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For a finite regular polytrope with fixed index, the scale ${\alpha }^2\propto K{\rho }_c^{(1-n)/n}$ implies $M^{n-1}{R}^{3-n}\propto K^n$. Thus, for $n\ne 3$, $R\propto K^{n/(3-n)}{M}^{(1-n)/(3-n)}$. For $n=3/2$, $R\propto K{M}^{-1/3}$. Fixed composition makes the cold electron degeneracy pressure coefficient approximately fixed, but a convective ideal gas has $K$ controlled by entropy. A stellar sequence with changing entropy therefore need not follow the fixed-$K$ negative slope.

For finite-radius regular polytropes of one fixed index $n$ and composition, let ${\xi }_1$ be the first zero of the Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. Hydrostatic scaling gives $GM/R=(n+1)(Y_n/{\xi }_1)\mathcal{R}{T}_c$. Fixed central temperature therefore implies $M\propto R$. This is not the fixed-entropy-constant sequence; if the index or composition changes, their prefactors cannot be discarded.

An idealized self-gravitating interior with constant mass density. Negligible surface pressure and spherical hydrostatic equilibrium give $M=4\pi \rho R^3/3$ and the uniform-density planetary pressure profile. It is the $n=0$ polytrope interpretation, rather than a literal substitution into $P=K{\rho }^{1+1/n}$. It approximates weak compression and misses real compositional stratification.

The classical Lane-Emden equation is the dimensionless equationfor a spherically symmetric polytrope in three dimensions. The Generalized Lane-Emden equation extends it to arbitrary dimensions and to functions without radial symmetry.

Write the polytropic law as $p=K{\rho }^{1+1/n}$, with constant $K$ and $n>0$. Its pressure potential, or polytropic enthalpy relative to zero density, isHydrostatic balance is $dH/dr=-d\psi /dr$. If the surface has zero density and potential ${\psi }_0$, integration gives $H={\psi }_0-\psi$. At the centre, $\Delta \psi \equiv {\psi }_0-{\psi }_c=(n+1)K{\rho }_c^{1/n}>0$. ThereforeInsert $\psi ={\psi }_0-\Delta \psi \theta$ and $\rho ={\rho }_c{\theta }^n$ into the spherical Poisson equation for Newtonian gravity:The central conditions are $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$; the latter follows from a regular spherically symmetric centre. This reproduces the supplied coordinate and potential scaling rather than introducing an unrelated normalization.

For the sound-speed task, the adiabatic sound speed of a perfect polytropic gas isA diatomic gas with active rotational but frozen vibrational degrees of freedom has $\gamma =7/5$ and $n=5/2$. With the requested surface-potential gauge ${\psi }_0=0$,The interior potential is then negative. Choosing zero potential at the surface is legitimate; for a finite isolated sphere it is a different additive gauge from setting potential to zero at infinity. Both choices cannot generally be imposed simultaneously.

For the mass-radius task, let ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. A finite-radius polytrope hasThusAt fixed central temperature, fixed composition and fixed polytropic index, $M\propto R$. These are homologous polytropes at fixed central temperature. They do not in general have constant $K$; imposing constant entropy instead gives a different polytropic mass-radius sequence. If $n$ or $\mu$ varies between stars, the displayed dimensionless structure factor and composition factor must be retained: equal $T_c$ alone is not a universal exact proportionality for arbitrary stellar models.

A spherical harmonic of angular degree $\ell$ has horizontal wavenumber $k_h=L/r$, with $L=\sqrt{\ell (\ell +1)}$. The local acoustic dispersion relation therefore gives the radial wavenumberIn the high-order regime a travelling stellar acoustic mode accumulates phase $\int k_{r}dr$ across its radial cavity. Reflection at the inner turning point and at the near-surface boundary fixes additional phases. A round trip must return the same complex amplitude, so WKB quantization givesHere $n$ counts radial oscillations, up to the choice of origin for radial order, and $\alpha$ incorporates both reflection phases. The outer acoustic reflection need not coincide exactly with the geometric stellar radius; replacing it by $R$ and absorbing its correction in $\alpha$ is the leading asymptotic approximation. The model neglects buoyancy, perturbations of self-gravity and rapid background variations on a wavelength.

Resolved observations of the Sun measure line-of-sight surface velocity through the Doppler effect, or brightness fluctuations. Projection on spherical harmonics identifies $\ell$ and azimuthal order $m$; a temporal Fourier transform identifies the mode angular frequencies $\omega$. The sequences of peaks of successive radial order form ridges in the frequency-degree diagram. Their ordering and comparison with a reference stellar oscillation model identify $n$; the surface image alone does not count inaccessible radial nodes. Fit the residual phase of the radial quantization relation across the observed mode set to estimate $\alpha$. A simultaneous fit with the sound-speed profile, or a calibrated reference model, is needed: $\alpha$ is not determined independently by one mode, and an integer shift of the numbering of $n$ shifts $\alpha$ oppositely. In a more accurate fit the surface phase depends on frequency. These steps constitute helioseismology.

Set $a(r)=c(r)/r$ and $w=\omega /L$. Assume $a$ decreases outwards in the region to be inverted. With the surface approximation above, the Duvall law isThus a suitably scaled mode set samples one function of $w$. Smooth the measured values before differentiating; inversion amplifies observational noise. At a finite outer reference point $r_s$, the already known contribution exterior to $r_s$ must be subtracted. For the remaining cavity, let $w_s=a(r_s)$ and $h(b)=-d\ln r/db$. Changing integration variable from radius to $b=a(r)$ givesThe boundary term vanishes because the square root vanishes at $b=w$. Now multiply the derivative by $w/\sqrt{a^2-w^2}$ and interchange the integrals. The kernel identityfollows by setting $w^2=b^2+(a^2-b^2){\sin }^2\theta$ and then integrating $1/[b^2{\cos }^2\theta +a^2{\sin }^2\theta ]$. It yields the Abel inversion of stellar acoustic travel times:Knowing the monotone relation $r(a)$ then gives $c=ar$. Observed turning points delimit the region recoverable without extrapolation; an unknown surface contribution or nonmonotone $a$ prevents this simple unique inversion.

There are genuine inconsistencies in the printed numerical example. Since $w$ and $w_0$ have dimensions of angular frequency, $F$ must have dimensions of time, but the displayed empirical right-hand side is dimensionless. Moreover its derivative supplies a factor $1/w_0$, whereas the requested inversion supplies $1/w_0^2$. The dimensionally repaired relation that produces the requested inversion integral isSubstitution in the boxed Abel transform inversion gives precisely the exponent coefficient $2a^2/(\pi w_0^2)$, with the square-root denominator inside the integral. This repairs the first normalization but does not repair the later algebra.

Indeed, put $x_s=w_s/a$ and $b=\sqrt{1-x_s^2}$. The required elementary integrals evaluate toThe finite-reference-point answer following from $F_c$ is thereforeIt still depends on $w_s$. The later printed profile suppresses that dependence, which requires the additional surface limit $w_s\to 0$, $r_s\to R$. Even in that limit it is not the result of this integral. To see this exactly, write $\mathcal{L}=\ln (R/r)$ and $y=a^2/(\pi w_0^2)$. The integral givesThe minus branch is selected by $a\to 0$ at the surface. Its decreasing-radius branch extends only to $\ln (R/r)=1/6$, where $a^2=\pi w_0^2/6$; it cannot describe $r=R/e$. The stated lower bound on the largest sampled $w$ does not fix this algebraic obstruction. In contrast, squaring and rearranging the printed closed profile would require $\mathcal{L}=18y-81y^2$. This differs identically from $2y-6y^2$, so the two printed steps cannot both follow from the same data.

The surface polytrope test can nevertheless be performed exactly, and clarifies the final discrepancy. For a plane-parallel polytropic atmosphere of index $m=3$, take $g=GM/R^2$ constant and $z=R-r$. The stellar hydrostatic equation and $p=K{\rho }^{4/3}$ give $4p/\rho =gz$, hence the surface sound speed of a polytropic star isThe exponent in the equilibrium polytropic equation of state need not equal the perturbation adiabatic exponent $\gamma$. For the corrected inversion above, expansion at $z/R\ll 1$ gives $c^2\simeq \pi w_0^{2}Rz/2$, so matching the surface polytrope fixes $w_0^2=\gamma GM/(2\pi R^3)$.

Alternatively, if the later printed closed profile is adopted as a separate imposed model, its surface expansion gives $c^2\simeq \pi w_0^{2}Rz/18$. It also has the correct square-root depth dependence, but requiresAt $r=R/e$ that imposed profile then givesThe claimed coefficient $\pi /3$ is inconsistent with this substitution; $1/e\ne \pi /3$. Thus the surface scaling is demonstrable, but the empirical normalization, the requested closed inversion and the final numerical coefficient cannot all be asserted as printed. These discrepancies are visible in the original PDF, not introduced by its TeX conversion.

The mass-radius curve of solar-composition substellar objects reflects the transition from weak compression to pressure ionization and electron degeneracy pressure, followed by sustained hydrogen burning. A schematic joining representative object classes is:

The illustration is not an age-specific numerical evolutionary model. In particular, ice giants contain much more heavy material than a solar-composition giant, so a single uniform-composition equation of state does not describe the entire joined curve.

Planet/brown-dwarf naming conventions and deuterium burning do not define a universal discontinuity in the mass-radius relation. Composition, age and irradiation move the curves; the hydrogen-burning transition changes the long-term energy source more fundamentally.

Hydrostatic equilibrium requiresFor the polytrope $p=K{\rho }^{1+1/n}$, the specific enthalpy isThus $h+\Phi$ is constant. Since $\rho =p=0$ at $r=R$, this constant is ${\Omega }^2{R}^2/2$, andfor $0\le r\le R$, with both fields zero outside the model star.

The coefficient $K$ in a polytropic equation of state $p=K{\rho }^{\Gamma }$. It is spatially constant for one simple polytrope, but need not stay constant between different equilibrium models. For a fixed-composition ideal gas with $\Gamma$ equal to its adiabatic index, $K$ parametrizes specific entropy: $s=c_v\ln K+\text{constant}$ with a suitable dimensional reference inside the logarithm.