Eddington Standard Model

The Eddington Standard Model is the first stellar interior that a physicist can carry in their head. Real stars are governed by four coupled differential equations — hydrostatic equilibrium, mass continuity, energy generation, and energy transport — closed by an equation of state, an opacity law, and nuclear reaction rates, and in general only a computer can integrate them. Eddington, writing before nuclear physics existed and before any of those inputs were known, cut through the whole tangle with one audacious postulate: that the fraction $\beta \equiv P_{\mathrm{gas}}/P$ of gas pressure to total pressure is the same number at every point in the star. That single assumption is not exact, but for main-sequence stars it is astonishingly good, and it does something magical to the mathematics — it turns the combined ideal-gas-plus-radiation mixture into a polytrope of fixed index $n=3$ . Once the star is an $n=3$ polytrope its mechanical structure decouples completely from its thermal structure and is described by a single dimensionless equation, the Lane–Emden equation of index three, whose universal solution supplies every structure constant the model needs. Feed those constants back into the definition of $\beta$ and Eddington's celebrated quartic drops out: a relation that fixes the radiation fraction $1-\beta$ from nothing but the stellar mass $M$ and the mean molecular weight $\mu$ . The quartic says something profound — radiation pressure is a negligible sliver in the Sun but grows as $M^2$ , taking over the star's support as the mass climbs, and hinting at an upper mass limit long before anyone could compute one. This article recalls the polytrope machinery from aq-astro-polytropic-models by name and asks what Eddington's one assumption buys: a hand-calculable star, the mass–luminosity relation, and the conceptual seed of the Chandrasekhar mass.

Key takeaways

The Eddington Standard Model assumes the gas-to-total pressure ratio $\beta = P_{\mathrm{gas}}/P$ is constant throughout a star; under that one assumption the combined gas-plus-radiation equation of state becomes a polytrope of index $n=3$ , $P=K\rho^{4/3}$ .
Constant $\beta$ forces $T\propto\rho^{1/3}$ , so both $P_{\mathrm{gas}}\propto\rho T$ and $P_{\mathrm{rad}}\propto T^4$ scale as $\rho^{4/3}$ — the effective adiabatic index is $\gamma=4/3$ , which is exactly $n=3$ .
The structure obeys the $n=3$ Lane–Emden equation, whose solution gives the surface constant $(-\xi^2\theta')_{\xi_1}=2.018$ at $\xi_1=6.897$ and a central-to-mean density ratio $\rho_c/\bar\rho=54.18$ .
Eddington's quartic fixes the radiation fraction from mass and composition alone: $1-\beta \approx 0.003\,\mu^4\beta^4\,(M/M_\odot)^2$ . For the Sun ( $\mu\approx0.62$ ) this gives $1-\beta\approx4\times10^{-4}$ , and it rises as $M^2$ .
Combined with radiative transport the model yields $L=(1-\beta)\,L_{\mathrm{Edd}}$ and the main-sequence mass–luminosity relation $L\propto M^3$ , a landmark success achieved without solving the full structure equations.
Key idea: if the gas-to-total pressure ratio $\beta$ is constant, radiation and gas pressure fuse into an effective $n=3$ polytrope $P=K\rho^{4/3}$ , and Eddington's quartic $1-\beta\approx0.003\,\mu^4\beta^4(M/M_\odot)^2$ fixes $\beta$ from the mass and composition alone.

The core insight

If the gas-to-total pressure ratio $\beta$ is constant, radiation and gas pressure combine into an effective polytrope $P = K\rho^{4/3}$ (index $n=3$ ), and Eddington's quartic $1-\beta\approx 0.003\,\mu^4\beta^4(M/M_\odot)^2$ fixes $\beta$ from the stellar mass and composition alone.

Where this lives today

Eddington's model is a century old, yet its logic runs straight into the front line of modern astrophysics. Its clearest descendant is the study of the most massive stars: the quartic's $1-\beta\to1$ behaviour is the seed of the observed upper mass limit near $150$ – $300\,M_\odot$ , above which radiation pressure and pulsational instability tear a star apart. Luminous blue variables such as Eta Carinae sit close to their Eddington limit and shed mass in giant eruptions precisely because $L\to L_{\mathrm{Edd}}$ — the same $L=(1-\beta)L_{\mathrm{Edd}}$ relation the model derives. The Eddington luminosity itself is now a universal yardstick far beyond stars: it caps the steady accretion rate of black holes and sets the reference against which super-Eddington accretion in ultraluminous X-ray sources and early quasars is measured. And the $n=3$ polytrope's mass-independent-of-density property remains the conceptual template for the Chandrasekhar mass that governs Type Ia supernovae and neutron-star limits. Modern stellar-evolution codes such as MESA no longer assume constant $\beta$ — they integrate the full equations with tabulated opacities — but they are routinely initialized on polytropes, and every student still meets the star's interior first through Eddington's one brilliant assumption.

The physical picture

Picture the pressure that holds a star up against its own gravity as a partnership between two very different agents. One is the jostling of gas particles — protons, electrons, ions — that provides $P_{\mathrm{gas}}$ ; the other is the momentum carried by the flood of photons streaming outward, which provides $P_{\mathrm{rad}}$ . At every radius the total support is $P = P_{\mathrm{gas}} + P_{\mathrm{rad}}$ , and the two partners split the load in some ratio $\beta:(1-\beta)$ . Eddington's bet is that this split is democratic and unchanging — the same $\beta$ deep in the core and out near the photosphere. That is not obvious, and it is not exactly true, but in a radiative star the way opacity and flux conspire keeps $\beta$ remarkably flat, and the payoff for pretending it is exactly constant is enormous. The magic of a constant $\beta$ is that it locks temperature to density. Gas pressure scales as $\rho T$ and radiation pressure as $T^4$ ; demanding their ratio be fixed forces $T\propto\rho^{1/3}$ . Substitute that back and both pieces — and therefore the total — become proportional to $\rho^{4/3}$ . The star has become a polytrope with $\gamma=4/3$ , i.e. index $n=3$ , and all of its mechanical structure now follows from one dimensionless equation. The first figure follows this logic and shows how the radiation partner, a mere sliver in the Sun, swells to nearly half the support in a very massive star.

Figure

The pressure partnership and why constant β forces an n=3 polytrope. Left: gas (amber) and radiation (red) shares of the total support — β = 0.9996 in the Sun but only β = 0.56 in a 100 solar-mass star, so radiation grows with mass. Right: constant β locks T ∝ ρ^(1/3), making both pressures scale as ρ^(4/3), the n = 3 polytrope

Why is $n=3$ such a privileged index? Because for an $n=3$ polytrope the total mass depends only on the polytropic constant $K$ and fundamental constants — not on the central density. Squeeze the star harder and it does not gain mass; the mass is pinned. That single mathematical fact is the seed of two of the deepest results in stellar physics. Turned toward degenerate matter it becomes the Chandrasekhar mass, the fixed ceiling above which no white dwarf can stand. Turned toward Eddington's ordinary gas-plus-radiation star it means that raising the mass cannot just raise the density — it must force radiation to shoulder more of the support, driving $1-\beta$ upward. The second figure is the quantitative heart of the model: Eddington's quartic, plotting how the radiation fraction grows with stellar mass.

Figure

Eddington's quartic 1−β ≈ 0.003 μ⁴ β⁴ (M/M⊙)² with μ = 0.62, on log–log axes. The radiation fraction is a negligible 4×10⁻⁴ at one solar mass, reaches a few percent near 10 M⊙, and approaches one-half by 100 M⊙ — the growth as M² foreshadows an upper mass limit

The core equations

The starting point is the total pressure of a mixture of ideal gas and blackbody radiation. Both are present in any hot stellar interior; the gas obeys the ideal law and the radiation supplies one-third of its energy density as pressure. Everything else in the model is a consequence of asking that their ratio be spatially constant.

P = P_{\mathrm{gas}} + P_{\mathrm{rad}} = \frac{\rho k_B T}{\mu m_H} + \frac{1}{3}a T^4 \qquad (1)

where: $P$ — total pressure; $\rho$ — mass density; $T$ — temperature; $\mu$ — mean molecular weight; $m_H$ — atomic mass unit; $k_B$ — Boltzmann constant; $a=4\sigma/c=7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ — radiation constant. Reading it: the first term is the kinetic pressure of the particles, the second the pressure of the photon gas. The defining assumption of the model is a single symbol held constant. Split the total pressure into its gas and radiation shares and demand that the split not vary with position.

\beta \equiv \frac{P_{\mathrm{gas}}}{P}, \qquad 1-\beta = \frac{P_{\mathrm{rad}}}{P}, \qquad \beta = \mathrm{const} \qquad (2)

where: $\beta$ — gas-to-total pressure ratio, taken independent of radius. Reading it: $\beta\to1$ is a gas-supported star, $\beta\to0$ a radiation-supported one; the whole model lives in between and pins $\beta$ from the mass. Eliminating the temperature between the two shares — the algebra is carried out in the first derivation below — collapses the mixture into a single power law. Because $\beta$ is constant the bracket is a constant, so this is genuinely a polytrope.

P = \left[\frac{3k_B^4}{a\,\mu^4 m_H^4}\,\frac{1-\beta}{\beta^4}\right]^{1/3}\rho^{4/3} \equiv K\rho^{4/3} \qquad (3)

where: $K$ — polytropic constant, fixed by $\beta$ and the composition $\mu$ ; the exponent $4/3$ means $\gamma=1+1/n=4/3$ , hence $n=3$ . Reading it: a hotter, more radiation-heavy star (smaller $\beta$ ) has a larger $K$ and is more distended. With $P=K\rho^{4/3}$ in hand, hydrostatic equilibrium and mass continuity fuse into the dimensionless Lane–Emden equation of index three — the machinery is owned by aq-astro-polytropic-models and recalled here by name.

\frac{1}{\xi^2}\frac{\mathrm{d}}{\mathrm{d}\xi}\!\left(\xi^2\frac{\mathrm{d}\theta}{\mathrm{d}\xi}\right) = -\theta^{\,3}, \qquad \theta(0)=1,\ \theta'(0)=0 \qquad (4)

where: $\xi$ — dimensionless radius, $r=\alpha\xi$ ; $\theta$ — dimensionless density variable, $\rho=\rho_c\theta^{\,3}$ ; $\rho_c$ — central density; $\alpha^2=(n+1)K\rho_c^{1/n-1}/(4\pi G)$ — the radial scale. Reading it: the surface is the first zero $\xi_1=6.897$ , and the tabulated constants $(-\xi^2\theta')_{\xi_1}=2.018$ , $\rho_c/\bar\rho=54.18$ describe every $n=3$ star. Feeding the $n=3$ mass back into $K(\beta)$ eliminates the central density and yields Eddington's quartic — the model's signature result. It fixes the radiation fraction from mass and composition with no free parameters.

1-\beta = \frac{\pi}{48}\,\frac{a\,G^3\,\mu^4 m_H^4}{k_B^4}\,\frac{M^2}{(-\xi^2\theta')^2_{\xi_1}}\,\beta^4 \;\approx\; 0.003\,\mu^4\beta^4\!\left(\frac{M}{M_\odot}\right)^2 \qquad (5)

where: $M$ — total stellar mass; $G$ — gravitational constant; $(-\xi^2\theta')_{\xi_1}=2.018$ — the $n=3$ surface constant. Reading it: the right side grows as $M^2$ , so more massive stars carry a larger radiation fraction — the physical content of the whole model. Closing the model with radiative transport gives the standard-model luminosity, which turns out to be a fixed fraction $1-\beta$ of the Eddington luminosity. This is the bridge from structure to the observable mass–luminosity relation.

L = \frac{4\pi c G M (1-\beta)}{\kappa} = (1-\beta)\,L_{\mathrm{Edd}}, \qquad L_{\mathrm{Edd}} = \frac{4\pi c G M}{\kappa} \qquad (6)

where: $L$ — luminosity; $c$ — speed of light; $\kappa$ — opacity (mass absorption coefficient); $L_{\mathrm{Edd}}$ — Eddington luminosity, at which radiation force balances gravity. Reading it: since $1-\beta$ rises with mass and $L_{\mathrm{Edd}}\propto M$ , the luminosity climbs steeply — roughly $L\propto M^3$ for intermediate masses.

Derivation AFrom a constant pressure ratio to an n=3 polytrope

The goal is to show that the single assumption of a spatially constant $\beta$ converts the two-component equation of state — ideal gas plus blackbody radiation — into a pure power law $P=K\rho^{4/3}$ , and hence into a polytrope of index $n=3$ . The strategy is to write each pressure share as a fixed multiple of the total, use the gas law to express temperature in terms of $P$ and $\rho$ , substitute that into the radiation term to remove $T$ entirely, and collect the surviving powers. The bracketed coefficient will be built only from $\beta$ and fundamental constants, so when $\beta$ is constant it is constant too — which is exactly the definition of a polytrope.

Step 1. Write both pieces of the pressure as fixed fractions of the total: $P_{\mathrm{gas}}=\beta P=\rho k_B T/(\mu m_H)$ and $P_{\mathrm{rad}}=(1-\beta)P=\tfrac{1}{3}aT^4$ . Why: these are two independent expressions for the same $P$ , and the constancy of $\beta$ is what makes the coefficients ahead pure constants rather than functions of radius.

Step 2. Solve the gas relation for temperature, $T = \beta P\mu m_H/(\rho k_B)$ . Why: temperature is the variable that couples the two pressure terms; expressing it through $P$ and $\rho$ is the move that will let us eliminate it and leave a relation between $P$ and $\rho$ alone.

Step 3. Substitute this $T$ into the radiation relation, giving $(1-\beta)P = \tfrac{a}{3}\big(\beta P\mu m_H/(\rho k_B)\big)^4$ . Why: the temperature has now vanished; a single power of $P$ on the left faces a fourth power of $P$ on the right, so the equation can be solved for $P$ as a function of $\rho$ .

Step 4. Cancel one power of $P$ and rearrange: $P^3 = \dfrac{3(1-\beta)}{a}\Big(\dfrac{\rho k_B}{\beta\mu m_H}\Big)^4$ , and take the cube root to obtain $P = \Big[\dfrac{3k_B^4}{a\mu^4 m_H^4}\dfrac{1-\beta}{\beta^4}\Big]^{1/3}\rho^{4/3}\equiv K\rho^{4/3}$ . Why: the exponent $4/3$ is forced, and with $\beta$ constant the bracket $K$ is a genuine constant — this is a polytrope with $\gamma=4/3$ , i.e. $n=1/(\gamma-1)=3$ .

Step 5. Insert $P=K\rho^{4/3}$ into the combined hydrostatic-plus-continuity relation $\dfrac{1}{r^2}\dfrac{\mathrm{d}}{\mathrm{d}r}\big(\dfrac{r^2}{\rho}\dfrac{\mathrm{d}P}{\mathrm{d}r}\big)=-4\pi G\rho$ , and non-dimensionalize with $\rho=\rho_c\theta^3$ , $r=\alpha\xi$ to reach the $n=3$ Lane–Emden equation with first zero $\xi_1=6.897$ . Why: this is the standard polytropic reduction owned by aq-astro-polytropic-models; the constant $\beta$ is the only ingredient that made the reduction applicable to a radiative star. *Verified: numerical integration of the $n=3$ Lane–Emden equation reproduces $\xi_1=6.897$ , $(-\xi^2\theta')_{\xi_1}=2.018$ , and $\rho_c/\bar\rho=\xi_1^3/[3(-\xi^2\theta')_{\xi_1}]=54.18$ .*

Go deeper · Derivation B · 1 minEddington's quartic for the radiation fraction

Derivation BEddington's quartic for the radiation fraction

The aim is to convert the composition-dependent polytropic constant $K(\beta)$ into a direct relation between the radiation fraction $1-\beta$ and the stellar mass $M$ . The key lever is a property unique to $n=3$ : the polytropic mass depends on $K$ alone and not on the central density $\rho_c$ . We write the $n=3$ mass in terms of $K$ , square it to isolate $K^3$ , then set that equal to the $K^3$ delivered by the equation of state. Everything except $\beta$ , $M$ , and $\mu$ cancels, leaving a quartic in $\beta$ whose numerical coefficient we pin against the Sun.

Step 1. The polytropic mass is $M = 4\pi\alpha^3\rho_c\,(-\xi^2\theta')_{\xi_1}$ ; for $n=3$ the combination $\alpha^3\rho_c = (K/\pi G)^{3/2}$ is independent of $\rho_c$ , so $M = 4\pi\,(K/\pi G)^{3/2}\,(-\xi^2\theta')_{\xi_1}$ . Why: this is the mathematical miracle of $n=3$ — the central density drops out, so a single mass corresponds to each value of $K$ , and hence to each $\beta$ .

Step 2. Square the mass relation to remove the fractional power: $K^3 = \dfrac{\pi G^3 M^2}{16\,(-\xi^2\theta')^2_{\xi_1}}$ . Why: the equation of state gives $K^3$ cleanly (the cube root in $K$ disappears when cubed), so matching powers of $K^3$ is the natural way to equate the two descriptions.

Step 3. From the effective polytrope, $K^3 = \dfrac{3k_B^4}{a\mu^4 m_H^4}\dfrac{1-\beta}{\beta^4}$ . Equate the two expressions for $K^3$ . Why: one $K^3$ came from the mechanics (mass and gravity), the other from the thermodynamics (gas plus radiation); setting them equal is what ties the star's mass to its radiation fraction.

Step 4. Solve for $1-\beta$ : $\ 1-\beta = \dfrac{\pi}{48}\dfrac{a\,G^3\mu^4 m_H^4}{k_B^4}\dfrac{M^2}{(-\xi^2\theta')^2_{\xi_1}}\,\beta^4$ . Inserting CODATA constants, $(-\xi^2\theta')_{\xi_1}=2.018$ and $M=M_\odot$ collapses the prefactor to $0.003$ , so $1-\beta\approx0.003\,\mu^4\beta^4(M/M_\odot)^2$ . Why: this is Eddington's quartic; the right side grows as $M^2$ , confirming that radiation takes over as mass rises, and the numerical coefficient reproduces the solar value below. *Verified: for $\mu=0.62$ , $M=M_\odot$ the quartic solves to $\beta=0.99956$ , i.e. $1-\beta=4.4\times10^{-4}$ ; the dimensionless prefactor evaluates to $0.00299$ , matching the textbook $\approx0.003$ . Dimensional analysis confirms the right side is dimensionless (the source's inverted form is not).*

Go deeper · Derivation C · 1 minThe standard-model luminosity and the Eddington limit

Derivation CThe standard-model luminosity and the Eddington limit

The goal is to close the model by adding energy transport and to show that Eddington's constant- $\beta$ star radiates a fixed fraction $1-\beta$ of the Eddington luminosity, thereby producing the mass–luminosity relation without solving the thermal structure. The idea is that in radiative equilibrium the radiation pressure gradient is set by the outward flux and the opacity; if $\beta$ is constant, that gradient is simply $(1-\beta)$ times the total pressure gradient, which hydrostatic equilibrium already fixes. Equating the two expressions cancels the density and even the radius, leaving luminosity proportional to mass.

Step 1. Radiative diffusion gives the radiation pressure gradient in terms of the local luminosity, $\dfrac{\mathrm{d}P_{\mathrm{rad}}}{\mathrm{d}r} = -\dfrac{\kappa\rho}{c}\dfrac{L_r}{4\pi r^2}$ . Why: this is the transport equation recalled from aq-astro-radiative-diffusion-rosseland-opacity — photons pushing outward against the opacity is what carries the flux, and it is the only place luminosity enters the mechanics.

Step 2. Because $\beta$ is constant, $P_{\mathrm{rad}}=(1-\beta)P$ , so $\dfrac{\mathrm{d}P_{\mathrm{rad}}}{\mathrm{d}r}=(1-\beta)\dfrac{\mathrm{d}P}{\mathrm{d}r}=-(1-\beta)\dfrac{GM_r\rho}{r^2}$ using hydrostatic equilibrium. Why: constancy of $\beta$ is what lets the radiation gradient inherit the full hydrostatic gradient — this is where the model's one assumption does the decisive work.

Step 3. Equate the two gradients; the factors $\rho$ and $r^2$ cancel, leaving $\dfrac{L_r}{M_r}=\dfrac{4\pi cG(1-\beta)}{\kappa}$ . With $\kappa$ taken constant this holds globally, so $L=(1-\beta)\dfrac{4\pi cGM}{\kappa}=(1-\beta)L_{\mathrm{Edd}}$ . Why: the star radiates exactly the fraction of its Eddington luminosity that radiation contributes to its pressure — a clean, memorable statement, and inserting $1-\beta\propto M^2\beta^4$ from the quartic gives $L\propto M^3$ at intermediate masses. *Verified: the cancellation of $\rho$ and $r^2$ is exact for constant $\beta$ and $\kappa$ ; combining $L=(1-\beta)L_{\mathrm{Edd}}$ with the quartic reproduces the observed slope of the main-sequence mass–luminosity relation, steepening further under a Kramers opacity law.*

Limiting cases

The behaviour of $\beta$ across the mass spectrum — and the special status of the $n=3$ mass — is the entire physical content of the model:

Case	Result	Meaning
Low-mass stars ( $M\ll M_\odot$ )	$1-\beta\to0,\ \beta\to1$	Gas pressure dominates; radiation is a negligible correction. Real low-mass stars are convective and closer to $n=3/2$ , but the standard model formally keeps $n=3$ .
Solar-mass star ( $M\sim M_\odot$ )	$1-\beta\approx4\times10^{-4},\ \beta\approx0.9996$	Radiation is a tiny but nonzero share; the constant- $\beta$ approximation is excellent and central conditions come out within a factor of two of detailed models.
Radiation–gas parity	$\beta=\tfrac12$ at $M\approx130\,M_\odot$	The two partners support the star equally only at extreme mass — a landmark on the way to radiation domination.
Very massive stars ( $M\gg M_\odot$ )	$1-\beta\to1,\ P\to P_{\mathrm{rad}}$	Radiation dominates; the star approaches the Eddington limit and the effective adiabatic index sinks toward the dynamical-instability value $4/3$ .
$n=3$ mass	$M=4\pi(K/\pi G)^{3/2}(-\xi^2\theta')_{\xi_1}$	Mass depends on $K$ and constants only, independent of $\rho_c$ — the same fact that fixes the Chandrasekhar mass.

Common trapConstant β is an assumption, and the n=3 index is a coincidence — not a claim about degeneracy

Two traps hide in this model. First, the constancy of $\beta$ is a postulate, not a theorem. It is an excellent approximation for radiative main-sequence stars, but it fails where convection, a strong opacity gradient, or degeneracy takes over — in those regions $\beta$ genuinely varies with radius and the $n=3$ polytrope is only a caricature. Second, the appearance of $n=3$ (i.e. $\gamma=4/3$ ) is a numerical coincidence with the index of a relativistically degenerate electron gas. It does not mean Eddington's gas is degenerate or relativistic — his star is ordinary non-degenerate ideal gas plus thermal radiation. The two systems share a mathematical index and therefore share the "mass independent of central density" property, which is why the Chandrasekhar mass and the Eddington model are cousins, but the underlying physics is entirely different.

Worked examples

1 · The radiation fraction of the Sun (fully worked)

Using Eddington's quartic, find the gas-to-total pressure ratio $\beta$ at the centre of the Sun, taking $M=M_\odot$ and a fully ionized mean molecular weight $\mu=0.62$ .

Step 1. Write the quartic in its handy form, $1-\beta = 0.003\,\mu^4\beta^4(M/M_\odot)^2$ . With $M=M_\odot$ the mass factor is 1, and $\mu^4 = 0.62^4 = 0.1478$ , so the coefficient is $0.003\times0.1478 = 4.43\times10^{-4}$ . The equation to solve is $1-\beta = 4.43\times10^{-4}\,\beta^4$ .

Step 2. Because $\beta$ is very close to 1, set $\beta^4\approx1$ on the right for a first pass: $1-\beta\approx4.43\times10^{-4}$ , so $\beta\approx0.99956$ . Re-inserting $\beta^4=(0.99956)^4=0.99823$ barely changes it, confirming $\mathbf{\beta\approx0.9996}$ and $\mathbf{1-\beta\approx4.4\times10^{-4}}$ .

Step 3. Interpret it: radiation contributes only about four parts in ten thousand of the Sun's central support — the Sun is overwhelmingly a gas-pressure star, which is exactly why simple ideal-gas models of the Sun work so well. But the quartic warns that this fraction scales as $M^2$ : a star ten times more massive has $1-\beta$ a hundred times larger, near $4\times10^{-2}$ , and radiation can no longer be ignored. β ≈ 0.9996 for the Sun; radiation is a 0.04 % correction that grows as M².

2 · Central conditions of the Sun from the n=3 polytrope (fully worked)

Estimate the central density, pressure, and temperature of the Sun as an $n=3$ polytrope, using $M_\odot=1.989\times10^{30}\ \mathrm{kg}$ , $R_\odot=6.96\times10^{8}\ \mathrm{m}$ , $\mu=0.62$ , and the tabulated constants.

Step 1. The mean density is $\bar\rho = 3M/(4\pi R^3) = 3(1.989\times10^{30})/[4\pi(6.96\times10^8)^3] = 1.41\times10^{3}\ \mathrm{kg\,m^{-3}}$ . The $n=3$ solution gives $\rho_c/\bar\rho = 54.18$ , so $\mathbf{\rho_c \approx 7.6\times10^{4}\ kg\,m^{-3}}$ (about 76 g cm⁻³).

Step 2. The central pressure of an $n=3$ polytrope is $P_c = W_3\,GM^2/R^4$ with $W_3 = 1/[4\pi(n+1)\theta'(\xi_1)^2] = 11.05$ . Numerically $P_c = 11.05\,(6.674\times10^{-11})(1.989\times10^{30})^2/(6.96\times10^8)^4 = \mathbf{1.25\times10^{16}\ Pa}$ .

Step 3. Get the temperature from the gas share of the pressure, $\beta P_c = \rho_c k_B T_c/(\mu m_H)$ with $\beta\approx0.9996$ from Example 1: $T_c = \beta P_c\,\mu m_H/(\rho_c k_B) = (0.9996)(1.25\times10^{16})(0.62)(1.66\times10^{-27})/[(7.6\times10^4)(1.38\times10^{-23})] = \mathbf{1.22\times10^{7}\ K}$ .

Step 4. Sanity-check against reality: detailed solar models give $\rho_c\approx1.5\times10^5\ \mathrm{kg\,m^{-3}}$ , $P_c\approx2.5\times10^{16}\ \mathrm{Pa}$ , $T_c\approx1.57\times10^{7}\ \mathrm{K}$ . The polytrope lands within a factor of about two on every quantity — remarkable for a model that never solved an energy equation, and more than enough to confirm that the Sun's core is hot enough for hydrogen fusion ( $T_c\sim10^7\ \mathrm{K}$ ). The n=3 Sun: ρc ≈ 7.6×10⁴ kg m⁻³, Pc ≈ 1.25×10¹⁶ Pa, Tc ≈ 1.2×10⁷ K — all correct to a factor of two.

3 · Where radiation catches up with gas

At what mass does radiation pressure equal gas pressure at the centre, i.e. $\beta=\tfrac12$ ? Take $\mu=0.62$ .

Step 1. Set $\beta=\tfrac12$ in the quartic: $1-\beta = \tfrac12$ and $\beta^4 = (\tfrac12)^4 = \tfrac{1}{16}$ , so $\tfrac12 = 0.003\,\mu^4(\tfrac{1}{16})(M/M_\odot)^2$ . Solve for the mass factor: $(M/M_\odot)^2 = 0.5\times16/(0.003\times0.1478) = 8/(4.43\times10^{-4}) = 1.8\times10^4$ .

Step 2. Take the square root: $\mathbf{M\approx130\,M_\odot}$ . Only in the realm of the most massive stars known does radiation pressure rival gas pressure at the centre. Above this the star runs toward the Eddington limit, its effective index slides to $4/3$ , and it becomes increasingly vulnerable to radiation-driven pulsational instability — the physics behind the observed upper mass cutoff of stars. Radiation equals gas pressure only near 130 M⊙, deep in the regime of dynamical instability.

Physics of the portrait

The Eddington Standard Model carries one name because one person, almost single-handedly, turned the interior of a star from a mystery into a calculation. Arthur Eddington did it with radiation pressure as his central character and a bold constant- $\beta$ simplification as his tool.

Origins · 1 minArthur Eddington and how this was discovered

Arthur Eddington

1882–1944 · the physics of stellar interiors and radiation pressure

Arthur Stanley Eddington was the towering figure of early-twentieth-century astrophysics — the man who first took seriously the idea that a star's interior is governed by physics we can write down. Beginning around 1916 he built a theory of stars in radiative equilibrium, insisting that in the deep interior energy flows outward as radiation and that the pressure of that radiation is a genuine mechanical support, comparable to gas pressure and dominant in massive stars. Working before the source of stellar energy was known (he could only guess it was subatomic) and before electronic computers, he needed a way to solve the structure by hand, and he found it in the assumption that the ratio $\beta$ of gas to total pressure is constant — reducing the star to an $n=3$ polytrope and producing the quartic equation that bears his name. From this framework he derived, in 1924, the mass–luminosity relation, showing that a star's brightness is set mainly by its mass, and he crystallized the whole programme in his 1926 masterwork The Internal Constitution of the Stars. Eddington was also the experimentalist's theorist: his 1919 eclipse expedition measured the bending of starlight and confirmed general relativity, making him world-famous. His constant- $\beta$ star, and the upper mass limit its logic implied, remain the standard pedagogical entry into stellar structure a century later.

History

1870

— Lane poses the problem of a self-gravitating gas sphere in equilibrium

1907

— Emden's Gaskugeln systematizes polytropic gas spheres and tabulates solutions

1916

— Eddington develops the theory of stars in radiative equilibrium with radiation pressure

1924

— Eddington derives the mass–luminosity relation for main-sequence stars

1926

— Eddington publishes The Internal Constitution of the Stars, presenting the standard model and the quartic

1939

— Chandrasekhar's Introduction to the Study of Stellar Structure tabulates the n=3 solution used today

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., §10.6 (polytropes and the Eddington standard model) · Eddington, The Internal Constitution of the Stars (1926), Ch. IV · Chandrasekhar, An Introduction to the Study of Stellar Structure (1939), Ch. IV & VI · Kippenhahn, Weigert & Weiss, Stellar Structure and Evolution, 2nd ed., Ch. 19 · Hansen, Kawaler & Trimble, Stellar Interiors, 2nd ed., Ch. 7.

Provenance. The polytrope reduction to the $n=3$ Lane–Emden equation is owned by aq-astro-polytropic-models and recalled here by name; radiation pressure $P_{\mathrm{rad}}=\tfrac13aT^4$ comes from aq-astro-radiation-pressure-stellar-interiors, the ideal-gas term from aq-astro-equation-of-state-stellar-matter, and the transport step in Derivation C from aq-astro-radiative-diffusion-rosseland-opacity. A verifier should recompute: (1) the elimination of $T$ under constant $\beta$ giving $P\propto\rho^{4/3}$ ; (2) the $n=3$ Lane–Emden constants $\xi_1=6.897$ , $(-\xi^2\theta')_{\xi_1}=2.018$ , $\rho_c/\bar\rho=54.18$ by numerical integration; (3) the corrected quartic $1-\beta=\tfrac{\pi}{48}(aG^3\mu^4m_H^4/k_B^4)M^2\beta^4/(-\xi^2\theta')^2_{\xi_1}$ , whose dimensionless prefactor is $0.003$ at $M=M_\odot$ — the source's inverted form $(3/a)(k_B/\mu m_H)^4\beta^4/\pi G^3$ is dimensionally wrong and was corrected here; and (4) the solar values $\beta=0.9996$ , $\rho_c=7.6\times10^4\ \mathrm{kg\,m^{-3}}$ , $P_c=1.25\times10^{16}\ \mathrm{Pa}$ , $T_c=1.22\times10^7\ \mathrm{K}$ . Constants $a=4\sigma/c$ , $k_B$ , $m_H$ , $G$ are CODATA/textbook standard.

Prerequisites. Hydrostatic Equilibrium in Stars · Mass Continuity Equation · Equation of State for Stellar Matter · Radiation Pressure in Stellar Interiors. Leads to. Radiative Temperature Gradient · Opacity and Radiative Transfer in Stars · Equations of Stellar Structure.

Ask the tutor: Ask me to walk through solving Eddington's quartic for any stellar mass, or to show why the $n=3$ polytrope's density-independent mass is the same fact behind the Chandrasekhar limit.

Is the constant-β assumption actually true in real stars?

Only approximately, and that is the honest heart of the model. In a genuinely radiative main-sequence star the opacity and flux do conspire to keep $\beta$ nearly flat, so the constant- $\beta$ star is a very good caricature — good enough to get central conditions right to a factor of two. But it is a postulate, not a law. Where a star becomes convective (its envelope, or the whole of a low-mass star), where the opacity changes sharply, or where matter degenerates, $\beta$ varies with radius and the $n=3$ polytrope breaks down. Modern codes therefore never assume constant $\beta$ ; they integrate the full structure equations. The value of Eddington's assumption is pedagogical and conceptual, not that it is exactly obeyed.

Why exactly does constant β pick out n=3 and not some other index?

It comes down to how gas and radiation pressure depend on temperature. Gas pressure goes as $\rho T$ and radiation pressure as $T^4$ . If their ratio $\beta$ must be the same everywhere, then $\rho T / T^4 = \rho/T^3$ must be constant, which forces $T\propto\rho^{1/3}$ . Substitute that back: gas pressure becomes $\rho\cdot\rho^{1/3}=\rho^{4/3}$ and radiation pressure becomes $(\rho^{1/3})^4=\rho^{4/3}$ — both, and hence the total, scale as $\rho^{4/3}$ . A polytrope with $P\propto\rho^{\gamma}$ has $\gamma=1+1/n$ , so $\gamma=4/3$ means $n=3$ . No other index is compatible with holding $\beta$ fixed for this particular pair of pressures.

What is the physical meaning of Eddington's quartic?

It answers a single question with no free parameters: given a star's mass and chemical composition, what fraction of its support comes from radiation? The answer, $1-\beta\approx0.003\,\mu^4\beta^4(M/M_\odot)^2$ , says that radiation's share is tiny in low-mass stars and grows as the square of the mass. For the Sun it is four parts in ten thousand; by ten solar masses it is a few percent; near a hundred solar masses radiation supplies almost half the pressure. The quartic thus predicts, from first principles and long before any computer, that there must be a heaviest possible star — one so massive that radiation would take over its support entirely and destabilize it. It is the model's most far-reaching consequence.

How can this hand model get the Sun's centre right to a factor of two?

Because the mechanical structure of a star — how pressure and density run from centre to surface — is far less sensitive to the detailed microphysics than the thermal structure is. Hydrostatic equilibrium is a strong constraint: a self-gravitating ball of a given mass and radius has a central pressure fixed almost entirely by $GM^2/R^4$ , whatever the star is made of. The $n=3$ polytrope captures that constraint faithfully, so its central pressure, density, and temperature are close to the truth. What it does not get right is the luminosity's fine detail, the convective zones, or the exact density concentration — those need the energy and transport equations. Getting three central numbers within a factor of two from one assumption is exactly why the model endures.

What is the connection between this model and the Chandrasekhar mass?

They share the mathematical backbone $n=3$ . For any $n=3$ polytrope the total mass depends only on the polytropic constant $K$ and fundamental constants — not on how densely the star is packed. In Eddington's star $K$ is set by $\beta$ and composition, giving the $M$ – $\beta$ relation of the quartic. In a white dwarf supported by relativistically degenerate electrons the equation of state is also $P\propto\rho^{4/3}$ , so it is also $n=3$ , but there $K$ is built from $h$ , $c$ , and the electron mass alone — leaving a single fixed mass, the Chandrasekhar limit near $1.4\,M_\odot$ . Eddington's model is the pedagogical bridge: it shows why any $\gamma=4/3$ structure has a special, density-independent mass scale, which is precisely the property that makes the Chandrasekhar mass a definite number.

What does L = (1−β)L_Edd tell us about massive stars?

It says a star's luminosity is exactly the fraction $1-\beta$ of the Eddington luminosity, the brightness at which outward radiation force balances inward gravity. In the Sun $1-\beta$ is minuscule, so the Sun shines at a tiny fraction of $L_{\mathrm{Edd}}$ and is completely stable. But in very massive stars $1-\beta$ approaches one, so their actual luminosity approaches the Eddington limit — the point at which radiation can lift the outer layers off the star. This is not a mathematical curiosity: luminous blue variables like Eta Carinae live near this limit and shed mass in violent eruptions, and the same $L_{\mathrm{Edd}}$ caps how fast black holes and neutron stars can accrete. The relation Eddington derived for stars became the yardstick for radiation-limited systems everywhere.