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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 chemically uniform, gas-pressure-supported stars with homologous dimensionless profiles, mass conservation and hydrostatic pressure support equation give ${\rho }_c\propto M/R^3$ and $T_c\propto M/R$. Suppose opacity scales as $\kappa \propto {\rho }^a{T}^b$ and specific nuclear energy generation as $ϵ\propto {\rho }^c{T}^d$. ThenThe second scaling follows from radiative diffusion in a star, $L\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Equating the two giveswhen the denominator is nonzero. Fixed composition and self-similar profiles are essential; these are relations between idealized models rather than universal stellar laws.

The hydrostatic pressure support equation in enclosed-mass coordinate is $dP/dm=-Gm/(4\pi r^4)$. Since $r(m)\le R$, integration from surface pressure zero to the centre givesIf mass density decreases outward, the mean density within $r$ is at least the global mean, so $r(m)\le R(m/M{)}^{1/3}$. This improves the bound towith equality for uniform mass density. Centrally concentrated planets need larger central pressure to support their inner mass.

Efficient convection keeps the interior at nearly uniform specific entropy. For the fully ionized monatomic gas, the heat capacity ratio is $5/3$, so the adiabatic process gives $P=K_{\rho }{\rho }^{5/3}$, with $K_{\rho }$ spatially constant. The ideal gas relation $P=\mathcal{R}\rho T/\mu$ then impliesThe constant is spatial, not temporal: loss of specific entropy during contraction changes it. This is a stellar polytrope with polytropic index $n=3/2$. We neglect radiation pressure and take the thin photosphere as an atmospheric matching layer rather than a significant fraction of the stellar mass.

For a fixed dimensionless profile of the Lane-Emden equation, the stellar homology relations have the formwhere $C_{\rho },C_P$ are positive dimensionless structure constants. More explicitly, write $\rho (r)={\rho }_{c}f(x)$ and $m(r)=Mh(x)$ with $x=r/R$ and fixed dimensionless profiles. Mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^{1}f(x)x^{2}dx$. Integrating the hydrostatic pressure support equation, with negligible surface pressure on the interior scale, gives $P_c=(GM{\rho }_c/R){\int }_0^{1}h(x)f(x)x^{-2}dx$. Both integrals are fixed positive numbers, proving the two central scalings. Substituting these relations into $K=P_c/T_c^{5/2}$ givesThus the same polytropic index fixes $K_0$ throughout this homologous sequence, while the dimensional stellar polytrope constant changes.

At the photosphere, whose optical depth is $2/3$, identify the temperature with the effective temperature $T_e$. The ideal gas relation and the interior adiabat give ${\rho }_{\mathrm{ph}}=\mu K{T}_e^{3/2}/\mathcal{R}$. With the prescribed opacity and stellar surface boundary condition,Eliminating $K$ therefore provesUsing the Stefan–Boltzmann law, with $\sigma =ac/4$, gives the luminosity

The gravitational energy of a stellar polytrope is $\Omega =-6G{M}^2/(7R)$ for $n=3/2$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, and hence the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. With no appreciable stellar nuclear fusion or accretion, energy conservation givesCombining the two expressions for $L$ yieldsMultiplication by $-7R^{-9/2}/2$ and integration proves the contraction law with density-linear fourth-power stellar opacity:When the accumulated contraction term dominates the initial-radius term, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Substitution into the central ideal gas relation giveswhich has the stated $t^{2/7}$ behavior when the time origin is negligible. This late-time approximation still requires the assumed fully convective star, fixed composition and opacity law to remain applicable.

Introduce the characteristic scalesThen $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }t$. The gas-only equation of state gives the important dimensionless identity $p=bt$. The PDF's CNO cycle approximation has temperature power 13; the converted TeX's power 1.3 is an OCR error.

Since $d/dr=R^{-1}d/dx$, the mass and hydrostatic pressure support equation immediately becomeThus $dq/dx=x^{2}b$ and $dp/dx=-bq/x^2$. The stellar structure equations for nuclear heating givesoFor radiative diffusion in a star, $\kappa \rho ={\kappa }_0{\rho }^2{T}^{-3}$, and thereforeThis gives the radiative gradient coefficientIn an efficiently convective region, the monatomic adiabatic temperature gradient is $2/5$. Hence $dt/dx=(2/5)(t/p)(dp/dx)=-2q/(5x^2)$. If the radiative gradient is shallower, radiation suffices; if it is steeper than the adiabatic gradient, the Schwarzschild criterion selects convection and the shallower adiabatic profile. Thus

The regular central conditions are $q(0)=l(0)=0$ with finite positive $p(0),t(0)$. In particular $q\sim b(0)x^3/3$ and $l\sim Eb(0{)}^{2}t(0{)}^{13}{x}^3/3$, so the pressure and temperature derivatives tend to zero at the centre. The normalized surface has $q(1)=l(1)=1$. In the idealized zero-pressure, zero-temperature surface used for stellar homology, $p(1)=t(1)=0$; a physical photosphere supplies small nonzero surface values that are neglected in these bulk homology relations.

The common dimensionless profiles and boundary conditions keep $D,E$ constant. With fixed ${\kappa }_0,{ϵ}_0$, $D$ constant yields $L{\mu }^{-7}{M}^{-5}=\mathrm{constant}$. Substitution into the nuclear coefficient then gives $R^{16}\propto {\mu }^6{M}^{10}$. Consequently the CNO homology with density-dependent inverse-cubic opacity isThe Stefan–Boltzmann law gives $T_e^4\propto L/R^2\propto {\mu }^{25/4}{M}^{15/4}$. Eliminating mass, rather than dropping composition dependence, gives the Hertzsprung-Russell diagram relationThe additive constant depends on the fixed physical and structural coefficients; logarithms refer to fixed choices of units.

For physical evolution time $s$, distinct from the dimensionless temperature $t(x)$, hold stellar mass fixed and put $u=3+5X$, $u_0=3+5X_0$. The mean molecular weight approximation and the homology relation give $L=L_0(u_0/u{)}^7$. Converting hydrogen mass to helium releases energy at rate $L$, so fuel conservation is $M{E}_{H}dX/ds=-L$. ThereforeIntegration with the initial composition proves the homogeneous fuel-depletion luminosity feedback lawIt is valid while hydrogen remains and the homology assumptions hold. Fuel exhaustion occurs when $u=3$, before the formal denominator vanishes; the apparent infinite luminosity is not a physical endpoint of this stellar model.

Outside the thin burning shell, the enclosed mass and luminosity are approximately $M_c$ and $L$. Divide the hydrostatic pressure support equation by radiative diffusion in a star with constant opacity:In the constant-opacity radiative-zero giant envelope approximation, the outer pressure and temperature constants are negligible. Integration givesThe radiation pressure is $a{T}^4/3$, so the gas pressure fraction $\beta$ is constant and satisfies $(1-\beta )\Lambda =a/3$. Thus alsoThe physical branch has positive gas pressure, hence $L<4\pi cG{M}_c/{\kappa }_0$, the constant-opacity Eddington luminosity for the core mass.

Using the gas part of the equation of state, define $H=\mu \beta \Lambda /\mathcal{R}$, so $\rho =H{T}^3$. Substitution into hydrostatic balance givesConsequently, in the same deep-envelope radiative-zero approximation,Precisely, a finite outer matching point $(R_s,T_s)$ gives $T=T_s+\mu \beta G{M}_c(1/r-1/R_s)/(4\mathcal{R})$. The displayed $1/r$ profile neglects that additive outer-boundary constant; it is appropriate near a compact core when the envelope extends to much smaller temperatures.

To estimate the thin-shell luminosity, write $K_T=\mu \beta G{M}_c/(4\mathcal{R})$, so $T=K_T/r$. The hydrogen-layer heating law givesThis steep radial dependence strongly concentrates the emission near $R_c$. Continuing the outer profile across the thin emitting layer and taking a distant upper limit givesThe same radiative coefficient implies $H=16\pi ac{K}_T/(3{\kappa }_{0}L)$. Hence the high-power shell burning in a constant-opacity giant envelope satisfiesTaking the cube root gives the requested form,Thus $C$ contains only the specified opacity and heating normalizations and physical constants. The explicit coefficient uses the outer radiative profile as a thin-shell estimate; resolving the variation of $L_r$ within the emitting layer can change that coefficient. Also $\beta$ depends on $L$, so the radiation-pressure-inclusive relation is implicit, not a pure explicit eighth-power core-mass law with independently fixed $\beta$.

Assume $K>0$ and a positive central mass density. For this polytrope of index one, the spherical hydrostatic pressure support equation becomesDifferentiate the second equation and use mass conservation. With $k^2=2\pi G/K$, this givesThe regular central solution, with $\rho (0)={\rho }_c$ and ${\rho }^{\prime }(0)=0$, isThis is also the index-one solution of the Lane-Emden equation. The mass density remains positive up to its first zero, so the free surface is at $kR=\pi$, givingTaking a later zero would include a region of negative mass density and would not describe a physical star.

Integrating the mass givesConsequentlyThe radius is independent of the central mass density, whereas the mass is proportional to it; this is the special polytropic mass-radius relation at index one.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.

In the thin radiative envelope, take enclosed mass $m(r)\simeq M$, luminosity $L_r\simeq L$, and constant mean molecular weight $\mu$. Combining radiative diffusion in a star with the hydrostatic pressure support equation givesThe ideal gas relation is $\rho =AP/T$, where $A=\mu m_u/k_B$, so the opacity law impliesHere $a_{\mathrm{rad}}$ is the radiation constant, distinct from the exponent $a$. Since $\nabla =(P/T)dT/dP$, separation gives $P^{a-1}dP=C{T}^{b-1}dT$. With photospheric $T=T_{\mathrm{eff}}$, the power-law opacity radiative envelope therefore hasThis expression assumes $a,b\ne 0$ and a positive bracket on the physical branch. If $b=0$, replace $(T^b-T_{\mathrm{eff}}^b)/b$ by $\log (T/T_{\mathrm{eff}})$; if $a=0$, the integrated relation is $\log (P/P_{\mathrm{phot}})=C(T^b-T_{\mathrm{eff}}^b)/b$, with the same logarithmic limit when $b=0$.

For a spherical planet, combine the hydrostatic pressure support equation with $dm/dr=4\pi r^2\rho$:Neglect surface pressure. Since $r(m)\le R$, the hydrostatic lower bound on planetary central pressure isFor a physically usual mass density decreasing outward, the mean interior density exceeds the global mean. Hence $r(m)\le R(m/M{)}^{1/3}$, giving the sharper minimum within this class,Uniform mass density attains the sharper bound; direct integration with $m(r)=M{r}^3/R^3$ gives $P_c=3G{M}^2/(8\pi R^4)$. Real planets are centrally concentrated through compression and dense cores, so their central pressure is higher.

Using $M_{\oplus }=5.97\times 10^{24}\mathrm{kg}$, $R_{\oplus }=6.37\times 10^6\mathrm{m}$, $M_J=1.90\times 10^{27}\mathrm{kg}$, and $R_J=7.15\times 10^7\mathrm{m}$, the universal bounds are approximately $5.75\times 10^{10}\mathrm{Pa}$ for Earth and $3.66\times 10^{11}\mathrm{Pa}$ for Jupiter. The uniform-density estimates are respectively $1.72\times 10^{11}\mathrm{Pa}$ and $1.10\times 10^{12}\mathrm{Pa}$, or $1.72$ and $11.0$ megabars. Either comparison gives a Jupiter minimum about 6.4 times the Earth minimum.

Assume stellar homology: fixed dimensionless profiles, uniform fixed composition, negligible radiation pressure, ideal gas support, and radiative transport throughout the model. Mass conservation gives ${\rho }_c\propto M/R^3$. The hydrostatic pressure support equation gives $P_c\propto G{M}^2/R^4$; combining with $P\propto \rho T$ gives $T_c\propto M/R$.

Integrating specific proton–proton chain energy generation over mass givesFor radiative diffusion in a star, $T_c/R\propto {\kappa }_c{\rho }_{c}L/(R^2{T}_c^3)$, hence $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. With the Kramers' opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Equating generated and transported luminosity gives $M^{1/2}\propto R^{13/2}$, so the radiative homology with proton-proton burning and Kramers opacity relation isThe proportionality constants are fixed only within the adopted homologous family, not for arbitrary evolving stars.

For the nonrelativistic white dwarf, combine the hydrostatic pressure support equation and enclosed mass equation:Eliminating $m$ yieldsIntroduce the Lane-Emden variables for a stellar polytrope,The resulting Lane-Emden equation isThe white dwarf surface is at the first zero ${\xi }_1$ of $\theta$, with $R=a{\xi }_1$.

Let $M(r)$ be the enclosed mass. A spherical shell of thickness $dr$ has inward gravitational acceleration $GM(r)/r^2$, so hydrostatic equilibrium requiresWith $\rho =A{r}^{2}P$, these equations reduce toFor a regular central value $P(0)=P_c>0$, the first equation integrates toThe exponential function is strictly positive at every finite $r$, so the pressure and density cannot reach zero at a finite stellar surface. The equations may describe an atmosphere whose density decreases at large radius, but they admit no nontrivial hydrostatic star of finite radius. Therefore

In a negligible-mass white dwarf envelope, take enclosed mass and luminosity constant, neglect radiation pressure, and use an ideal gas with the Kramers' opacity law. Dividing hydrostatic pressure support equation by radiative diffusion in a star gives $PdP\propto (M/L)T^{15/2}dT$. Neglecting the surface integration constant produces the displayed pressure and mass density laws. It is a deep-envelope approximation, not a finite-temperature photospheric boundary condition.

The radiative temperature gradient needed to carry a spherical stellar luminosity follows by dividing radiative diffusion in a star by the hydrostatic pressure support equation. It is a logarithmic pressure gradient, not $dT/dr$ itself. Comparing it with the adiabatic temperature gradient by the Schwarzschild criterion determines whether purely radiative transport is locally unstable at uniform composition.