7430\times4.47\times10^{-8} = 3.3\times10^{-4}\ \text{m}\]
\[N_D = \frac{4}{3}\pi(10^{14})(3.3\times10^{-4})^3 = 1.5\times10^4\]

\(\lambda_D/L = 0.033 \ll 1\) ✓ and \(N_D \gg 1\) ✓. A candle flame passes both — the free electrons in it do behave collectively. What it is not is fully ionised: the ionised fraction is around \(10^{-10}\), so neutral collisions dominate the dynamics and the third condition (\(\omega_{pe} \gg \nu\)) is where it actually fails. This is the standard trap in these questions: passing the \(\lambda_D\) and \(N_D\) tests does not make something a good plasma, it makes it a system in which Debye shielding is meaningful.

The interesting failure: copper

Conduction electrons in copper have a spectacular density, \(8.5\times10^{28}\) m⁻³, and a characteristic energy set by the Fermi level, \(E_F \approx 7\) eV. The Debye (here Thomas–Fermi) length comes out at \(6.7\times10^{-11}\) m — smaller than the atomic spacing — and

\[N_D = 0.11 < 1\]

Fewer than one electron sits inside a screening sphere. The whole derivation of \(\lambda_D\) assumed a smooth statistical cloud of shielders responding in a Boltzmann fashion; with \(N_D < 1\) that picture is meaningless, and so is the continuum description built on it. Copper is a strongly coupled, degenerate system, correctly described by solid-state physics, not plasma physics. It is also Fermi-degenerate, so Boltzmann statistics were the wrong tool from the start.

Compare the mean interparticle spacing: \(n^{-1/3} = 2.3\times10^{-10}\) m, larger than \(\lambda_D\). That comparison is the cleanest way to see the failure — the shielding cloud is smaller than the gap between the particles supposed to form it.

What the survey teaches

  1. Density and temperature alone decide nothing. Only the combinations \(\lambda_D\) and \(N_D\) do, which is why they, and not \(n\) and \(T\), are the parameters of the subject.
  2. The ideal-plasma boundary is a coupling boundary. \(N_D \gg 1\) means the potential energy between neighbours is small compared with their kinetic energy. Cross it and you get liquids, crystals, and white-dwarf interiors.
  3. ICF lives near the edge. At \(N_D \sim 500\) the compressed core is still nominally ideal, but the margin is thin, and during the colder early stages of compression it is genuinely strongly coupled — which is why ICF modelling needs equation-of-state physics that a tokamak modeller never touches.

Ideal plasma · Debye shielding · Plasma frequency · Coulomb collisions · Debye shielding lab — run any row yourself