Phys216 Practical Astrophysics - Astrophysics Research Institute

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PHYS216 Practical Astrophysics Module Leader: Course Lecturer: Dr Matt Darnley Dr Chris Copperwheat Astrophysics Research Institute Liverpool Telescope / ARI Liverpool John Moores University Liverpool John Moores University [email protected] [email protected] Notes online All course materials available on Vital These Powerpoint slides: ● http://www.astro.ljmu.ac.uk/~cmc/#teaching Other good resources online: ● http://star-www.st-and.ac.uk/~fv/webnotes/ ● http://www.vikdhillon.staff.shef.ac.uk/teaching/phy21 7/instruments/phy217_inst_course.html 2 Great Circles and Small Circles Great Circle • The intersection of a plane containing the centre of a sphere and its surface, e.g. ABCD and BEDF. • P and Q are poles of plane ABCD. Small Circle • A circle which does not include the centre of the sphere, e.g. WXYZ. 3 On the Earth… • All lines of Longitude are great circles • All lines of Latitude are small circles except the equator. Lines of equal LATITUDE Lines of equal LONGITUDE Q. Is the tropic of Cancer a small or a great circle? 4 Spherical Trigonometry Spherical triangle Formed from three arcs of Great Circles. Distances (a; b; c) are measured as angles. Sum of the 3 angles A + B + C > 180º (this is non-Euclidean geometry appropriate to curved space). e.g. if • arc AB is part of the Earth's equator • arc CB is the Greenwich Meridian (long=0º) • arc CA is longitude=90º then A = B = C = 90º so A + B + C = 270º in this case. As the triangle gets small relative to the size of the sphere then A + B + C -> 180º 5 Spherical sine rule: Spherical cosine rule: How do these rules relate to Euclidean (plane) geometry? If the length of the sides are very small (compared to the radius of the sphere), then sin a ~ a etc. (small angle approximation). The spherical sine rule then becomes: The Euclidean cosine rule can also be recovered by the same method. 6 Celestial Sphere • Celestial Meridian – a great circle which passes through Zenith and the North & South celestial poles. It is perpendicular to the horizon. If you stand facing North, the meridian is a line that passes from north on the horizon, directly over your head, to south on the horizon behind you 7 Celestial Sphere As the earth rotates, stars (like the sun) rise in the east, pass over the meridian (transit), and set in the west. The hour angle tells you how long it will be before the star transits (or how much time has passed since it transited!) • Hour Angle - angle between a star's current position and the meridian (measured WESTWARD in hours, where 1 hour is equivalent to 15 degrees – because 24 hours = 360 degrees). MERIDIAN An object transits or culminates when HA ~ 3 hr passing through the meridian. It has an HA = 0 hr when culminating. Its HA then increases as it moves towards the west. At HA = 23h it is just one hour short S E W of culminating again. 8 Celestial Sphere • Celestial Equator - Projection of the Earth's equator out onto the Celestial Sphere.  • Celestial Poles - we have two of them, a North and South pole! • Declination - angular distance of a star above the Celestial Equator. Analogous to LATITUDE, hence the Celestial Equator comprises all points at zero Declination. 9 Observing the sky from the surface of the rotating Earth Apparent direction of stars Horizon - you can’t see below this, so some stars are too far south to be observed from e.g. Europe. Zenith - directly overhead Altitude - angular height of star above the horizon Zenith distance/angle angle between zenith and direction to star. Altitude of the pole above the horizon = latitude of the observer, Φ (phi) 10 The Celestial Sphere Star X Path of Star X on the sky is a small circle (red) parallel to the celestial equator (blue). It rises in the East, transits the meridian, and sets in the West. Star X rises and sets where its small circle intersects the observer’s horizon (black). General rule - for an observer in the NORTH: • If a northern hemisphere star’s small circle does not intersect the observer’s horizon, the star never SETS and the star is said to be circumpolar • If a southern hemisphere star’s small circle does not intersect the observer’s horizon, the star never RISES 11 Circumpolar Stars These objects (at high declination) never set, and depend on the observer's latitude Φ For a star that just grazes the horizon at its lowest elevation, declination δ is given by: δ = 90º – Φ i.e. if δ > 90º - Φ then the star is circumpolar The Latitude of Liverpool is 53º N. Therefore stars with δ > 37º never set! At the north pole, essentially all visible stars are circumpolar; at the equator, none are! At what declination do stars never rise in Liverpool? 12 Simplest design of Telescope It moves up and down, and rotates from left to right… Altitude Azimuth 13 Altitude-Azimuth system The Alt-Az system uses observer based coordinates. Zenith – the point on the sky directly above the observer. Altitude, a - angle between the observer's horizon and the object, measured on a great circle through the object and the zenith. Zenith distance, z - angular distance between zenith and object, z = 90º - a Azimuth, A - angle measured along the horizon, Eastward from North, to the great circle used for altitude measurement. a S E A Alt-Az coordinates of a star are specific to the time and the observer's location. 14 The Celestial Sphere We really need a fixed coordinate system so we can catalogue the positions of the stars... What about Hour Angle and Declination? Last week we learned that: • Celestial Equator - Projection of the Earth's equator out onto the Celestial Sphere. • Celestial Poles – we have two of them, a North and South Pole! • Hour Angle - angle between a star's current position and the meridian (measured WESTWARD in hours) • Declination - angular distance of a star above the Celestial Equator. 15 Equatorial Coordinates - 1 HA and Declination – nice – but is this a fixed coordinate system? Declination is a fixed coordinate on the celestial sphere. Its analogous to latitude on the Earth. It tells you how high above (or below) the Celestial Equator your target is. (nb. the celestial equator is an extension of the earth’s equator out into space.) Hour Angle is not a fixed coordinate system. The stars rise and set like the sun; the hour angle tells you how long it’ll be before the target transits (reaches its highest point on the sky) or since it has transited MERIDIAN HA ~ 3 hr (remember: HA increases to the west) E S W Together HA and Dec are useful for determining whether an object is currently observable, and how long before (or since) it transited - but it's still not a fixed coordinate system. Hour Angle is time-dependent – its varies continuously! 16 Equatorial Coordinates - 2 For our fixed coordinate system we need to use something other than HA! Declination is still fine – its still a fixed coordinate on the celestial sphere, analogous to latitude on Earth. Right Ascension is like HA. However, it is referenced to a fixed point on the celestial sphere (rather than the celestial meridian) called the First Point of Aries (). RA is analogous to longitude on Earth; lines of RA are great circles which pass through the poles, and whose planes pass through the centre of the celestial sphere (and Earth).  Ecliptic Equatorial Lines of equal RA 17 Origin of the Equatorial System Right Ascension is referenced to a fixed point on the celestial sphere, called the First Point of Aries (FPoA),  The zero-point of Right Ascension is the great circle which passes through the FPoA and the poles. The RA and Dec of the FPoA are 0,0. RA has units in hrs/mins/sec: 24 hr = 360º, so 1 hr is equivalent to 15º (same as HA). 18 Equatorial vs Ecliptic Planes Equatorial plane – A projection of the Earth's equator out onto the celestial sphere.  Ecliptic plane – The apparent path of the Sun on the celestial sphere. It is equivalent to the plane of the Earth’s orbit around the Sun. The angle between the equatorial plane and the ecliptic plane is 23.5º Ecliptic coordinates are sometimes used for objects in the Solar System. Most planets (apart from Mercury) and most asteroids have orbits with small inclinations to the ecliptic. 19 First Point of Aries and the Vernal Equinox • The FPoA represents one of the two points on the Celestial Sphere where the Ecliptic Plane and the Equatorial Plane cross one another. • The First Point of Aries is the point in space beyond the sun on the Vernal Equinox, March 21st / 22nd each year. 'Ver' from the Latin for Spring 20 First Point of Aries and the Vernal Equinox Vernal (or Spring) Equinox & Autumnal Equinox – When the sun crosses the equator. The Vernal Equinox – when the Sun, Earth and First Point of Aries (FPoA) are in line –corresponds to when the Sun moves above the equator as it moves around the ecliptic. 21 RA and Dec The positions of astronomical sources are usually (but not always) quoted in Right Ascension and Declination – RA and Dec. Remember: • RA increases TO THE LEFT! • Dec (+ve) increases up. • North is up, East is to the LEFT! • 1 hour is split into 60 minutes • 1 minute is split into 60 seconds (Equatorial coords) N E ------ W S • Example coordinates: Supernova SN1987A: 05h 35m 28.0s , -69º 16´ 12´´ Barred spiral galaxy, NGC 55: 00h 14m 53.6s , -39º 11´ 48´´ Ring Nebula, M57: 18h 53m 35.1s , +33º 01´ 45´´ Targets with an angular distance that is > 90º from the latitude of the observer can’t be reached 22 Converting to Alt-Az Coordinates In order to successfully point an Altitude-Azimuth mounted telescope at a given object, one must first convert that object's position from equatorial coordinates (RA and Dec) to the Alt-Az system. The altitude portion of this transformation is also important when determining if a particular telescope can observe an object (i.e. is the object higher than the observatory wall, a nearby tree or volcano?) and for calculating the airmass (see later). Altitude 23 Converting to Alt-Az Coordinates To convert between equatorial coordinates (HA and Dec) and “horizon” (Alt-Az) for star X, we use a spherical triangle XPZ, where Z is the zenith, P is the North Celestial Pole, and X is the star. me HA rid i an Note: Φ is the latitude of the observer (remember, angular height of Polaris above the horizon is roughly equal to Φ) N The sides of the triangle: PZ is the observer's co-latitude = 90°- Φ. ZX is the zenith distance of X = 90°- a. PX is the North Polar Distance of X = 90°- δ. The angles of the triangle: The angle at P is HA, the local Hour Angle of star X. The angle at Z is 360°- A, where A is the azimuth of star X. The angle at X is q, the parallactic angle. Tip: if you don’t get this, try looking down on Z from above! 24 Converting to Alt-Az Coordinates Given: the latitude of the observer, Φ, the hour angle, HA and declination, δ we can calculate azimuth A and altitude a. By the cosine rule: cos(90°-a) = cos(90°-δ) cos(90°-Φ) + sin(90°-δ) sin(90°-Φ) cos(HA) which simplifies to: sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) … This gives us the altitude a. HA By the sine rule: sin(360°-A) / sin(90°-δ) = sin(HA) / sin(90°-a) which simplifies to: - sin(A) / cos(δ) = sin(HA) / cos(a) then: sin(A) = - sin(HA) cos(δ) / cos(a) … This gives us the azimuth A (IF δ > 0). 25 Converting to Alt-Az Coordinates Given: the latitude of the observer, Φ, the hour angle, HA and Declination, δ we can calculate azimuth A and altitude a. By the cosine rule: cos(90°-a) = cos(90°-δ) cos(90°-Φ) + sin(90°-δ) sin(90°-Φ) cos(HA) which simplifies to: sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) … This gives us the altitude a. Alternatively, by the cosine rule again: cos(90°-δ) = cos(90°-Φ) cos(90°-a) + sin(90°-Φ) sin(90°-a) cos(360°-A) which simplifies to sin(δ) = sin(Φ) sin(a) + cos(Φ) cos(a) cos(360°-A) Rearrange to find A: cos(360°-A) = cos(A) = [ sin(δ) - sin(Φ) sin(a) ] / cos(Φ) cos(a) … This again gives us the azimuth A (for +ve and –ve declination) 26 Converting to Alt-Az Coordinates Strictly speaking: • Altitude, a (the angular height above the horizon) sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(HA) • Azimuth, A (angular rotation clockwise, i.e. from North towards East): cos(360°-A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) (0 hrs < HA < 12 hrs target is setting) cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) (12 hrs < HA < 24 hrs target is rising) Because, e.g., cos-1(cos(315º)) = 45º 27 Converting to Alt-Az Coordinates (an example…) Planetary Nebula, M76, equatorial coordinates: Convert to degrees. α = 1h 39m 10s → 1h 39.17m → 1.653h → ( ÷24 and *360) → 24.795º δ = 51º 19´ 30´´ → 51º 19.5´ → 51.325º We will observe at 3am when the target has transited: Hour angle, HA = 2.20 hrs To convert hour angle to degrees - multiply by 15 (nb. 24 hrs is equivalent to 360º). Therefore: HA = 33.00º Altitude, a: sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) Observatory in La a = 56.624º Palma, latitude = 56º 37´ 29´´ Φ = 28.7624º Azimuth, A (0 < HA < 12 hrs): cos(360°-A) = [ sin(δ) - sin(Φ) sin(a) ] / cos(Φ) cos(a) A = 321.781º or -38.219º (NB: a -ve angle, or very large +ve = 321º 46´ 50´´ angle, means target is West of North) 28 Making sense of these numbers… M76, equatorial coordinates: α = 1h 39m 10s δ = 51º 19´ 30´´ When we want to observe it: Hour Angle, HA = 2.2 hrs Altitude, a = 56.624º Azimuth, A = -38.219º or 321.781º HA ~ 2.2 hr 56º 38º Observatory in La Palma, latitude Φ = 28.7624º • M76 is in the North – the object’s declination is GREATER THAN the latitude of our observatory • M76 therefore rises in the EAST and moves to the left. • Our target is already >2 hours over, i.e. its in the west; it therefore makes sense that Azimuth is a large (or negative) angle, since A is measured East of North. 29 The Pole Star (a special case) The “pole star” Polaris has a declination of δ ≈ +90º. Since sin 90º = 1 and cos 90º = 0, its Altitude is given by: Altitude, a: sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) sin(a) = sin(Φ) a=Φ The Altitude, a, of Polaris depends only on the Latitude, Φ, of the observer. THIS IS NOT TRUE FOR THE OTHER STARS! 30 An example to try for yourself! Convert the Equatorial Coordinates of the Crab Nebula, M1, to Alt-Az coords. Coordinates of the Target: α = 1h 39m 10s Coordinates of the Observatory: • HA: -2.0 hrs (or +22 hrs) δ = 51º 19´ 30´´ 19.8207º N, 155.4681º W Crab Nebula, observed by the LT with IO:O on, 30 Oct 2013. 31 Precession Unfortunately no coordinate system can be permanently fixed. This is because the Earth's rotation axis precesses slowly, with a period of 25,600 years. This happens because the Earth is not quite spherical, but is oblate and tilted. Therefore, the direction of the Sun's gravity does not pass directly through the Earth's centre of rotation. The overall effect is that the position of the Vernal Equinox on the sky moves with respect to the stars, at approx 50´´ (arc-seconds) per year. Because of precession, we must define an appropriate equinox for the RA and Dec catalogue positions. Standard equinoxes are defined every 50 years, e.g. 1950, 2000 etc. Current catalogues use J2000 coordinates. 32 Precession Converting coordinates between two Equinoxes, or updating to current Epoch: And: Where: αT and δT are the RA and Dec (both in degrees) of an object at time interval T (in years) after the catalogue equinox, E. αE and δE are the catalogue coordinates, RA and Dec (for equinox E). θ (the precession constant) = 50.4´´ per year (multiply by 25,600 yrs to get 360º!) ε is the angle between the equatorial and ecliptic planes, precisely 23º 27´ 8´´ *** NOTE *** Equinox defines a standardised coordinate system (1950, 2000) Epoch can be any time (its usually when you want to observe) 33 Converting between Equinox 1950 and Equinox 2000 Convert Equinox 1950 coords to degrees: α1950 = 1h 39m 10s → 1h 39.17m → 1.653h → (÷24 and *360) → 24.795º δ1950 = 51º 19´ 30´´ → 51º 19.5´ → 51.325º Precess 1950 degrees to 2000 degrees and convert back to RA and Dec α2000 = 24.795º + [0.014º . 50yrs . (cos23.5º + sin23.5º . sin24.795º . tan51.325º)] = 25.583º = 1.7055h ≈ 1h 42m 20s δ2000 = 51.325º + [0.014º . 50yrs . (sin23.5º . cos24.795º)] = 51.578 ≈ 51º 34´ 40´´ Remember: Precession constant is 50.4´´ = 0.014º / year. Angle between ecliptic and equatorial planes Is 23.5º) 34 Converting Equinox 2000 to the current Epoch Convert Equinox 2000 coords to degrees: α2000 = 1h 42m 20s → 1h 42.33m → 1.706h → 25.583º δ2000 = 51º 34´ 40´´ → 51º 34.67´ → 51.577º Time between 2000 baseline (1 Jan) and date we want to observe = 16.5 years. Therefore, precess 2000 coords forward by this period: α2016.5 = 25.583º + [0.014º . 16.5yrs . (cos23.5º + sin23.5º.sin25.583º.tan51.577º)] = 25.845º = 1.723h ≈ 1h 43m 23s δ2016.5 = 51.577º + [0.014º . 16.5yrs . (sin23.5º.cos25.583º) = 51.660º = 51º 39´ 36´´ 35 Converting current Epoch to Alt-Az Convert Epoch 2015.5 declination to decimal degrees: α2016.5 = 1h 43m 23s → 25.845º δ2016.5 = 51º 39' 36'' → 51.660º Finally, convert Epoch 2016.5 coords to Alt-Az system for La Palma, Φ = 28.760º (assume target transits at 1.00 am; I get access to the telescope at midnight, so hour angle, HA = -1 hrs, or -15º) sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) = sin(51.660). sin(28.760) + cos(51.660).cos(28.760).cos(-15) Altitude, a = 64.509º 12 < HA < 24 hrs: cos(A) = [ sin(δ) - sin(Φ) sin(a) ] / cos(Φ) cos(a) ] = [ sin(51.660) – sin(28.760).sin(64.509) ] / cos (28.760).cos(64.509) Azimuth, A = 21.913º 36 One for you to try…. • Step 1: precess Equinox 1950 coords to Epoch 2016.5 • Step 2: convert these new Equatorial (RA-Dec) coords to Alt-Az coordinates (assume HA = 0, i.e. object is transiting!) Object: Betelgeuse Coords: α1950 = 5h 52m 28s , δ1950 = +7º 23´ 58´´ • Obs latitude: Φ = 28.76 • Precession const: θ = 0.014º , eclip/eq angle ε = 23.5º You’ll need the following: Precession: Convert RA/Dec to Alt-Az: Altitude, a: sin(a) = sin(δ) sin(Φ) + cos(δ) cos(Φ) cos(HA) Azimuth, A: cos(A) = [ sin(δ) - sin(Φ) sin(a) ] / cos(Φ) cos(a) 37 Coordinates of Solar System Objects All the objects considered so far have been "fixed stars", which keep almost constant values of Right Ascension and Declination. But bodies within the Solar System move a lot within the equatorial coordinate system. The most important one to consider is the Sun. The Sun's declination can be found by measuring its altitude when it's on the meridian (at midday). -Through the year, it varies between +23°26' and -23°26' – WHY? The Sun's Right Ascension can be found by measuring the Local Sidereal Time of meridian transit – see next lecture... The Sun's RA increases by approximately 4 minutes a day… The path apparently followed by the Sun is called the Ecliptic. 38 Parallax • We can use the motion of the earth around the sun as a tool for measuring distances to nearby stars. • Foreground stars appear to move against the background of distant stars and galaxies as the earth orbits the sun. Annual parallax θ – the maximum displacement of a star from its mean position; measured when the earth is on either side of the sun (six months apart) • A star with an annual parallax of 1 arcsec is at a distance of 1 parsec Distance, D = 1/θ E.g.Proxima Centauri, Annual parallax = 0.772 arcsec, D = 1/0.772 = 1.30 pc 39 Calendars & Julian Date Gregorian calendar, used universally for civil purposes Julian calendar, its predecessor in the western world The two calendars have identical month names and number of days in each month, and differ only in the rule for leap years. The Julian calendar has a leap year every fourth year, while the Gregorian calendar has a leap year every fourth year except century years not exactly divisible by 400. Change from Julian to Gregorian occurred gradually in 16th and 17th centuries. The Julian date (JD) is simply a continuous count of days and fractions of days since Greenwich noon on 1st January, 4713 BCE (on the Julian calendar). Almost 2.5 million days have transpired since this date – so it’ll be a big number!. Julian dates are widely used as time variables within astronomical software. Note, however, that the Julian Day Count has nothing to do with Julius Caesar, who introduced the Julian calendar. It is named after Julius Scaliger who invented the concept. Sometimes using the Modified Julian Date (MJD) is more convenient. This is simply the Julian Date, JD – 2400000.5. 40 Calculating Julian Date The Julian Day Number of any date on the Gregorian Calendar is calculated as follows: 1) Express the date as y m d, where y is the year, m is the month number (Jan = 1, Feb = 2, etc.), and d is the day in the month. 2) If the month is January or February, subtract 1 from the year to get a new y, and add 12 to the month to get a new m. (This is because we consider January and February as being the 13th and 14th month of the previous year). 3) Dropping the fractional part of all results of all calculations (except JD), let A = rounddown(y/100) B = 2 – A + rounddown(A/4) C = rounddown(365.25 * Y) D = rounddown(30.6001 * (m + 1)) JD= B + C + D + d + 1720994.5 e.g. JD on 2014 July 31 – A = rounddown(2014/100) = 20 B = 2 – 20 + roundown(20/4) = -13 C = rounddown(365.25 * 2014) = 735613 D = rounddown(30.6001 * (7+1)) = 244 JD = -13 + 735613 + 244 + 31 + 1720994.5 = 2456869.5000 (at 0:00 hrs) This is the Julian Day Number for the beginning of the date in question at 0 hours (Greenwich time). Note that this always gives you a half day extra because the Julian Day begins at noon – which is handy for astronomy... 41 Universal Time For a variety of reasons, astronomers often need to assign a “time” to an observation. For this they use Universal Time. Universal Time is the name by which Greenwich Mean Time (GMT) became known for scientific purposes in 1928. UT is based on the daily rotation of the Earth. However, the Earth’s rotation is somewhat irregular and can therefore no longer be used as a precise system of time. There are several versions of UT. The most import are: UT1 - conceptually it is the mean solar time at 0° longitude (Greenwich). In other words, the Sun transits at noon, UT1. It is derived from observations of distant quasars as they transit the Greenwich meridian. UTC – Coordinated Universal Time, is the time given by broadcast time signals since 1972, and is derived from atomic clocks. UTC is kept to within 1 second of UT1 by adding or deleting a leap second as needed! Remember: UT = GMT 42 Solar vs Sidereal Time Our 'common sense' notion of a day is based on the time it takes from one transit of the Sun (when the Sun is overhead, i.e. midday) to the next - this is a Solar Day. For astronomers it makes more sense to define this time in terms of distant stars. Pick a distant star (shouldn't matter which!) and measure the time between one transit and the next – this defines a Sidereal Day … but a sidereal day isn’t quite as long as a normal day!. A sidereal day is about 23 hours, 56 minutes, 4 seconds in length. Why? 43 4 mins/day = 2 hrs/month = 24 hrs/year Earth must rotate almost 1º more ( ≈ 360/365) to get the Sun to transit. Takes approx 4 mins to rotate through 1º Hence: • a Sidereal Day is 4 mins shorter than the (mean) Solar Day • the Local Sidereal Time (LST) gets 4 mins later at a given clock time every day. Things to remember: 1. LST is the Hour Angle of the Vernal Equinox, , (see next-but-one slide), so... 2. The RA of a star = its Hour Angle relative to . 3. At meridian transit of any star, LST = RA 4. LST tells us which RA is currently going through transit 5. LST - RA of an object = Hour Angle of the object 44 When is an object observable? For optical and infrared wavelengths we need to observe at night (not so for radio astronomers, who may have to stay awake 24 hours when observing!). On March 21st, the Sun is at the Vernal Equinox, i.e. on March 21st the RA of the Sun = 00h. Also, on March 21st at noon, LST is exactly 12 hrs ahead of local time At transit, RA = LST, and the Sun transits at midday, so…. At midday on March 21st, LST = 0 hrs At midnight on March 21st, LST = 12 hrs – this means that targets at RA = 12hrs are transiting Each month sidereal time moves 2 hours ahead of clock time (solar time) At midday on April 21st, LST = 2 hrs At midnight on April 21st, LST = 14 hrs … and so on, in an annual cycle. 45 When is an object observable? • At 00:00 hrs LST on March 21st, a person in Greenwich facing due south would be staring right at the meridian-transiting Sun… • On March 21st, as the earth rotates, the HA of  and the LST both increase together. • No matter what day it is, LST = always equals the Hour angle of the Vernal Equinox LST = 15 hrs Local time = 1 am   To Vernal Equinox,  LST = 12 hrs Local time = 10 pm LST = 0 hrs Local time = noon LST = 12 hrs Local time = midnight 46 When is an object observable? Example: The Hyades (open cluster) has RA ≈ 04h 30m, Dec ≈ +15º Ideally want to observe it on a night when LST = 04h 30m at midnight • LST = 12 hrs at midnight on March 21st (Sources with RA = 12 hr transiting…) • 04h 30m is 16.5 hours later than 12 hrs • LST moves on by 2 hours/month w.r.t solar time • 16.5 hours difference = 8.25 months • 8.25 months after March 21st is …  Late November is  the best time to  observe the Hyades. When would be the best time to observe the Hyades if we had access to the Telescope tonight? 47 What is the approximate Local Sidereal Time? LST depends on Time and Location… Its 22:05 PDT on 1st June 2015 at the Mount Laguna Observatory, near San Diego, California. What is the LST and thus the RA of transiting sources? KEY: on March 21st, RA~12hrs transits at ~midnight (local time) 1st June is 2-and-a-bit months later, so add 2 hours per month: RA ~ 16 hrs transits at midnight But we’re observing about 2 hours earlier, so RAs that are 2 hrs less transit RA ~ 14 hrs transits at about 10pm at the end of May We are therefore looking for an LST of about 14 hrs In other words… Targets with RA ~ 14 hrs will be passing through the meridian at approx. 10 pm on 1st June. 48 Calculating Local Sidereal Time To the nearest hour (good enough for a small telescope, maybe) Convert local time at the Observatory to UT/GMT UT = tloc + Δt Where tloc is the local time in decimal hours, and Δt is the time difference between local and GMT/UT. Calculate the approximate LST at the Observatory LST ~ (UT - 12) + Δd * (4/60) – λ * (4/60) LST = GMST – λ * (4/60) GMST is the Greenwich Mean Sidereal Time In the above eqn ’12’ corrects for the 12 hr time difference between LST and UT on 21 March. UT is the Universal Time, Δd is the number of days AFTER the Vernal Equinox (noon on 21 March, when LST = 0 hrs), and λ is the longitude WEST, in decimal degrees. The factors 4/60 convert both Δd and λ to decimal hours. Your answer will therefore be in decimal hours. Q. What is the approximate LST (to within a few minutes) at the Armagh Observatory, λ = 6.6500º W, at 19.00 BST on 28 March? 49 Calculating Local Sidereal Time more precisely The precise formula for calculating LST must take into account Nutation and Precession (see e.g. the Astronomical Almanac published by the US and UK Nautical Almanac Offices: aa.usno.navy.mil/faq/docs/GAST.php). 1. Convert UT Date and Time to a precise Julian date • JD - see slides at start of this lecture … 2. Calculate the precise number of days, D, since 1 January, 2000 at 12 hrs UT. • D = (JD – 2451545) 3. Calculate GMST • GMST = 18.697374558 + (24.06570982 * D) (this number will probably be large; reduce it to within 24hrs by subtracting some multiple of 24) 4. Correct for Longitude of observatory (add if E of Greenwich, subtract if W) • LST = GMST +/- λ * (4/60) 50 Calculating Alt-Az from RA, Dec, and Sidereal Time So how do I point my telescope at M3 ? Need to know: • RA (α) and Dec (δ) • Latitude of the observatory, Ф • Local Sidereal Time, LST Remember: to calculate Alt and Az, you ONLY need HA, δ , and Φ. Need to remember: • HA is the time since the target transited • LST is equivalent to the RA that is transiting • Therefore: HA = LST - RA Example: 1. Target is M3: RA: 13h 42m 11.6s Dec: +28° 22′ 38.2″ (assume current epoch) 2. LST is 14hrs 03 min on Mount Laguna (latitude, Ф = 32.8400º) 3. Now, calculate (i) HA from the RA and LST and (ii) the Altitude and Azimuth 51 Other Things Which Affect Sky Positions - 1 (Things you need to know before writing telescope control software!) 1. Nutation In addition to Precession (see last week’s notes) we have Nutation. This is a 9 arcsec wobble of the polar axis along the precession path - caused by the Moon’s gravitational pull on the oblate Earth. Main period = 18.66 years. R = Rotation of earth P = Precession N = Nutation 52 Other Things Which Affect Sky Positions - 2 (Things you need to know before writing telescope control software!) 2. Refraction Refraction in the Earth's atmosphere displaces a star's apparent position towards the zenith. R ≈ tan z where R is in arcminutes and z, the zenith distance, is in degrees. (Note that this equation is only really accurate for z << 90º, since tan90 = ∞ ) 53 How does refraction affect the sun’s appearance at sunrise/sunset? Due to refraction, the Sun appears to set 2 minutes AFTER it actually does set! To work this out you need a more precise empirical formula: R = cot ( 90-z + 7.31/[90-z+4.4] ) At z = 90º: R = cot (7.31/4.4) = 34.4 arcmin. R ≈ 0.5 deg. If it takes 6 hrs for the sun to move from zenith to the horizon, i.e. through 90 deg, it takes 6 hrs x 0.5/90 = 0.033 hrs = 2 minutes to move 0.5 deg. 54 Other Things Which Affect Sky Positions - 3 (Things you need to know before writing telescope control software!) 3. Height above sea level Observer's height above sea level means that the observed horizon is lower on the celestial sphere, so the star's apparent elevation increases. Measured angle of elevation, θ’, above the observed horizon = θ + a where displacement, α, in arcmins is given by: α = 1.78 √h (h = height above sea level, in metres) Q. Which is perpendicular to the radius of the Earth, the Celestial or the Observed Horizon? 55 Other Things Which Affect Sky Positions - 4 (Things you need to know before writing telescope control software!) 4. Stellar Aberration Caused by velocity of the Earth around the Sun ( ≈ 30 km/s). Need to point the telescope slightly ahead in the direction of motion. The amount depends on the time of year and the direction of the star. Maximum effect ≈ 20 arcsec LEFT: The angle at which the rain appears to be falling depends on the speed of the falling rain and the speed at which the person is running: sin θ = vman / vrain. RIGHT: For a star near the ecliptic pole, or for a star in the plane of the ecliptic and at right angles to the direction of motion of the Earth around the sun: sin θ = vearth / c vearth = 30 km/s and the speed of light, c = 300,000 km/s. Therefore, θ = 0.0057 deg = 20 arcsec 56 Angular Separations and converging lines of RA Stars 1 & 2: RA: 10h and 12h Dec: 0º Stars 3 & 4: RA: 10h and 12h Dec: +60º Even though stars 1 & 2 are 2 hrs apart in RA, and stars 3 & 4 are also 2 hrs apart in RA, the angular separation of stars 1 & 2 is NOT the same as for stars 3 & 4 because lines of right ascension converge towards the poles! 3 1 4 2 57 Angular Separations and converging lines of RA Stars 1 & 2: RA: 10h and 12h Dec: 0º Stars 3 & 4: RA: 10h and 12h Dec: +60º 3 4 Angular separation of stars 1 & 2 in degrees: 2 hrs = 360º x 2/24 hrs x cos δ = 360º x 2/24 hrs x cos 0 = 30º Stars 3 & 4: 2 hrs = 360º x 2/24 hrs x cos δ = 360º x 2/24 hrs x cos 60 = 15º 1 2 58 Small Angular Separations It is often important to know the angular separation, θ, of 2 objects on the sky (e.g. binary star components). This is especially useful if you want to know whether you can see two or more objects in one telescope pointing (NB need to know field-of-view of the telescope), or for calculating the proper motion of an object. For two objects, A and B, with coordinates (RA and Dec) αA , δA and αB , δB Δδ = δA - δB Δα = ( αA – αB ) cos δmean Where δmean is the mean declination of both objects, in degrees (only valid if θ < 1º). For Δα, the cos δmean term is required because lines of equal RA converge towards the poles. Hence, angular separation, q , is given by θ = √ (Δα2 + Δδ2) 59 Small Angular Separations (an example) Δδ = 40´´ δmean = +20º16´45´´ = +20º16.67´ = +20.28º Δα = 3.6 seconds of time A 20:17:00 Star A: 18h 29m 49.6s +20º17´05´´ Star B: 18h 29m 46.0s +20º16´25´´   B Key: 1 sec of time = 15´´ . cos δmean Angular separation, θ , is given by: 20:16:00 Therefore: 3.6 sec of time = 3.6 × 15´´ × cos20.28º Δα = 51´´  18:29:50 48 46 θ = √ (Δα2 + Δδ2) = √ (40*40 + 51*51) = 65´´ 60 Large Angular Separations To calculate the angular separation, θ, of 2 objects with a large separation (θ > 1º) or in the general case, the following formula can be used: 61 Tangential or “Proper” Motions Some nearby stars have significant ‘real’ motion in the plane of the sky (i.e. the star really is moving relative to the Earth). These motions may be up to a few arc-seconds per year and must be taken into account when using catalogue positions. There are separate catalogues of High Proper Motion objects for this purpose. For example, the brightest star in the sky, Sirius, has the following catalogue position and proper motion, μ: Correct the RA and Dec coordinates separately So in order to observe such an object in the summer of 2015, say, one must (a) precess the catalogue coordinates to that epoch, (b) correct for 15.5 years of proper motion, and then (c) convert to Altitude and Azimuth! 62 Asteroids and Comets Asteroids and comets can have very high proper motions (arcseconds per second!) Ephemeris – a table of coordinates over a range of dates 2 July EXAMPLE Near-Earth Asteroid NEO 2012-DA On 1 July at 12.00 UT its coordinates are: 18h 29m 49.6s +20º17´05´´  It's proper motions are μa cos δ = 1.0 arcsec/minute μd = 2.0 arcsec/minute 1 July  Q. What are its coordinates on 2 July at 12.00 UT? 63 Luminosity and Flux Luminosity (L) - the total Power from a source of radiation (in units of energy/second=Watts) emitted in all directions over all wavelengths. Flux - the luminosity, or Energy per second, emitted per unit area of the source (f), or detected per unit area by the observer (F), over all wavelengths (Wm-2) Total luminosity, L, given by: L = 4 π R★2 f where f = surface flux and R★ = stellar radius. At a distance D from the source, if the measured Flux (power received per unit detector area) = F then: L = 4 π D2 F therefore: D2 F = R★2 f or F / f = R★2 / D2 This is the Inverse Square Law for radiation Remember! Surface Area of a Sphere = 4 π R2 ; if R doubles, surface area quadruples… 64 The Magnitude System The Greek astronomer Hipparchos is usually credited with the origin of the magnitude scale. He assigned the brightest stars he could see with his eye a magnitude of 1 and the faintest a magnitude of 6. However, in terms of the amount of energy received, a sixth magnitude star is not 6 times fainter than a first magnitude star; its is approx 100 times fainter, due to the eye's non-linear response to light. This led the English astronomer Norman Pogson to formalize the magnitude system in 1856. He proposed that a sixth magnitude star should be precisely 100 times fainter than a first magnitude star, so that each magnitude corresponds to a change in brightness of 1001/5 = 2.512 The bottom line: Magnitude is proportional to the log10 of Flux. Remember: The GREATER the magnitude, the FAINTER the object! 65 The Magnitude System Pogson's equation is: F1 / F2 = 2.512 -(m1 - m2) Betelgeuse m = 0.4 Bellatrix m = 1.6 Where F1 and F2 are the fluxes of two stars, and m1 and m2 are their corresponding magnitudes. The minus sign is in there because magnitudes are smaller for brighter stars. If we take the logarithm of both sides we find Saiph m = 2.1 Rigel m = 0.1 log10(F1 / F2) = - ( m1-m2 ) * log10(2.512) Log10(2.512) = log10( 1001/5) = 0.4. We rearrange to get the magnitude equation m1 - m2 = -2.5 log10( F1 / F2 ) 66 The Magnitude System Relative magnitudes are given by: m1 - m2 = -2.5 log10( F1 / F2 ) Betelgeuse m = 0.4 i.e. difference in magnitude between two stars is given by the ratio of fluxes. Bellatrix m = 1.6 If one star has a magnitude of zero, then the above equation defines the Apparent magnitude, m, e.g.: m = -2.5 log10( F / F0 ) Here F0 is the flux from the zeroth magnitude star, Vega, the “primary” standard. This equation can also be re-written: m = -2.5 log10F + Z where Z is the zero-point (described later). In other words, if we can measure the Flux of Vega, and the Flux of another star, we can calculate the apparent mag of that star Saiph m = 2.1 Rigel m = 0.1 Q. If star A is 100x brighter than star B, what’s the magnitude difference? Q. How much brighter is Rigel than Bellatrix? 67 The Magnitude System Apparent magnitude, m, is given by (see previous slide): m = -2.5 log10( F / F0 ) - where F0 is the flux of the “zeroth mag” star, Vega m1 - m2 = -2.5 log10( F1 / F2 ) Absolute magnitude, M, is the apparent magnitude a star would have if it was at a distance of 10 parsecs (pc). It is a measure of the intrinsic brightness of a star Remember, F is proportional to 1 / d2 so: m - M = -2.5 log10(102 / d2) = -2.5 log10(d-2) - 2.5 log10(102) - where d is in parsecs Therefore the Distance Modulus: m - M = 5 log d - 5 For example: • if the distance modulus, m - M = 0, d = 10 pc • if the distance modulus, m - M = 5, d = 100 pc • if the distance modulus, m - M = 10, d = 1000 pc etc… Q1. What is the Absolute magnitude, M, of • Sirius, the brightest star in the sky: m = -1.5 mag, d=2.6 pc • The Sun, which is actually the brightest star in the sky: m = -26.7 mag, d = 5.10-6 pc Q2. How much brighter would Sirius be if both stars were at a distance of 10 pc? 68 Optical imaging through Filters Stars have different magnitudes at different wavelengths, i.e. when viewed through different filters/in different wavelength bands, or “wavebands” Top-left: RATCam on the Liverpool Telescope Above: RATCam’s filter wheel Bottom-left: the CCD detector mounted in RATCam 69 Filter sets U B V R I Z J H K L M N Q 3600 Å 4300 Å 5500 Å 6500 Å 8200 Å 9000 Å 1.25 μm 1.65 μm 2.20 μm 3.7 μm 4.7 μm 10.5 μm 20.9 μm Left – the standard optical filter profiles (Johnson system). Below – IR filters plotted against atmospheric transmission in the near-IR. Two narrow-band filters are also plotted (in green). NB. 1 Angstrom (Å) = 10-10 m; 9000 Å = 0.9 10-6 m = 0.9 mm The wavelengths listed above correspond to the centre of the filter’s transmission. Filter bandwidths are typically 20% (i.e δλ ~ 0.2λ) in the optical; 10% in the IR. The near- and mid-IR bands roughly correspond to atmospheric “windows”. H was defined after J and K – its not true that IR astronomers struggle with their alphabet! 70 Johnson-Morgan-Cousins vs Sloan The most widely used Photometric System (i.e set of filter definitions) is the Johnson-Morgan-Cousin UBVRI system, which is an extension of the Ultraviolet-Blue-Visible system defined by Johnson and Morgan in the 1950s to include redder filters, Red and Infrared, by Cousins a decade later. The central wavelengths and bandwidths have been modified slightly over the years – most notably by Bessel in the 1990s to better match the performance of CCDs. Bessel optical filters are used today in many telescopes – including the LT The magnitude of an object, when seen through a given filter, is referred to as mB, mV, mR, etc., or simply by B, V, R. For example, Bellatrix has apparent magnitudes U = 0.54 mag B = 1.42 mag V = 1.64 mag K = 2.38 mag Bellatrix is a red star; its brighter at longer wavelengths 71 Johnson-Morgan-Cousins vs Sloan However, in 1996… The Sloan Digital Sky Survey (SDSS) introduced a new set of optical filters, which are referred to as u’ g’ r’ i’ z’. These filters have broader bandwidths than the J-M-C set, they have higher transmission (they let through more light – so fainter objects like distant galaxies can be seen), and their bandwidths don’t overlap in wavelength. They are ideal for measuring the red-shifts of galaxies, for example – see right. Sloan optical filters are now used in many other telescopes – including the LT The SDSS map of Galaxies out to redshift z=0.15 between -1.5º < δ < 1.5º. Each dot is a galaxy containing perhaps 100 billion stars… 72 The Magnitude System The Apparent magnitude, m - specific to the waveband through which it is observed. For example: Betelgeuse has U = 4.3 mag, B = 2.7, V = 0.42 mag, J = -3.0 mag, K = -4.4 mag Betelgeuse is a very red star! The higher the magnitude, the fainter the star. Vega has U = 0 mag, B = 0 mag, V = 0 mag, J = 0 mag, etc.. Debris disk around Vega (HST image) 73 The Magnitude System Apparent magnitude: m = - 2.5 log10( F / F0 ) where F0 is the flux of Vega. If you know the Flux of Vega, F0 , in each filter If you measure the Flux of a star on your CCD, through the same filters ….. You can work out the apparent magnitude of that star. Q1. Calculate the Apparent U,B and V mags, mU, mB, mV of Rigel Q2. Calculate its Absolute Magnitudes, MU, MB, MV (assume a distance, d = 250 pc) Rigel (the bright blue star in Orion): • FU,Rigel = 3.47*10-9 W m-2 , FU,Vega = 2.09*10-9 W m-2 • FB,Rigel = 4.58*10-9 W m-2 , FB,Vega = 4.98*10-9 W m-2 • FV,Rigel = 4.30*10-9 W m-2 , FV,Vega = 4.80*10-9 W m-2 In ancient Egypt, Rigel’s name was… Seba-en-Sah, which means Foot Star or maybe Toe Star! 74 Flux vs. Flux Density The Flux is specific to the waveband, but also the photometric system, used. For example, the Johnson-Morgan-Cousin’s U-band flux of a star is slightly different from the Sloan u’-band flux. This is because the filters have a different central wavelength and band-pass (width), so they let a different amount of light through. Therefore, its often better to quote the Flux Density, i.e. the Flux per unit wavelength, Fλ (in W m-2 nm-1 or W m-2 Angstrom-1) or Flux per unit frequency, Fν (in W m-2 Hz-1). You can approximate the Flux Density by simply dividing the flux by the “width” of the filter, in nm, Angstrom, or even Hz. Rigel E.g. Rigel (the bright blue star in Orion): - FB = 4.58*10-9 W m-2 B-band filter: Δλ = 72 nm, Δν = 1.17*1014 Hz Therefore: - Fλ,B = 6.36*10 -11 W m-2 nm-1 - Fλ,B = 3.91*10 -23 W m-2 Hz-1 Flux density in the centre of the B-band www.astropixels.com 75 Spectral Energy Distribution Plot Flux Density against Wavelength to get a Spectral Energy Distribution (or simply ‘spectrum’) for a star. The curve is called a Blackbody spectrum and is defined by the Planck function; the spectrum peaks at a different wavelength depending on the temperature of the star. Cool stars are brighter in the IR (λ > 1 μm) than in the optical (λ < 1 μm). F The sun has a surface temperature of 5,800 K; it’s a yellow star. Rigel (blue) : 11,000 K Betelgeuse (red) : 3,500 K 76 An aside: Rigel vs. the Lightbulb (!?) You may have noticed (from the last few slides) that even bright stars seem to produce very little Flux. For example, Rigel’s B-band flux is FB,Rigel = 4.6x10-9 W m-2. How does this compare to a 60W light-bulb? 1. First, 60W is the power consumed by the bulb! An incandescent light bulb is perhaps 10% efficient, so the total flux radiated (across all wavelengths) may be only 6W! 2. This energy is radiated over a broad spectral (wavelength) range. In fact, burning at 3,000 K (i.e. like Betelgeuse), most of an incandescent bulb’s energy is radiated in the IR! Only about 10% is radiated in the optical (between 300 nm and 800 nm), and only ~20% of this optical wavelength range is covered by our B-band filter. The Luminosity of our bulb in the B-band is therefore: LB,bulb = 6 x 0.1 x 0.2 = 0.12 W 3. Finally, if we assume our bulb is 1 km away (so that our bulb looks like a star). B filter L = 4 π D2 F … so … FB,bulb = 0.12 / (4 π 10002 ) = 9.5x10-9 W m-2 Our bulb is now almost as faint as Rigel, when viewed through a B-band filter at a distance of ~1 km 77 Colour and Colour Index The colour of an object is defined in terms of the ratio of fluxes in different wavebands. This corresponds to a difference in magnitudes in two different bands, e.g. mB - mV = (B - V ), where (B - V ) is referred to as a the `colour index'. [ Remember: m1 - m2 = -2.5 log10( F1 / F2 ) ] Any colour index can be constructed; (U - R), (V - I), (J - K) etc. For example: Betelgeuse Betelgeuse: B – V = 2.7 – 0.42 = +2.28 (optical) J – K = -3.0 – (-4.4) = +1.4 (near-IR) Rigel: B – V = 0.09 – 0.12 = -0.03 J – K = 0.206 – 0.213 = -0.07 Note: colours can be negative, or even zero, like Vega! Rigel 78 Colour and Colour Index (B - V ) is the most frequently used optical colour index, and is a measure of the effective temperature, Teff , of a star. (The effective temperature is the temperature of a blackbody that would emit the same amount of radiation – typically this is similar to the temperature near the surface of a star; the temperature at the core is usually much higher!) For example: For Vega: mB = mV = 0.0 (by definition) hence (B - V)vega = 0.0 - Teff ≈ 9,900 K F B - V = -2.5 log10( FB / FV ) [email protected]μm [email protected]μm Sun For the Sun: mB = -26.14, mV = -26.78 hence (B - V)sun = 0.64 - Teff ≈ 5,700 K For Betelgeuse: mB = 2.70, mV = 0.42 hence (B – V)bet = 2.28 - Teff ≈ 3,600 K Betelgeuse 79 Empirical Formula for Colour Index The observed relationship between (B - V) and Teff for Main Sequence stars (stars in the main/stable part of their evolution, like the sun) is given by: Note: For Main Sequence stars (like the Sun) each Teff corresponds to a single value of (B - V) . We can’t use B or V (or any other magnitude) alone to measure Teff. However, the stars colour does give you its temperature. 80 Teff and Bolometric Luminosity (the Herzsprung-Russell Diagram) Teff is related to the total (or bolometric) luminosity, L = 4πR*2 σ Teff4 Where σ is the Stephan-Boltzman constant L is the intrinsic or absolute (not apparent!) brightness of the star: it represents the total outflow of radiation per second from the star (at all wavelengths). The Temperature and Luminosity can be plotted on a Hertzsprung-Russell diagram. Most stars fall on the Main Sequence, a tight correlation between Teff and L. 81 Bolometric Magnitude (and Bolometric Corrections) We can determine the total (or bolometric) magnitude (apparent or absolute) via a bolometric correction. We define: mbol = mV - BC and hence Mbol = MV - BC BC is ~zero for a star with a Teff = 5700 K, i.e. like our sun. Where the bolometric correction, BC, is a function of (B - V) or, equivalently, Teff. (We assume there is no extinction, see next section.) Note that very hot and very cool objects have large BCs, because their spectral energy distribution peaks a long way from the V band (at 5500 Angstrom).… 82 Bolometric Magnitude (and Bolometric Corrections) The Sun has MV = 4.82 and BC = 0.07, Hence: Mbol = 4.82 – 0.07 = 4.75 (absolute bolometric magnitude). Most of the sun’s energy is radiated in the optical. Our eyes have evolved to be sensitive over the same wavelength range. The sun doesn’t radiate much light in the ultraviolet or infrared, so your eyes aren’t sensitive in these parts of the electromagnetic spectrum. 83 Bolometric Magnitude (and Bolometric Corrections) What about using other colour indices, e.g. (U - B)? The relationship between (U - B) and Teff is not monotonic, which suggests that the spectral energy distribution (or spectrum, for short) deviates from a simple black body in the U-band or between U and B. (This is the Balmer discontinuity at 364 nm, which is due to the ionisation of hydrogen out of the n = 2 level: this affects A0 - F0 stars in particular). A single value of U-B can be equivalent to multiple values of B-V ! 84 Affects of interstellar dust Light from distant stars is both absorbed and scattered by interstellar dust (absorbed light is re-emitted in the far-IR; scattered light is actually absorbed and re-emitted in a different direction!). Both affects cause extinction, and mean that objects appear fainter than they should. For small dust grains: scattering cross-section, sscat propto λ-4 absorption cross-section, sabs propto λ-1 Tiny interstellar dust particles (image: A Davis, U Chicago) The overall effect is that the observed extinction cross-section is: sext propto λ-a , where a in the range 1-2. Red light (longer wavelengths) is less extinguished than Blue light, so objects appear redder than they should when viewed through dust - hence the term reddening (see e.g. the example to the right). 85 Extinction and Reddening AV = absorption in the V (visible) band, in magnitudes – the visual extinction Absorption in other bands is different because of the dependency on wavelength, e.g. AU = 1.53 AV - absorption in the UV is more than in the visible AB = 1.32 AV - absorption in the BLUE is (a bit) more than in the visible AK = 0.11 AV - absorption in the IR is much less than in the visible! The variation in extinction with wavelength means that, in the IR, we can often see through clouds of dust that are opaque in the optical! B68 - optical B68 - IR 86 Colour “excess” Extinction changes the colour of a star! The Colour Excess, E(B - V), also referred to as the reddening, is the additional (B - V) colour caused by this wavelength-dependent extinction, so: E(B - V) = AB - AV = 1.32AV - AV = (1.32 - 1) AV E(B - V) = 0.32 AV AV = 3.1 E(B - V) In other words, the greater the extinction, AV (the more dust between you and the star), the greater the affect on the B-V colour! Nice blue star… mB = 7.1 mag mV = 7.6 mag Nice blue star with a cloud in front! AV of cloud = 3.0 mag (B-V) colour = 7.1 – 7.6 = -0.5 mag E(B-V) = 0.32 AV = 0.96 mag (B-V) is now = -0.5 + 0.96 = 0.46 mag a negative number; star is brighter in the blue (B) band! A positive number; star is brighter in V and is looking a bit reddish… 87 Extinction and Reddening From effective temperature, Teff , can calculate the intrinsic colour, (B - V)0 , compare this to the observed colour, and thus calculate the colour excess. Remember from a few slides back? E( B - V ) = ( B – V )observed - ( B – V )0 From E( B - V ) we can calculate extinction, AV and correct the V-band magnitude. For example: G2V star has Teff = 5520 K; mB = 15.3 (observed); mV = 14.1 (observed). • • • • • Its intrinsic (B-V) colour (based on Teff ) Its observed colour is Colour excess, E( B - V ) = 1.2 – 0.68 Visual extinction, AV = 3.1 * E( B - V ) Extinction-corrected V-band mag of the star → → → → → (B – V)0 = 0.68 mag (B – V)observed = 1.2 mag E(B – V) = 0.52 mag AV = 1.61 mag mV = 12.5 mag Finally, when using the Distance Modulus equation it is important to account for extinction, because it affects the apparent magnitude of the star. The modified equation becomes: (mV,observed - AV ) - MV = 5 log d - 5 88 Atmospheric Absorption (and Airmass corrections) At night the sky ain’t blue – however, it still absorbs and scatters star-light! The altitude of your target therefore affects how much light gets through to the telescope at a given wavelength… The distance traveled through the atmosphere is given by: m0 mz d ≈ h/cos z = h sec z ‘sec z’ is known as the airmass. - at an airmass of 1: - at an airmass of 2: z=0º z= 60º Its preferable not to observe at airmass > 2. Attenuation is proportional to sec z, and: mz = m0 - C sec z C is a constant which depends on λ (but changes from site to site, and time to time). Before magnitudes can be properly used in data analysis etc, they must all be 'corrected' to the same airmass - generally, an airmass of 1.0, which corresponds to an observation made directly overhead. 89 Calibrating Stellar Magnitudes Pogson’s Ratio gives: m1 – m2 = -2.5 log10 ( f1 / f2 ) where f1 and f2 are measured fluxes. Note that ( f1 / f2 ) is dimensionless – from the ratio of integrated fluxes of two stars, star 1 and star 2, one could easily calculate the difference in their magnitudes, m1 – m2 . Typically one would measure the total, integrated counts in an aperture on a CCD image minus the counts in the background sky (which won’t be complete dark, i.e. devoid of counts). 90 Calibrating Stellar Magnitudes If one of the two stars was Vega, then: m1 - mvega = -2.5 log10 ( f1 / fvega ) = 2.5 log10 ( fvega ) - 2.5 log10 ( f1 ) But mvega = 0 in all photometric bands (B,V,R,I, etc.) by definition, so: m1 = 2.5 log10 ( fvega ) - 2.5 log10 ( f1 ) The quantity “2.5 log10 ( fvega )” is known as the Zero-point, Z . Similarly, “ -2.5 log10 ( f1 )” is the instrumental magnitude, minst, of Star 1. Vega Star 1 91 Calibrating Stellar Magnitudes So, our equation relates the true absolute magnitude of the star to its instrumental magnitude and the Zero-point, m1 = Z + minst provided that the instrumental magnitude and Z were measured with the same instrument, and in the same manner. Generally speaking, for calibration purposes, remember that: m = Z - 2.5 log10 f This means that, in theory, if we measure the flux, fstd , of a star with a known magnitude, mstd , we can calculate Z. Once we know Z, then if we measure the flux of a second star, f★ , we can calculate its magnitude, m★ . In other words: fstd and mstd give you Z Z and f★ give you m★ fstd and f★ must be in the same units (usually integrated counts per second) 92 Standard Star Catalogues There are a number of standard star catalogues in use by professional astronomers. For example: A portion of the Landolt Catalogue of optical standard stars presented on the European Southern Observatory’s (ESO’s) website …etc. Note that the table above gives only the V-band magnitude. Magnitudes in other filters can be calculated from their quoted colours. 93 Zero-point, Z, versus Airmass (Altitude) Now for the bad news: Even on a completely clear night Z changes with Airmass. Remember – Airmass is a measure of how low an object is in the sky: AM = 1 means the target is at an Altitude of 90º (i.e. directly overhead); AM = 2 means its at an Altitude of 30º, or at a zenith angle, z, of 60º. The flux from a star gets fainter as the airmass increases. Therefore, Z decreases as the airmass increases (because m = Z - 2.5 log10 f ) 94 How does Zero-point vary with Airmass We derive Z at AM=1 by observing a standard star at different times in the night, when the star is at different altitudes (different AMs). We then plot Z against AM. If we fit a straight line through the data, then the slope of the line gives the atmospheric extinction in magnitudes per airmass, while the intercept gives Z at AM=0, i.e. above the atmosphere (in space!). From the equation of the straight line we can thus predict Z at AM=1. Remember: mstd = Z - 2.5 log10 ( fstd ) Z where mstd is the catalogue magnitude of the standard star, and fstd is the flux (in integrated counts per second) that is measured from each observation (at each AM). Airmass = sec z (where z is the zenith angle) 95 Calculating a stars magnitude from Z (and a measure of atmospheric extinction) Once we know Z at AM=1, and know how atmospheric extinction varies with AM, we can calculate the apparent magnitude of any other star observed with the same system (telescope, instrument, filter) under the same observing conditions. For Example: • On a particular night, through a particular telescope and instrument, using a particular filter, Z is determined at AM=1 from the “ZP plot” (see previous slide). Assume Z1 = 11.35. • The atmospheric extinction, the slope of the ZP plot (see previous slide) allows us to correct our observations for airmass. Assume atmos. extinction = -0.36 mag/airmass (note: –ve !!) • If a star is observed at AM = 1.28 with an integrated flux, f = 3599 counts/sec, then its apparent magnitude is: m★ = Z1 – 2.5 log f★ + airmass correction = 11.35 – 2.5 log(3599) + (-0.36 x 0.28) = 2.36 f = 215,950 counts in 60 sec, or 3599 counts/sec 96 More on Zero-Points… • Must establish the zero-point for each filter AND for each night. • However, Zero-points calculated for AM=1 are often made available at professional telescopes, so that observers can estimate how much signal they are likely to detect from their targets. They may need to know if a certain observation is feasible! • Z at AM=1 should be the same from night-to-night, provided observing conditions (and the instrument itself) don’t change. Zero-points for IR filters in WFCAM at UKIRT, quoted for AM=1 and “photometric” conditions. • A star with the same magnitude as the zero-point would produce only one count per second in an image • Can you see this from the ZP equation (and Pogson’s relationship)? m★ - Z = - 2.5 log10 (f★) • You’d want to observe stars that are either much brighter than the zero-point, or observe 97 [integrate] for a lot longer than a second to gather more counts/photons! Zero-points and bad weather! Clear Sometimes, telescopes will monitor Z so that they know how the weather is changing Cloudy On the CLEAR night, the zero-point, Zclear ≈ 22.8 mag On the CLOUDY night, sometimes Zcloudy ≈ 21.0 mag How does the cloud affect the flux reaching the telescope? We know that: mstd = Z – 2.5 log (fstd) But mstd is fixed! Therefore: Zclear – 2.5 log ( fstd,clear ) = Zcloudy – 2.5 log ( fstd,cloudy ) 98 Noise Why can't a detector detect any source, however faint? The reason is - Noise. All measured signals are accompanied by noise: random variations about the mean. Main sources of noise in most astronomical images is photon noise: • Photon noise - photons are produced and arrive at the detector at random. The probability of arrival is given by the Poisson Distribution, which describes all random events. For MANY random events, this becomes the Gaussian or Normal Distribution. Gaussian distributions about the expected value, μ, with a variance, σ. 99 Noise Both thermal and photon noise have distributions with: standard deviation  √N Where N is the number of events (photons, thermal electrons, etc.), and √N is the uncertainty in N. e.g. A person catching kittens in a bucket! If 1,000,000 kittens land in the bucket, that person would in theory count 1,000,000 ± 1,000 kittens. 999,397… 999,398! Note that these statistical arguments only apply if there is a very large number of kittens! 100 Noise 1. Adding 2 or more images – how does that affect Noise? Lets assume we measure a signal A, with uncertainty X, on a CCD detector. If we add together multiple measurements of signal A: - for 2 measurements added: signal = 2.A, but uncertainty = √2.X - for 3 measurements added: signal = 3.A, but uncertainty = √3.X … etc. When N measurements are added: signal = N.A, uncertainty = √N.X 2. Taking a longer exposure – how does that affect Noise? Integrating the signal over time is equivalent to ADDING N measurements. signal is proportional to t and noise is proportional to √t Hence, the Signal-to-Noise ratio, SNR is proportional to t / √t or SNR is proportional to √t For example, if you integrate for 4-times as long, the SNR should increase by a factor of only 2. Alternatively, if a source is 4-times brighter (1.5 mag brighter), the SNR should be double. 101 Noise 3. Integrating signal over more bandwidth - how does that affect Noise? Adding up signal over adjacent frequencies will also improve SNR. The signal is proportional to Δν or Δλ (i.e. filter bandwidth) The noise is proprtional to √(Δν) or √(Δλ) Therefore, the SNR is proportional to √(Δν) or √(Δλ) Narrow “H2” filter A broad-band filter is often 10x wider than a narrow-band filter; a star viewed through the former will have ≈ 3x the S/N than if observed through the latter (all else being equal)! Broad K filter The image on the left is a 60 second exposure taken through a narrow-band H2 filter, while the image on the right is a 6 second exposure taken through a broad, K-band filter ( ΔλK ≈ 10x ΔλH2 ). The SNR in the K image should be √10–times better, but the exposure time in the H2 image is 10-times longer, so this cancels out the affect of the wider filter bandwidth. 102 Other sources of Noise The signal from the star is accompanied by noise. However, there are other sources of noise which may contribute to the overall signal-to-noise ratio: Background moonlight, twilight, and even light from distant (unresolved) stars and galaxies. All of these mean that the background in an image is almost never completely devoid of light; the signal in blank areas of an image is rarely at zero! We can subtract the background signal but NOT the random noise that is associated with it. For example, if the total signal detected, T = S + B, then: T ± ΔT = (S ± ΔS) + (B ± ΔB) = S + B ± √(ΔS2 + ΔB2) If we then attempt to subtract the background B, we get: (T ± ΔT) - (B ± ΔB) = S + B ± √(ΔS2 + ΔB2) - (B ± ΔB) = S ± √(ΔS2 + 2ΔB2) We have made the (noise) situation worse! We cannot avoid this, so we need to minimise background signals (and the accompanying √N noise) as much as possible. 103 Other sources of Noise Dark Current. Signal associated with the thermal excitation of electrons from the conduction to the valence band in the CCD semiconductor material. Cooling the detector reduces this “false” signal. Read noise. When the signal on a CCD or IR detector is read out, there is an inherent uncertainty associated with this measurement. In low background situations (e.g. narrow-band imaging or high-resolution spectroscopy), readnoise can be the main source of noise in your data. Left: Readout electronics used with CCD detectors. Right: Noise in a blank image, largely produced by readout but also due to dark current. 104 Noise Even the “dark sky” is noisy! Note that the average number of counts along the white line – as displayed in the pop-up graph – is about 25. The root-mean-square (rms) noise level is about 5 counts ( = √25). Almost all of the noise in this image is statistical noise from the “bright” sky background! Q. Is a 9.5 mag star (observed at zenith) detectable in this image? Assume exposure time is 10 sec, the zero-point, Z = 11.35 Assume also that the flux is spread evenly over 4 pixels. Remember: apparent magnitude, m★ = Z – 2.5 log f★ 105 Background Noise So, object signal is accompanied by background (or sky) signal, and both sources of signal have associated noise. Simply subtracting the sky signal from an object signal does not improve the SNR – in fact, it makes it worse! Baffles Its therefore imperative that the sky signal is minimised as much as possible: a low sky signal means a lower sky noise contribution. • Baffles can be used to stop stray light from entering the system. • It’s a good idea to observe in the dark! • In the Infrared, all of the optics inside an instrument must be cooled – to reduce background IR signal which would otherwise be produced by the warm optics. Baffles on the WHT primary and secondary mirrors cut down unwanted stray background light entering the system 106 Calculating Signal-to-Noise Ratio (SNR) The basic equation for all SNR calculations for CCDs and similar array detectors is the Array Equation: where: Sobj = signal from the object (electrons, e-, per resolution element per second) Ssky = signal from sky or background (same units) D = thermal signal, also known as “dark current” (e- per second) N = number of pixels per resolution element R = rms read or readout noise (e- per pixel) When imaging, by “resolution element” we effectively mean the area covered by the star FIG 107 Calculating Signal-to-Noise Ratio (SNR) Object Signal, Sobj , is a function of several elements: Sobj = (Pλ x Δλ x A) x E where: Pλ is the flux density of photons at each wavelength received at the telescope per unit area per second (i.e. number of photons /m2/μm/sec, or /m2/Å/sec). Δλ is the filter band-width (in μm or Å ). A is the telescope collecting area (the area of the primary mirror, in m2). E is the system efficiency. This is a combination of the “quantum efficiency” of the detector, the reflectivity of the telescope mirrors, and the throughput of the overall camera/optics system. A  CCD P filter E 108 Calculating Signal-to-Noise Ratio (SNR) Sky Signal, Ssky , is the flux density from the sky background, which varies from site to site and from night to night. Obviously, most of the things that affect the object signal (E, A, Δλ - see the previous slide) also affect the sky signal. Sky brightness is usually quoted in magnitudes per square arc-second and is very different at different wavelengths, e.g. the night sky is relatively dark at optical wavelengths (obviously!) but it glows brightly in the infrared. In fact, in the near-IR the sky background is usually much brighter than any astronomical object we may be trying to measure. Infrared images, before and after subtraction of the bright sky background signal (data courtesy U. Oregon) Raw H-band Sky-subtracted H-band 109 SNR – Background Limited case From the Array Equation - if Sskyt >> N.R2, and treating dark current, D, as part of Ssky, then we can write: This is the sky limited or background limited case, where SNR is proportional to √t . What SNR do astronomers typically aim for? • SNR = 10 means that fluxes measured to an accuracy of 10% • SNR = 100 means that fluxes measured to an accuracy of 1% SNR = 3 is usually the accepted limit for a simple detection of a signal. This is often known as a `3-sigma' (3σ) detection, where σ is the standard deviation (assumes a Gaussian distribution). 110 SNR – How does it vary with time? In the background-limited regime: If in 2 seconds we measure: • Sobj = 125 counts • Ssky = 1000 counts SNR = (125 x √2) / √(125 + 1000) = 5.27 in 2 seconds Q. SNR of 5.3 is quite low – how long would it take to get SNR ~ 20 ? SNR is proportional to √t. Therefore: SNRt1 / SNRt2 = √(t1 / t2) or t1 = (SNRt1 / SNRt2)2 x t2 If it takes 2 seconds to get a SNR of 3.72, to get SNR = 20 it would take: Time = (20 / 5.27)2 x 2 = 28.8 sec 111 Spectroscopy We have seen that, by taking images in different filters, the variation in flux or (more correctly) flux density with frequency or wavelength can be plotted. We can, for example, use this information to figure out the colour, or effective temperature, of a star. But a spectrum constructed in this way doesn’t have very much detail in it. We may only get a flux measurement every 100 nm (1000 Å)! It is for this reason that the field of Spectroscopy has been developed! B V U Photometry R v I  v Z v Spectroscopy 112 Spectroscopy Sir Isaac Newton’s famous experiment (circa 1660), in which he split sunlight into the colours of the rainbow by passing it through a glass. He is widely recognised as the father of spectroscopy. Newton was the first to use the word “spectrum” to describe the rainbow of light produced in this experiment 113 Slit-less vs Long-slit Spectroscopy Spectroscopy provides wavelength information, but usually at the expense of some spatial information Image Spectra Slit-less spectroscopy Long Long-slit slit spectroscopy spectroscopy 114 Spectroscopy Astronomers use spectroscopy to: • Measure accurate wavelengths of emission and absorption lines (lines at discrete wavelengths are “red-shifted” or “blue-shifted” by the Doppler shift). • Measure the widths of those lines (to get velocities, or pressures). • Measure strengths (equivalent widths) of those lines (to estimate the density and/or temperature of the gas, or to get abundances of atoms/molecules). • Measure the spectral energy distribution of the continuum radiation (which, as we have seen, can lead to an estimate of the effective temperature). Star’s continuum emission (smooth slope) Absorption lines 115 Spectroscopy Emission lines: • Produced for example by warm gas surrounding a star. The atoms and molecules in this gas are thermally excited, which means that electrons are raised to higher energy states. These electrons fall back down to lower energies, emitting light at very specific wavelengths. Emission from oxygen atoms Emission from hydrogen atoms heated to thousands of degrees This is a red star – notice how the spectrum steadily rises at longer wavelengths 116 Spectroscopy Absorption lines: • Often produced by cooler gas surrounding a star or in the “line-of-sight” (i.e. between the star and the earth). The atoms in this gas “absorb” light from the star and re-radiate it at different wavelengths. The absorption causes dips at very specific wavelengths in the spectrum. The star’s continuum is much brighter this time! Relatively cold hydrogen atoms absorbing light from the star Broad absorption caused by O2 and H2O molecules in the earth’s atmosphere 117 Spectroscopy – Resolving lines If the dispersion of your spectrograph is large enough (in other words, if the light is dispersed, or spread out, sufficiently), the “shape” of an emission or absorption line can be “resolved”. An unresolved emission line. The line shape is more-or-less Gaussian. In this case the width is simply equal to the resolution of the spectrometer. A resolved emission line. The shape may or may not be “Gaussian”. It depends on what affects the “width” of the line! 118 Line broadening - 1 Emission and absorption lines are broadened by three main mechanisms Natural broadening is caused by the uncertainty in the energy levels between the two states in an electronic transition due to the Heisenberg Uncertainty Principle, ΔE = ( h/2π ) / Δt, where Δt is the finite life-time of the state. Hydrogen electronic states/transitions 119 Line broadening - 2 Emission and absorption lines are broadened by three main mechanisms • Pressure broadening can be caused by collisions between particles. Collisions will further reduce the lifetime of an excited state, therefore increasing the uncertainty in photon energy and hence the width of the line. The higher the pressure the more frequent and energetic the interactions, therefore the broader the line. 120 Line broadening - 3 Emission and absorption lines are broadened by three main mechanisms • Doppler broadening is caused by the motion of the emitting particles of a source with respect to the frame of rest of that source. These motions can include thermal motion, turbulent motion, pulsation, rotation, infall and outflow. Wavelengths are Doppler shifted according to Δλ / λ = vr /c, where vr is the radial component of the velocity vector. 121 Line broadening The observed width of a line is usually a combination of these three effects, added to (or “convolved” with) the resolution of the instrument. The resolution of the spectrometer will further broaden or “smooth” emission (or absorption) lines. Lines that are close together in wavelength may even start to “merge together”, or be blended. 3 … further smoothed by the resolution of the spectrometer 2 … Natural, Doppler, and Pressure broadening 1 … Un-broadened, “infinitely narrow”, emission lines (which you would never see!). 122 Complex Line widths The width of a spectral line is typically characterized by a measurement or estimation of the full width at half maximum (FWHM) intensity. This is often measured by fitting a Gaussian distribution to the observed line. Sometimes a complex line structure – resulting from bulk motions of the gas (e.g. in the ejecta from a supernova or a planetary nebula) – can not be approximated by a single Gaussian. In such instances lines can be separated using deconvolution techniques. FWHM A single Gaussian profile nicely fits the spectral line on the left: the FWHM of this line can be easily measured. However, two Gaussians are needed to fit the line on the right. Perhaps this emission line comes from two discrete regions, at different distances along the line-of-sight, and at different radial velocities (so one is Doppler shifted with respect to the other!) 123 Line Equivalent Widths Finally, sometimes one would like to measure the intensity of a line. This can be difficult if the target is observed through thin cloud, or at high airmass. Astronomers therefore use the equivalent width, W, of a line as a measure of its strength. W is equivalent to the integrated area under a spectral line divided by the continuum flux density, Fc . W can be measured for an emission line or an absorption line! Traditionally, positive numbers are used for absorption lines, and negative numbers for emission lines (this is because most stars exhibit absorption rather than emission lines). Note that W should not be affected by extinction caused by clouds or airmass > 1 – provided the target (and line) are still detected. 124 Wavelength Coverage No telescope yet conceived can detect photons with wavelengths covering the entire EM spectrum. For example the physics of detecting (or intercepting) very high energy photons (gamma, X-rays) is fundamentally different from the physics of detecting low energy radio waves, or even optical photons. This problem is somewhat amplified by the Earth's atmosphere which does a very good job of absorbing IR and high energy photons (although this has a good side from a biological/life on Earth point of view). 125 UV IR 126 Absorption of the Sun’s Spectrum by the Atmosphere 127 128 Why use telescopes? • Light collection power - Think about the flux intercepted by the eye (a few millimetres in diameter) and the largest ground based optical telescopes (some over 10m in effective diameter). • Integration - Integrations of many hours are possible using ground based telescopes, even days or weeks for space based ones. However, the eye has a sampling frequency of around 25Hz (although this itself is a vast area of research and not so straight forward). • Magnification and Resolving power - The eye is rarely seeing-limited; due to its small aperture the eye is almost exclusively diffraction-limited, giving very low resolution. 129 Reflectors vs Refractors The first astronomical telescopes, such as those designed by Galileo, used lenses and are therefore called refractors. Refractors - Pros: • Rugged - optical alignment not subject to much disturbance after initial setting up. • Inner glass surfaces sealed in, so rarely need cleaning. • Air current and temperature effects minimised due to enclosed tube. Refractors - Cons: • Lenses must be supported around the edge. • Larger lenses are very heavy, therefore making them very expensive. • Various forms of aberration. 130 Reflectors vs. Refractors Most modern astronomical telescopes use mirrors rather than lenses and thus are called reflectors. The largest single-piece optical telescope in the world is currently the Large Binocular Telescope : 2x 8.4 m mirrors. Segmented mirrors (see e.g. the Keck mirror below-left) are the future..? Reflectors - Pros: Above - LBT 8.4m mirror. Below – Keck 10m mirror • No chromatic aberration. • By using paraboloidal shapes, spherical aberration can be removed. • Easier to manufacture • Can be supported from behind Reflectors – Cons: • Open to the elements • Coma and astigmatism still a problem 131 Reflectors Most large telescopes are reflectors (though their instruments may still use lenses) Naysmith Focus Gemini 8 meter Telescope Instruments are usually mounted at the Cassegrain, Prime or Naysmith Focus 132 Magnification & Field-of-View Magnifying power = Focal ratio of objective lens (or mirror) Focal length of objective, fo Focal length of eyepiece, fe = Focal length of lens, fo Diameter of lens, D The focal ratio is indicative of the maximum field-of-view available through the telescope system (primary/secondary mirror); lower f-ratios give larger f-o-v. 133 Liverpool Telescope The f-ratio of the Liverpool Telescope primary mirror is f/3. Primary diameter, D = 2.0 m. Therefore, the mirror focuses light to a point 6 m above the primary. However…. the secondary mirror changes the overall f-ratio of the telescope! The effective f-ratio of the LT is f/10 Therefore, the focal length is ≈ 20 m That’s the distance from the primary mirror, up to the secondary, and back down to the Cassegrain where the instruments are mounted. 134 Magnification Unfortunately a small fuzzy blob when magnified just becomes a large fuzzy blob (remember seeing?)! Magnified objects are also fainter, since you spread the light out over more pixels on your detector. Magnification is not always the primary need when selecting a telescopes for astronomical research. 135 What defines the resolution of a telescope? There is a fundamental maximum to the resolution of any optical system (telescope, microscope, your eye!) which is due to diffraction. An optical system with the ability to produce images with an angular resolution (the angular separation between two objects, or the “width” of a star measured at half the intensity) that is as good as the instrument's theoretical limit is said to be diffraction limited. This limit is defined by the Rayleigh Criterion 136 The Rayleigh (or Diffraction) Limit The angular resolution of an optical system, θ, can be estimated using the Rayleigh criterion, which relates the size of the aperture, D (e.g. the diameter of the telescope primary mirror) and the wavelength of the observation, λ (550 nm, say, for the optical) to θ : sin θ = 1.22 λ / D. The Gemini Telescope has an 8 meter diameter mirror, and it operates in the optical (550 nm) and the infrared (5 mm) Image: Vik Dhillon At which wavelength would you expect to get better spatial resolution? The Hubble Space Telescope has only a 2.4 m aperture (primary mirror). How does the diffraction limit of HST compare to Gemini (in the optical)? 137 But Gemini’s resolution is hugely affected by the atmosphere (which isn’t an issue for HST!) The blurring or twinkling of stars by the atmosphere is referred to as Seeing. As light rays enter the denser atmosphere from the vacuum of space they are refracted. Variations in the density, due to temperature gradients or turbulence, cause the refraction to alter as the light passes through the air… Understanding atmospheric 138 physics can ultimately affect where you build your telescope! What else affects the seeing? • The seeing changes with wavelength and the altitude of the object being observed. • At the Liverpool Telescope we estimate the seeing by measuring the FWHM of stars in images. We then calculate the seeing at zenith in the V-band using the following empirical relationships: Seeing, θ is proportional to 1 / λ 0.45 Seeing, θ is proportional to 1 / cos(z) 0.5 Where z is the zenith distance For example: If θ = 1.0 arcsec at a zenith distance of 45º, what is it at zenith? θz / θ45 = sqrt (cos45 / cos0) θz = 0.84 arcsec 139 Aberrations Spherical aberration - A spherically shaped lens will produce a curved focal plane with blurred off-axis images. In the example below, the further the rays are from the optical axis, the closer to the lens they cross the axis when focused. 140 Aberrations Coma (left) and Astigmatism (right) - Off-axis images further distorted. Coma refers to the “comet-like” appearance of images suffering from Coma. Coma: Rays parallel with the optical axis are focused to a point, those not parallel with the axis are spread out… Astigmatism: Rays in perpendicular planes have different focii 141 Aberrations Chromatic aberration - Refractive index in a lens is a function of wavelength (see lab section of the course). Different foci for different wavelengths give rise to distorted colour “halos" around stars (nb. refractive index is wavelength-dependent). A long focal length (thin lenses) helps reduce this problem, as do compound lenses made of different materials. 142 Aberrations Instrument designers produce “spot diagrams" with ray-tracing software to determine how distorted images will be. The aim is to make the effects of distortion negligible compared to the intrinsic image quality (e.g. seeing). Spot diagrams for the Subaru/CISCO instrument, at different wavelengths and off-axis angles. Aberrations can to some degree be reduced by using coatings and multiple-lens systems. 143 Telescope Mountings - Equatorial There are two main types of telescope mounting: Equatorial and Altitude-Azimuth (Alt-Az). Equatorial telescopes: • Must be set-up/designed for a particular latitude, such that the declination axis points directly to the north (or south) pole. • Only needs to track stars through one axis. • Simple design, but heavy. • Object tracking is straight forward; the telescope only needs to track in RA (along one axis); Dec is fixed! Model of the UK Infrared Telescope. The red “tube” is an instrument sitting on the primary mirror! 144 Telescope Mountings – Alt-Az Two main types of telescope mounting: Equatorial and Altitude-Azimuth. Alt-Az telescopes: • Simple “up-down" and “revolving" mount. • Must track through both the Alt and Az axis, PLUS Cassegrain de-rotation. • Tracking can be tricky • Zenith “blind spot" due to near-infinite de-rotator speeds. • Simple engineering • Used on most modern large telescopes. Computer model of the proposed Thirty Meter Telescope (TMT) 145 The Eye The most important detector for hundreds of years was the Human Eye! There are 2 types of detectors (retinal receptor cells) in the eye, rods and cones Rods - detect monochrome only - they contain rhodopsin, a complex protein pigment in layers 20 nm thick. When a photon is absorbed, a piece of the protein breaks off leaving behind a colourless substance (opsin). An associated change in potential causes an electrical signal in the optic nerve. Peak sensitivity of rods occurs at wavelengths around 500 nm (for obvious reasons!). Rhodopsin regenerates slowly (see below). Cones - come in 3 types and are sensitive to 3 distinct wavelength ranges which the eye interprets as blue (narrow, with a peak near 419 nm), green (broader, with a peak near 531 nm) and red (also broad, with a peak near 558 nm, which is actually more like yellow). 146 Wavelength range and dark adaptation Eye sensitive to wavelengths between 315 nm (UV) and 1000 nm (near-IR) Wavelengths shorter than 315 nm (UV) are absorbed by the cornea (potentially causing injury) and do not reach the retina. Retinal sensitivity sometimes extends (with very low sensitivity) to 1000 to 1050 nm (into the near-IR). Note that if your eyes were sensitive to much longer wavelengths you would look through a sort of `infrared fog', since you would see heat energy everywhere. In the dark, rods dominate since their sensitivity is about 100 times greater than for cones - hence at night we rely on monochrome vision. In the dark, rhodopsin gradually reforms on a time-scale of 30 mins. Red light is much less likely to reduce dark adaption (hence red lights are often used in `dark' rooms, and red torch filters are used on observing trips!). 147 Naked Eye Astronomy The Rayleigh (diffraction) limit* of the fully open iris (D = 5-7mm) is ≈ 25´´ . But the receptor cell spacing is much coarser: 1´ - 2´ is therefore the max resolution you can expect from your eye. The eye's response to light intensity is logarithmic: this is the origin of the magnitude scale. The faintest stars which are naked eye visible on a moonless night have mV ≈ 6. Q. Can you resolve (distinguish) Maia from Alcyone with the naked eye? Maia (V = 3.9m) RA: 3h 45m 50.0s Dec: 24º 22´ 10´´ Alcyone (V = 2.9m) RA: 3h 47m 29.0s Dec: 24º 06´ 20´´ David Malin/AAT *The angular resolution, θ , can be estimated using the Rayleigh criterion, which relates the size of the aperture, D and wavelength of the observation, λ, to θ : sin θ = 1.22 λ / D. 148 Photographic Plates Photons cause chemical changes in photographic emulsions. The use of hypersensitizing can increase the efficiency, though only to perhaps 10% (i.e. only 1 in 10 photons detected!). • Limitations: Non-linear response. Scattering degrades resolution. • Advantages: Large area and relatively cheap. Scanning machines have been used extensively to digitize photographic plates, e.g. the Digitised Sky Survey Catalogue – still widely used by Astronomers! Photo courtesy ESO The DSS largely contains scanned photographic plates obtained at the Palomar Observatory (North) and the UK Schmidt Telescope (South) 149 Photographic Plates Left: Hubble at the Palomar Schmidt; Right: His 1926 plate of M33 with at least one Classical Nova marked While once widely (even exclusively) used by professional astronomers, now only a handful of amateurs still regularly use plates. However, vast archives of almost a century of plate observations still exist and are still frequently referred to. 150 CCDs (Charge-Coupled Devices) Photons incident on a semiconductor (usually doped silicon) produce electron-hole pairs (below-left). The electrons are trapped in potential wells produced by attached electrodes, and accumulate until read out by charge coupling the electrodes. By cycling the voltages on each electrode in sequence, we can shuffle the charge along to an output electrode (below-right). This process is known as charge coupling. The resulting analogue signal is amplified and digitised. Note: IR photons with λ > 1.1 μm don’t have enough energy to create a free electron! Above: a photon with sufficient energy excites an electron from the valence to the conduction band, thereby producing an electron-hole pair. Electrons collect in potential wells produced by the gate electrodes (G); charge is shifted to the right by shifting the applied voltage on each gate. An amplifier at the end of the row converts the charge to a voltage… 151 CCDs (Charge-Coupled Devices) CCDs are of course now commonly used in digital cameras. They were adopted for use by astronomers in the mid-1980s because of their much greater efficiency and sensitivity compared to photographic film. Whilst CCDs are commonly used in every-day items, “science grade” optical CCDs are still extremely expensive, and IR detectors (strictly-speaking, these are not CCDs, since they don’t shuffle charge) can be a factor of ten times more expensive again. (Left) The array of four 2k x 4k pixel detectors of the INT WFC (plus the finder chip). (Right) WFC3 being installed on-board the HST, the camera contains two 2k x 4k pixel detectors and a single 1k x 1k detector. Note the size of the camera compared to the detectors . 152 CCDs – Quantum Efficiency Efficiency of conversion of photons into electrons - the Quantum Efficiency - in the CCD material is around 75-90% (cf 2% for photographic film). However, QE does change with wavelength and temperature, and even depending on whether the “front or back” of the CCD is illuminated! LEFT: The QE response of the detectors used in the Andor cameras at the LT when cooled to -100ºC. Note how QE drops off sharply in the UV (photons reflected) and near-IR (photons pass through the device). Photons falling onto the “front” of the detector must traverse the gate electrodes (see two slides back); the gate electrodes can reflect or absorb photons, thereby reducing the QE. It may therefore be desirable to illuminate the back of the CCD. Standard optical CCDs are typically sensitive to radiation in the wavelength range λ = 400 - 1000 nm. 153 CCDs - Cryogenics Scientific grade CCDs (e.g. those used on telescopes) need to be cooled. Expansion of with liquid Nitrogen (T~70K) or even liquid Helium (T~ 4K) can be used. The CCD is mounted inside in a cryostat or dewar. Cooling reduces noise associated with “dark current” - signal associated with thermal excitation of electrons in the semiconductor material. The associated noise is proportional to √dark current (see next section). Closed-cycle means that the fluid (liquid or gas) that exchanges the heat is kept within a closed circuit. RIGHT: The IO:O cryostat – the gold cylinder - mounted on the bottom (the Cassegrain focus) of the Liverpool Telescope. So how does a CCC work? Remember those laws of Thermodynamics!? PV = nRT ?? 154 CCDs - Cryogenics This is the layout of the Closed-Cycle Cooling system being used at the LT to cool the detectors in IO and FRODOspec. CryoTiger Compressor RETURN (30 psi) Air flow Cold Head on the instrument SUPPLY Gas Compressor (300 psi) Gate valve to vacuum pump Heat Shield Cold Plate Getter Heat Exchanger Copper heat strap The Cryotiger pumps compressed PT13 liquid into the cold-head on the instrument, where it expands adiabatically, thereby extracting heat from the copper heat strap. “Warm” PT13 gas returns to the cryotiger, where it is passes through the compressor and heat exchanger, before again being pumped back through the cold-head. Note that everything inside the coldhead is kept in a vacuum, and the heat shield protects the cold internals from radiative heating from the outer jacket. 155 CCDs - Cryogenics “Amateur" astronomical CCD cameras - such as those used in your labs and during the field trip - use thermo-electric cooling, via the Peltier effect, to cool the detectors by typically 20 – 30º below the ambient temperature. The Peltier effect uses Thermo-electric cooling (conversion of temperature differences to voltage) to extract heat energy from the system. Usually a hot and cold plate are connected by doped semi-conductive thermo-couples. When a DC voltage is applied across the device (note the +ve and –ve terminals) the hot plate heats up and the cold plate cools down… The device effectively transfers heat from one side to the other. Advantages of this cooling device: lack of moving parts (reliable, low maintenance) and does not involve cryogenic liquids (like PT13 or N2). Note: in the Peltier effect, heat is transferred from one material to another as electrons flow to an environment with higher energy, E, but also higher entropy, S, and therefore lower free energy (F = E-ST) 156 CCDs – Gain & Bias GAIN: In an ideal world, the number of “counts" recorded in a particular pixel in the image from a CCD camera would be equal to the number of photons detected. However the electronics that lie between the CCD camera and the resultant image can slightly complicate matters. When designing the electronics (the Analogue-to-Digital converter, or ADC) one must decide upon the length of numbers used, typically 16-bit unsigned integers (giving a range from 0 → [216 – 1] or 0 → 65535). This number is unlikely to relate directly to the maximum number of electrons that can be stored in a given pixel. Hence, the gain allows one to convert between “counts" and photons/electrons. Gain = electrons per pixel / counts per pixel A gain of 8 means that each count is equivalent to 8 photoelectrons BIAS: The bias is a small voltage applied to raise the output signal above zero at all times. This ensures that the signal + random noise is always positive, since the Analogue-to-Digital conversion electronics cannot deal with negative analogue input. 157 CCDs – Cosmic Rays CCDs are sensitive to the passage of cosmic rays, which produce many free electrons in the semiconductor creating bright pixels or 'spikes', which have nothing to do with the signal we are trying to measure (unless you're a gamma-ray astronomer!). Cosmic rays are the bane of optical astronomers’ lives and have to be “removed” from an image before any meaningful science can be done. Before After IO:O images of NGC 7479, before and after cosmics have been removed 158