# Introduction: Astronomical units, jargon, coordinates, etc.

The goal of this course is to lead you on a journey through the universe on a large scale: our main subjects of study will be galaxies, both individuals and herds. However, in order to make sense of what is to come, you'll need to have a good understanding of some basic astronomical concepts, such as the magnitude scale, the units we use to describe masses and distances and luminosities, and so forth; and you'll also need to know a bit about stars, since they make up the galaxies we'll be studying.

Therefore, we should go over these ideas at the start, to make sure that no one is missing that little bit of knowledge which might be crucial to understanding the third stage in the cosmological distance ladder ...

#### Units

Astronomers don't use the SI system of meters, kilograms and seconds very frequently (why not?). Instead, we adopt units which are similar in scale to the objects we study. You should know the meaning of all the following terms.

• distance
• Angstrom (Å)
• astronomical unit (AU)
• light-year (ly)
• parsec (pc)
• And one which often serves as a proxy for distance

• redshift (z)

• mass
• Jupiter mass (MJ)
• solar mass (M)

• energy
• erg
• keV

• luminosity
• erg/s
• solar luminosity (L)

#### Coordinate systems

Astronomers have devised several coordinate systems for describing the location of objects in the sky. The most widely used involves Right Ascension and Declination.

As with latitude, Declination is measured away from the celestial equator. But there is again no obvious choice for the starting point of the other set of coordinates. Where should we start counting Right Ascension? The rather arbitrary choice made by astronomers long ago was to pick the point at which the Sun appears to cross the celestial equator from South to North as it moves through the sky during the course of a year. We call that point the "vernal equinox".

There are several ways to express a location. The star Sirius, for example, can be described as at

Right Ascension 101.287 degrees, Declination -16.716 degrees

We can also express the Declination in Degrees:ArcMinutes:ArcSeconds, just as we do for latitude; and, as usual, there are 360 degrees around a full circle. For Right Ascension, astronomers always use the convention of Hours:Minutes:Seconds. There are 24 hours of RA around a circle in the sky, because it takes 24 hours for the Sun to move all the way from sunrise to the next sunrise.

Right Ascension 06:45:09, Declination -16:42:58

meaning
• the Right Ascension is 6 hours, 45 minutes, 09 seconds
• the Declination is -16 degrees, 42 arcminutes, 58 arcseconds

What's the difference between an "arcminute" and a "minute"?

• one degree is divided into 60 arcminutes
• one arcminutes is divided into 60 arcseconds. Therefore, there are 3600 arcseconds in one degree

• one hour of time is divided into 60 minutes of time
• one minutes of time is divided into 60 seconds of time. Therefore, there are 3600 seconds of time in one hour of time

#### Exercises

1. How many degrees are there all the way around the celestial equator? How many hours are there around the celestial equator?
2. How many degrees make up one hour of Right Ascension?
3. How many degrees are there in one minute of Right Ascension?
4. How many arcminutes are there in one minute of Right Ascension?
5. How many arcseconds are there in one second of Right Ascension?

It is possible to convert from (RA, Dec) to (alt, az), or vice versa. One needs to know two factors:

• the location of the observer on Earth
• the time of the observation
The calculations involve some spherical trigonometry. One can find the details in any good book on celestial mathematics, such as
• Textbook on Spherical Astronomy by W. M. Smart
• Computational Spherical Astronomy by L. G. Taff
• Spherical Astronomy by R. M. Green
• Practical Astronomy with your Calculator by P. J. Duffet-Smith
• Astronomical Formulae for Calculators by J. Meeus

In these modern times, it's usually easiest to use one of the many fine planetarium programs on a computer to do this work.

#### Ecliptic (Solar System) coordinates

For objects within our solar system -- planets, asteroids, comets -- it often helps to use a coordinate system centered on the Sun, with its equator running along the plane of the planetary orbits. We call this the ecliptic coordinate system.

• Ecliptic Latitude is measured north from the plane of the solar system; the North Ecliptic Pole is in Coma Bernices, near RA = 18:00 and Dec = +66:34.
• Ecliptic Longitude is measured east from the Sun, and increases as one moves eastwards away from the vernal equinox.

The ecliptic coordinate system is convenient when dealing with objects in the solar system: they are concentrated towards the ecliptic equator:

Ecliptic coordinates can also be important when you want to avoid the solar system. Telescopes in space, such as the Hubble Space Telescope or the Chandra X-ray Telescope, cannot point close to the Sun (or else they might suffer damage to their detectors). For some purposes, astronomers want to make very, very long exposures: days or even weeks long. During such a long exposure, the Earth may move a significant fraction of its entire orbit, which can cause a target originally far from the Sun ....

... to move closer to the Sun, from the telescope's point of view.

Therefore, astronomers sometimes choose their very deep fields

to be not too far away from the ecliptic poles, so that the Sun is always many degrees away from these locations.

#### Galactic Coordinates

One more set of coordinates comes into play if one studies the distribution of stars within our Milky Way Galaxy, or the distribution of other galaxies in the far reaches of space.

• Galactic Latitude is measured north from the plane of the Milky Way; the North Galactic Pole is in Coma Bernices, near RA = 12:52 and Dec = +26:19.
• Galactic Longitude is measured east from the direction towards the center of the Milky Way, which is in Sagittarius, near RA = 17:45 and Dec = -29:22. It increases as one moves northwards in Dec away from the galactic center (i.e. higher in the sky as seen from Rochester).

On a warm July evening in Rochester, the Milky Way stretches overhead, with the galactic center just above the southern horizon.

If you make a map of the sky in galactic coordinates, the Milky Way runs right across the middle. The section we see in the summer sky from Rochester is in mostly the left half of this map.

An infrared map of the sky in galactic coordinates made by the COBE satellite is dominated by emission from dust in the Milky Way.

What is the faint blue band in this COBE image?

#### Converting from one set of coordinates to another

All these coordinate systems use two values to describe the position on the surface of a sphere. It is possible to convert from one system -- say, RA and Dec -- to any other system -- say, galactic longitude and latitude -- using spherical trigometry. In general, you need to know three angular quantities in order to make the conversion:

• how much is the equator of one system tilted relative to the equator of the other?
• what is the position in the first coordinate system at which the two equators are most offset?
• what is the difference in longitude-like coordinate between the zero-point of the two systems?

In the bad old days, you would have to use these three angles in a few long equations in order to convert the position from one system to another. But these days, it's usually easier to use a computer to do the job. There are webs sites to do the trick interactively, and lots of software libraries if you need to convert many items.

Note, by the way, that even when one has settled on a particular coordinate system, one may still display the information in different ways. For example, compare this view of the sky in equatorial coordinates and a Mercator projection

with this view, again in equatorial coordinates, but using a Hammer-Aitoff projection:

#### The magnitude scale

Optical astronomers, and, to some extent, infrared astronomers, are stuck with the historical artifact known as the magnitude scale. We describe the brightness of stars not in a linear sense -- "this one sends 1200 photons per second into our detector, that one sends only 492 photons per second" -- but in a logarithmic sense. Specifically, we have adopted the convention suggested long ago by Pogson that, if two stars have intensities of light I1 and I2, then the difference in their magnitudes is

Note that this definition says nothing about the zero-point of a magnitude: it provides only the DIFFERENCE between two stars. Exactly where to set the zero-point of the magnitude scale is a matter of some debate, and eventually comes down to an arbitrary choice. We'll deal with it later.

One can convert from a difference of magnitudes to a ratio of intensities like so:

Note that even a relatively small difference in magnitudes can mean a very large ratio of brightness. For example, Sirius, the brightest star in the sky, has a magnitude of about -1.4. If you look at a chart of its constellation (Canis Major), it is drawn as the biggest circle:

The stars in the little triangle to the upper-left of Sirius are about magnitude 5.

Q:  How many times brighter is Sirius (mag -1.4)
than a star of magnitude 5?

Wow! That's much brighter, as you will see at once if you look at a photograph of the area:

Q: The average diameter of the dark-adapted pupil in
a human eye is about 6 millimeters; the average
person can see a star of magnitude 6 on a clear, dark night.
If the same person were to look through typical 7x35 binoculars,
how faint a star might he be able to detect?

#### The painful side of magnitudes: adding and subtracting

There are drawbacks to the magnitude system. One of the big ones is the work one must do when trying to figure out the result of adding or subtracting two stellar sources, rather than multiplying or dividing them. Suppose there are two stars, A and B, with magnitudes m(A) and m(B), which appear so close together that their light blends into a single source. What is the magnitude of the resulting blend?

m(A + B)  =?    m(A) + m(B)               NO!

The proper way to do this calculation is to convert the magnitudes back into intensities, add together the intensities, and then convert back into magnitudes. There's no way around it.

#### Why do we continue to use magnitudes?

Why do we continue to use this system? There are several of reasons:

1. The objects we study cover a huge range in apparent brightness: the brightest star visible with the naked eye is over 600 times brighter than the faintest one. The Sun is over 6 TRILLION times brighter than the faintest star visible to the naked eye. It's awkward to deal with numbers this large.
2. The magnitude system is logarithmic, which turns the huge range in brightness ratios into a much smaller range in magnitude differences: the difference between the Sun and the faintest star visible to the naked eye is only 32 magnitudes.
3. Over the past few hundred years, astronomers have built up a vast literature of catalogs and measurements in the magnitude system.
4. Astronomers have figured out how to use magnitudes in some practical ways which turn out to be easier to compute than the corresponding brightness ratios.

Astronomers who study objects outside the optical wavelengths -- in the radio, ultraviolet, or X-ray regimes -- do not have any historical measurements to incorporate into their work: these fields are all very recent, dating to the 1930s or later. In those regimes, measurements are almost always quoted in "more rational" systems: units which are linear with intensity (rather than logarithmic) and which become larger for brighter objects. In the radio, for example, sources are typically measured in janskys, where

A source of strength 5 Janskys is 5 times brighter than a source of 1 Jansky, just as one would expect.

#### Zeropoints for magnitude scales

As you may have already noticed, the magnitude scale is defined in relative terms: star A is 5 magnitudes brighter than star B. But how is the zero point of the scale set? In other words, how do astronomers decide that stars A and B should have magnitudes 5 and 10 instead of 6 and 11, or 20 and 25?

There are two basic methods of setting the zero point of a magnitude scale, both of which are used at the present time.

object-based methods
This one is simple to do in practice: just decide that some particular source has magnitude zero (or 5, or 3.8, or whatever fiducial value you prefer). Then measure all other sources relative to that special source.

Vega is commonly used as a fundamental standard, and in some magnitude systems, it has either a magnitude of zero (or very close to it), or a color of zero. For example, the Johnson UBV system is defined by Johnson and Morgan (1953) so that the colors of stars of spectral class A0V such as Vega are zero -- that is, (B-V) = 0 -- and the zero point is set, not by Vega, actually, but by a set of stars near the North Pole (see Table 3 in Johnson and Morgan 1951). The magnitude of Vega itself turns out to be very nearly zero: V = 0.03, in fact.

The advantage of this system is that it's easy for everyone to share, as long as you pick a good fundamental standard or standards. But if you happen to choose a standard which varies, or which can't be seen from some parts of the world, you're in trouble. Moreover, it's not clear how to translate the magnitude value for any star into physical units, such as the flux of energy in some particular range of wavelengths.

flux-based methods
Some people prefer to begin with a specified amount of flux and use it to define the zero point of their magnitude scale. It certainly makes it easy to convert from a magnitude to physical units, but it has problems of its own, as we shall see in a moment.

The SDSS u'g'r'i'z' magnitude system, for example, is designed by Fukugita et al. (1996) so that the following relationship holds between the flux per unit frequency and the magnitude AB:

So, for example, an object of magnitude zero must provide a flux (above the Earth's atmosphere) of about 3.63 x 10-20 ergs per second per square cm per Hz.

The big problem with flux-based systems is that it's very, very difficult to measure accurately and precisely the tiny amount of energy we can gather from any star other than the Sun, thanks in large part to the fact that we live at the bottom of a very thick atmosphere.

Whatever magnitude system you happen to use, be sure that you know where you can find the definition of its zero-point. You may need to use that zero-point to convert the magnitude of an object into energy, or photons, or some other sort of quantity.

For example, suppose you are trying to figure out the proper length to expose your detector; you want to achieve a signal-to-noise ratio of 3000, which means you need to collect at least nine million photons.

Q:  You look at Vega, a star of apparent magnitude V=0,
with a telescope of diameter d = 6 inches = 15 cm.
to the V passband, which has

effective wavelength  =   5514  Angstroms
effective frequency   =   5.50 x 10^(14) Hz
equivalent width      =   9.05 x 10^(13) Hz

How many photons enter the telescope each second?

What should your exposure time be?

Notice, by the way, that astronomers may describe the spectral energy output of a star or galaxy in two slightly different ways.

flux per unit wavelength or "flam"
This describes the amount of energy per second per square centimeter of area, over some interval of wavelength. If you see a graph of a spectrum with wavelength on the bottom, then it's probably showing flams. I am much more accustomed to dealing with them, just because there is a more simple match to the typical observed optical spectrum.
flux per unit frequency or "fnu"
In this case, the spectrum is given in terms of energy per second per square centimeter, over some interval of frequency. If you see a graph with frequency on the x-axis, it's probably in these units. I believe that this may be more common for discussions involving very high (X-rays) or very low (radio waves) frequencies.

The units favored by radio astronomers and (some) high-energy observers, Janskys, are examples of "fnu"-type units. They express flux density in terms of a frequency interval.

For example, consider a perfect blackbody. The shape of the spectrum is slightly different, depending on how you measure it.

The spectrum of a real object can look quite different, too, depending on which way you choose to display it; consider this A0 V star, with data taken from Pickles (1998). Note that the horizontal axis is logarithmic in both cases.

No matter how you express it, the bottom line is that an object emits the same amount of energy within some range of wavelengths as it does between the equivalent range of frequencies. That means

You can work out the relationship between the two ways of expressing spectral energy density:

#### Apparent and absolute magnitudes

When we look up into the sky at night, we see some stars which appear very bright, and others which are so faint that we can barely detect them.

The star Sirius, for example, has a magnitude of about -1.5; a bit more than one degree away, the star HD 49980 shines relatively feebly at magnitude 5.8.

Q:  How many times brighter does Sirius appear?

Q:  Does this mean that Sirius is a much more powerful star,
one which emits hundreds of times as much energy
as HD 49980?

No!

The reason, of course, is that two factors determine the apparent brightness of a star in our sky.

• The amount of light it emits
• Its distance from the Earth

In this particular instance, the apparent magnitude of these two stars, based on their apparent brightness is quite misleading.

apparent magnitude: a measure of the brightness a star appears to have as we observe it in the night sky from Earth

It turns out that Sirius is one of the closest stars to the Sun; it is only 2.64 parsecs away. A parsec is a unit of distance equal to about 3.3 light years, or 3.1 x 1016 meters; we'll discuss this unit later. On the other hand, HD 49980 is very distant, located about 500 parsecs from the Sun.

Q:  How many times more distant is HD 49980?

Q:  If we were to pick up Sirius and move it
so that it was at the same distance as HD 49980,
how many times fainter would it become?

Q:  Which star would look brighter if they
were sitting side-by-side at the same distance?

Since we often want to compare the intrinsic properties of stars, we'd like to have some measure of brightness which is connected directly with luminosity; a type of magnitude which does not depend on distance. Astronomer convert apparent to absolute magnitudes to compare stars fairly, as if they were all side-by-side at a standard distance.

absolute magnitude: the magnitude a star would have, if it were moved to a distance of 10 parsecs from the Sun.

The ordinary convention is to write apparent magnitudes with a lower-case letter m, and absolute magnitudes with an upper-case M. One can derive a formula which connects the apparent and absolute magnitudes of a star, using the inverse square law. If we express the distance d in parsecs, then

Q:  The distance to Sirius is 2.64 pc, and its apparent
magnitude is m = -1.5.  What is the absolute magnitude
of Sirius?

Q:  The distance to HD 49980 is 500 pc, and its apparent
magnitude is m = 5.8.  What is the absolute magnitude
of HD 49980?

Stars with small absolute magnitudes are truly luminous beasts, radiating huge amounts of energy into space each second. Stars with large absolute magnitudes are relatively feeble creatures, dimly illuminating their immediate surroundings but little else.

#### The distance modulus

The difference between the apparent and absolute magnitude of a star, (m - M), is called its distance modulus.

As the equation above shows, it is a simple function of the distance to the star. In practice, astronomers sometimes prefer to specify the distance to a star by its distance modulus, rather than by the distance itself. For example, Look at an extract from the abstract to this paper on the distance to stars in the galaxy NGC 2403:

Why use distance modulus instead of distance? I can think of two reasons, though they really boil down to the same thing.

1. Often, one determines the distance to a distant object by assuming that it is identical to some nearby object whose distance is known, and comparing the apparent brightness of the two objects. In this case, the quantities actually observed are the two apparent magnitudes; so using distance modulus is natural.

Q:  Joe Astro is studying the Andromeda galaxy.
He notices a star in that galaxy which appears
similar in many ways (color, temperature, etc.)
to the Sun.  The apparent magnitude of this
solar twin is m = 30.0.

What is the distance modulus to the Andromeda
Galaxy?

2. When planning an observing run, astronomers need to figure out how long their exposures should be. This depends on the apparent magnitude of their targets. If (as is often the case) an astronomer has a good idea of the intrinsic properties of target stars -- their absolute magnitudes -- he can most quickly and easily calculate their apparent magnitudes if he knows the distance modulus.

The basic idea is that, from a practical, observational point of view, it is often more useful to have a catalog of distance modulus values instead of real distances.

A second advantage appears when astronomers use distance modulus as a relative measure between two objects. For example, the Large Magellanic Cloud (LMC), the nearest galaxy to our own Milky Way, is often used as a stepping-stone to other, more distant galaxies. Our current estimate of the distance to the LMC is about 50,000 pc = 50 kilo-parsecs (kpc). Suppose that you measure the distances to the LMC and several other galaxies by observing a particular sort of star in each galaxy. You might find

apparent mag    distance modulus    distance
galaxy     of star       relative to LMC       (kpc)
------------------------------------------------------
LMC         16.5           0.0               50
M31         24.3           7.8             1800
M81         26.8          10.3             5740
------------------------------------------------------

Ten years from now, astronomers discover a systematic error in measurements to the LMC; instead of being 50 kpc away from us, it actually turns out to be 60 kpc away.

• if you published your results for M31 and M81 using the absolute distances for each one, in kpc, then your work immediately becomes invalid. In order to use your results, astronomers will have to re-calculate all your values using the new distance to the LMC
• but if you published your results in distance modulus, relative to the LMC, your results are still as correct as ever. Astronomers will still know exactly how much fainter a star in M31 will appear, relative to an identical star in the LMC