advanced35 min studyLesson 892 in Physics

Equations of state inside stars

Count ions and electrons consistently and distinguish classical gas, radiation, and degenerate-electron pressure domains.

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What you’ll understand

  • Compute mean molecular weights from declared mass fractions and ionization.
  • Compare ideal-gas and radiation pressure with consistent SI units.
  • State the temperature and momentum orderings behind degenerate-electron limits.

The same mass density can produce very different pressure depending on ionization, composition, temperature, and electron degeneracy. A stellar equation of state must also supply internal energy and thermodynamic derivatives, because transport and stability depend on more than pressure alone. We first treat a fully ionized, weakly coupled mixture with classical ions and electrons, then identify where radiation and quantum statistics change the result. All numerical constants in the exercise are supplied rounded toy inputs; they do not describe a measured stellar interior.

Mass fractions are not particle fractions

For species i, specify mass fraction Xi, nuclear mass number Ai, and ionic charge Zi in units of the elementary charge. The ion density is its share of the mass divided by the mass per nucleus. Full ionization supplies Zi electrons per nucleus. Summing both ions and electrons defines the dimensionless mean molecular weight mu, whereas summing only electrons defines mu electron. Neither symbol is a mass in kilograms. Charge neutrality determines electron number without creating additional baryonic mass.

For pure ionized hydrogen there are two pressure-carrying particles per baryon, giving mu equal to one half. For ionized helium there are three particles per four baryons, giving mu equal to four thirds. Replacing hydrogen by helium therefore reduces classical gas pressure at fixed density and temperature. For metals, the frequently used approximation of one electron per two baryons neglects a smaller ion contribution and depends on the chosen mixture; an exact species sum avoids silently assigning every metal the same atomic mass.

1μ=∑iXi(1+Zi)Ai,1μe=∑iXiZiAi,∑iXi=1\frac1\mu=\sum_i\frac{X_i(1+Z_i)}{A_i},\qquad \frac1{\mu_e}=\sum_i\frac{X_iZ_i}{A_i},\qquad \sum_iX_i=1
The sums apply to a fully ionized mixture using nuclear masses approximately Ai mᵤ. With partial ionization, use actual mean ionic charges and a self-consistent ionization calculation. μ and μe are dimensionless and differ from chemical potentials.↔ Scroll wide equations horizontally.

Add pressure sources within their shared domain

The classical ideal gas contributes particle number density times Boltzmann constant times temperature. Locally thermalized photons add one third of their energy density as radiation pressure. Their relative importance increases rapidly with temperature and decreases with material density. The sum is useful only when the material remains classical and weakly coupled and the radiation field is sufficiently close to local thermal equilibrium. It is not an equation of state for an arbitrary optically thin, anisotropic radiation field.

An equation of state table can include partial ionization, Coulomb corrections, pair creation, and other effects absent from the simple sum. Partial ionization stores heat in changing ionization rather than only random motion, altering heat capacities and adiabatic exponents. Radiation pressure likewise changes the relation between an adiabatic temperature change and compression. Using a monatomic ideal-gas adiabatic index everywhere merely because ideal-gas pressure was included would miss these mechanisms. Thermodynamic derivatives must be computed from the same pressure and energy model.

P=Pgas+Prad=ρkBTμmu+aT43,PradPgas=aμmuT33ρkBP=P_{\rm gas}+P_{\rm rad}=\frac{\rho k_BT}{\mu m_u}+\frac{aT^4}{3},\qquad \frac{P_{\rm rad}}{P_{\rm gas}}=\frac{a\mu m_uT^3}{3\rho k_B}
ρ is kg m⁻³, T is K, a is J m⁻³ K⁻⁴, and both pressures are Pa. Classical, fully ionized gas and locally isotropic thermal radiation are assumed. Degeneracy and nonideal interaction corrections are omitted.↔ Scroll wide equations horizontally.

Electron degeneracy is not zero internal motion

When electron states are densely occupied, the exclusion principle produces a filled momentum distribution up to the Fermi momentum in the cold limit. Electrons still have momentum and exert pressure even as their temperature approaches zero. At temperatures far below the Fermi temperature and momenta small compared with electron rest mass times light speed, the electron pressure scales as electron density to the five-thirds power. The ions may remain classical even while the electrons are degenerate.

As the Fermi momentum becomes much larger than electron rest mass times light speed, the limiting exponent changes to four thirds. Intermediate degeneracy and relativity require the appropriate Fermi integrals rather than switching formulas at an arbitrary density. Composition matters through electron number per unit mass. The cold electron law approximates white-dwarf support, but finite temperature, Coulomb effects, electron captures, and relativistic gravity eventually modify simple inferences about radii or limiting masses. A pressure law alone does not specify the complete evolutionary state.

ne=ρμemu,pF=ℏ(3π2ne)1/3,Pe,NR=ℏ2(3π2)2/35mene5/3n_e=\frac{\rho}{\mu_em_u},\quad p_F=\hbar(3\pi^2n_e)^{1/3},\quad P_{e,{\rm NR}}=\frac{\hbar^2(3\pi^2)^{2/3}}{5m_e}n_e^{5/3}
The displayed pressure is the zero-temperature nonrelativistic limit, valid approximately for kBT ≪ pF²/(2me) and pF ≪ mec. Its units are Pa. In the ultrarelativistic cold limit the density exponent becomes 4/3; the displayed nonrelativistic coefficient must not be retained.↔ Scroll wide equations horizontally.

Count a hydrogen–helium toy mixture

Use fully ionized X=3/4 hydrogen and Y=1/4 helium, with no metals. Find μ and μe. At ρ=10³ kg m⁻³ and T=10⁷ K, use supplied kB/mᵤ=8 × 10³ J kg⁻¹ K⁻¹ and a=7.5 × 10⁻¹⁶ J m⁻³ K⁻⁴ to compare the classical gas and radiation terms. Assume classical weak coupling for this calculation.

  1. The total-particle sum is 1/μ=2(3/4)+(3/4)(1/4)=27/161/\mu=2(3/4)+(3/4)(1/4)=27/16, so μ=16/27. The electron sum is 3/4+1/8=7/8, giving μe=8/7.

  2. The gas term is Pgas=103(8×103)107(27/16)=1.35×1014 PaP_{\rm gas}=10^3(8\times10^3)10^7(27/16)=1.35\times10^{14}\ {\rm Pa}. This is the declared classical approximation.

  3. The radiation term is 2.5 × 10¹² Pa, so its ratio to gas pressure is 1/54, approximately 1.85 × 10⁻². Both contributions are positive.

The result

μ=16/27 and μe=8/7; gas pressure dominates the two supplied terms, with a small radiation correction.

Explore the model

Interactive lab

Interactive lab · function

Distinguish powers with a controlled surrogate

Reset and set BOTH Coefficient a to 0.5 and Position x to 1; predict y=0.5. Then Reset and set BOTH a to 0.5 and x to 2; predict y=2. This exact quadratic factor-of-four comparison helps distinguish a power law from a linear law. It does not implement the stellar density exponents 5/3 or 4/3, ionization, a Fermi integral, or any stellar equation of state.

Distinguish powers with a controlled surrogateThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA density exponent must come from the statistical and relativistic regime rather than from the appearance of a curve.

Check your understanding

Which conditions justify the displayed nonrelativistic cold electron-pressure law?

Try it yourself

Compare pure fully ionized hydrogen with pure fully ionized helium at the same mass density and temperature. Find the helium-to-hydrogen classical gas-pressure ratio. Then, separately assuming cold nonrelativistic degenerate electrons in both samples, find their electron-pressure ratio.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For the gas count all ions and electrons; for degeneracy compare electron densities and raise their ratio to the five-thirds power.

Compare with the solution

Hydrogen has μ=1/2 and μe=1, whereas helium has μ=4/3 and μe=2. Thus the classical gas-pressure ratio is (1/2)/(4/3)=3/8(1/2)/(4/3)=3/8. Electron density is halved, so the cold electron-pressure ratio is 2−5/3≈0.3152^{-5/3}\approx0.315. These are comparisons in two separately declared statistical regimes; they are not additional gas and degenerate pressures to add for the same electron population.

Take it with you

  • Composition and ionization determine how many ions and electrons occupy a given mass density.
  • Gas plus radiation pressure needs consistent internal energy and thermodynamic derivatives.
  • Cold electron degeneracy has distinct nonrelativistic and ultrarelativistic density exponents.

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