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An ultrarelativistic degenerate electron gas has $P=K{\rho }^{4/3}$, so it forms an index-three stellar polytrope. Insert $P_c=c_{3}G{M}^2/R^4$ and ${\rho }_c=3a_{3}M/(4\pi R^3)$ into that equation of state: the radius cancels, leaving the displayed mass. For fully ionized helium, $n_e=\rho /(2m_p)$ and $K=(3/(2\pi ){)}^{1/3}hc/(16m_p^{4/3})$. This is the ideal limiting Chandrasekhar mass, rather than a claim that every ordinary white dwarf has that mass.

For a spherical stellar polytrope of finite radius and zero surface pressure, the specific enthalpy is $(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =-GM/R$, hence $(n+1)\int PdV=-G{M}^2/R-2\Omega$. Eliminating the pressure integral with the stellar virial theorem proves the displayed formula. For a gas adiabatic stellar polytrope, $n=1/(\gamma -1)$ and $U=G{M}^2/[(5\gamma -6)R]$; a general polytropic index need not have this relation to the specific-heat ratio.

Match an $n=3/2$ stellar polytrope to the grey-atmosphere convection onset with seventeenth-power opacity. At each instant the convective entropy coefficient obeys $P=K{T}^{5/2}$, $K\propto M^{-1/2}{R}^{-3/2}$ at fixed composition, while the boundary gives $K\propto g^{1/2}{T}_e^{-21/2}$. Thus $T_e^{21/2}\propto M{R}^{1/2}$ and the first displayed luminosity law follows. The stellar virial theorem for a monatomic gas gives total energy $-3G{M}^2/(7R)$, so Kelvin-Helmholtz contraction has $\dot{R}=-7L{R}^2/(3G{M}^2)$. Integrating at fixed mass gives $R^{-67/21}-R_0^{-67/21}\propto t$. The second displayed relation holds after the initial-radius term is negligible. The entropy coefficient is spatially constant but changes during contraction.

Match a fully convective stellar polytrope of index $3/2$ to the grey-atmosphere convection onset with eighth-power temperature opacity. At the fixed matching optical depth, $T_b\propto T_e$ and $P_b\propto g^{2/3}{T}_e^{-16/3}$. The ideal-gas adiabat $P=K{\rho }^{5/3}$ gives $K\propto T_b^{5/3}{P}_b^{-2/3}\propto T_e^{47/9}{g}^{-4/9}$. The polytropic mass-radius relation gives $K\propto R{M}^{1/3}$; with $g\propto M/R^2$, this yields $T_e^{47}\propto R{M}^7$. The Stefan–Boltzmann law then gives the displayed luminosity. Composition and opacity normalization are held fixed between models.

For a homologous stellar polytrope of index $3/2$, take $P=K{\rho }^{5/3}+\mathcal{R}\rho T/\mu$ and zero surface pressure. The stellar virial theorem, with $\Omega =-6G{M}^2/(7R)$, givesThe normalized density profile fixes $A$ independently of mass and radius at fixed composition. Writing ${\overline{T}}_M=\alpha /R-\delta /R^2$ shows that its maximum is ${\alpha }^2/(4\delta )=3\mu G^2{M}^{4/3}/(49\mathcal{R}A)$, at $R=7A{M}^{-1/3}/(3G)$. Below this radius contraction increases the fraction of support provided by electron degeneracy pressure and lowers the thermal temperature. A fixed normalized profile also fixes the ratio of central to mass-averaged temperature, giving a hydrogen-burning minimum mass.

For a spherical stellar polytrope, the axial moment of inertia is $I=(8\pi /3){\int }_0^R\rho (r)r^{4}dr$. With Lane-Emden variables for a stellar polytrope, first zero ${\xi }_1$ and Lane-Emden surface mass constant ${\omega }_n$, its dimensionless form isA polytrope of index zero gives $2/5$, while a polytrope of index one gives $(2/3)(1-6/{\pi }^2)$. The latter is smaller because more of the mass lies near the center. The scalar second mass moment $\int r^{2}dm$ is $3I/2$, not the axial moment of inertia.

A fully convective, monatomic ideal gas is approximately isentropic through its deep interior. Its adiabatic index is $5/3$, so $P=K{\rho }^{5/3}$ and $T\propto K{\rho }^{2/3}$. The $n=3/2$ stellar polytrope scaling, or direct substitution of the central homology scalings into $K=P_c/{\rho }_c^{5/3}$, gives $K\propto R{M}^{1/3}$.

Near the photosphere, hydrostatic balance expressed in optical depth gives $P_{\mathrm{ph}}\sim g/{\kappa }_{\mathrm{ph}}$, with $g\propto M/R^2$. Treat $T_{\mathrm{ph}}$ as proportional to $T_{\mathrm{eff}}$ and match to the convective adiabat, neglecting a varying surface entropy jump in this homology approximation. Since $\rho \propto P/T$ and ${\kappa }_{\mathrm{ph}}\propto \rho T^{10}\propto P{T}^9$,The adiabat gives $P\propto K^{-3/2}{T}^{5/2}$, so the surface condition yieldsThe Stefan–Boltzmann law consequently gives $L_{\mathrm{surf}}\propto M^{4/7}{R}^{16/7}$. The same specific nuclear law as before still gives $L_{\mathrm{nuc}}\propto M^7{R}^{-8}$. Equating these scalings produces $R^{72/7}\propto M^{45/7}$. Thus the fully convective homology with tenth-power surface opacity isThen $T_{\mathrm{eff}}\propto M^{3/16}$, and the theoretical HR slope isIts steepness expresses the relatively weak change of effective temperature with mass. The graph using $-\log T_{\mathrm{eff}}$ has the negative of this slope.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

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.

A stellar polytrope replaces a detailed equation of state and thermal structure by $P=K{\rho }^{1+1/n}$ for $n>0$, with spatially constant $K$ and polytropic index $n$. Together with spherical hydrostatic equilibrium and mass conservation, it gives a tractable family of stellar density profiles. Introduce the Lane-Emden variables for a stellar polytropeCombining $dP/dr=-Gm\rho /r^2$ with $dm/dr=4\pi r^2\rho$ eliminates $m$ to give $r^{-2}d[r^2{\rho }^{-1}(dP/dr)]/dr=-4\pi G\rho$. Substitution of the pressure law and the dimensionless variables gives the Lane-Emden equationA regular finite-radius model for $0\le n<5$ ends at the first zero ${\xi }_1$. Its radius and mass areThese formulas supply central-to-mean density ratios and stellar homology scalings without solving a full evolutionary model. Eliminating ${\rho }_c$, for $0<n<5$ and $n\ne 3$, gives the polytropic mass-radius relation

An efficient-convection monatomic gas has $n=3/2$. The same index describes a cold, nonrelativistic electron degeneracy pressure law; at fixed composition it gives $R\propto M^{-1/3}$ for a white dwarf. Ultrarelativistic electron degeneracy pressure has $n=3$, for which the mass is independent of ${\rho }_c$ and scales as $(K/G{)}^{3/2}$. This is the origin of the Chandrasekhar limit. A radiation-dominated, approximately constant-entropy interior also approaches $P\propto {\rho }^{4/3}$ and $n=3$. The polytrope of index zero is an incompressible comparison model, useful for illustrating the role of central concentration.

The gravitational energy of a stellar polytrope, $\Omega =-3G{M}^2/[(5-n)R]$, and the stellar virial theorem make polytropes useful for contraction-energy estimates. They also provide structural models for perturbation and stability calculations. A polytrope is a structural approximation, not a complete stellar evolution calculation. One must still specify stellar nuclear fusion, opacity, energy transport, composition and boundary conditions to determine the luminosity and history. Moreover, the structural exponent $1+1/n$ need not equal the adiabatic exponent relevant to a rapid perturbation. Real stars can have different effective indices in their cores and envelopes, as well as ionization zones or varying electron degeneracy pressure.

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the Electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic Electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete Electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once Electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

Let $\Omega =-{\int }_0^{M}Gmdm/r$ be the gravitational energy. Multiply the stellar hydrostatic equation by $4\pi r^3$ and integrate through the star. Integration by parts and zero surface pressure givewhile its hydrostatic right-hand side is $-{\int }_0^{R}4\pi Gmr\rho dr=\Omega$. Hence the stellar virial theorem is

There is a genuine reference-sheet error to resolve. A monatomic ideal gas with adiabatic exponent $5/3$ has specific internal energy of an ideal gas $u=P/[\rho (5/3-1)]=3P/(2\rho )$, not the printed $3P/\rho$ on PDF page 2. The latter would give total internal energy $-\Omega$ and zero total energy, inconsistent with radiating contraction. The following derivation uses the consistent monatomic value.

A fully convective star is approximately isentropic and has an adiabatic stellar polytrope with index $3/2$. The given polytropic gravitational-energy formula yields $\Omega =-6G{M}^2/(7R)$. The virial relation gives total internal energy $U=-\Omega /2$, so total energy is $U+\Omega =-3G{M}^2/(7R)$. At constant mass with no nuclear heating, Kelvin-Helmholtz contraction obeys $d(U+\Omega )/ds=-L$, whenceHere $s$ denotes physical time.

In the plane-parallel atmosphere, the inward optical depth satisfies $d\tau =-\kappa \rho dr$, so hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. Neglecting radiation pressure, $\rho =\mu P/(\mathcal{R}T)$ and the specified opacity becomes $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{16}$. ThereforeThe grey atmosphere relation and the boundary $P(0)=0$ integrate toTo apply the Schwarzschild criterion, put $w=2+3\tau$. Logarithmic differentiation givesThis increases inward, starting from zero. Marginal convection has ${\nabla }_{\mathrm{rad}}=2/5$, so $w_b^3=136/5$. Since $T_b^{12}/T_e^{12}=w_b^3/64$, the grey-atmosphere convection onset with seventeenth-power opacity isThe purely radiative atmosphere is used only up to this boundary.

The convective relation $P=K{T}^{5/2}$ is spatially uniform in its entropy coefficient $K$ at each instant; $K$ can change as the star radiates and contracts. At the convection boundary $w_b$ is a fixed number, so $P_b\propto g^{1/2}{T}_e^{-8}$ at fixed composition and opacity normalization. ThusOn the other hand, the central scalings for the $n=3/2$ stellar polytrope give $P_c\propto G{M}^2/R^4$ and $T_c=\mu P_c/(\mathcal{R}{\rho }_c)\propto \mu GM/(\mathcal{R}R)$. Hence, at fixed composition,Equating the interior and atmosphere expressions yields $T_e^{21/2}\propto M{R}^{1/2}$. Applying the Stefan–Boltzmann law then proves the Hayashi contraction with seventeenth-power opacity relation

At fixed mass write $L=A{M}^{8/21}{R}^{\alpha }$, with constant $A$ and $\alpha =46/21$. The contraction equation is $\dot{R}=-C{R}^{\alpha +2}$, where $C=7A{M}^{8/21}/(3G{M}^2)$. Direct integration givesMultiplying through by $R^{\alpha +1}$ and substituting the current luminosity givesFor the stated very large initial radius, or after that initial-radius term becomes negligible,This is the luminosity-age relation within the contraction model, rather than an assertion that a physical star begins at literally infinite radius.

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.

Efficient convection makes the interior follow a nearly adiabatic process and maintains uniform specific entropy. For the fully ionized monatomic ideal gas, the specific-heat ratio is $5/3$, so the adiabatic temperature gradient gives $d\log T/d\log P=2/5$. Integrating at a fixed time givesThe coefficient is spatially constant because the composition and specific entropy are uniform. It can change as the star loses energy. Equivalently the gas is an adiabatic stellar polytrope with $P=K_{\rho }{\rho }^{5/3}$ and polytropic index $n=3/2$; $K_{\rho }$ and the pressure-temperature coefficient $K$ are different quantities.

This fixes the dimensionless Lane-Emden equation profile throughout a stellar homology sequence. Writing $\rho (r)={\rho }_{c}f(r/R)$ in the mass integral gives ${\rho }_c=C_{\rho }M/R^3$, with a fixed positive shape constant $C_{\rho }$. Integrating the stellar hydrostatic equation through the same profile, with negligible surface pressure compared with central pressure, gives $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

At the approximate photosphere, take $T=T_e$ and use the ideal gas relation $\rho =\mu P/(\mathcal{R}T)$. The prescribed opacity givesThe stellar surface boundary condition and surface gravity of a star therefore implyThis is the temperature-four special case of convective contraction with power-law opacity.

Using the Stefan–Boltzmann law, the luminosity isFor $n=3/2$, the supplied gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, so the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. During Kelvin-Helmholtz contraction, with constant mass and no significant nuclear source, $L=-dE/dt$. ThusMultiplying by $-(7/2)R^{-9/2}$ and integrating givesThe sign makes the radius decrease as time increases.

When the accumulated contraction term dominates the initial-radius term, this yields $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Combining it with the central ideal gas temperature gives the late-time central-temperature scalingReplacing $t-t_0$ by $t$ gives the stated late-time form. The asymptotic step also requires a time long enough to forget the initial radius, not merely a choice of zero for the age.

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.