Radiation Pressure in Stellar Interiors

Deep inside a star, matter floats in an almost perfectly isotropic sea of thermal photons. Those photons are not passive: each carries momentum $p=E/c$ , and their ceaseless bombardment of any surface exerts a real pressure, exactly the way gas particles do. Because the radiation field is very close to a blackbody in local thermodynamic equilibrium (LTE), that pressure takes one universal form, $P_{\mathrm{rad}}=\tfrac13 aT^4$ — independent of composition, fixed by temperature alone. This article follows that single expression outward, from the geometry of an isotropic photon gas to the largest question stellar physics asks: how massive can a star be? Three consequences organize everything. First, the total support is $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$ , and because gas pressure grows only as $\rho T$ while radiation grows as $T^4$ , radiation quietly takes over the hot, tenuous cores of massive stars — a race between exponents the gas always loses. Second, the gradient of $P_{\mathrm{rad}}$ is a genuine outward force, photons handing momentum to matter through opacity; when it rivals gravity the star reaches the Eddington limit and blows off its outer layers. Third, a radiation-dominated gas is soft — its adiabatic index falls to $\Gamma_1=4/3$ , exactly the value at which a self-gravitating body is marginally bound, so the photons that support a massive star also destabilize it. We derive $P_{\mathrm{rad}}=\tfrac13 aT^4$ two ways, split the pressure with $\beta$ , build Eddington's standard model and luminosity limit, and show why $\Gamma_1\to4/3$ caps stellar masses — results that feed directly into the radiative temperature gradient and the criteria for convection.

Key takeaways

Radiation pressure is the momentum flux of a blackbody photon gas in LTE and equals $P_{\mathrm{rad}}=\tfrac13 aT^4$ , where $a=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$ is the radiation constant.
The factor $\tfrac13$ is pure geometry — the isotropic average $\langle\cos^2\theta\rangle=1/3$ — so any isotropic relativistic gas obeys $P=\tfrac13 u$ .
Because $P_{\mathrm{rad}}\propto T^4$ while $P_{\mathrm{gas}}\propto\rho T$ , radiation dominates the hot low-density cores of massive stars; the split is tracked by $\beta=P_{\mathrm{gas}}/P$ .
Eddington's standard model treats $\beta$ as constant, making the star an $n=3$ polytrope, $P=K\rho^{4/3}$ , with $1-\beta$ fixed by mass through the Eddington quartic.
Radiation caps luminosity at the Eddington limit $L_{\mathrm{Edd}}=4\pi cGM/\kappa\approx3.8\times10^4\,(M/M_\odot)\,L_\odot$ for electron-scattering opacity.
A radiation-dominated gas has $\Gamma_1\to4/3$ , the threshold of dynamical stability, which together with $L_{\mathrm{Edd}}$ sets an upper bound on stellar mass.
Key idea: photons in LTE form a blackbody gas whose pressure is $P_{\mathrm{rad}}=\tfrac13 aT^4$ ; because this grows as $T^4$ while gas pressure grows only as $\rho T$ , radiation pressure quietly dominates the hot interiors of massive stars and sets an upper bound on stellar luminosity and mass.

Where this lives today

The Eddington limit is the working currency of high-energy astrophysics, far beyond the massive stars where Eddington found it. Accreting black holes and neutron stars radiate at or near $L_{\mathrm{Edd}}$ , and it sets the fastest rate at which a black hole seed can grow — the central puzzle of how billion-solar-mass quasars existed when the universe was under a billion years old, apparently demanding super-Eddington episodes. In massive stars, the modern refinement is that opacity is not the clean electron-scattering value: iron-group "opacity bumps" raise $\kappa$ in specific temperature zones, pushing the local Eddington factor past unity, launching the line-driven winds of O and Wolf–Rayet stars, and powering the eruptions of luminous blue variables such as $\eta$ Carinae. The softening $\Gamma_1\to4/3$ is equally alive: pair-instability supernovae in the most massive metal-poor stars are the endpoint of this fragility, and gravitational-wave detections near the "pair-instability mass gap" are testing where it bites. And the same $\tfrac13 aT^4$ Eddington wrote for a stellar core is the pressure of the photon gas that set the expansion rate through the early universe's radiation-dominated era.

The physical picture

Think of the photon gas exactly as you think of an ordinary gas: particles bounce off a wall, transfer momentum, and pressure is the momentum deposited per unit area per unit time. A photon differs only in being relativistic — it carries momentum $p=E/c$ and moves at speed $c$ regardless of energy. In LTE the field is a blackbody with energy density $u=aT^4$ , and the pressure it exerts is the isotropic momentum flux of that field. The whole content of $P_{\mathrm{rad}}=\tfrac13 u$ is angular geometry: the momentum crossing a surface picks up one $\cos\theta$ from the flux and a second from projecting onto the normal, and averaging $\cos^2\theta$ over a full sphere gives $1/3$ . The first figure makes that counting concrete.

Figure

Photons carry momentum p = E/c and strike a surface from every direction; the normal momentum flux picks up two factors of cos θ, whose isotropic average ⟨cos²θ⟩ = 1/3 gives P_rad = ⅓u = ⅓aT⁴. With a = 7.566×10⁻¹⁶ J m⁻³ K⁻⁴ and the solar-centre temperature T = 1.57×10⁷ K this evaluates to P_rad ≈ 1.5×10¹³ Pa

The dominance of radiation in massive stars is that race between exponents made visible. Gas pressure carries one power of temperature, $P_{\mathrm{gas}}\propto\rho T$ ; radiation carries four, $P_{\mathrm{rad}}\propto T^4$ . So the ratio $P_{\mathrm{rad}}/P_{\mathrm{gas}}\propto T^3/\rho$ climbs steeply toward hotter, more diffuse cores, and a virial estimate turns that into $\propto M^2$ — radiation is a heavyweight's burden. In the Sun it is a $\sim5\times10^{-4}$ correction at the centre; by $30\,M_\odot$ photons supply roughly a sixth of the central pressure, and near $140\,M_\odot$ they carry half. The second figure plots that rise.

Figure

The radiation-pressure fraction 1 − β at a star's centre versus mass, from the Eddington quartic 1 − β = 0.003 (M/M⊙)² μ⁴ β⁴ with μ = 0.61. It rises as roughly M²: 0.0004 at 1 M⊙, 0.04 at 10 M⊙, 0.28 at 50 M⊙, and 0.43 at 100 M⊙, reaching parity β = 0.5 near 140 M⊙

Radiation pressure also has a darker role. When it dominates, the gas responds to compression like a $\gamma=4/3$ fluid — the index at which the total energy of a self-gravitating polytrope is marginally zero. A star there barely resists a nudge, so massive stars are driven toward instability: the photons that hold them up have softened their equation of state to the edge of stability.

The core equations

The central result is the pressure of a blackbody photon gas. It follows from the isotropic momentum flux (Derivation A) and depends on nothing but temperature — which makes it universal across compositions.

P_{\mathrm{rad}} = \frac{1}{3}aT^4 \qquad (1)

where: $P_{\mathrm{rad}}$ — radiation (photon) pressure; $a$ — radiation constant, $7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$ ; $T$ — local temperature (LTE). Reading: pressure is one-third of the radiation energy density and rises as the fourth power of temperature — the steepest term in the whole pressure budget. The radiation constant is fixed by fundamental constants through the Planck spectrum, and is emphatically not the Stefan–Boltzmann constant $\sigma$ ; the two differ by a factor $c/4$ .

u = aT^4, \qquad a = \frac{4\sigma}{c} = \frac{\pi^2 k_B^4}{15\hbar^3 c^3} \qquad (2)

where: $u$ — radiation energy density; $\sigma$ — Stefan–Boltzmann constant, $5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ ; $k_B$ — Boltzmann constant; $\hbar$ — reduced Planck constant. Reading: $\sigma$ governs surface flux ( $F=\sigma T^4$ ), $a$ the volume energy density; radiation pressure uses $a$ . The pressure that enters hydrostatic equilibrium is the sum of the two gases sharing the same volume and temperature — each computed at the local $T$ , their sum balancing gravity.

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

where: $\rho$ — mass density; $\mu$ — mean molecular weight (particles per $m_H$ ); $m_H$ — atomic mass unit. Reading: the first term is the ideal-gas law for a plasma of mean particle mass $\mu m_H$ ; the second, the photon contribution, is negligible in the Sun but decisive in massive cores. The balance between the two is captured by a single dimensionless number that reappears throughout stellar-structure theory.

\beta \equiv \frac{P_{\mathrm{gas}}}{P}, \qquad 1-\beta = \frac{P_{\mathrm{rad}}}{P} \qquad (4)

where: $\beta$ — fraction of total pressure supplied by gas, $0\le\beta\le1$ . Reading: $\beta\to1$ is a gas star (the Sun), $\beta\to0$ a radiation star; real massive stars live in between. Radiation pressure is not only a static term — its gradient is a net force. Deep in a star the outward flux is driven by that gradient through the opacity.

\frac{\mathrm{d}P_{\mathrm{rad}}}{\mathrm{d}r} = -\frac{\kappa\rho}{c}\,F \qquad (5)

where: $\kappa$ — opacity (mass absorption coefficient, $\mathrm{m^2\,kg^{-1}}$ ); $F=L(r)/4\pi r^2$ — radiative flux; $c$ — speed of light. Reading: photons streaming outward deposit momentum at the rate $\kappa\rho F/c$ per unit volume — a force density equal to minus the radiation pressure gradient. Setting that photon force equal to gravity gives the maximum luminosity a mass $M$ can radiate and stay bound — the Eddington limit (Derivation C).

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

where: $L_{\mathrm{Edd}}$ — Eddington luminosity; $M$ — enclosed mass; $G$ — gravitational constant. Reading: for electron-scattering opacity $\kappa_{\mathrm{es}}\approx0.034\,\mathrm{m^2\,kg^{-1}}$ this is $\approx3.8\times10^4\,(M/M_\odot)\,L_\odot$ ; exceed it and hydrostatic equilibrium fails. If $\beta$ is taken constant through the star — Eddington's key idealization — the equation of state collapses to an $n=3$ polytrope, the backbone of the standard model.

P = K\rho^{4/3}, \qquad K = \left[\frac{3}{a}\,\frac{1-\beta}{\beta^4}\left(\frac{k_B}{\mu m_H}\right)^{4}\right]^{1/3} \qquad (7)

where: $K$ — polytropic constant set by $\beta$ and composition. Reading: constant $\beta$ forces $P\propto\rho^{4/3}$ , so the interior obeys the $n=3$ Lane–Emden equation and is known once $\beta$ is. Matching the polytrope's mass to $\beta$ yields Eddington's quartic, radiation fraction rising as $M^2$ (Derivation B).

1-\beta = 0.003\left(\frac{M}{M_\odot}\right)^{2}\mu^4\beta^4 \qquad (8)

where: $M/M_\odot$ — mass in solar units; the coefficient $0.003\approx(18.1)^{-2}$ comes from the $n=3$ polytrope mass. Reading: solve this transcendental equation for $\beta$ given $M$ and $\mu$ — the curve in the second figure. Finally, mixing gas and radiation interpolates the adiabatic exponent between the gas value $5/3$ and the radiation value $4/3$ .

\Gamma_1 = \beta + \frac{(4-3\beta)^2(\gamma-1)}{\beta + 12(\gamma-1)(1-\beta)} \qquad (9)

where: $\Gamma_1$ — first adiabatic exponent $(\partial\ln P/\partial\ln\rho)_S$ ; $\gamma$ — gas adiabatic index, $5/3$ for a monatomic ideal gas. Reading: $\beta\to1$ gives $5/3$ , $\beta\to0$ gives $4/3$ ; at $\beta=0.5$ it is $1.43$ , already near the marginal value.

Derivation ARadiation pressure from the isotropic photon flux and the Planck spectrum

The goal is to obtain $P_{\mathrm{rad}}=\tfrac13 aT^4$ twice — first from pure geometry, treating the photon gas as an isotropic swarm of momentum-carriers, then from the Planck distribution itself — so that the factor $\tfrac13$ and the fourth-power law both rest on explicit foundations rather than assertion. The two routes must agree exactly, and their agreement is the check that the energy density $u=aT^4$ and the pressure $\tfrac13 aT^4$ describe one field.

Step 1. Pressure is the flux of the normal component of momentum across a surface. A photon at angle $\theta$ to the normal delivers momentum $p=E/c$ ; the rate at which such photons cross the surface carries one factor $\cos\theta$ (fewer arrive per area when oblique), and the normal component of the momentum carries a second. Why: each projection — flux across the surface, momentum along the normal — is a cosine, and this is the origin of the two factors that average to $\tfrac13$ .

Step 2. Average over an isotropic field. Integrating $\cos^2\theta$ over all incoming directions on one hemisphere and normalizing over the full solid angle gives

\langle\cos^2\theta\rangle = \frac{1}{4\pi}\int_0^{2\pi}\!\!\int_0^{\pi}\cos^2\theta\,\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi = \frac{1}{3}.

Multiplying the energy density $u$ by this angular factor gives $P_{\mathrm{rad}}=\tfrac13 u$ . Why: isotropy is the entire physical input, so the same $\tfrac13$ governs any isotropic relativistic gas — the CMB as much as a stellar core — independent of the spectrum.

Step 3. Now do the spectral integral to identify $u=aT^4$ . The Planck energy density per unit frequency is

u_\nu = \frac{8\pi h\nu^3}{c^3}\,\frac{1}{e^{h\nu/k_BT}-1},

and substituting $x=h\nu/k_BT$ turns $\int_0^\infty u_\nu\,\mathrm{d}\nu$ into $(8\pi k_B^4 T^4/h^3c^3)\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x$ . Why: the frequency dependence factors out as a pure number times $T^4$ — the $T^4$ scaling is inevitable once $k_BT$ is the only energy scale.

Step 4. Evaluate the standard integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x = \pi^4/15$ , giving

\int_0^\infty u_\nu\,\mathrm{d}\nu = \frac{8\pi^5 k_B^4}{15 h^3 c^3}\,T^4 = aT^4, \qquad P_{\mathrm{rad}} = \frac{1}{3}aT^4.

The prefactor is $a=8\pi^5 k_B^4/15h^3c^3=\pi^2 k_B^4/15\hbar^3c^3=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}=4\sigma/c$ . Why: the geometric $\tfrac13$ and the spectral $aT^4$ meet in one expression, so each route confirms the other and the result is doubly grounded.

Go deeper · Derivation B · 1 minThe gas–radiation split and the Eddington quartic

Derivation BThe gas–radiation split and the Eddington quartic

The aim is to turn the constant- $\beta$ idealization into a concrete equation of state and then a relation between radiation fraction and stellar mass, so that "massive stars are radiation-supported" becomes the statement $1-\beta=0.003(M/M_\odot)^2\mu^4\beta^4$ . We eliminate temperature between gas and radiation pressures at fixed $\beta$ , recognize the $n=3$ polytrope, and match its mass eigenvalue to close the loop.

Step 1. Write both partial pressures in terms of the total. From $P_{\mathrm{rad}}=(1-\beta)P=\tfrac13 aT^4$ solve for temperature, $T=[3(1-\beta)P/a]^{1/4}$ , and use $P_{\mathrm{gas}}=\beta P=\rho k_BT/\mu m_H$ for the density. Why: with $\beta$ fixed, $T$ and $\rho$ are no longer independent, and trading $T$ for $P$ and $\rho$ is the only route to a pressure–density relation.

Step 2. Substitute the temperature into the gas law and collect powers. Eliminating $T$ gives

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

This is a polytrope with $P\propto\rho^{1+1/n}$ and $n=3$ . Why: constant $\beta$ makes the composite gas behave as a single $\Gamma=4/3$ fluid, so the model is an $n=3$ Lane–Emden star and inherits that solution's structure.

Step 3. Match the polytrope's mass. For an $n=3$ polytrope the mass is an eigenvalue independent of central density: $M = 4\pi\,(K/\pi G)^{3/2}\,(-\xi^2\theta')_1$ , with the Lane–Emden surface value $(-\xi^2\theta')_1\approx2.018$ . Because $K$ contains $(1-\beta)/\beta^4$ , inverting for the mass yields Eddington's relation $M = 18.1\,\mu^{-2}\sqrt{(1-\beta)/\beta^4}\,M_\odot$ . Why: the $n=3$ polytrope has a fixed mass for a given $K$ , so the star cannot choose its mass and $\beta$ independently — that rigidity ties one to the other.

Step 4. Rearrange to the quartic. Squaring Eddington's relation gives

1-\beta = \frac{1}{18.1^2}\left(\frac{M}{M_\odot}\right)^2\mu^4\beta^4 = 0.003\left(\frac{M}{M_\odot}\right)^2\mu^4\beta^4.

For $\mu=0.61$ this yields $1-\beta\approx4\times10^{-4}$ at $1\,M_\odot$ , $0.04$ at $10\,M_\odot$ , $0.28$ at $50\,M_\odot$ , and $0.43$ at $100\,M_\odot$ , with parity $\beta=0.5$ near $140\,M_\odot$ . Why: the $M^2$ is the punchline — radiation's share grows quadratically with mass, a burden specific to heavyweight stars, and the coefficient makes the crossover a concrete number.

Go deeper · Derivation C · 1 minThe Eddington luminosity and the stellar mass limit

Derivation CThe Eddington luminosity and the stellar mass limit

The purpose is to convert the radiation force density of equation (5) into a hard ceiling on luminosity, and then argue that this ceiling — together with the softening $\Gamma_1\to4/3$ — is what forbids arbitrarily massive stars. We balance the outward photon force against gravity, insert the electron-scattering opacity, and compare $L_{\mathrm{Edd}}$ with the steep main-sequence mass–luminosity relation.

Step 1. Balance forces per unit volume. The outward radiation force density is $\kappa\rho F/c=\kappa\rho L/4\pi r^2c$ ; the inward gravitational force density is $GM\rho/r^2$ . Setting them equal,

\frac{\kappa\rho L}{4\pi r^2 c} = \frac{GM\rho}{r^2},

the density and $r^2$ cancel, leaving a condition on $L$ alone. Why: both forces scale identically with $\rho$ and $r^{-2}$ , so the balance is geometry-free — a global statement about how much luminosity a given mass can push against.

Step 2. Solve for the limiting luminosity, $L_{\mathrm{Edd}}=4\pi cGM/\kappa$ . Insert the electron-scattering (Thomson) opacity for ionized H-rich gas, $\kappa_{\mathrm{es}}=0.02(1+X)\,\mathrm{m^2\,kg^{-1}}\approx0.034\,\mathrm{m^2\,kg^{-1}}$ at $X=0.7$ . Why: electron scattering is the floor opacity of an ionized plasma, giving the cleanest, most conservative limit; line and bound-free opacities only lower the ceiling.

Step 3. Evaluate in solar units. With $\kappa_{\mathrm{es}}=0.034\,\mathrm{m^2\,kg^{-1}}$ ,

L_{\mathrm{Edd}} = \frac{4\pi cGM}{\kappa_{\mathrm{es}}} \approx 3.8\times10^4\,\frac{M}{M_\odot}\,L_\odot,

so a $30\,M_\odot$ star may radiate up to $\sim1.1\times10^6\,L_\odot$ . (For pure ionized hydrogen, $\kappa=0.04$ , the coefficient drops to $3.2\times10^4$ .) Why: pinning the number lets us compare it to what stars actually emit — a limit is only interesting if real stars approach it.

Step 4. Close the mass argument. Main-sequence luminosity climbs as $L\propto M^{3.5}$ while $L_{\mathrm{Edd}}\propto M$ , so $L/L_{\mathrm{Edd}}\propto M^{2.5}$ rises without bound and sufficiently massive stars hit their own limit — while $\Gamma_1\to4/3$ makes them dynamically fragile. Together these cap stellar masses at a few hundred $M_\odot$ . Why: neither effect alone is a clean wall, but the luminosity ceiling and the vanishing binding energy bite in the same mass range — which is why the stellar mass function falls off above $\sim150$ – $300\,M_\odot$ .

Limiting cases

The gas–radiation split is governed by $\beta=P_{\mathrm{gas}}/P$ , and the important regimes are the two extremes plus the constant- $\beta$ idealization that connects them.

Case	Result	Meaning
Gas-dominated ( $\beta\to1$ )	$P\approx\dfrac{\rho k_B T}{\mu m_H}$	Low-mass stars like the Sun: radiation is a $\lesssim10^{-3}$ correction at the centre; $\Gamma_1\approx5/3$ ; structure set by the ideal-gas EOS.
Radiation-dominated ( $\beta\to0$ )	$P\approx\dfrac{1}{3}aT^4$	Hypothetical supermassive object: photons supply nearly all support; $\Gamma_1\to4/3$ , marginal for gravitational stability.
Eddington standard model (const. $\beta$ )	$P=\left[\dfrac{3}{a}\dfrac{1-\beta}{\beta^4}\left(\dfrac{k_B}{\mu m_H}\right)^4\right]^{1/3}\rho^{4/3}$	$n=3$ polytrope; the quartic fixes $1-\beta$ from $M$ : $\approx0.04$ at $10\,M_\odot$ , $\approx0.43$ at $100\,M_\odot$ .

Common trapConfusing radiation pressure with radiation energy density, and the constant a with σ

The single most common slip is writing $P_{\mathrm{rad}}=aT^4$ . It is not: the pressure is one-third of the energy density, $P_{\mathrm{rad}}=\tfrac13 aT^4=\tfrac13 u$ , and the missing factor is the isotropic average $\langle\cos^2\theta\rangle=1/3$ , not a typo. The second trap is the constant itself. The radiation constant $a=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$ governs energy density inside a cavity, while $\sigma=5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ governs the flux leaving a surface; they are related by $a=4\sigma/c$ and differ by nineteen orders of magnitude, so mixing them is a wrong answer, not a rounding error. A third, subtler slip is treating $\Gamma_1$ as a simple ratio; the correct mixed-fluid form (equation 9) is $\beta$ plus a fraction, and only that stays between $4/3$ and $5/3$ for all $\beta$ .

Worked examples

1 · Radiation pressure at the centre of the Sun (fully worked)

Central conditions: $T_c=1.57\times10^7\,\mathrm{K}$ , $\rho_c=1.48\times10^5\,\mathrm{kg\,m^{-3}}$ , $\mu=0.61$ . Find $P_{\mathrm{rad}}$ , $P_{\mathrm{gas}}$ , and $1-\beta$ .

Step 1. Radiation pressure. With $(1.57\times10^7)^4=6.08\times10^{28}$ , $P_{\mathrm{rad}}=\tfrac13 aT_c^4=\tfrac13(7.566\times10^{-16})(6.08\times10^{28})=\mathbf{1.5\times10^{13}\ Pa}$ .

Step 2. Gas pressure. $P_{\mathrm{gas}}=\rho_c k_B T_c/\mu m_H=(1.48\times10^5)(1.381\times10^{-23})(1.57\times10^7)/[(0.61)(1.673\times10^{-27})]=\mathbf{3.1\times10^{16}\ Pa}$ .

Step 3. Radiation fraction. $1-\beta=P_{\mathrm{rad}}/(P_{\mathrm{gas}}+P_{\mathrm{rad}})=1.5\times10^{13}/(3.1\times10^{16}+1.5\times10^{13})\approx\mathbf{4.9\times10^{-4}}$ . Radiation supplies about five parts in ten thousand of the Sun's central support — negligible for its structure, which is why solar models drop it to leading order.

2 · The Eddington luminosity of a 30 M⊙ O-star

Estimate $L_{\mathrm{Edd}}$ for a $30\,M_\odot$ star with $\kappa_{\mathrm{es}}=0.034\,\mathrm{m^2\,kg^{-1}}$ , and compare to an observed $L\approx1\times10^5\,L_\odot$ .

Step 1. With $M=30\,M_\odot=5.97\times10^{31}\,\mathrm{kg}$ and prefactor $4\pi cG=0.251$ , $L_{\mathrm{Edd}}=0.251\times5.97\times10^{31}/0.034=\mathbf{4.4\times10^{32}\ W}$ .

Step 2. In solar units ( $L_\odot=3.83\times10^{26}\,\mathrm{W}$ ): $L_{\mathrm{Edd}}=4.4\times10^{32}/3.83\times10^{26}=\mathbf{1.1\times10^6\ L_\odot}$ , matching the rule $3.8\times10^4\,(M/M_\odot)$ .

Step 3. The Eddington ratio is $\Gamma_{\mathrm{Edd}}=L/L_{\mathrm{Edd}}=10^5/1.1\times10^6\approx\mathbf{0.09}$ — the star radiates at $\sim9\%$ of its limit. Comfortable globally, but iron-group line opacity can raise the effective $\kappa$ tenfold in cooler layers, pushing local $\Gamma_{\mathrm{Edd}}$ toward unity and driving the powerful winds O-stars are famous for.

3 · Solving the Eddington quartic for a 50 M⊙ star

Use $1-\beta=0.003(M/M_\odot)^2\mu^4\beta^4$ with $M=50\,M_\odot$ , $\mu=0.61$ to find $\beta$ , then $\Gamma_1$ .

Step 1. Coefficient: $C=0.003\times50^2\times0.61^4=0.003\times2500\times0.1385=\mathbf{1.04}$ , so $1-\beta=1.04\,\beta^4$ .

Step 2. Iterate. Try $\beta=0.72$ : right side $1.04\times0.72^4=0.280$ , left side $0.280$ — they match, so $\beta\approx\mathbf{0.72}$ , $1-\beta\approx\mathbf{0.28}$ . Radiation carries about a quarter of the central pressure in a $50\,M_\odot$ star.

Step 3. Adiabatic exponent. From equation (9) with $\gamma=5/3$ : $\Gamma_1=0.72+\dfrac{(4-2.16)^2(2/3)}{0.72+12(2/3)(0.28)}=0.72+\dfrac{2.26}{2.96}=\mathbf{1.48}$ . The fluid is already softened from $5/3$ toward the dangerous $4/3$ — the star is stable but its binding energy is measurably reduced, foreshadowing the instability of still heavier objects.

Physics of the portrait

Radiation pressure runs from thermodynamics and electromagnetism straight into the architecture of stars, and one figure more than any other made that connection the working core of astrophysics.

Origins · 1 minArthur Eddington and how this was discovered

Arthur Eddington

1882–1944, English astrophysicist · the theory of stellar interiors and radiative equilibrium

Arthur Stanley Eddington took the abstract fact that photons carry momentum and made it a load-bearing beam of stellar structure. Radiation pressure itself was older — Maxwell's electromagnetism predicted that light presses on matter, Boltzmann in 1884 used $P=\tfrac13 u$ to derive the Stefan $T^4$ law, and Lebedev, Nichols and Hull measured the feeble push around 1901 — but it was Eddington, from about 1916 through his 1926 masterwork The Internal Constitution of the Stars, who realized that in a stellar interior this pressure is anything but feeble. Assuming radiative equilibrium, with energy carried outward by photons diffusing through an opaque gas, he showed the gas–radiation balance could be captured by a single nearly constant parameter $\beta$ — his "standard model," an $n=3$ polytrope. From it he extracted the quartic relating $\beta$ to mass and, most consequentially, the luminosity limit that now bears his name: no star can shine brighter than $L_{\mathrm{Edd}}=4\pi cGM/\kappa$ without blowing itself apart. He also found the mass–luminosity relation before anyone knew the energy source, remarking that stars drew on some sub-atomic reservoir he could not name. His limit still governs everything from massive-star winds to the accretion of black holes.

History

1873

— Maxwell predicts from electromagnetic theory that light exerts a pressure, with P = u/3

1884

— Boltzmann derives the Stefan T⁴ law thermodynamically using radiation pressure

1901

— Lebedev, and Nichols and Hull, measure radiation pressure in the laboratory

1916

— Eddington begins his radiative-equilibrium models of stellar interiors

1926

— Eddington's The Internal Constitution of the Stars gives the standard model, the quartic, and the luminosity limit

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., §10.2 and §11.1 (radiation pressure, the standard model, the Eddington limit) · Kippenhahn, Weigert & Weiss, Stellar Structure and Evolution, 2nd ed., Ch. 13 · Prialnik, An Introduction to the Theory of Stellar Structure and Evolution, Ch. 3 · Hansen, Kawaler & Trimble, Stellar Interiors, §3 · Ryden, Introduction to Cosmology, §2, and Dodelson, Modern Cosmology, §2 (the photon gas and $P=\tfrac13 u$ ).

Provenance. Equations follow Carroll & Ostlie Ch. 10–11 and Kippenhahn, Weigert & Weiss Ch. 13; the blackbody integral is cross-checked against Ryden and Dodelson. Every number here was recomputed: (1) $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\pi^4/15$ and hence $a=8\pi^5k_B^4/15h^3c^3=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}=4\sigma/c$ ; (2) $L_{\mathrm{Edd}}=3.8\times10^4\,(M/M_\odot)\,L_\odot$ for $\kappa_{\mathrm{es}}=0.034\,\mathrm{m^2\,kg^{-1}}$ ( $3.2\times10^4$ for pure-H $\kappa=0.04$ ); (3) the Eddington quartic $1-\beta=0.003(M/M_\odot)^2\mu^4\beta^4$ giving $1-\beta=\{0.0004,\,0.04,\,0.28,\,0.43\}$ at $M=\{1,10,50,100\}\,M_\odot$ for $\mu=0.61$ , with parity at $\approx140\,M_\odot$ ; (4) the corrected mixed-fluid $\Gamma_1=\beta+(4-3\beta)^2(\gamma-1)/[\beta+12(\gamma-1)(1-\beta)]$ , which returns $5/3$ , $1.43$ , $4/3$ at $\beta=1,\,0.5,\,0$ ; (5) the solar-centre $1-\beta\approx5\times10^{-4}$ .

Builds on. Equation of State for Stellar Matter · Hydrostatic Equilibrium in Stars · Opacity and Radiative Transfer in Stars. Leads to. Radiative Temperature Gradient · Schwarzschild and Ledoux Criteria for Convection · Virial Theorem for Stars.

Ask the tutor: ask me to solve the Eddington quartic for any stellar mass, or to show why $\Gamma_1=4/3$ makes a radiation-dominated star marginally bound.

Why exactly one-third of the energy density, when a monatomic gas gives two-thirds?

The difference is relativistic kinematics. For any gas, pressure is $\tfrac13\langle n\,p\,v\rangle$ after the isotropic average, with $p$ the momentum and $v$ the speed. Photons have $E=pc$ and $v=c$ , so $pv=E$ and $P=\tfrac13\sum E/V=\tfrac13 u$ . A non-relativistic particle has $E=\tfrac12 mv^2$ but $pv=mv^2=2E$ , so $P=\tfrac23 u$ . The geometric $\tfrac13$ is the same in both; what changes is whether $pv$ equals the energy (photons) or twice it (slow particles).

Is radiation pressure a force, or just a term in the pressure?

Both, and the distinction matters. As a scalar it is a term in the equation of state, $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$ , that helps hold the star up. But its gradient is a genuine body force: $-\mathrm{d}P_{\mathrm{rad}}/\mathrm{d}r=\kappa\rho F/c$ is the rate at which streaming photons deposit momentum in matter. In hydrostatic equilibrium that force is already in the pressure balance, but when luminosity is large it can locally overwhelm gravity — precisely the Eddington limit and the origin of radiatively driven winds.

Why does the radiation fraction depend on mass as M² rather than on temperature directly?

Because temperature and density are not free in a star — they are set by mass and radius through hydrostatic balance. The ratio $P_{\mathrm{rad}}/P_{\mathrm{gas}}\propto T^3/\rho$ depends on local conditions, but the virial theorem gives $T_c\propto M/R$ and $\rho_c\propto M/R^3$ , so $T_c^3/\rho_c\propto(M/R)^3/(M/R^3)=M^2$ and the radius cancels. That cancellation is why the radiation fraction is a clean function of mass and composition, and why the Eddington quartic expresses $\beta$ in terms of $M$ alone.

Does the Eddington limit really forbid more luminous stars?

It forbids more luminous hydrostatic stars under the stated opacity. A star cannot sit in static equilibrium above $L_{\mathrm{Edd}}$ , but nature has workarounds: the excess drives mass loss, or the object sheds material in eruptions. In accretion disks, super-Eddington flow is possible because the geometry is not spherical and photons escape sideways while matter falls in. So the limit is best read as "the luminosity at which spherical hydrostatic equilibrium fails," not an absolute ceiling on brightness.

What does Γ₁ → 4/3 physically mean for stability?

$\Gamma_1$ measures the stiffness of the gas: compress it and the pressure rises by $\Gamma_1$ powers of the compression. For a self-gravitating body the total energy is $E\propto(\Gamma_1-4/3)$ times a positive structural integral, so at $\Gamma_1=4/3$ the total energy passes through zero and the restoring response to a global compression vanishes to leading order — the star is marginally stable. Radiation drives $\Gamma_1$ toward this value, and any extra softening (ionization, pair production, general relativity) can tip a very massive star into collapse or pulsation.

Why is the constant a, not σ, in the radiation pressure formula?

Because $\sigma$ describes energy leaving a surface while $a$ describes energy filling a volume, and pressure is about the volume-filling field inside the star. The Stefan–Boltzmann law $F=\sigma T^4$ is a flux through an area; the interior energy density is $u=aT^4$ , and pressure is one-third of that. They are tied by $a=4\sigma/c$ : the factor $c$ converts an energy density to a one-sided flux, and the factor $4$ combines the isotropic averaging. Using $\sigma$ where $a$ belongs misplaces the answer by nearly twenty orders of magnitude.