Accretion discs

Builds on: Sections 10.2, 10.4, 4.6 and 8.6
Reading time: 35 min
Exercises: 3
Olympiad problems: 1

In the constellation Virgo, a moderate telescope shows a star of about the 13th magnitude that is not a star at all. It is the quasar 3C 273, and its light has been travelling for about 2.0 billion years. To look so bright from so far it must give out more light than a few trillion Suns, and since it flickers within days, the light comes from a region only a few light-days across. The source is a disc of gas spiralling into a black hole of several hundred million solar masses. In this section we see how such a disc works and why it is so hot.

10.5.1Why gas forms a disc

Gas that falls towards a compact object almost never heads straight at it. It has some sideways motion, and therefore angular momentum, which it keeps as it falls, so it cannot reach the centre and instead goes into orbit. Streams of gas on different orbits collide and share their motion, and their angular momentum, until they all move round in the same direction in the same plane, a flat (Section 8.6).

Each ring of the disc moves on a nearly circular orbit, faster close in, as Kepler's third law requires. Neighbouring rings rub against each other. This friction slows the inner rings, which then drop a little closer, and speeds up the outer ones, which move out. Angular momentum is carried outwards and mass inwards, so the gas spirals slowly towards the centre. The rubbing heats the gas, and the disc glows.

10.5.2The energy of a disc

Before you readMake a guess first

Gas spirals slowly inwards through a disc, from far away to a circular orbit at radius $r$ . What fraction of the potential energy it has lost has it given out as radiation?

On a circular orbit of radius $r$ a mass $m$ has the kinetic energy $G M m / (2 r)$ and the total energy $- G M m / (2 r)$ (3.4). Gas that comes from far away, where its energy was close to zero, and settles onto this orbit has therefore given out $G M m / (2 r)$ , half of the potential energy it lost. The gas keeps spiralling in until it reaches the inner edge of the disc, at radius $R_{in}$ , so a disc fed at a steady rate $\dot{M}$ gives out the luminosity $G M \dot{M} / (2 R_{in})$ . This is half the accretion luminosity of (10.11). Onto a star with a surface, the other half is released where the disc meets the star, and into a black hole it is swallowed.

Round a black hole the inner edge is the innermost stable circular orbit, $R_{in} = 3 R_{s} = 6 G M / c^{2}$ (Section 10.2), which gives $L = \dot{M} c^{2} / 12$ , an efficiency of 8.3%. Einstein's theory gives 5.7% for a black hole that does not spin, so the simple estimate is a little too high and gives the right order of size.

To find the temperature, we look at one ring of the disc, between $r$ and $r + d r$ . The gas passing through it in each second gives out the difference of the energies $G M \dot{M} / (2 r)$ at its two edges, which is $G M \dot{M} d r / (2 r^{2})$ . The ring has two faces, top and bottom, of total area $2 \times 2 π r d r$ . If each face radiates as a blackbody, $4 π r d r σ T^{4} = G M \dot{M} d r / (2 r^{2})$ .

ResultThe temperature of an accretion disc(10.13)

T = {(\frac{G M \dot{M}}{8 π σ r^{3}})}^{1 / 4} \propto r^{- 3 / 4}

$T$	the temperature of the surface of the disc at radius $r$
$M$	the mass of the central object
$\dot{M}$	the rate at which mass flows through the disc
$σ$	the Stefan-Boltzmann constant

This simple model assumes that each ring radiates the energy released inside it, as a blackbody from both faces. The full theory of thin discs keeps

T \propto r^{- 3 / 4}

far out, with a temperature larger there by a factor of

3^{1 / 4}

, because friction also carries energy outwards from the inner rings. It also gives a cooler inner edge. For estimates the simple model is enough, and it is the one that the IOAA 2024 problem below uses.

The temperature falls as $r^{- 3 / 4}$ , so a ring ten times further out is 5.6 times cooler. The disc is hottest near its inner edge, and the light from there dominates the total. Its spectrum is the sum of the blackbodies of all the rings, each at its own temperature, so it is much broader than the spectrum of a star.

Figure 10.5An accretion disc round a black hole

Mass of the black holeAccretion rate

Inner edge
Peak of the inner edge
Luminosity of the disc

The figure shows, on the left, a disc round a black hole seen face on, from its inner edge I at

3 R_{s}

out to a thousand times further, with brighter shading for hotter gas on a logarithmic scale of radius. On the right, above, the temperature falls as

r^{- 3 / 4}

from its value at the inner edge, and below, the heavy curve is the spectrum of the whole disc out to a thousand inner radii, each curve scaled to its own peak. It is broader than the dashed spectrum of a single blackbody at the temperature of the inner edge. The dotted lines mark the bands of X-rays, ultraviolet, visible and infrared light. The figure can be set to , , or .

The inner edge sits at a radius proportional to $M$ , so for a given rate of feeding (10.13) gives an inner temperature proportional to $M^{- 1 / 2}$ . The disc round a black hole of 10 solar masses fed with $10^{- 8}$ solar masses a year reaches $5.4 \times 10^{6}$ K at its inner edge and peaks at 0.54 nm, in X-rays, which is why black holes in binaries were found by X-ray satellites. The far larger discs round the black holes of quasars are much cooler.

Example 10.9The disc of a quasar

A black hole of $10^{8}$ solar masses takes in one solar mass of gas a year. Find the luminosity of its disc, the temperature at the inner edge and the wavelength at which the inner edge shines most strongly.

Solution

The rate is $\dot{M} = 1.988 \times 10^{30} / (3.156 \times 10^{7})$ kg/s, and the luminosity is $\dot{M} c^{2} / 12 = 4.7 \times 10^{38}$ W. The inner edge is at $6 G M / c^{2}$ , which is 5.9 au, and (10.13) gives $T = 1.7 \times 10^{5}$ K there. By Wien's law (4.11) the peak is at $2.898 \times 10^{- 3} / (1.7 \times 10^{5})$ m, which is 17 nm.

AnswerThe disc gives out

4.7 \times 10^{38}

1.2 \times 10^{12}

solar luminosities and 0.38 of its Eddington luminosity (10.12), with the inner edge at

1.7 \times 10^{5}

K, shining most strongly at 17 nm in the far ultraviolet. Most of this light is absorbed by gas in our own Galaxy, but the cooler parts of the disc further out make quasars bright in ordinary ultraviolet and blue light.

Example 10.10How big a quasar can befinish it

A quasar doubles in brightness in 10 days. Light from its far side reaches us later than light from its near side, so a region larger than the distance light travels in 10 days could not change all at once. Find the greatest size of the region, in astronomical units.

The start

The region can be no larger than $c \times 10$ days, which is $2.998 \times 10^{8} \times 8.64 \times 10^{5}$ m.

Your stepNow turn this distance into astronomical units.

Common errorsSection 10.5

Gas on its way in through a disc radiates only half of the potential energy it loses. The other half goes into the speed of its orbit.
The temperature of a disc falls as $r^{- 3 / 4}$ . It is not the same everywhere, so a disc does not have the spectrum of a single blackbody.
Each ring of the disc radiates from two faces, top and bottom, which gives the factor $8 π$ in (10.13).
A heavier black hole fed at the same rate has a cooler disc, because its inner edge is further out. Discs round quasars shine in ultraviolet light and those round stellar black holes in X-rays.

ExercisesSection 10.53 questions

10.5.1
By what factor is a ring of an accretion disc 10 times further from the centre cooler than the first ring?
10.5.2
Two black holes, of 10 and $10^{8}$ solar masses, take in gas at the same rate. Which disc has the hotter inner edge?
10.5.3
In the simple model of this section, what percentage of the rest energy of the gas does a disc round a black hole give out, if its inner edge is at $3 R_{s}$ ?
%

Olympiad problemsPast IOAA problems, restated1 problem

2024The physics of accretion35 points

Adapted from IOAA 2024 (Rio de Janeiro), theory problem T9. We have restated it in our own words and kept the original data. In part (f) the official solution rounds the temperature to 4.8 × 10⁴ K and gives 1.3 × 10⁹ solar masses. Without that rounding the mass is 1.2 × 10⁹, a difference of only 1% that falls either side of 1.25. The solution was last checked on 23 October 2026.

A thin, steady disc of gas orbits a compact object of mass $M$ and radius $R$ , from an outer radius $R_{max}$ to an inner radius $R_{min}$ , with $R_{min} / R_{max}$ much less than 1. Ignore the gravity of the disc itself.

(a) An element of gas of mass $Δ m$ moves on a nearly circular orbit and spirals slowly inwards, its lost energy turned into heat by friction. Find the energy per unit mass, $Δ E / Δ m$ , that it releases between $R_{max}$ and a radius $r$ much less than $R_{max}$ . (6 points)

(b) Mass enters the disc at the rate $\dot{M}$ , and all the energy released becomes radiation. Find the total luminosity of the disc. (5 points)

(d) The energy released in each ring is radiated from its surface as a blackbody. Find the temperature of the ring. (10 points)

(e) The central object is a black hole of 3 solar masses fed at $10^{- 9}$ solar masses a year, with $R_{min} = 3 R_{s}$ . Find the luminosity of the disc and the peak wavelength of its inner part. Ignore the gravitational redshift. (3 points)

(f) Another black hole is fed at one solar mass a year, and the inner part of its disc peaks at $6 \times 10^{- 8}$ m. Estimate its mass. (3 points)

I solved it without the solution

Answer(a) $G M / (2 r)$ . (b) $G M \dot{M} / (2 R_{min})$ . (c) $G M \dot{M} / (2 r^{2})$ . (d) $T = (G M \dot{M} / (8 π σ r^{3}))^{1 / 4}$ . (e) $4.7 \times 10^{29}$ W, with a peak at $5.2 \times 10^{- 10}$ m. (f) About $1.2 \times 10^{9}$ solar masses, a quasar.