#### Photometric Systems

We have already seen that astronomers use filters to isolate parts of the spectrum, and so measure monochromatic flux. The amount of the spectrum that a filter allows through is known as the bandpass. Filters are usually categorized into narrow-band filters, which have a bandpass of order 10 nm and typically isolate a spectral line, and broad-band filters, which have a bandpass of order 100 nm. The central wavelength of the filter bandpass is known as the effective wavelength. Most modern filters are constructed of different coloured glasses, often in conjunction with thin-film coatings to help define the bandpasses and minimise reflection at the surfaces.

Photometry of a source in a set of filters provides crude spectral information about the source. Well-defined sets of filters are known as photometric systems. A photometric system with too many filters, each with a very narrow bandpass, would make it difficult to detect sufficient photons from a source, and strong absorption/emission features in the spectrum might adversely affect some of the bandpasses. Conversely, a photometric system with too few filters, each with a very wide bandpass, would provide insufficient spectral information.

The most widely used photometric system today is the UBVRI system, also known as the Johnson-Morgan-Cousins system (see Prof Vik Dhillon's notes for an excellent discussion of the history of this system). The filters cover the whole range of optical wavelengths - UBV covers the "Ultraviolet", "Blue" and "Visual" ranges, whilst RI cover the "Red" and "Infrared" range. The UBVRI filter set is shown in figure 20 below.

Figure 20: Filter profiles of the UBVRI filter set. Also plotted is the transmission of the atmosphere (dotted line) and the quantum efficiency of a typical CCD (dashed line). Credit: Vik Dhillon.

Figure 21: A photograph of the UBVRI filters. The U and I filters look black since they transmit light which is largely beyond the range of wavelengths detectable by the eye.

As discussed in PHY104, it is common practice to refer to apparent magnitudes in a particular filter by the name of the filter. If the apparent magnitude of a star in the V-band is $$m_V = 15.5$$, this is often simply written as $$V=15.5$$. The difference in magnitude between two filters is called a colour index. If a star's magnitudes in two filters are $$B=16.0$$ and $$V=15.5$$, the star has a colour index of $$B-V = 0.5$$.

Although the UBVRI system is very widely used, it is not the only photometric system. Another system you may well come across is that used by the Sloan Digital Sky Survey (SDSS). This uses five filters, named u'g'r'i'z'. The SDSS filter set has filters with higher transmissions than UBVRI, making them good for faint sources. The SDSS filters also have minimal overlap between filters. It is likely that this filter set will become more common, and eventual dominant over time. The SDSS filter set is shown in figure 22.

#### Magnitude Systems

Recall that the definition of apparent magnitude is
$m= -2.5 \log_{10} F + c.$
It can be seen that the magnitude scale depends upon our choice of the constant $$c$$. The most logical way of choosing this scale is to choose a zero-point, i.e to choose a star that has a magnitude of 0. We can then measure the magnitudes of all other stars with respect to this one.

The A0V star Vega was chosen as this so-called primary standard because it indeed does have a magnitude close to zero as determined by Hipparchos' original magnitude scale, it is easily observable in the northern hemisphere for more than 6 months of the year, it is non-variable, relatively nearby (and hence unreddened by interstellar dust), and it has a reasonably flat and smooth optical spectrum. However, because Vega is too bright to observe with modern telescopes and instruments without saturating their detectors, and because it is not always observable, an all-sky network of fainter secondary standards has also been defined, where the magnitude of each star relative to Vega has been carefully calibrated. Over the years, refinements in the definition, number and measurement accuracy of the primary and secondary standards has resulted in the apparent magnitude of Vega now being 0.03 in the V-band, and it is also thought that Vega may be slightly variable, but for the purposes of this course we can ignore this few per cent offset and assume it is 0 in all bands.

Filter $$\lambda_{\rm eff}$$ (nm) $$\Delta \lambda$$ (nm) $$m_{\rm vega}$$ $$F_{\nu} {\rm (W\,m}^{-2}{\rm Hz}^{-1}{\rm )}$$ $$F_{\lambda} {\rm (W\,m}^{-2}{\rm nm}^{-1}{\rm )}$$ $$N_{\lambda} {\rm (photons\,s}^{-1}{\rm cm}^{-2}{\rm \unicode{xC5}}^{-1}{\rm )}$$
U 360 50 0 $$1.81 \times 10^{-23}$$ $$4.19 \times 10^{-11}$$ 759
B 430 72 0 $$4.26 \times 10^{-23}$$ $$6.91 \times 10^{-11}$$ 1496
V 550 86 0 $$3.64 \times 10^{-23}$$ $$3.61 \times 10^{-11}$$ 1000
R 650 133 0 $$3.08 \times 10^{-23}$$ $$2.19 \times 10^{-11}$$ 717
I 820 140 0 $$2.55 \times 10^{-23}$$ $$1.14 \times 10^{-11}$$ 471

Table 1: Characteristics of the UBVRI photometric system. $$m_{\rm vega}$$ refers to the magnitude of the star Vega in the filter. By definition, the magnitude of Vega is actually $$V=0.03$$ and all the colours (e.g B-V) are zero. Magnitudes defined this way are referred to as the Vega magnitude system. The final three columns give the flux expected at the top of the Earth's atmosphere from a zero magnitude source, in various units.

Once we know that the magnitude of Vega is defined as zero, this allows us to calculate the value of c, and also some physical characteristics of the photometric system. Table 1 uses the definition of Vega as having $$m=0$$ to list the characteristics of the UBVRI system. All values are taken from Chris Benn's ING Signal program. From left-to-right are tabulated the filter name, the effective wavelength ($$\lambda_{\rm eff}$$), the bandpass ($$\Delta \lambda$$; FWHM), the approximate apparent magnitude of Vega ($$m_{\rm vega}$$), the band-averaged monochromatic fluxes in both frequency ($$F_{\nu}$$) and wavelength ($$F_{\lambda}$$) units of a V=0 A0V star, and the photon flux ($$N_{\lambda}$$) in units of photons s-1 cm-2-1 (the latter units are used for convenience as they result in more easily remembered values).

Figure 22: Filter profiles of the SDSS filter set. Also plotted is the transmission of the atmosphere (dotted line) and the quantum efficiency of a typical CCD (dashed line). Credit: Vik Dhillon

#### Calibrating Photometry

Previously, we saw how to extract the sky-subtracted signal from an object, measured in counts, from an image. Once this has been done, it is useful to convert to a magnitude in a photometric system. To see why this is important imagine two astronomers observing the same star, through the same filter, but with very different telescopes. Clearly an astronomer with a large telescope is going to measure more counts than the unfortunate astronomer who has a small telescope. It is therefore not very meaningful to share our results with others in units of counts.

Converting a measurement in counts into a calibrated magnitude involves five steps:

1. Divide the number of counts by the exposure time, to get a measure of flux in counts per second.
2. Calculate the instrumental magnitude, from the counts per second
3. Determine the extinction coefficient, and correct the instrumental magnitude to the above-atmosphere value.
4. Repeat the above steps for a standard star and use the resulting above-atmosphere instrumental magnitude of the standard star to calculate the zero point.
5. Use the zero point to transform the above-atmosphere instrumental magnitude of the target star to the required photometric system.
The above steps are described in more detail below.

Instrumental Magnitudes

Let us call the sky-subtracted signal from our target object, in counts, $$N_t$$. We convert this to an instrumental magnitude, using the formula
$m_{\rm inst} = -2.5 \log_{10} \left( N_t / t_{\exp} \right),$
where $$t_{\rm exp}$$ is the exposure time of the image in seconds. The instrumental magnitude depends on the characteristics of the telescope, instrument, filter and detector used to obtain the data. It is therefore meaningless to compare two instrumental magnitudes taken under different situations, without first putting them on a calibrated scale.

The relationship between instrumental magnitudes and calibrated magnitudes can be understood as follows. The counts per second $$N_t / t_{\exp}$$ is proportional to the flux, $$F_\lambda$$. Hence
$m_{\rm inst} = -2.5 \log_{10} \left( \kappa F_\lambda \right) = -2.5 \log_{10} F_\lambda + c',$
therefore, instrumental magnitudes are offset from calibrated magnitudes by a constant:
$m_{\rm calib}= m_{\rm inst} + m_{\rm zp},$
where the constant, $$m_{\rm zp}$$, is known as the zero point. The zero point depends upon the telescope and filter used. We can understand this because if we used a larger telescope to observe a star, the instrumental magnitude would change, but the calibrated magnitude must not!

A handy tip to remember about zero-points is this; an object with a calibrated magnitude equal to the zero point gives one count-per-second at the telescope. For example, suppose the zeropoint of a telescope/filter combination is $$m_{\rm zp} = 19.0$$. If we observe a star with $$m_{\rm calib}= 19.0$$, then it follows that, for this star $$m_{\rm inst}=0$$. By definition then, this star gives one count-per-second.

Extinction

The next step is to convert the instrumental magnitude, which is measured on the surface of the Earth, to the instrumental magnitude that would be observed above the atmosphere. This is necessary because the Earth's atmosphere absorbs light from the star, and the amount of light absorbed depends on the angle of the star above the horizon. If we didn't correct this effect, observations of objects at different times of night would give different magnitudes! We'll see how to correct for extinction below, but first we need to work out how extinction depends on the angle of the star above the horizon.

We can derive a simple equation for the extinction correction by assuming the atmosphere is a series of thin plane-parallel layers. Figure 23 shows such a layer, of thickness $$dx$$ at an altitude $$x$$. The path length through the layer for light from a star at a zenith distance $$z$$ is equal to $$dx \,/ \cos z = dx \sec z$$. The term $$\sec z$$ is known as the airmass, and is sometimes given the symbol $$X$$. At the zenith, $$X= \sec z = 1$$, and this increases to a value of 2 at a zenith distance of 60o. Note that the approximation of the atmosphere as being plane-parallel breaks down at larger zenith distances. Due to the amount of extinction, it's a bad idea to observe astronomical objects this low, so we shall not worry about the curvature of the atmosphere here.

Figure 23: A thin, plane-parallel layer in the Earth's atmosphere. As the zenith distance of the star increases, the path length through the atmosphere increases, and hence the absorption increases. Credit: Vik Dhillon

If the monochromatic flux from an object incident on the layer is $$F_\lambda$$ then the flux absorbed by the layer $$dF_\lambda$$ will be proportional to both $$F_\lambda$$ and the path length through the layer. Therefore:
$dF_\lambda = - \alpha_\lambda F_\lambda \sec z \, dx,$
where the constant of proportionality, $$\alpha_\lambda$$ is known as the absorption coefficient, with units of m-1. The absorption coefficient is a function of the composition and density of the atmosphere, and hence the altitude of the layer, $$x$$. We can re-arrange this equation (and drop the $$\lambda$$ subscripts for clarity) to give
$\frac{dF}{F} = - \sec z \, \alpha \, dx.$
Integrating the equation above for $$x$$ values from the top of the atmosphere, $$t$$, to the bottom $$b$$, we obtain
$\int_t^b \frac{dF}{F} = - \sec z \int_t^b \alpha dx.$
Hence
$\frac{F_b}{F_t} = \frac{F}{F_0} = \exp \left( -\sec z \int_t^b \alpha dx \right),$
where for clarity we have renamed the above-atmosphere flux $$F_t = F_0$$ and the flux measured at the ground by the observer $$F_b = F$$. Remembering that the difference between two magnitudes is $$m_1 - m_2 = -2.5 \log_{10} (F_1/F_2)$$, we can write (via some magic with the change of base formula),
$m-m_0 = -2.5\log_{10} (F/F_0) = 2.5 \sec z \log_{10}(e) \int_t^b \alpha dx .$
We define the extinction coefficient, $$k$$, as:
$k = 2.5 \log_{10}(e) \int_t^b \alpha dx,$
to finally obtain:
$m = m_0 + k \sec z = m_0 + kX,$
where $$m_0$$ is the magnitude of a star observed with no extinction (i.e above the atmosphere) and $$m$$ is the magnitude of a star observed at the Earth's surface at zenith distance $$z$$. As an example, if the extinction coefficient from a site in the V-band is $$k=0.15$$ magnitudes/airmass then a star would appear 0.15 magnitudes fainter at the zenith than it would appear above the atmosphere, and 0.3 magnitudes fainter than above the atmosphere when at a zenith distance of 60o.

The dominant source of extinction in the atmosphere is Rayleigh scattering by air molecules. This mechanism is proportional to $$\lambda^{-4}$$, which means that extinction is much higher in the blue than in the red. The extinction can also vary from night to night depending on the conditions in the atmosphere, e.g. dust blown over from the Sahara can increase the extinction on La Palma during the summer by up to 1 magnitude. Table 2 lists the extinction coefficients on a typical (undusty) night on La Palma in UBVRI. For reference, the night sky brightness on La Palma when the Moon illumination is 0% (Dark), 50% (Grey) and 100% (Bright) illuminated is also listed. All values in table 2 have been taken from Chris Benn's ING signal program.

Filter $$\lambda_{\rm eff}$$ (nm) $$k$$ (mags/airmass) $$m_{\rm sky}$$ (mags/arcsec2)
Dark Grey Bright
U 360 0.55 22.0 20.0 17.7
B 430 0.25 22.7 20.7 18.4
V 550 0.15 21.9 19.9 17.6
R 650 0.09 21.0 19.7 17.5
I 820 0.06 20.0 18.9 16.7

Table 2: typical extinction and sky brightness values in the UBVRI photometric system at a high-quality astronomical site.

Correcting for extinction

To measure the extinction on a particular night, it is necessary to measure the signal from a non-variable star at a number of different zenith distances. Inspecting the equation $$m = m_0 + k \sec z$$, it can be seen that the extinction would be given by the gradient of a plot of the instrumental magnitude of the star versus $$\sec z$$ and the $$y$$-intercept would give the above-atmosphere instrumental magnitude. Although such a plot would give the most accurate answer, it is also possible to obtain an estimate of $$k$$ from just two measurements of the instrumental magnitude of a star at two different zenith distances: subtracting $$m_{z_1} = m_0 + k \sec z_1$$ from $$m_{z_2} = m_0 + k \sec z_2$$ eliminates $$m_0$$, allowing $$k$$ to be derived.

Note that no explicit extinction correction is required when performing relative photometry. This is because the target and comparison stars are always observed at the same airmass and hence suffer the same extinction. Hence, when the target signal is divided by the comparison star signal to correct for transparency variations, the variation due to extinction present in the comparison star is removed from the target star.

Aside: Secondary Extinction

For very accurate photometry, the wide bandpass of broad-band filters has to be taken into account when correcting for extinction. Because extinction is so strongly colour dependent, a blue object actually loses more light to the atmosphere than a red one. This is true even within the relatively narrow wavelength range of the UBVRI filters. The solution is to introduce a colour-dependent secondary extinction coefficient, $$k_2$$, which modifies the above extinction correction equation to:
$m = m_0 + k \sec z + k_2 C \sec z,$
where $$C$$ is the colour index, e.g. when correcting a V-band magnitude for extinction, $$C = B - V$$. The secondary extinction coefficient is usually of order a hundredth of a magnitude, so it will be ignored for the remainder of this course.

Calibrated magnitude

Now we can find the above-atmosphere instrumental magnitude of any object. The final step is to observe a primary or secondary photometric standard star to convert to calibrated magnitudes. If the above-atmosphere instrumental magnitude of our standard star is $$m_{{\rm std},0,i}$$, then:
$m_{\rm zp} = m_{\rm std} - m_{{\rm std},0,i}.$
The calibrated magnitude of our standard star $$m_{\rm std}$$ can be looked up in a catalog. The calibrated magnitude of our target star, $$m$$, can then be found using:
$m = m_{\rm zp} + m_{0,i},$
where $$m_{0,i}$$ is the above-atmosphere instrumental magnitude of our target star.

Each filter in a photometric system will have a different zero point. Once the zero point has been measured for a particular telescope, instrument, filter and detector combination, it should remain unchanged, although dirt and the degradation of the coatings on the optics will cause minor changes to the zero point on long timescales. To determine the zero points for the UBVRI system, the photometric standards measured by Landolt can be used.

For very accurate photometry, the wide bandpass of broad-band filters also has to be taken into account when converting instrumental magnitudes to standard values. This is because the telescope, instrument, filter and detector used by an observer will always have a slightly different response to light as a function of wavelength than those used by the astronomers who originally defined the magnitudes of the photometric standard stars. The biggest discrepancy is often in the profile of the filter. Figure 24 shows two V filters used at the Kitt Peak National Observatory in Arizona. The WIYN V filter has a sharper increase in transmission on the blue side of the profile than the Bessell V filter, allowing more light to enter from hot, blue stars than from cool, red stars. This creates a systematic error that makes blue stars seem a bit brighter than red stars when observed through the WIYN V filter.

Figure 24: A plot by Michael Richmond showing two different V filters and the spectra of hot and cold stars, demonstrating why correcting for colour terms is necessary when performing high-accuracy photometry (see text for details).

Fortunately, it is straightforward to correct for this systematic error by observing a field with many standard stars possessing a range of colours. The advantage of observing a single field is that all of the stars will then be at the same airmass and hence extinction effects are cancelled out. If this is not possible, you can use many observations of standard stars in different fields. We have already seen that:
$m_{\rm zp} = m_{\rm std} - m_{{\rm std},0,i}.$
Hence, if the telescope, instrument, filter and detector combination being used matches that of the photometric system perfectly, we can write:
$m_{\rm std} - m_{{\rm std},0,i} - m_{\rm zp} = 0.$
In practice, however, the observer's equipment is never identical to that used to define the photometric system, resulting in the above equation being modified to:
$m_{\rm std} - m_{{\rm std},0,i} - m_{\rm zp} = c C,$
where $$c$$ is the colour term, and $$C$$ is the colour index (e.g $$B-V$$). Hence, the colour term is equal to the gradient in a plot of $$m_{\rm std} - m_{{\rm std},0,i} - m_{\rm zp}$$ against $$C$$, i.e a plot of the difference between the catalogue magnitudes of the standard stars and their calibrated magnitudes as a function of colour; the $$y$$ intercept is set to zero by using a zero point calculated from a star of $$B - V = 0$$. An example of such a plot is shown in figure 25, which has a gradient of 0.072. You can see from this plot that use of colour terms are necessary to obtain accuracies of order 0.01 magnitudes (i.e 1% in flux). If your requirements are less accurate, you can ignore colour terms, as we will do for the remainder of this course.

Figure 25: A plot of the difference between the catalogue magnitudes of a set of standard stars and their calibrated magnitudes (y axis) as a function of colour (x axis). The gradient of the line is equal to the colour term.