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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.

For completely degenerate, nonrelativistic electrons in fully ionized hydrogen and helium with hydrogen mass fraction $X$, $n_e=(1+X)\rho /(2m_p)$. Filling the spin-degenerate momentum states to $p_F$ gives $n_e=8\pi p_F^3/(3h^3)$ and $P_e=(1/3){\int }_0^{p_F}(p^2/m_e)(8\pi p^2/h^3)dp=8\pi p_F^5/(15m_e{h}^3)$. Eliminating $p_F$ gives $P_e=K{\rho }^{5/3}$, whereIt is a polytropic index $3/2$ relation. Complete degeneracy suppresses thermal dependence, while the nonrelativistic assumption requires $p_F\ll m_{e}c$.

A cold nonrelativistic white dwarf supported by electron degeneracy pressure has a polytropic equation of state $P=K_{\mathrm{NR}}{\rho }^{5/3}$, so its polytropic index is $3/2$. The Lane-Emden mass formula and polytropic mass-radius relation give $R\propto K_{\mathrm{NR}}{M}^{-1/3}$ at fixed composition. As density rises, relativistic corrections soften the exponent toward $4/3$ and lead toward the Chandrasekhar mass regime.

A stellar polytrope approximates a spherical self-gravitating star by the structural relationwhere $K$ is constant and $n$ is the polytropic index. Combining hydrostatic equilibrium with mass conservation givesThe Lane-Emden variables for a stellar polytrope separate the dimensional scales from the profile:For $n>0$ these substitutions give the Lane-Emden equationThe central conditions express finite central mass density and spherical symmetry. If ${\xi }_1$ is the first zero, a zero-pressure surface hasthe Lane-Emden mass formula. For $0<n<5$ this surface is at finite radius. For example, the polytrope of index one has $\theta =\sin \xi /\xi$ and ${\xi }_1=\pi$, giving the density profile shown in the overview figure.

An isentropic ideal monatomic gas has $p\propto {\rho }^{5/3}$ and hence $n=3/2$, a useful approximation for a fully convective ideal-gas star. Nonrelativistic electron degeneracy pressure has the same density exponent. The exponent $4/3$, giving $n=3$, occurs for ultrarelativistic electron degeneracy and in appropriate radiation-dominated stellar models. The structural index should not be confused with the adiabatic index controlling a perturbation unless the model is actually isentropic.

Polytropes give useful mass-radius scalings and central concentration with very few parameters. They do not by themselves specify nuclear fusion, energy transport, or the star's thermal evolution. Nor does a structural profile alone establish dynamical stability; that requires the response of pressure to the perturbation.

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.

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.

Put $\mathcal{R}=k_B/m_u$, with $m_u$ the atomic mass constant and $a_r$ the radiation energy-density constant. The ideal gas and radiation pressure contributions obey $P_g=\rho \mathcal{R}T/\mu$ and $P_r=a_r{T}^4/3$. With the constant stellar gas-pressure fraction, eliminate $T=\mu \beta P/(\mathcal{R}\rho )$ to getHenceSince $\mu$ and $\beta$ are spatially constant, so is $K$. Comparing with $P=K{\rho }^{1+1/n}$ gives polytropic index $n=3$. This structural stellar polytrope relation does not assert that every perturbed fluid parcel has stellar adiabatic exponent $4/3$.

For polytropic index three, the Lane-Emden mass formula cancels the central mass density:Substituting the preceding constant $K$ givesFor fixed composition this is $M\propto \sqrt{1-\beta }/{\beta }^2$, as required. Squaring and rearranging yields the Eddington quartic relation, $(1-\beta )/{\beta }^4\propto {\mu }^4{M}^2$. The larger masses in this model have smaller gas fractions and relatively more radiation support.

For a fixed polytropic index $n>0$ and fixed equation-of-state constant $K$, let ${\xi }_1$ be the first zero of the regular Lane-Emden equation solution. A finite-radius model requires such a zero; for the usual nonnegative indices this holds for $n<5$. The surface radius and mass follow from $r=\alpha \xi$ and $\rho ={\rho }_c{\theta }^n$:The last equality integrates the Lane-Emden equation; define the positive Lane-Emden surface mass constant ${\omega }_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. Substituting $\alpha =C_1{\rho }_c^{(1-n)/(2n)}$ givesThe central-density exponents cancel when the required powers are taken. Thus the polytropic mass-radius relation isIt is independent of ${\rho }_c$, with $n,K$ and composition held fixed. For $n=1$ the radius is fixed; for $n=3$ the mass is fixed. These limiting powers need no division by a vanishing exponent.

For $n=0$, the literal pressure-density power and the supplied expression for $\alpha$ are singular. Interpret this case separately as an incompressible planetary interior with constant density. Then $M=4\pi \rho R^3/3$, or $R\propto M^{1/3}$ at fixed density. A weakly compressed rocky body, or a rough uniform-density approximation to Earth, is an example; realistic terrestrial planets are stratified and compressible.

For $n=3/2$, $P\propto {\rho }^{5/3}$. This describes a cold nonrelativistic degenerate electron gas of fixed composition, or a fully convective monatomic ideal gas at fixed entropy. Examples are a nonrelativistic white dwarf, a sufficiently cooled partly degenerate brown dwarf as an approximation, and an approximately fully convective low-mass star for the ideal-gas version. The fixed-$K$ scaling is $R\propto M^{-1/3}$. It must not be applied to an entire main-sequence stellar sequence with different entropies; the entropy dependence of a polytropic mass-radius relation explains the distinction.

Regularity at the centre requires finite ${\rho }_c,P_c$ and $M(0)=0$. Then $M(r)=4\pi {\rho }_c{r}^3/3+o(r^3)$, so $P^{\prime }(0)=0$. At a free stellar surface, $P(R)=0$; more generally it matches the external pressure. For $P=K{\rho }^2$ with $K>0$ and vacuum outside, this also gives $\rho (R)=0$.

Inside the star, cancel positive $\rho$ to obtain $2K{\rho }^{\prime }=-GM/r^2$. Differentiate and use mass conservation:Thus the Lane-Emden equation of polytropic index one isFor example, the regular solution is $\rho ={\rho }_c\sin x/x$, with its central value interpreted by continuity, and the first vacuum surface is at $x=\pi$. This also verifies the stated central and surface boundary conditions.

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

Take $\gamma >1$ for the ideal gas. At fixed cylindrical radius, put $\mathcal{R}=k/(\mu m_p)$, where $k$ is the Boltzmann constant, $\mu$ the mean molecular weight and $m_p$ the proton mass. Also write $a=9\alpha \Omega /4$. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law requireConsequently, with $K=A/a$,Integrating from the midplane yieldsThe polytropic vertical structure in stellar gravity therefore has, for $|z|\le H$,whereThis is a polytropic equation of state $P\propto {\rho }^{1+1/\beta }$ with polytropic index $\beta$. For $|z|>H$, set $\rho =P=0$; temperature has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential disk scale height.

When $s=0$, part i gives the polytropic index $n=3$. Stellar homology givesbecause gas pressure is the fraction $\beta$ of the pressure required by hydrostatic equilibrium. The Kramers' opacity law therefore scales asWith constant $\eta$, the opacity relation from part i gives

For an $n=3$ polytrope, $M\propto (K/G{)}^{3/2}$. The explicit $K(\beta )$ of part i obeysso the Eddington quartic relation isat fixed composition. Substitution gives

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.