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A band insulator has an integer number of completely filled energy bands, with the Fermi energy inside a positive band gap. At zero temperature, no arbitrarily low-energy charged excitation is then available.

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.

The Fermi sea is the zero-temperature many-fermion state in which every one-particle state below the Fermi energy is occupied and every state above it is empty.

Integrating the hydrostatic pressure support equation with the radiative-zero white-dwarf envelope law yields $T=4GM(1/r-1/R)/(17\mathcal{R})$. Matching $T=T_c$ at $r_b$ gives $s=R-r_b$ as displayed. A strongly degenerate bulk core has thermal energy small compared with its Fermi energy, making $q\ll 1$ and hence $s\ll R$. The inequality $T_c\gg T_{\mathrm{eff}}$ alone does not establish geometrical thinness.

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.

Apply the preceding temperature profile at the envelope base. Introduce the dimensionless parameterThen $R/r_b-1=q$, so the geometrical thickness of a white-dwarf radiative envelope isThe smallness to be established is therefore $\mathcal{R}{T}_c\ll GM/R$, up to the numerical factor $17/4$.

That ordering follows from the cold, strongly electron-degenerate white dwarf model. Throughout the mass-bearing core, $k{T}_c\ll E_F$, where $E_F$ is the Fermi energy. For nonrelativistic electron degeneracy pressure,The stellar hydrostatic equation gives a characteristic core $P_{\mathrm{deg}}/\rho$ of order $GM/R$. Consequentlyfor fixed composition with order-one molecular-weight ratios, and $s\ll R$. Equivalently, the matched base mass density is much lower than a characteristic core mass density, since ${\rho }_b/{\rho }_{\mathrm{core}}\sim [\mathcal{R}{T}_c/(GM/R){]}^{3/2}$. The degeneracy ordering applies in the bulk core; at the matching point degeneracy is becoming marginal.

The inequalities $T_c\gg T_{\mathrm{eff}}$ and $k{T}_c\ll E_F$ express different facts. The first justifies ignoring surface-temperature terms, while the second makes the envelope geometrically thin. If the core were hot enough that $q$ were not small, the exact expression $s/R=q/(1+q)$ would not justify a thin-layer approximation.

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

At zero temperature the Fermi-Dirac distribution is the indicator of $0\le \epsilon <E_F$. HenceSolving for the Fermi energy gives