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Consider an ideal degenerate electron gas at zero temperature. Counting two spin states per phase space cell gives number density $n_e=p_F^3/(3{\pi }^2{\hslash }^3)$. The isotropic momentum flux gives pressureHere $m_e$ is electron mass, $c$ is the speed of light, $\hslash =h/(2\pi )$ with Planck constant $h$, and special relativity gives $v(p)=pc/\sqrt{p^2+m_e^2{c}^2}$. The factor $1/3$ averages the directional momentum flux, and the spin factor is already included. This normalization makes $P_{\ast }$ a pressure, not an energy per electron.

For $x\ll 1$, expand the denominator: $I(x)\sim x^5/5$ and $P\simeq n_e{p}_F^2/(5m_e)$. For $x\gg 1$, the integrand approaches $t^3$: $I(x)\sim x^4/4$ and $P\simeq n_e{p}_{F}c/4$. At fixed composition, mass density is $\rho ={\mu }_e{m}_u{n}_e$, with ${\mu }_e$ the mean molecular weight per electron and $m_u$ the unified atomic mass unit. Since $n_e\propto x^3$, the logarithmic stiffness isPressure keeps rising; its fractional gain per fractional compression becomes smaller. Relativistic electrons approach a speed ceiling while their momenta keep increasing. Thus the high-density equation of state loses the extra compression leverage of a nonrelativistic white dwarf. The Chandrasekhar mass from central polytropic constants supplies the corresponding global mass calculation.

For a calculated local example with ${\mu }_e=2$, $x=1$ gives $\rho \simeq 1.95\times 10^9\mathrm{kg}{\mathrm{m}}^{-3}$, $I(1)=0.1537384$ and $P\simeq 7.38\times 10^{21}\mathrm{Pa}$. Its stiffness is $1.53314$, intermediate between the two limits. These are ideal equation of state values at one point, not a measured central mass density or a whole white dwarf model. Finite temperature and interactions are excluded.

For an isotropic degenerate electron gas, pressure is one third of the momentum flux. In the ultrarelativistic limit each occupied mode has speed $c$, so the supplied momentum-space state count givesA fully ionized helium nucleus provides two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Eliminating the Fermi momentum $p_0=(3h^3\rho /(16\pi m_p){)}^{1/3}$ therefore givesThe two-spin-state degeneracy is already included in the supplied state count; no further factor 2 is needed.

If electron degeneracy pressure supplies the star's pressure, this is an index-three stellar polytrope. Its central pressure and density obeyThe radius cancels, leaving $M^{2/3}=K(3a_3/(4\pi ){)}^{4/3}/(c_{3}G)$. Thus the Chandrasekhar mass from central polytropic constants isThe dependence on microscopic constants reflects the balance of fermionic pressure and gravity. The absence of radius dependence is special to the limiting $4/3$ pressure exponent. Ordinary partially relativistic white dwarfs can have a range of masses below this Chandrasekhar limit; the formula does not force every white dwarf to have this mass. As a degenerate stellar core grows toward the limiting mass, compression cannot provide an arbitrarily larger supporting pressure relative to gravity. Above it the ideal cold, nonrotating electron-supported configuration cannot remain in equilibrium; nuclear ignition, collapse or other changes in the support mechanism must intervene. Which outcome occurs depends on the core's composition and evolutionary state.