Measuring on images and spectra

Builds on: Sections D1, D4 and S4
Reading time: 50 min
Exercises: 6
Olympiad problems: 1

A little over a century ago Henrietta Leavitt found the law that links the periods of Cepheids to their luminosities by measuring their images, one by one, on glass photographic plates of the Small Magellanic Cloud. Astronomers still measure pictures, though the plate has become a grid of pixels, and data-analysis problems ask for the same. They hand out an image or a spectrum and expect positions, widths and brightnesses to be read off it with a ruler or by counting pixels, and then turned into angles, magnitudes and speeds. In this section we learn to make those measurements and to carry their uncertainties into the answer.

D9.1From the picture to the sky

Before you readMake a guess first

Two stars 0.50° apart on the sky appear 1.2 cm apart on a photograph. What is the scale of the photograph?

Every image has a scale, the angle on the sky that one millimetre or one pixel covers. When the focal length $F$ of the telescope is known, the scale follows from (5.3), and for a camera the pixel scale is 206 265″ times the size of a pixel divided by the focal length. For the camera of IOAA 2009, with pixels 25 µm across on a telescope of focal length 1.20 m, the pixel scale is $206 265^{″} \times 25 \times 10^{- 6} / 1.20 = {4.30}^{″}$ . When the focal length is not known, two stars of known separation give the scale instead. A ruler is read to half a millimetre, as Section D1 explains, so the scale is best found from two stars far apart, for which that reading error is a small fraction of their distance.

D9.2The size and centre of a star's image

Before you readMake a guess first

A star's profile peaks at 900 counts on a sky of 100 counts. At what level of counts is its full width at half maximum measured?

The width of a star's image tells us the , the blurring by the air that Section 5.4 describes, and it is measured as the full width at half maximum. We first find the level of the sky from pixels with no star in them, then the peak of the star, and the half maximum lies halfway between the two. The points where the profile crosses that level on either side of the peak, found by drawing a straight line between neighbouring pixels, are the two ends of the width. Multiplying the width in pixels by the pixel scale gives the seeing in arcseconds.

Figure D.9Measuring the width of a star's imageinteractive

Left cursorRight cursor

Guide lines

Width between the cursors
Check

The figure shows the counts in a row of fifteen pixels across the image of a star, from a camera with pixels of 1.5″, and a thin line joins the tops of the columns. The sliders set two cursors, and the table gives the width between them and says whether they sit where the profile crosses half its height above the sky. The control row, or tapping , draws lines at the sky, the half maximum and the peak. Tapping or moves the cursors there.

Here the sky is 120 counts and the brightest pixel holds 1850, so the half maximum is at 985 counts. The profile crosses it 3.06 pixels apart, which with 1.5″ pixels is a seeing of 4.6″, as on a poor night. Each pixel averages the light over its own width, and the straight lines between pixels cut across the curve, so the measured width comes out a little larger than the true width of the image, which here is 2.9 pixels. This matters only when a star covers just a few pixels.

The centre of a star's image is found as a mean of the pixel positions, each weighted by its counts above the sky. It can be found much more precisely than the image is wide, as (5.13) shows.

Quick checkThe centre of a star

Three neighbouring pixels at positions 4, 5 and 6 hold 300, 900 and 600 counts above the sky. Where is the centre of the star?

pixels

D9.3The brightness of a star

Before you readMake a guess first

A square aperture of 25 pixels round a star holds 52 000 counts in all, and the sky gives 400 counts in every pixel. How many of the counts come from the star?

To measure a star's brightness on an image we add up the counts in an aperture, a small square or circle round the star, and take away the sky, which is the mean sky per pixel, measured where there are no stars, times the number of pixels in the aperture. The counts of the star per second of exposure give its instrumental magnitude $m_{I}$ , which depends on the telescope and camera as well as on the star. Stars of known magnitude in the same image link the instrumental magnitudes to true ones through a zero point, after allowing for the dimming by the air, which grows with the airmass $X$ , the thickness of air the light crosses compared with the thickness straight up.

ResultAperture photometry(D.12)

\begin{aligned} m_{I} & = - 2.5 \log \frac{\sum C - N {\bar{C}}_{sky}}{t} \\ m_{I} & = m + K X - Z \end{aligned}

$\sum C$	the counts added up over the $N$ pixels of the aperture
${\bar{C}}_{sky}$	the mean counts of the sky in one pixel
$t$	the exposure time
$m$	the true magnitude, above the air
$K, X$	the extinction coefficient and the airmass, about $1 / \cos z$ at a zenith distance $z$
$Z$	the zero point of the camera

The aperture should be large enough to hold nearly all of the star's light, a few times the full width at half maximum across, and the same for every star in the image.

Example D.18A zero point and the brightness of the sky

In IOAA 2009 an image taken through a V filter, 450 s long, at a zenith distance of 25°, held five stars. The extinction was 0.3 magnitudes per airmass and the pixel scale 4.30″. Stars 1, 3 and 4, with catalogue magnitudes 9.03, 6.22 and 8.02, had instrumental magnitudes −3.02, −5.85 and −4.04, and stars 2 and 5 had −2.13 and −0.66. The sky gave 4.42 counts per pixel. Find the zero point, the magnitudes of stars 2 and 5, and the brightness of the sky in magnitudes per square arcsecond.

Solution

The airmass is $X = 1 / \cos 25 ° = 1.103$ , so $K X = 0.3 \times 1.103 = 0.331$ . By (D.12), $Z = m + K X - m_{I}$ , which gives 12.38, 12.40 and 12.39 for stars 1, 3 and 4, with a mean of 12.39. Then $m = m_{I} - K X + Z$ gives 9.93 for star 2 and 11.40 for star 5.

The sky gives $4.42 / (450 \times {4.30}^{2})$ counts per second in each square arcsecond, since a pixel covers ${4.30}^{2} = 18.5$ square arcseconds. Its magnitude per square arcsecond is

\begin{aligned} m_{sky} & = - 2.5 \log \frac{4.42}{450 \times {4.30}^{2}} + Z \\ = 8.19 + 12.39 = 20.6 . \end{aligned}

AnswerThe zero point is 12.39, stars 2 and 5 have magnitudes 9.93 and 11.40, and the sky has a brightness of 20.6 magnitudes per square arcsecond, as in the official solution. Like the official solution we leave out the extinction for the sky, whose light comes from the air itself.

D9.4Wavelengths on a spectrum

Before you readMake a guess first

Two comparison lines whose wavelengths differ by 21 Å lie 42 pixels apart on a spectrum. What is the dispersion of the spectrum?

A spectrograph spreads light out by wavelength along one axis of the image. The dispersion, the change of wavelength per pixel or per millimetre, is found from lines of known wavelength, usually from a lamp whose spectrum is recorded beside the star's. The wavelength of any other feature then follows by counting pixels from the nearest known line, and over a short stretch of spectrum a straight-line scale is good enough. The centre of a line is taken as the middle of the line at half its depth, or half its height for an emission line, or as a weighted mean like the centre of a star. Its shift from the rest wavelength gives the speed along the line of sight, $v = c Δ λ / λ$ by (S.30).

When the slit of a spectrograph lies along a rotating galaxy, one side of the galaxy approaches and the other recedes, so its emission lines come out tilted across the spectrum. Reading off the shift at each point along the slit gives the rotation curve, as the original of the problem at the end of this section asked. A slit spectrum gives only the part of the speed along the line of sight. For a thin round disc whose image is an ellipse with axes $a$ and $b$ , the disc is tilted by an angle $i$ from face-on, with $\cos i = b / a$ , and the speed of rotation is the measured part divided by $\sin i$ .

Example D.19The speed of a galaxy from its spectrumfinish it

On the spectrum of a galaxy, two comparison lines of wavelengths 4921.9 Å and 5015.7 Å lie 400 pixels apart. The galaxy's Hβ line, whose rest wavelength is 4861.3 Å, lies 212 pixels from the first comparison line, towards the second. How fast is the galaxy receding?

The start

The dispersion is $(5015.7 - 4921.9) / 400 = 0.2345$ Å per pixel, so the line lies at $4921.9 + 212 \times 0.2345 = 4971.6$ Å. Its shift from the rest wavelength is $4971.6 - 4861.3 = 110.3$ Å.

Your stepNow find the speed of recession.

km/s

Common errorsSection D9

The half maximum of a star's profile lies halfway between the sky and the peak. Taken as half of the peak's counts measured from zero, it gives a width that is too large whenever the sky is bright.
The sky has to be taken away from every pixel of an aperture, so the amount to take away is the sky per pixel times the number of pixels in the aperture.
A slit spectrum gives only the part of a speed of rotation that lies along the line of sight. For a disc tilted by $i$ , the speed in the disc is that part divided by $\sin i$ .
A scale found from two stars close together is uncertain, because the reading error of the ruler is a large fraction of their separation. Stars far apart give a better scale.
The dispersion of a spectrum changes slowly along it, so wavelengths are best found from the nearest known lines and not from one scale across the whole spectrum.

ExercisesSection D96 questions

D9.1
A camera with pixels 9 µm across sits at the focus of a telescope of focal length 1000 mm. What is its pixel scale?
arcseconds
D9.2
A star's image is 3.2 pixels wide at half maximum, on a camera whose pixels cover 0.8″. What is the seeing?
arcseconds
D9.3
A star gives 42 000 counts above the sky in a 60-second exposure. What is its instrumental magnitude?
D9.4
The Hα line, of rest wavelength 6562.8 Å, appears at 6578.2 Å in the spectrum of a galaxy. How fast is the galaxy receding?
km/s
D9.5
A star's profile peaks at 2400 counts on a sky of 200 counts. At what count is its full width at half maximum measured?
D9.6
The image of a round galaxy disc is an ellipse 6 cm long and 3 cm wide. A slit spectrum along its long axis shows a speed along the line of sight of 150 km/s. What is the speed of rotation?

Olympiad problemsPast IOAA problems, restated1 problem

2011Weighing the spiral galaxy NGC 708325 points

Adapted from IOAA 2011 (Katowice), data-analysis problem 2. We have restated it in our own words and kept the original data. The original problem gave a photograph and a spectrum to measure, and we give the measurements that the official solution made on them. The solution was last checked on 20 October 2026.

The spiral galaxy NGC 7083 lies 40 Mpc away. The original problem gave its photograph and part of its spectrum, taken with the slit of the spectrograph along the long axis of the galaxy's image. Along the slit one pixel covers 0.82″. Two bright emission lines, with rest wavelengths 6564 Å and 6584 Å, run across the spectrum, tilted because one side of the galaxy approaches us and the other recedes. The official solution measured the following on the images. The two lines are 22 ± 2 pixels apart along the wavelength axis. Measured 10 pixels from the core on either side, the two arms of the 6564 Å line are 7 ± 1 pixels apart along the wavelength axis. On the photograph the image of the galaxy is 11.5 ± 0.5 cm long and 6.5 ± 0.5 cm wide.

(a) Find the speed of rotation of the galaxy 10 pixels from its core, taking the disc to be round. (b) Taking the central bulge to be a sphere, estimate its mass within that distance. The whole problem, which also asked for the rotation curve to be plotted, was worth 25 marks.

I solved it without the solution

AnswerThe galaxy rotates at about 176 km/s, and its bulge holds $(1.1 \pm 0.4) \times 10^{10}$ solar masses within 1.6 kpc of the centre. The official solution follows the same method, but in the step for the tilt it writes 216.7 km/s where its own numbers give 177.5 km/s, and so it finds $1.7 \times 10^{10}$ solar masses.