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The Fermi level is the chemical potential controlling electron occupations. For an ideal Fermi gas at zero temperature it equals the Fermi energy; at nonzero temperature the chemical potential need not equal that zero-temperature energy.

Fermi momentum is the momentum at the boundary of occupied one-particle states in a zero-temperature Fermi gas. For spin-one-half particles in three dimensions, counting the two spin states gives $p_F=\hslash (3{\pi }^{2}n{)}^{1/3}$, where $n$ is number density.

For a two-dimensional Fermi gas of $N$ spin-one-half particles with Fermi energy $E_F$, the constant density of states makes the low-temperature internal energy correction and heat capacity at constant areaThus its low-temperature heat capacity obeys a power law with exponent one.

Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while electron degeneracy pressure supplies support. By the Pauli exclusion principle, the two electron spin states fill a momentum sphere up to Fermi momentum $p_F$. Counting states giveswhere ${\mu }_e$ is mean molecular weight per electron. The momentum flux of this isotropic Fermi gas isWriting $x=p_F/(m_{e}c)$ and performing the integral gives the equation of state of a cold electron gasIn the nonrelativistic limit $p_F\ll m_{e}c$, $v(p)\simeq p/m_e$, soIn the ultrarelativistic limit $p_F\gg m_{e}c$, $v(p)\simeq c$, soThese are polytropic equations of state with indices $3/2$ and $3$, respectively. The relativistic softening underlies the Chandrasekhar mass limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.

For an ultrarelativistic particle, $\epsilon =pc$. Including the two spin states of an electron, counting wavevectors in a sphere gives the density of statesThe grand canonical partition function of the ideal Fermi gas is thereforeSince the grand potential satisfies $pV=k_{B}T\log \mathcal{Z}$, integration by parts givesThusThis is the equation of state of an ideal ultrarelativistic gas.

For density of states proportional to ${ϵ}^a$, $a>-1$, and positive temperature, set $q=\mu /(kT)$, $n(x)=(1+e^{x-q}{)}^{-1}$, $b=a+1$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(1-n)dx$. Differentiation and integration by parts giveThe inequality is strict Cauchy-Schwarz inequality, since $x$ is not constant under this positive weight. Thus fixed entropy per particle forces fixed $\mu /T$. This is the injectivity step behind the adiabatic scaling of an ideal Fermi gas.