Surface brightness and photon counts

Builds on: Sections 4.1, 4.2, 5.3 and S4
Reading time: 30 min
Exercises: 4
Olympiad problems: 1

Seen from Mars, the Sun covers less than half the area of sky that it covers from Earth, and it gives each square metre there only 43 per cent as much light, as Section 4.1 found. Yet every small patch of its disc looks just as dazzling from Mars as from Earth. Galaxies behave in the same way. A galaxy twice as far away gives a quarter of the light, but it also covers a quarter of the area, so each square arcsecond of it is as bright as before. In this section we measure how bright an object is for each piece of sky it covers, see why that brightness does not depend on distance, and then count the photons that a star sends into an eye or a telescope.

4.5.1Surface brightness

Before you readMake a guess first

A galaxy is moved to three times its distance from us. What happens to the brightness of each square arcsecond of its image?

A star is so small that its light seems to come from a single point. A planet, a nebula or a galaxy covers an area of sky, and how easy it is to see depends on how much light falls on each piece of that area as well as on its total. The flux from each square arcsecond of an object is its surface brightness. Astronomers write it as a magnitude, the magnitude that one square arcsecond of the object would have on its own, in magnitudes per square arcsecond.

An object of total magnitude $m$ spread evenly over an area $Ω$ square arcseconds gives each square arcsecond a fraction $1 / Ω$ of its flux. By (4.2), one square arcsecond is fainter than the whole by $2.5 \log_{10} Ω$ magnitudes.

ResultSurface brightness(4.8)

μ = m + 2.5 \log_{10} Ω

$μ$	the surface brightness, in magnitudes per square arcsecond
$m$	the total magnitude of the object
$Ω$	the area of sky it covers, in square arcseconds

The formula gives the mean surface brightness over the object. Real galaxies are brighter in the middle than at the edges, so their surface brightness changes across the image. A dark country sky has a surface brightness of about 21 to 22 magnitudes per square arcsecond in the V band, and the faint outer parts of a galaxy are fainter still.

The guess above shows why $μ$ does not depend on distance. At $n$ times the distance the flux falls by $n^{2}$ , which adds $5 \log_{10} n$ to $m$ . The angular size also falls by $n$ , so $Ω$ falls by $n^{2}$ , which takes $5 \log_{10} n$ away from $2.5 \log_{10} Ω$ . The two changes cancel. This holds whenever no dust dims the light and the distances are small enough for the expansion of the Universe to be ignored, which Chapter 13 takes up.

Quick checkThe surface brightness of the Sun

The Sun has magnitude −26.74 and a disc 1918″ across. What is its surface brightness?

mag per square arcsecond

A telescope cannot raise the surface brightness of what it shows. It gathers more light than the eye, but it also spreads the image over a larger angle, and the two effects cancel at best. When the exit pupil of Section 5.3 is as wide as the pupil of the eye, a nebula looks as bright for each piece of its area as it does to the naked eye, only larger. At higher magnifications the exit pupil shrinks and the nebula looks dimmer. Stars behave differently, because a point stays a point at any magnification, and so a telescope shows faint stars that the eye cannot see.

Example 4.9A small galaxyfinish it

A galaxy has total magnitude 10.0 and looks like an even disc 60″ across. What is its mean surface brightness?

The start

The disc covers $Ω = π \times 30^{2} = 2827$ square arcseconds.

Your stepNow use (4.8).

mag per square arcsecond

4.5.2Counting photons

Light arrives as photons, each carrying an energy $E = h c / λ$ (Section S4). A flux $F$ is a stream of energy, so dividing it by the energy of one photon gives the number of photons that cross each square metre each second. A telescope or an eye of collecting area $A$ catches $A$ times as many.

ResultPhotons from a source(4.9)

N = \frac{F A λ}{h c}

$N$	the number of photons collected each second
$F$	the flux, in W m⁻²
$A$	the collecting area of the eye or telescope, in m²
$λ$	the wavelength, in m, taken as one wavelength for the whole flux
$h, c$	the Planck constant, $6.626 \times 10^{- 34}$ J s, and the speed of light, $2.998 \times 10^{8}$ m s⁻¹

Real light is spread over many wavelengths, so the rule is an estimate that uses one wavelength in the middle of the band, such as 550 nm for the V band.

Example 4.10Photons of sunlight

Sunlight brings 1361 W m⁻² above the air. If all of it were at 550 nm, how many photons would cross each square metre each second?

Solution

One photon of wavelength 550 nm carries

\begin{aligned} E & = \frac{h c}{λ} = \frac{6.626 \times 10^{- 34} \times 2.998 \times 10^{8}}{550 \times 10^{- 9}} \\ = 3.61 \times 10^{- 19} J . \end{aligned}

The number of photons is therefore $1361 / (3.61 \times 10^{- 19}) = 3.8 \times 10^{21}$ each second.

AnswerAbout

3.8 \times 10^{21}

photons of sunlight cross each square metre every second.

The same rule counts the photons that a faint star sends into the eye. Section S4 restates a problem from IOAA 2009 that does this for a star of magnitude 6, the faintest the naked eye can see, and finds a few thousand photons each second.

Common errorsSection 4.5

Surface brightness does not fall with distance, though the total flux does. A distant galaxy is hard to see because its image is small, while each part of it is as bright as it would be close by.
In $μ = m + 2.5 \log_{10} Ω$ the sign is a plus. One square arcsecond of an object is fainter than the whole object, so its magnitude is larger.
In $E = h c / λ$ the wavelength must be in metres. A wavelength left in nanometres gives an energy $10^{9}$ times too small.

ExercisesSection 4.54 questions

4.5.1
A nebula of total magnitude 12.0 covers an area of 100 square arcseconds evenly. What is its surface brightness?
mag per square arcsecond
4.5.2
What is the energy of one photon of wavelength 500 nm?
J
4.5.3
A detector receives $1.00 \times 10^{- 12}$ W of light at 550 nm. How many photons does it receive each second?
photons
4.5.4
Two identical galaxies lie 10 Mpc and 40 Mpc away. How do their surface brightnesses compare, ignoring dust and the expansion of the Universe?

Olympiad problemsPast IOAA problems, restated1 problem

2011How much of the sky do stars cover?10 points

Adapted from IOAA 2011 (Katowice), theory problem S9. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

The distances between stars are huge compared with their sizes, so a star cluster or a galaxy without gas and dust hides almost nothing behind it. We look towards a galaxy whose surface brightness is 18.0 magnitudes per square arcsecond. Assume that its stars are all like the Sun, and estimate what fraction of the sky in that direction is covered by the discs of its stars.

I solved it without the solution

AnswerThe stars cover only about $4 \times 10^{- 12}$ of the sky in that direction.