Accretion and the Eddington limit

Builds on: Sections 10.1, 10.2, 8.5, 8.6 and 4.1
Reading time: 45 min
Exercises: 3
Olympiad problems: 1

In 1962 a small rocket carried an X-ray detector above the atmosphere for a few minutes, and it found that the brightest source of X-rays in the sky, apart from the Sun, lies in Scorpius. Scorpius X-1, as it is called, turned out to be a binary of an ordinary faint star and a neutron star. Gas from the star spirals down onto the neutron star and lands at a large fraction of the speed of light, and the heat of the landing makes it shine in X-rays with tens of thousands of times the Sun's power. In this section we see how much energy falling gas can release, and why there is a limit to how bright such a source can be.

10.4.1Energy from falling

Before you readMake a guess first

Which gives out more energy for each kilogram of hydrogen: fusing it into helium inside a star, or dropping it onto a neutron star?

The capture of gas by a star or a compact object is called accretion. Gas that falls from far away onto the surface of a body of mass $M$ and radius $R$ loses the potential energy $G M m / R$ (S.20). It first gains this as kinetic energy, and when it strikes the surface it is stopped and heated, so the energy comes out as radiation. If gas falls in at a steady rate $\dot{M}$ , kilograms each second, the power given out is the energy lost by each kilogram times the rate.

ResultThe accretion luminosity(10.11)

L_{acc} = \frac{G M \dot{M}}{R} = η \dot{M} c^{2}, η = \frac{G M}{R c^{2}} = \frac{R_{s}}{2 R}

$L_{acc}$	the luminosity of the accreting gas
$M, R$	the mass and radius of the body the gas lands on
$\dot{M}$	the rate at which mass falls in
$η$	the efficiency, the fraction of the rest energy $m c^{2}$ of the gas that comes out as radiation

This assumes that all the energy of the fall is radiated and that the gas starts from rest far away. The efficiency depends only on how compact the body is.

For a white dwarf like Sirius B the efficiency is $2.7 \times 10^{- 4}$ , less than the 0.7% of hydrogen fusion. For a neutron star of 1.4 solar masses and radius 12 km it is 0.17, so each kilogram that lands releases $1.55 \times 10^{16}$ J, about 25 times as much as fusion. A black hole has no surface to land on, and gas falls through the horizon carrying much of its energy with it. Gas that spirals in through a disc radiates until it reaches the innermost stable orbit at $3 R_{s}$ (Section 10.2), which releases 5.7% of its rest energy for a black hole that does not spin and up to 42% for one that spins as fast as it can. Accretion onto neutron stars and black holes is the most efficient way of making light that nature has, apart from the meeting of matter with antimatter.

Quick checkAn X-ray binary

A neutron star of 1.4 solar masses and radius 12 km takes in $10^{- 9}$ solar masses of gas a year. What is its accretion luminosity, in solar luminosities?

solar luminosities

10.4.2The Eddington limit

Light carries momentum, so radiation that streams out from a bright source pushes on the gas that is falling in. In ionised hydrogen the push acts on the electrons, which scatter the light, and each electron presents a target of area $σ_{T} = 6.65 \times 10^{- 29}$ m², called the Thomson cross-section. The electrons cannot leave the protons behind, because the electric pull between them is enormous, so the push on each electron acts on a whole atom of hydrogen. Gravity pulls on the proton, which has almost all the mass.

At a distance $r$ the light of a source of luminosity $L$ carries the momentum $L / (4 π r^{2} c)$ through each square metre each second, so the push on one electron is $σ_{T} L / (4 π r^{2} c)$ . The pull of gravity on the proton is $G M m_{p} / r^{2}$ . Both fall off as $1 / r^{2}$ , so whichever is stronger at one distance is stronger at every distance. The luminosity at which they balance is the Eddington luminosity.

ResultThe Eddington luminosity(10.12)

L_{Edd} = \frac{4 π G M m_{p} c}{σ_{T}} = 1.26 \times 10^{31} W \times \frac{M}{M_{⊙}}

$L_{Edd}$	the Eddington luminosity
$M$	the mass of the shining body
$m_{p}$	the mass of the proton
$σ_{T}$	the Thomson cross-section of the electron, $6.65 \times 10^{- 29}$ m²

Above this luminosity the radiation pushes infalling gas away, so a source that is fed by falling gas cannot shine brighter for long. The limit is

3.3 \times 10^{4}

solar luminosities for each solar mass. It assumes ionised hydrogen and light that streams out evenly in all directions.

Figure 10.4The Eddington limit

MassLuminosity

Radiation over gravity
Infalling gas
Smallest mass for this luminosity

The figure shows, on the left, a graph of luminosity against mass, both on logarithmic scales, with the heavy Eddington line and the shaded region above it where radiation blows infalling gas away. The open circle S is the Sun, far below its limit, and X is the object of chosen mass and luminosity. On the right, a parcel of gas is drawn at the distances

r

1.5 r

and

2 r

from X, with the pull of gravity as the heavy arrow inwards and the push of the radiation as the thin arrow outwards. Both arrows shrink as

1 / r^{2}

, so their ratio stays the same. The figure can be set to , , or .

The Sun shines at only $3.0 \times 10^{- 5}$ of its Eddington luminosity. The luminosity of a star grows much faster than its mass, so the heaviest stars come closest to their limit. The heaviest stars known, of more than 100 solar masses, shine at a large fraction of their Eddington luminosity, and stars much heavier would blow off their own outer layers. This is one reason why no star of more than a few hundred solar masses is known.

Example 10.7How fast a neutron star can feed

Find the Eddington luminosity of a neutron star of 1.4 solar masses and radius 12 km, and the greatest rate at which it can take in gas.

Solution

By (10.12), $L_{Edd} = 1.4 \times 1.26 \times 10^{31} = 1.76 \times 10^{31}$ W. Setting the accretion luminosity of (10.11) equal to this gives

\begin{aligned} \dot{M} & = \frac{L_{Edd} R}{G M} = \frac{1.76 \times 10^{31} \times 1.2 \times 10^{4}}{6.674 \times 10^{- 11} \times 2.783 \times 10^{30}} \\ = 1.14 \times 10^{15} kg/s . \end{aligned}

AnswerThe neutron star can shine at up to

1.76 \times 10^{31}

W and take in at most

1.8 \times 10^{- 8}

solar masses a year. The brightest X-ray binaries, Scorpius X-1 among them, shine close to this limit.

Example 10.8The smallest mass of a quasarfinish it

A quasar, the bright centre of a distant galaxy, gives out $10^{40}$ W, which is $2.6 \times 10^{13}$ solar luminosities. If it is powered by gas falling onto a black hole, what is the smallest mass that the black hole can have?

The start

The luminosity can be no more than the Eddington luminosity of the black hole, so $M$ must be at least $L / (1.26 \times 10^{31} W)$ solar masses, by (10.12).

Your stepNow find the smallest mass.

× 10⁸ solar masses

Common errorsSection 10.4

The Eddington limit does not depend on distance, since gravity and the push of radiation both fall off as $1 / r^{2}$ .
The push of radiation acts on the electrons and the pull of gravity on the protons. Using the mass of the electron in $L_{Edd}$ gives a limit almost 2000 times too small.
The accretion luminosity uses the radius of the body the gas lands on, not the distance from which it falls, when the gas starts far away.
A source can exceed its Eddington luminosity for a short time, as a nova or a supernova does, but it cannot then be fed by steadily falling gas.

ExercisesSection 10.43 questions

10.4.1
What is the Eddington luminosity of a black hole of 10 solar masses, in units of $10^{5}$ solar luminosities?
× 10⁵ solar luminosities
10.4.2
A cloud of ionised hydrogen surrounds a bright star. If the cloud is moved twice as far from the star, what happens to the ratio of the push of radiation to the pull of gravity on it?
10.4.3
What is the efficiency of accretion onto Sirius B, of 1.02 solar masses and radius 5640 km, in units of $10^{- 4}$ ?
× 10⁻⁴

Olympiad problemsPast IOAA problems, restated1 problem

2022The limits of accretion20 points

Adapted from IOAA 2022 (Kutaisi), theory problem 10. We have restated it in our own words and kept the original data. In part (f) the official solution keeps the angular speed of the orbit fixed while the separation changes, which gives a change in separation four times too small. Using Kepler's third law, as we do, gives the same conclusion with the right size. The solution was last checked on 23 October 2026.

A compact object of mass $M$ draws in hydrogen evenly from all sides. The infalling gas heats up and radiates, and the radiation pushes on it with the force $F = σ_{T} I / c$ on each electron, where $I$ is the intensity of the light and

σ_{T} = \frac{8 π}{3} {(\frac{e^{2}}{4 π ε_{0} m_{e} c^{2}})}^{2}

is the Thomson cross-section. Although the force acts on the electrons, it acts on whole atoms.

(a) Find the Eddington luminosity $L_{E}$ , the greatest luminosity the accreting gas can have. (3 points)

(b) An object accretes gas with a luminosity equal to the Sun's. What is its smallest possible mass, in solar masses? (4 points)

(c) The gas starts far away and its energy of falling is turned into radiation. Find the accretion luminosity in terms of $M$ , the accretion rate $\dot{M}$ and the radius $R$ of the object. (2 points)

(d) Show that the greatest steady rate of accretion does not depend on the mass of the object. (2 points)

(e) Find this rate, in solar masses a year, for an object of radius $1.2 \times 10^{10}$ m. (2 points)

(f) The object, of mass $M_{1}$ , is now in a binary with a star of mass $M_{2}$ at a distance $a$ , and gains gas from it. Find the condition on the masses for the separation to grow. Ignore the spin of the stars. (7 points)

I solved it without the solution

Answer(a) $L_{E} = 4 π G M m_{H} c / σ_{T}$ . (b) $3.04 \times 10^{- 5}$ solar masses. (c) $L_{acc} = G M \dot{M} / R$ . (d) ${\dot{M}}_{max} = 4 π m_{H} c R / σ_{T}$ . (e) 0.018 solar masses a year. (f) The separation grows when the object gaining mass is the heavier one, as in (8.8).