The Eddington luminosity is the point at which the radiation coming out of an object pushes on the surrounding gas as hard as the object’s gravity pulls it in. Above that luminosity, material near the source is pushed away instead of falling in. It sets a rough ceiling on how bright an accreting object can be, a rough ceiling on how massive a star can get, and a rough floor on how long a black hole needs to grow. The formula is short enough to memorise, which is a shame, because the derivation takes about ten lines and every one of those lines contains an assumption you can later break on purpose.
Set up the simplest possible situation. A compact object of mass $M$ sits at the origin and radiates isotropically with luminosity $L$. Around it is fully ionised hydrogen, which means free protons and free electrons and nothing else. The gas is optically thin, so a photon travelling outward has a small chance of interacting and mostly just leaves. Ask what happens to one electron-proton pair sitting at radius $r$.
Start with the radiation. Luminosity is energy per unit time, and if the source is isotropic that energy is spread evenly over a sphere of area $4\pi r^2$, so the energy flux at radius $r$ is $$F=\frac{L}{4\pi r^{2}}.$$ That’s energy per unit area per unit time. What we actually need is momentum, because forces come from momentum transfer.
A photon of energy $E$ carries momentum $p=E/c$. That follows from the relativistic energy-momentum relation $E^2=p^2c^2+m^2c^4$ with $m=0$, and it’s also what classical electromagnetism gives you for the ratio of momentum density to energy density in a plane wave. So a stream carrying energy flux $F$ also carries momentum flux $F/c$, and momentum flux has units of pressure. This quantity, $F/c$, is the radiation pressure available to push on anything that gets in the way.1It helps to see $p=E/c$ from more than one direction. In classical electromagnetism it comes from the Poynting vector and the Maxwell stress tensor: a plane wave has momentum density $S/c^2$ and energy density $S/c$, so their ratio is $1/c$. In quantum terms each photon carries $E=h\nu$ and $p=h\nu/c=h/\lambda$. Both give the same momentum flux, which is why the classical derivation of the Eddington limit needs no quantum mechanics at all.
Now we need to know how much of that momentum a single electron actually receives, which means we need a cross section. Calling a cross section an area is useful, but it shouldn’t be taken literally. The electron isn’t a small disk of area $6.65\times10^{-25}\ \mathrm{cm^2}$. A cross section packages an interaction probability into the dimensions of an area: if a beam delivers a certain number of photons per square centimetre per second, multiplying that flux by $\sigma$ gives the rate at which one electron scatters them. The area belongs to the interaction, and it tells you nothing about how big the electron is. With that understood, a particle with momentum-transfer cross section $\sigma$ sitting in a stream of momentum flux $F/c$ receives net momentum at a rate $\sigma F/c$, and a rate of momentum gain is a force.
The relevant process here is Thomson scattering, the elastic scattering of low-energy photons off free electrons, and it’s worth deriving rather than looking up because the derivation tells you exactly when it stops being valid. Take an electromagnetic wave with electric field amplitude $E_0$ passing a free electron. The field accelerates the electron with $a=eE_0\cos\omega t/m_e$. An accelerating non-relativistic charge radiates according to the Larmor formula, $P=2e^2a^2/3c^3$, and time-averaging $\cos^2$ gives a factor of $1/2$, so $$\langle P\rangle=\frac{2e^{2}}{3c^{3}}\cdot\frac{e^{2}E_0^{2}}{2m_e^{2}}=\frac{e^{4}E_0^{2}}{3m_e^{2}c^{3}}.$$ The incident flux is the time-averaged Poynting vector, $\langle S\rangle=cE_0^2/8\pi$. The cross section is the ratio of power scattered to flux incident, $$\sigma_T=\frac{\langle P\rangle}{\langle S\rangle}=\frac{8\pi}{3}\left(\frac{e^{2}}{m_ec^{2}}\right)^{2}=\frac{8\pi}{3}r_e^{2}=6.65\times10^{-25}\ \mathrm{cm^{2}},$$ where $r_e=e^2/m_ec^2=2.82\times10^{-13}$ cm is the classical electron radius.
That derivation hands us something useful for free. The cross section scales as $1/m^2$, so the same calculation applied to a proton gives a cross section smaller by $(m_e/m_p)^2=(1/1836)^2\approx3\times10^{-7}$. Protons scatter light about three million times less effectively than electrons. So when we say the radiation force acts on the electrons and ignore the protons, that’s a conclusion with a number behind it.
So the outward radiation force on one free electron is $$f_{\rm rad}=\sigma_T\,\frac{F}{c}=\frac{\sigma_T L}{4\pi r^{2}c}.$$
There’s something slightly suspicious about what comes next. The radiation force we just wrote contains an electron cross section. The gravitational force is about to contain a proton mass. Why are we allowed to balance two forces that act on two different particles?
The answer is electrostatics. Gravity pulls on mass, and essentially all the mass is in the protons since $m_p/m_e\approx1836$, while radiation pushes on the electrons. If the electrons started to drift outward relative to the protons, that would immediately create a charge separation, which creates an electric field, which pulls the electrons back and drags the protons out. The question is how much separation is needed, and the answer is almost none, because the Coulomb interaction is absurdly stronger than gravity. For a single proton-electron pair, $$\frac{e^{2}}{G m_p m_e}=2.3\times10^{39}.$$ A fractional charge imbalance far too small to measure supplies an electric field far larger than anything gravity or radiation is doing. The plasma therefore moves as a neutral fluid, and the correct bookkeeping is to treat a proton-electron pair as the unit: it feels the radiation force of one electron and the gravitational pull of one proton.2You can also make this concrete. To transmit a gravitational force $GMm_p/r^2$ to an electron by electrostatic means, you need a field $E$ with $eE=GMm_p/r^2$, which for a spherical geometry corresponds to a net enclosed charge of $Q=GMm_p/e$. For a solar mass that’s about $10^{21}$ elementary charges, which sounds like a lot until you notice that a single gram of ionised hydrogen contains $6\times10^{23}$ electrons. The required fractional imbalance is tiny, and the plasma establishes it essentially instantly.
The inward gravitational force per pair is then $$f_{\rm grav}=\frac{GM(m_p+m_e)}{r^{2}}\simeq\frac{GMm_p}{r^{2}}.$$ Setting the two forces equal gives $$\frac{\sigma_T L}{4\pi r^{2}c}=\frac{GMm_p}{r^{2}},$$ and the radius cancels from both sides, leaving $$L_{\rm Edd}=\frac{4\pi GMm_pc}{\sigma_T}.$$
That cancellation is easy to read past, and it’s the reason the Eddington limit comes out as a statement about luminosity instead of a statement about some particular radius. Both forces fall off as $1/r^2$: gravity because it’s an inverse square law, and radiation because the flux from a point source dilutes as $1/r^2$. Moving farther away weakens gravity and radiation pressure in exactly the same proportion. Plotted against radius on log axes, both are straight lines of slope $-2$, which means they’re parallel and never cross. The balance is either satisfied at every radius or at no radius. There’s no critical distance inside which radiation wins. There’s a critical luminosity, and above it radiation wins at every radius at once.
Putting numbers in, $$L_{\rm Edd}=1.26\times10^{38}\left(\frac{M}{M_\odot}\right)\ \mathrm{erg\,s^{-1}}=3.3\times10^{4}\left(\frac{M}{M_\odot}\right)L_\odot.$$ The Sun’s actual luminosity is about $3\times10^{-5}$ of this value, so it sits very far from the balance point and nothing in its outer layers is close to being pushed off. Massive stars are a different matter, and we’ll come back to that.
Before breaking anything, rewrite the result in a form that makes the assumptions easier to isolate. Define the opacity $\kappa$ as cross section per unit mass, so that the force per gram of material is $\kappa F/c$. For pure ionised hydrogen there’s one electron per proton, so $\kappa_{\rm es}=\sigma_T/m_p=0.40\ \mathrm{cm^2\,g^{-1}}$, and the general statement becomes $$L_{\rm Edd}=\frac{4\pi GMc}{\kappa}.$$ The composition and the physics of the interaction are now quarantined inside a single symbol. Many of the corrections below can be understood either as changing $\kappa$, or as breaking one of the assumptions that let us write this simple force balance in the first place.
Check the units of that expression once, because it shows why $\kappa$ has to be an area per unit mass and not just an area: $$[G][M][c]\,[\kappa]^{-1}=\left(\frac{\mathrm{cm^{3}}}{\mathrm{g\,s^{2}}}\right)(\mathrm{g})\left(\frac{\mathrm{cm}}{\mathrm{s}}\right)\left(\frac{\mathrm{g}}{\mathrm{cm^{2}}}\right)=\frac{\mathrm{g\,cm^{2}}}{\mathrm{s^{3}}}=\mathrm{erg\,s^{-1}}.$$ Notice that the mass in $GM$ and the mass hidden inside $\kappa$ don’t cancel. That’s gravity pulling on the gas and radiation pushing on its cross section, showing up in the units.
Composition is the easiest thing to fix. What matters for scattering is the number of free electrons per gram, and hydrogen is unusual because it supplies one electron per nucleon while everything else supplies roughly one per two nucleons. For a fully ionised mix with hydrogen mass fraction $X$ and helium mass fraction $Y=1-X$, the electron density is $n_e=(\rho/m_p)(X+Y/2)=(\rho/m_p)(1+X)/2$, so $$\kappa_{\rm es}=\frac{\sigma_T}{m_p}\frac{1+X}{2}\simeq0.20\,(1+X)\ \mathrm{cm^{2}\,g^{-1}}.$$ For solar composition, $X=0.7$, this gives $0.34\ \mathrm{cm^2\,g^{-1}}$ and $L_{\rm Edd}=1.5\times10^{38}(M/M_\odot)\ \mathrm{erg\,s^{-1}}$. For hydrogen-free material the opacity halves and the Eddington limit doubles, purely because there are half as many electrons per gram to push on.3The commonly quoted $1.26\times10^{38}(M/M_\odot)$ assumes pure hydrogen. Papers on accreting neutron stars and X-ray bursts often use the solar or the helium value instead, so a factor of two discrepancy between two sources usually means they made different composition assumptions rather than that one of them is wrong. So check which $\kappa$ a quoted Eddington luminosity is built on before comparing two of them.
There’s a distinction buried in the derivation that causes a lot of confusion later, so let’s name it now. The force balance we wrote is local. It compares two forces on one parcel of gas at one place. It only turned into a statement about the total luminosity of the source because spherical symmetry let us write the local flux as $L/4\pi r^2$ everywhere at once. So $L_{\rm Edd}$ is a global luminosity scale derived from a local condition, and the translation between the two depends entirely on the geometry. When a source is described as super-Eddington in the literature, it’s often worth asking which of the two is meant, because a source can exceed the spherical Eddington luminosity as measured by a distant observer while the radiation force on every individual parcel of gas remains sub-Eddington.
That distinction is the first structural assumption to break. The derivation put a point source at the centre of a spherical shell of gas, so radiation and gravity were both radial and could be compared directly. Real accretion usually happens through a disc, and radiation escaping from an accretion flow doesn’t come out isotropically. If the gas arrives along the equatorial plane while the radiation leaves through low-density funnels along the rotation axis, the two never have to fight over the same material, and the luminosity inferred by a distant observer can exceed $L_{\rm Edd}$ while the accretion flow itself stays locally sub-Eddington.
The second structural assumption is that the situation can be treated through an instantaneous force balance. The derivation compares two forces acting on the gas at a given moment, and by itself says nothing about how the gas responds with time. A transient can therefore exceed the limit for as long as it takes the surrounding material to accelerate, expand, or otherwise adjust. Classical novae and X-ray bursts on neutron stars both show episodes consistent with locally super-Eddington fluxes, and the associated observational signature is photospheric radius expansion, in which the envelope is lifted, expands, and settles back as the luminosity drops.
The third assumption, and the one that fails most often in practice, is that electron scattering is the only relevant opacity. For soft photons in a hot, fully ionised plasma, electron scattering provides a useful baseline: it’s what remains when the gas is too hot to have any bound electrons left to absorb anything. Bound-free, free-free, line and dust interactions can all add substantially to it, and since $L_{\rm Edd}\propto1/\kappa$, every addition lowers the true limit below the value the standard formula gives. But even this baseline isn’t universal, and we’ll see shortly that once photon energies approach $m_ec^2$ the Thomson approximation itself fails and the scattering opacity falls below $\kappa_{\rm es}$.
Spectral lines are the clearest example of an addition. A bound electron in an ion can absorb a photon whose energy matches a transition, and the cross section at line centre can exceed the Thomson cross section by many orders of magnitude. In the ultraviolet spectrum of a hot star there are very large numbers of such transitions from iron-group ions, and although each covers a narrow frequency range, together they intercept a substantial fraction of the flux. The effective opacity for a hot star’s wind can be tens to hundreds of times $\kappa_{\rm es}$, which is how O stars can drive strong winds while their continuum Eddington ratios remain well below unity. The star is sub-Eddington for electron scattering and super-Eddington for lines.4This is the basis of the Castor, Abbott and Klein theory of line-driven winds from 1975, which parametrises the extra push through a force multiplier applied to the electron-scattering radiation force. The multiplier depends on the ionisation state and on the velocity gradient in the wind, because a wind that’s accelerating Doppler-shifts each line into fresh continuum photons and keeps the lines from saturating. The velocity gradient is therefore part of the opacity, which is an unusual feature of the problem.
Dust is the more extreme addition. A grain absorbs across a broad continuum rather than in narrow lines, and for gas with an ordinary interstellar dust-to-gas ratio the ultraviolet opacity per gram of gas is of order a few hundred $\mathrm{cm^2\,g^{-1}}$, which is roughly a thousand times $\kappa_{\rm es}$. In dusty gas the relevant Eddington limit therefore drops by about three orders of magnitude, and objects that look comfortably sub-Eddington by the standard formula can be strongly super-Eddington for the material around them. This is why radiation pressure on dust matters for massive star formation and for feedback in galactic nuclei, and why the electron-scattering formula is a poor guide in those settings.5The estimate comes from the standard interstellar extinction per hydrogen column, $A_V/N_H\approx5\times10^{-22}$ mag cm$^2$, converted to an opacity per gram using a mean mass per hydrogen nucleus of about $1.4m_p$, which gives roughly $200\ \mathrm{cm^2\,g^{-1}}$ in the visual and two to three times that in the ultraviolet. It scales linearly with the dust-to-gas ratio, so it drops in low-metallicity environments and vanishes wherever grains have been destroyed.
The fourth assumption is that the photons are soft. The Thomson derivation treated the electron as a non-relativistic charge oscillating in a classical wave, which requires the photon energy to be small compared to $m_ec^2=511$ keV. Above that, the photon recoils the electron appreciably, the scattering becomes inelastic, and the Klein-Nishina formula replaces the Thomson result. The cross section falls, roughly as $\ln(2x)/x$ for $x=h\nu/m_ec^2\gg1$. Hard photons are therefore less effective at pushing gas, and this is the one correction that pushes the limit up rather than down: a source radiating mainly above a few hundred keV can exceed the naive Eddington luminosity simply because its photons couple weakly to electrons.
The fifth assumption is Newtonian gravity, which is uncomfortable given that the objects people apply this to are usually black holes and neutron stars. In the Schwarzschild metric two things change for a static observer at radius $r$. The local gravitational acceleration picks up a factor $(1-r_s/r)^{-1/2}$, and the flux from a source of luminosity $L_\infty$ as measured at infinity picks up a factor $(1-r_s/r)^{-1}$ from gravitational redshift and time dilation together. Balancing them gives $$L_{\rm Edd}^{\infty}=\frac{4\pi GMc}{\kappa}\sqrt{1-\frac{r_s}{r}}.$$ The two corrections partly cancel, which is why the Newtonian answer survives as well as it does, but not completely. At $3r_s$ the limit is $82$ percent of the flat-space value and at $1.5r_s$ it’s $58$ percent. Note also that the answer now depends on $r$, so the radius no longer cancels and the earlier statement about parallel lines fails.
The sixth assumption is that the gas is ordinary matter with protons in it. Within the same simple force-balance picture, an electron-positron pair plasma changes the bookkeeping completely: there are no protons carrying most of the mass, so gravity acts on a mass of order $m_e$ per scattering particle rather than $m_p$, while the scattering cross section remains of order $\sigma_T$. The corresponding Eddington scale therefore drops by roughly $m_e/m_p\approx1/1836$. Real pair plasmas involve additional physics such as pair creation, annihilation and more complicated radiative transfer, so this factor is what you get by pushing the original argument into a new regime, and the real problem carries more than that. Pair plasmas are thought to be relevant in the inner regions of some accretion flows and in gamma-ray burst fireballs, where corrections of this size are not small.
The seventh assumption is that the gas is smooth. The derivation implicitly treats the medium as uniform, so every gram sees the same flux. If the gas is clumpy, radiation escapes preferentially through the low-density channels between the clumps and deposits less momentum than a uniform calculation predicts. The effective opacity of a porous medium is lower than that of a smooth one with the same mean density, which raises the effective limit. This idea, usually called porosity, is one of the mechanisms proposed for the eruptions of luminous blue variables, which appear to radiate above their nominal limits for extended periods.
With the limit itself understood, two more quantities follow, and these are what the Eddington luminosity actually gets used for in practice. If an accreting object converts a fraction $\eta$ of the rest mass energy of the infalling material into radiation, so that $L=\eta\dot Mc^2$, then the accretion rate corresponding to the Eddington luminosity is $$\dot M_{\rm Edd}=\frac{L_{\rm Edd}}{\eta c^{2}}=\frac{4\pi GM}{\eta\kappa c}\approx2.2\times10^{-8}\left(\frac{0.1}{\eta}\right)\left(\frac{M}{M_\odot}\right)M_\odot\,\mathrm{yr^{-1}}.$$ The efficiency $\eta$ is about $0.057$ for a non-rotating black hole and up to about $0.42$ for a maximally rotating one, so this number carries a factor of several of uncertainty depending on spin.
The second derived quantity needs one piece of care that’s easy to miss. If $\dot M$ is the rest mass supplied to the hole and a fraction $\eta$ of it is radiated away, then the mass actually retained is $\dot M_{\rm BH}=(1-\eta)\dot M$. Carrying that through, the e-folding time for the black hole mass under Eddington-limited accretion is $$t_S=\frac{\eta}{1-\eta}\,\frac{\kappa c}{4\pi G}\approx50\ \mathrm{Myr}\quad(\eta=0.1),$$ rather than the $45$ Myr you get by dropping the $(1-\eta)$. The distinction is small at low efficiency and large at high efficiency: for a maximally spinning hole with $\eta=0.42$ it changes the answer from $189$ to $326$ Myr.6The $45$ Myr form is the one usually quoted, and for order-of-magnitude arguments the difference rarely matters. Check which version a paper is using when the argument turns on a factor of two, which it sometimes does in discussions of early black hole growth.
That number is behind a well-known problem. Growing a $10\ M_\odot$ seed into a $10^9\ M_\odot$ quasar takes about $18$ e-foldings, which at $50$ Myr each is roughly $900$ Myr of continuous Eddington-limited accretion with no idle periods at all. Quasars of that mass are observed less than a billion years after the Big Bang, so the budget is uncomfortably tight, which is why heavier seeds and episodes of super-Eddington accretion are both actively discussed.
The limit bites hardest in massive stars. Main sequence luminosity rises steeply with mass, faster than linearly, while the Eddington luminosity rises exactly linearly, so the ratio $\Gamma=L/L_{\rm Edd}$ climbs with mass. Using standard zero-age main sequence luminosities, a $10\ M_\odot$ star has $\Gamma\approx0.03$ and a $100\ M_\odot$ star roughly $0.3$ on the continuum opacity alone. Add line opacity and the effective ratio approaches unity, which is connected both to the large mass loss rates of the most massive stars and to the existence of an upper limit on stellar masses. For these objects the Eddington limit is doing real work on the structure of the star.
References and Footnotes
- 1It helps to see $p=E/c$ from more than one direction. In classical electromagnetism it comes from the Poynting vector and the Maxwell stress tensor: a plane wave has momentum density $S/c^2$ and energy density $S/c$, so their ratio is $1/c$. In quantum terms each photon carries $E=h\nu$ and $p=h\nu/c=h/\lambda$. Both give the same momentum flux, which is why the classical derivation of the Eddington limit needs no quantum mechanics at all. ↩︎
- 2You can also make this concrete. To transmit a gravitational force $GMm_p/r^2$ to an electron by electrostatic means, you need a field $E$ with $eE=GMm_p/r^2$, which for a spherical geometry corresponds to a net enclosed charge of $Q=GMm_p/e$. For a solar mass that’s about $10^{21}$ elementary charges, which sounds like a lot until you notice that a single gram of ionised hydrogen contains $6\times10^{23}$ electrons. The required fractional imbalance is tiny, and the plasma establishes it essentially instantly. ↩︎
- 3The commonly quoted $1.26\times10^{38}(M/M_\odot)$ assumes pure hydrogen. Papers on accreting neutron stars and X-ray bursts often use the solar or the helium value instead, so a factor of two discrepancy between two sources usually means they made different composition assumptions rather than that one of them is wrong. So check which $\kappa$ a quoted Eddington luminosity is built on before comparing two of them. ↩︎
- 4This is the basis of the Castor, Abbott and Klein theory of line-driven winds from 1975, which parametrises the extra push through a force multiplier applied to the electron-scattering radiation force. The multiplier depends on the ionisation state and on the velocity gradient in the wind, because a wind that’s accelerating Doppler-shifts each line into fresh continuum photons and keeps the lines from saturating. The velocity gradient is therefore part of the opacity, which is an unusual feature of the problem. ↩︎
- 5The estimate comes from the standard interstellar extinction per hydrogen column, $A_V/N_H\approx5\times10^{-22}$ mag cm$^2$, converted to an opacity per gram using a mean mass per hydrogen nucleus of about $1.4m_p$, which gives roughly $200\ \mathrm{cm^2\,g^{-1}}$ in the visual and two to three times that in the ultraviolet. It scales linearly with the dust-to-gas ratio, so it drops in low-metallicity environments and vanishes wherever grains have been destroyed. ↩︎
- 6The $45$ Myr form is the one usually quoted, and for order-of-magnitude arguments the difference rarely matters. Check which version a paper is using when the argument turns on a factor of two, which it sometimes does in discussions of early black hole growth. ↩︎
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