Now it’s time time actually run phot. Get back into the phot parameter file again and enter the name of the first input image, the coordinate list and the results file. Get your demonstrator to check your parameter files (this may save tears later on), and then run it by typing :g. Do this for each of the reduced images, (making sure you change the results file name each time.)

You should have four files full of photometric information. All we really want is the summed, sky-subtracted pixel values for each star. Ask your demonstrator how to extract the column containing the pixel values, and transfer them across to the Excel on a Windows PC. Run > ../extract_params.pl OUTPUT.FILE and transfer the output files across to Excel on the Windows PC.

The magnitude system is a holdover from the ancient Greek method for categorising stars. In this system the brightest star in the northern sky, Vega, was classed “magnitude 0”. Fainter stars had higher magnitudes, with each increment in magnitude roughly corresponding to a factor of two reduction in brightness. With the advent of electronic detectors it was found that the ancient Greeks weren’t all that good determining “a factor of two in brightness”. However it turned out that they were very good at being consistent in their inaccuracies — each magnitude corresponded very closely to a factor of 2.51 difference in brightness. (More precisely, 5 magnitudes turned out to be factor of 100 in brightness, and \(100^{1/5}=10^{0.4}\approx 2.51\)). In other words, the magnitude \(m\) of an object is related to its brightness \(f\) by

where \(c\) is a constant which we’re going to determine. In practice, this equation is usually rearranged to give:

Derive this form. For two sources, the difference in their magnitudes is proportional to the ratio of their fluxes:

To figure out the magnitudes of our stars we need to calibrate the arbitrary flux units (‘counts’) we’ve measured into something more physical. In essence, we need “\(c\)” from the above equation. This constant depends on the size of the telescope (big telescopes collect more light, so you get more counts), the efficiency of the CCD (more efficient detectors get more counts), the filter (even the best filters loose some of the light), and the atmosphere (haze or dust absorbs or scatters the starlight before it reaches the telescope). Astronomers use images of standard stars to perform this calibration. These are stars of known (non-variable) brightness, whose magnitudes and locations on the sky are catalogued in long lists.

In our case we don’t need a separate standard star image, because we know the exact brightness of one of the stars in M93. This is the star indicated in the image in the previous section, which, hopefully, you placed at the top of your list. The apparent magnitudes for this star in our four bands are:

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  \(B=10.23\) mag

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  \(V=10.14\) mag

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  \(R=10.15\) mag

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  \(I=10.18\) mag.

Use your measured counts for this star, along with these apparent magnitudes, to determine the calibration constant for each filter. Each constant is valid for all the other stars in that image. Now calculate the apparent magnitudes of all your stars in each image. You can use the calibration constants and Equation (\ref{eq:fluxcalib}) to calculate the apparent magnitudes of all of your stars, or you can just take the ratio of fluxes and use Equation (\ref{eq:appmagfluxratio}).

From apparent magnitude we can calculate absolute magnitude (\(M\)). Absolute magnitude is defined as the magnitude a source would have if it were at a distance of 10 parsecs. You can find absolute magnitudes using this equation:

where \(d\) is the distance to the star in parsecs. This, of course, relies on the assumption that all the stars are at the same (known) distance. Can you figure out how this relation is derived?

Since magnitudes are logarithmic, the distance scaling factor (\((d/10pc)^{2}\)) becomes an additive constant (\(5+5\log_{10}d\)) between apparent and absolute magnitudes. This factor is called the “distance modulus”. Using the known distance (1104 pc) to the cluster, calculate the absolute magnitudes of your stars for all filters.

You now have magnitudes for a group of stars, using two or more filters. You can use them to create a Hertzsprung-Russell diagram, a plot of ‘colour’ against brightness.

‘Colour’ in astronomy is not colour like the colours of the rainbow. It is defined as the ratio of how bright a star is when viewed through two different filters. Remember that ratio of brightnesses is the same as the difference in magnitudes (just from the usual logarithm laws). If, for example, the magnitude of a star observed through the green filter is \(V\), and its magnitude when observed through a blue filter is \(B\), then its colour could be measured as \(B-V\).

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  If the object is relatively brighter in green light, compared to blue light, then its \(B-V\) will be a positive number. Astronomers would say it had a “red” colour, even though it is actually the blue and green light which is being considered.

• •

  If the object is brighter in blue light than green, its \(B-V\) will be negative. It would then have a “blue” colour.

For all the objects in your CCD frames, compute colours, and plot these colours against one of the absolute magnitudes (eg. \(M_{V}\) against \(B-V\)). This is an HR diagram!

Can you see a line of stars; a main sequence? If you see a main sequence, do you see stars in any other regions in the diagram? Compare your diagram to the colour-magnitude diagram in Figure x and see if you can figure out what types of stars are present in this cluster. Note that this HR diagram covers a wider range of absolute magnitude and colour than your plot, so be careful when identifying stars which are off the main sequence.

How does your HR diagram compare to the example in Figure x? We mentioned that Figure x covers a wider range of colours and magnitudes, can you think of why this might be? Why are you missing especially blue and red stars?

Extra question: Do you get the same or similar HR diagram if you use a different set of magnitudes? (ie. \(R\) and \(I\) instead of \(B\) and \(V\)) If there is any discrepancy, can you explain why? You will need to think carefully about this: what exactly does colour measure? If stars are just black-bodies, when would you expect colours to be the same or different?

To work out how bright our stars really are, we must first convert from magnitudes into energies. The star Vega is apparent magnitude zero in all filters (by definition), so we can use its known brightness at different wavelengths (Table \ref{tab:vegafluxes}) to calibrate our magnitudes to SI units. These brightnesses are in units of flux per unit wavelength; ie. the energy hitting a square metre every second with wavelengths in a 1nm range. The units of luminosity of a star is just energy per second per unit wavelength; ie. it is the total energy the star radiates at this wavelength, not just the amount that hits a square metre area at the Earth.

Using Table \ref{tab:vegafluxes} and the definition of magnitudes discussed in the preceding section, work out the flux your stars are producing at the different wavelengths. This tells you how much radiation from the stars are reaching the Earth. Using the distances to the clusters listed in Table \ref{tab:clusters} and the inverse square law, work out the total luminosity of the stars. How do they compare with the sun’s luminosity? Are these giant stars, or are we seeing things like the Sun? Or are we seeing things fainter still, like brown dwarfs, red dwarfs or white dwarfs?

The very brightest stars known may have luminosities as much as a few thousand times that of the Sun. These are the rare supergiants; some red (very old stars bloating out in their dying moments) and some blue (very young, very hot violent stars). Do you have any of these? If so, are they red or blue?