Section 8. How bright? Magnitudes and Telescopes 1

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Section 8. How bright? Magnitudes and Telescopes 1 More than 2000 years ago, the Greek astronomer Hipparchus put together the first star catalog (as far as we know). Since he wanted to indicate somehow the relative brightness of different stars, he invented a system of "stellar magnitudes" which we still use (more or less) today. In this system he divided up the stars that are visible to the naked eye (under the very best of conditions!) into six classes, from first magnitude = brightest to sixth magnitude = faintest. Later, when astronomers learned how to measure precisely how much light we are receiving from each star, it was found that Hipparchus magnitudes could be described mathematically in a very simple way. That is: if we compare the light received by any two stars that differ by one magnitude, the brighter star will be 2.512 times brighter than the fainter one. Thus if we compare a sixth magnitude star to a fifth magnitude star, the fifth magnitude star will be 2.512 times brighter. If we compare a fourth magnitude star to a fifth magnitude star, which is brighter? By how much? times. Compare the 4th magnitude star to the sixth magnitude star. The fourth magnitude star is 2.512 times brighter than the fifth magnitude star which is 2.512 times brighter than the sixth magnitude star. So fourth magnitude star = 2.512 x fifth magnitude star = 2.512 x (2.512 times sixth magnitude star) = (2.512x2.512) times sixth magnitude star, or about 6.3 times brighter. How much brighter than a sixth magnitude star is a 3rd magnitude star? 3rd = 2.512x4th = 2.512x(2.512)x5th = ( x( x( x6th))) = x6th A convenient fact to remember is that 5 magnitudes difference corresponds to a factor of 100 (= (2.512) 5 ) in brightness: If star A is five magnitudes brighter than star B, then we are getting 100 times as much light from star A as from star B. Another way to write this is brightness of star A = (2.512) (magnitude of B - magnitude of A) x brightness of star B. This is mathematically the same thing as the following equation using logarithms: log (2.512) (magnitude of B - magnitude of A) = log (brightness of A/brightness of B) or, since log(2.512) = 0.400 and 1/0.4 = 2.5 magnitude of B = magnitude of A + 2.5 log (brightness of A / brightness of B) This is the version of the definition of magnitude that most astronomers remember, because if one is comfortable with logarithms this is a very convenient formula to work with. For example, from this formula, we can easily verify that 5 magnitudes corresponds to 100x in brightness: if brightness of A / brightness of B = 100, then magnitude of B = magnitude of A + 2.5x2 = magnitude of A + 5. One can perhaps visualize stellar magnitudes better with some specific examples; the table on the next page gives some examples.

Section 8. How bright? Magnitudes and Telescopes 2 Apparent Magnitude 100W bulb at (distance) Street lamp at (distance) # of (2W) Xmas tree lights at 10 miles Number of stars this bright or brighter 1 7 miles 25 miles 100 12 A streetlight 1 block (50 yards) from you has an apparent brightness that is nearly a million times that of a first magnitude star. 2 11 miles 40 miles 40 40 3 17 miles 63 miles 16 4 27 miles 100 miles 6 458 On a good night around here, we can see stars to 4th magnitude (about 750 stars visible above the horizon). 5 43 miles 157 miles 2.5 1,508 6 68 miles 250 miles 1 4,900 The human eye is capable of seeing stars to 6th magnitude, given a dark observing site and a clear sky. On such a night, at such a location, you would see about 2,500 stars above the horizon. That means that your fist would block about a dozen stars, on average, anywhere in the sky. 7 22,000 8 72,000 9 222,000 10 722,000 11 680 miles 2500 miles 237,000 11th magnitude stars can be observed quite easily with a small telescope -- if the sky is not too bright. Numbers of stars brighter than a given magnitude for magnitude 6-11 are approximate. Hipparchus' magnitudes are what we now call apparent magnitudes: they describe how bright a star appears to us. If all stars were at the same distance from us, then this would also be a measure of how bright the stars were intrinsically (of themselves). However, stars we see in the night sky are at quite a range of different distances, so a big part of what determines how bright a star appears to us -- its apparent magnitude -- is how far away it is. If we imagine bringing all stars to the same distance from us, then we could compare their "true" intrinsic brightness. Astronomers define absolute magnitude as the magnitude a star would appear if it were moved to a distance of ten parsecs (= about 33 lightyears). If two identical stars (i.e. with the same absolute magnitudes) are located at different distances from us, the one that is farther away will appear fainter. To find out how much fainter, we invoke the "inverse square law": the amount of light we receive from a luminous object decreases as the

Section 8. How bright? Magnitudes and Telescopes 3 square of the distance (if there is nothing getting in the way). Thus, for example, two 100 W light bulbs, one of them twice as far away from us: that one will appear (1/2) 2 = 1/4 as bright. EXERCISE 8.1. Pollux has an apparent magnitude of +1.16; Rigel has an apparent magnitude of +0.14. Which star is brighter? The difference in magnitude between Rigel and Pollux is about 1 magnitude. That means that we are getting about times more light into our eyes from than from. EXERCISE 8.2: On a reasonably good night in Ames you may be able to see stars down to magnitude 3.5. These stars are about magnitudes fainter than Sirius (-1.5), or about times as much light comes into our eyes from Sirius as from these magnitude 3.5 stars. Suppose we have two stars that are identical, but at different distances. The closer one is first magnitude. The second one is 10 times farther away. What is its apparent magnitude? From the inverse square law, the apparent brightness of star 2 = (1/10) 2 times apparent brightness of star1 = times apparent brightest of star 1. Thus star 1 appears times as bright as star 2. From the equations above or from the table, deduce the apparent magnitude of star 2. Very bright stars in our sky must be either very close (Sirius) or very bright intrinsically (Betelgeuse, Rigel, and in fact most of the other "named" stars we will study). There are not very many stars very close to us, because the chances of finding a star in a small box (say, within a couple of light-years of the sun) is very small compared with the probability of finding a star in a large box (say, within 1000 light years of the sun.) Stars that are intrinsically very bright are also quite rare, because in order to be bright a star must be using up its fuel very quickly. These ideas are used extensively in mapping out the population of stars in our galaxy. (By "population of stars" astronomers mean, basically, how many stars similar to the sun, how many red giants, how many white dwarfs etc. do you expect to find in a given "box" with a volume of a few million cubic light-years). EXERCISE 8.3 Look at the list of the twenty brightest stars at the end of this section Of these 20, how many (total!) are brighter than (apparent) magnitude (-1)? how many (total!) are brighter than (apparent) magnitude (0)? how many (total!) are brighter than (apparent) magnitude (+1)? Note: The table lists A, B, and C components of double or triple star systems. For how many of the A components is the B component as bright as the faintest of the A components of the "20 brightest stars"?

Section 8. How bright? Magnitudes and Telescopes 4 EXERCISE 8.4. Look at the list of the 20 nearest stars at the end of this section. How many of these could be seen from a city like Ames on a good night (i.e., how many are brighter than apparent magnitude 4)? How many are naked-eye objects (brighter than apparent magnitude 6)? EXERCISE 8.5. These two figures show the part of the sky around Sagittarius -- on one, stars to magnitude 3.5, on the other, stars to magnitude 6. Which is which? Label the diagrams accordingly. About how many times more stars can you see if you can see all the way to 6th magnitude vs. if you only can see to magnitude 3.5? (circle one:) about (2, 4, 10, 100) times as many. Hint: If you wanted to know exactly how many more stars there are in the right hand diagram compared to the left-hand diagram, you would count them. But that is tedious! Look at the two diagrams carefully. Pick a small area that looks about average in terms of how close together the stars are in both diagrams. The lower left-hand corner would not be a good choice -- why not? Count the stars in those areas, and compare. Then try another "typical" area, and compare again. How many more would you expect to find? Well, from magnitude 3.5 to magnitude 6 is 2.5 magnitudes, or about 10 times in brightness. That means that if all stars were equally bright, we would be seeing stars about 10 times farther away in the right-hand figure. The volume of space that we are looking at in the right-hand figure is thus about ( 10) 2 or times the volume of space we are looking at in the left-hand figure. If the stars are scattered uniformly through space around us, we would expect to see about times as many stars in the right hand figure as in the left-hand figure. Does this agree with what you found? If it doesn't, can you think of any reason why not?

Section 8. How bright? Magnitudes and Telescopes 5 The twenty brightest stars Star name Spectral type 1 Apparent magnitude Sirius A -1.46 Canopus F -0.72 alpha Centauri G -0.01 Arcturus K -0.06 Vega A +0.04 Capella G +0.05 Rigel B +0.14 Procyon F +0.37 Betelgeuse M +0.41 Achernar B +0.51 beta Centauri B +0.63 Altair A +0.77 alpha Crucis B +1.39 Aldebaran K +0.86 Spica B +0.91 Antares M +0.92 Pollux K +1.16 Fomalhaut A +1.19 Deneb A +1.26 beta Crucis B +1.28 How many of these stars are you learning to find in your constellation work for recitation? Can you think of any good reason why we aren't asking youto learn to find the others? Hint: You might want to come back to this question after we have discussed coordinate systems and apparent daily motion; you might also want to look at the table of the brightest stars at the back of the book or use Voyager to find out where these stars are in the sky. It is possible to set the magnitude limit in Voyager so that only the brightest stars show up. 1 The spectral type is an indication of the color. O and B spectral types will appear blue; A, F, and G stars will be white or slightly yellow (the Sun is a G type star); and K and M stars will be pale orange or pink.

Section 8. How bright? Magnitudes and Telescopes 6 Telescopes -- what they are good for, how to select one, and related questions Telescopes serve two functions for astronomy: (a) they collect more light than the naked eye, so they allow us to see fainter objects; and (b) they allow us to magnify an image so as to see more detail. The size of the opening of a telescope (its aperture) is the most important attribute; this determines how much starlight can enter into the telescope. The amount of light collected depends on the area of the telescope, or the square of its diameter. The pupil of the human eye has a diameter of about 1 mm. Thus a telescope with a 1m aperture (about 40 inches in diameter) can collect about (1m/10-3 m) 2 = 10 6 times the light the naked eye can collect. Since the human eye can (under ideal conditions) see stars to magnitude 6, a 1m telescope can see stars up to a million times fainter, or, since 2.5log(10 6 ) = 6x2.5 = 15, in principle a 1m telescope should allow you to see stars to magnitude 21. Exercise 8.6. How much more light than the human eye can the Mt. Palomar 5m telescope collect? About how faint stars would you expect to be able to see with the Mt. Palomar telescope? The diameter of a telescope also determines how much detail you can resolve. Ignoring effects of the atmosphere, an ideal parabolic mirror can resolve detail down to α= 2.1x10 5 λ/d where α is the smallest angle on the sky that can be resolved, in arcseconds, λis the wavelength of the light and d is the diameter of the mirror; λ and d must both be given in the same units (for example, cm) for this equation to be correct. Visible light is centered around λ= 5.5x10-7 m (5000 Å or 500 nanometers, nm) so for visible light a good estimate is α in arcsec 12/(d in cm) Exercise 8.7. What is the theoretical resolution achievable by the E. W. Fick observatory's Mather telescope, which has a diameter of about 60 cm? A note on magnification: (a) Magnification depends on the eyepiece used and on the main mirror or lens, the "objective". The magnification you get = focal length of objective / focal length of eyepiece. (b) Since there is a limit to the resolution that can be produced by a given telescope on a given night in a given location -- such that there is a minimum angle on the sky corresponding to the smallest detail that can be distinguished -- there is no point in magnifying the image much more than so that the smallest detail can be comfortably seen by the human eye. Thus, for example, on a night when the "seeing" is one arc-sec, a magnification of x60 will take a 1arcminute feature and make it into a 1 arcminute feature -- just possible for the human eye to resolve. A magnification x120 will make it easier for the observer to be sure that he is seeing the smallest details, but will also spread the light out over 4 times the area (twice the diameter), making it harder to see faint features. For most astronomical objects, we are limited in what we can see by the faintness, so it is best to provide only the minimum necessary magnification. Whether a telescope can achieve its ideal resolution (given by the equation above) in practice depends on a number of factors: How accurately the mirror is made; how much the telescope's shape is distorted as it moves to track the star (or as the wind blows on it); and how much the atmosphere "scrambles" the starlight. Until very recently, most astronomers believed that the atmosphere limited telescopes to a resolution no better than about 1 arcsecond. (The human eye can resolve about 1 arcminute, so even the best telescope only gives us an improvement equivalent to magnification x60, by this estimate). However, it has recently been recognized that distortions in the telescope caused by the changing angle between the mirror and gravity, temperature differences, and other mechanical factors are responsible for about half of the smearing of the image that we see. The other half is due to the atmosphere: air cells at different temperature transmit light at slightly different speeds, causing the light to bend by slightly

Section 8. How bright? Magnitudes and Telescopes 7 different amounts, and spreading out the image. For small telescopes and the human eye, which look through one cell at a time, this atmospheric effect causes stars to appear to twinkle -- shift direction and vary slightly in brightness. For bigger telescopes, which look through several atmospheric cells at a time, the image becomes very patchy. The tendency of the atmosphere to smear out the image is called "seeing" by astronomers. Good "seeing" is less than one arcsec; if an astronomer says the "seeing" for his last observing run was 2 or 3 arcsec, express sympathy! With high speed computers it has become possible to build telescopes with "adaptive optics": that is, the mirror's shape is adjusted slightly by computer-controlled pistons on the back side to compensate for the effects of gravity, winds, and the distortions of "seeing". Ground based telescopes that are now being built are expected to give 0.1 to 0.2 arcsec resolution on the best nights. For comparison, the original specifications for the Hubble Space Telescope were that it should have >90% of the light within <0.08 arcsec -- not much different from the most optimistic estimte for the new generation of ground-based telescopes. Ground-based or space-based telescopes -- which are best? Pound for pound, space-based telescopes cost a lot more than ground-based ones. So anytime that a project can be done from the ground, that is the way that it should be done. Space-based telescopes have three potential advantages over ground-based ones: (1) They avoid the two major effects of the atmosphere: (a) "seeing" (as described above) and (b) absorption of all the radiation that falls outside of the visible light and radio wave "windows". Ultraviolet, X ray and gamma ray radiation, and most of the infrared radiation from distant stars and galaxies can only be detected by telescopes placed above the Earth's atmosphere. (2) They can be placed where observations may be made continuously for more than just the "night" time on Earth. (3) They are not subject to contamination of the starlight by "light pollution" (Section 9). The Hubble Space Telescope was designed to take advantage of both the improved resolution and the expanded wavelength coverage possible from space, by including ultraviolet detectors as well as detectors for visible light. Pre-repair, Hubble data could still be used for high-resolution studies in some special cases, because the optical problem was so simple that it could be corrected for by a computer calculation. However, when the problem caused the light from several stars to be blended together, this could not always be sorted out by calculations afterwards. Most of the work done on the Hubble telescope up to Dec. 93 therefore emphasized the use of the ultraviolet sensitivity of the telescope. Orbits that can be reached by the shuttle are about 1.5 hour orbits; for a few objects this may still allow for nearly continuous observations, but for most objects, this limits Hubble's observations to 1/2 hour or so at a time. The International Ultraviolet Explorer satellite, launched in 1978 (for a planned 3 year mission that has now gone over 15 years of successful operation!) is in a "synchronous" (24 hour) orbit that allows longer observations of many more objects. A telescope on the back side of the Moon would be able to track stars through an entire lunar night, about 2 weeks. If we can control light pollution on Earth, then ground-based telescopes will continue to be important for astronomy. If we cannot control light pollution, then we may ultimately have to do nearly all astronomical studies from space, at much higher cost and inconvenience. Light pollution is described and discussed in the next section.

Section 8. How bright? Magnitudes and Telescopes 8 Telescope mounts: A telescope mount serves several functions: it supports the tube and mirror or lens, it allows the tube to be moved to aim it at the object to be observed, and it allows the tube to move smoothly to "track" the star as it appears to move across the sky as the Earth rotates. There are two basic types of mounts: equatorial and altitude-azimuth (alt-az) mounts. The first is the traditional type of mount, because it allows a single motor to drive the telesope to compensate for Earth's rotation: To North Celestial Pole To * Clock drive moves telescope around the polar axis to compensate for Earth's rotation around the same axis Earth's axis of rotation (from South Pole through North Pole) This type of mount can be easily recognized: the polar axis points towards the celestial pole, a point in the sky along the N-S line (meridian) and at an angle to the horizon equal to the observer's latitude. angle equal to observer's latitude Polar axis OBSERVING EXERCISE: Which of the telescopes on the roof-top observing deck have equatorial mounts? The advantages of an equatorial mount are that it only requires one motor with the fine control needed to track an object you are observing, and that it does not require a computer to control it if the motor is reliable. The main disadvantage is that the telesope is supported by a bracket that is at an angle with respect to gravity, so it needs to be stronger and more massive than it would be if it were vertical. The alt-az mounting is the simplest and strongest mount to build, and is used on the largest of the new generation of telescopes as well as on many smaller telescopes, including those being built by amateurs for their own use. The disadvantages are that they need two precision motors, and that a computer control is necessary to keep the star centered in the field of view.