PHYS216 Practical Astrophysics Lecture 2 – Coordinate Systems 1

1. PHYS216 Practical AstrophysicsLecture 2 – Coordinate Systems 1 Module Leader: Dr Matt Darnley Course Lecturer: Dr Chris Davis

2. Simplest design of Telescope • It moves up and down, and rotates from left to right… Altitude Azimuth

3. Altitude-Azimuth MountsJodrell Bank (radio) & Liverpool telescope (optical)

4. Altitude-Azimuth system Observer based coordinates: Zenith – directly overhead. Altitude, h (or a) - angle between the observer's horizon and the object. Zenith distance, z- angular distance between zenith and object, z = 90o - h Azimuth, A - angle along the horizon, Eastward from North. Alt-Az coordinates are specific to the timeand the observer's location. h S E A

5. The Celestial Sphere Need a fixed coordinate system so we can catalogue the positions of the stars... • From last time… • Celestial Equator- Projection of the Earth's equator onto the Celestial Sphere. • Celestial Poles – North and South pole! • Hour Angle - angle between star's current position and the meridian (measured W, in hrs) • Declination - angular distance above (+ve) or below (-ve) the Celestial Equator.

6. Equatorial Coordinates - 1 HA and Declination – NOT a fixed coordinate system Declination - analogous to latitude on the Earth. Hour Angle – time before or since a star has transited. (remember: HA increases to the west) Together HA and Dec are useful for determining whether an object is currently observable. But… Hour Angle is time-dependent – its varies continuously!

7. Equatorial Coordinates - 2 HA and Right Ascension (RA) IS a fixed coordinate system Declination – as before, angular distance above/below celestial equator. Right Ascension - is measured like HA, but… - is referenced to a fixed point on the celestial sphere - the First Point of Aries (g) - is analogous to longitude on Earth; lines of R.A. are great circles through the poles g Ecliptic Equatorial Lines of equal RA

8. Zero-point of the Equatorial System Right Ascension reference - the First Point of Aries (FPoA), g The zero-point of RA - the great circle (line of longitude) which passes through the FPoA and the poles. The RA and Dec of the FPoA are 0,0. • R.A. units - hrs/mins/sec. 24 hr = 360o, so 1 hr = 15o (same as HA).

9. Zero-point of the Equatorial System First Point of Aries: RA: 0h 0m 0s Dec: 0o 0’ 0. This point in space looks pretty boring … because its not a specific object. It was defined, thousands of years ago, based on the passage of the seasons. Because of precession, the FPoA is now in the constellation of Pisces (also explained later)! Optical Image from the Digital Sky Survey • First Point of Aries: • Point where the Ecliptic plane crosses the Equatorial plane (see next slide)

10. Equatorial vs EclipticPlanes Equatorial plane – Projection of the Earth's equator onto the celestial sphere. Ecliptic plane – Apparent path of the Sun on the celestial sphere; 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 g Ecliptic coordinates are sometimes used for Solar System objects.

11. First Point of Aries and the Vernal Equinox • FPoA: one of the two points on the Celestial Sphere where the Ecliptic Planeand the Equatorial Plane cross one another. • FPoA: point in space beyond the sun on the Vernal Equinox, March 21st / 22nd each year. Note that Ver is Latin for Spring!

12. 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.

13. On the EQUATOR, the sun is directly overhead twice a year, at noon on the two equinoxes (March and September)….

14. RA and Dec(Equatorial coords) Positions: 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 • Example coordinates: • Orion Nebula, M42: • 05h 35m 17.0s , +5o 23’ 30” • Barred spiral galaxy, NGC 55: • 00h 14m 53.6s , -39o 11’ 48” • Ring Nebula, M57: 18h 53m 35.1s , +33o 01’ 45” N E ------ W S Which target can’t be observed from the Liverpool Telescope in La Palma (Lat 28.7o N)

15. Converting to Alt-Az Coordinates • To point at Alt-Az telescope must first convert from equatorial coordinates (RA and Dec) to the Alt-Az system. • Altitude tells you: • Is a target observable? • How much “air mass” must you observe through? Altitude

16. Converting to Alt-Az Coordinates Convert Equatorial coordinates (HA and Dec) to Horizon coordinates (Alt-Az) Use a spherical triangle XPZ, where Z is Zenith, P is North Celestial Pole, X is position of the star. Note:f is the latitude of the observer 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. meridian N The angles of the triangle: The angle at P is H, 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!

17. Converting to Alt-Az Coordinates • Given: the latitude of the observer, f, the hour angle, H 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(H) • which simplifies to: • sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(H) • This gives us the altitude a. • By the sine rule: • sin(360°-A)/sin(90°-δ) = sin(H)/sin(90°-a) • which simplifies to: • sin(A)/cos(δ) = sin(H)/cos(a) • then: • sin(A) = - sin(H) cos(δ) / cos(a) • This gives us the azimuth A.

18. Converting to Alt-Az Coordinates Given: the latitude of the observer, f, the hour angle, H 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(H) which simplifies to: sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(H) 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(A) Rearrange to find A: cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) This again gives us the azimuth A.

19. Converting to Alt-Az Coordinates(an example…) • Planetary Nebula, M76, equatorial coordinates: • Convert to degrees. a = 1h 39m 10s1h 39.17m 1.653h (÷24 and ×360)  24.795o d = 51o 19’ 30”  51o 19.5’ 51.325o

20. Converting to Alt-Az Coordinates(an example…) • Planetary Nebula, M76, equatorial coordinates: • Convert to degrees. a = 1h 39m 10s1h 39.17m 1.653h (÷24 and ×360)  24.795o d = 51o 19’ 30”  51o 19.5’ 51.325o • We will observe at 3am when the target has transitted: Hour angle, H = 2.20 hrs • To convert hour angle to degrees - multiply by 15 (nb. 24 hrs is equivalent to 360o). • Therefore: H = 33.00o

21. Converting to Alt-Az Coordinates(an example…) • Planetary Nebula, M76, equatorial coordinates: • Convert to degrees. a = 1h 39m 10s1h 39.17m 1.653h (÷24 and ×360)  24.795o d = 51o 19’ 30”  51o 19.5’ 51.325o • We will observe at 3am when the target has transitted: Hour angle, H = 2.20 hrs • To convert hour angle to degrees - multiply by 15 (nb. 24 hrs is equivalent to 360o). • Therefore: H = 33.00o Altitude, a: sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(H) a = 56.624o Azimuth, A: sin(A) = - sin(H) cos(δ) / cos(a) A = -38.219o or 321.781o Observatory in La Palma, latitude f= 28.7624 (NB: a –ve angle, or very large +ve angle, means target is West of North)

22. Converting to Alt-Az Coordinates(an example…) Remember this cartoon?? • M76, equatorial coordinates: a = 1h 39m 10s • = 51o 19’ 30” 56o • When we want to observe it: • Hour Angle, H = 2.2 hrs • Altitude, a = 56.624o • Azimuth, A = -38.219o or 321.781o • 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. 38o Observatory in La Palma, f= 28.7624

23. Converting to Alt-Az Coordinates(The Pole Star, a special case…) The “pole star” Polaris has a declination of d ≈ +90o. Since sin 90=1 and cos 90 = 0, its altitude above the horizon is given by: Altitude, a: sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(H) sin(a) = sin(φ) a = φ • Hence the • Altitude - a • of Polaris depends only on the • Latitude – f • of the observer (because it is a fixed point in space). • THIS IS NOT TRUE FOR THE OTHER STARS!

24. Polaris – not one star, but two, err, three… And not quite at the north ecliptic pole, either… Declination: 89o 15’

25. An example to try for yourself! Convert the Equatorial Coordinates of the Crab Nebula, M1, to Alt-Azcoords. • a = 1h 39m 10s , d = 51o 19’ 30” • Observatory Latitude: 19.8207o N, 155.4681o W • HA: -2.0 hours Crab Nebula, observed by the LT with IO:O on 30 Oct 2013.

27. Precession The Earth's rotation axis precessesslowly, with a period of 25,600 years. 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 position of the Vernal Equinox moves at approx 50” per year. Because of precession, we must define an appropriate equinox for the RA and Dec catalogue positions, e.g. 1950, 2000 etc.

28. Precession • Converting coordinates between two Eqinoxes, or updating to current Epoch: • And: • Where: • aT and dT are the RA and Dec (in degrees) of an object at time interval T (in years) after the catalogue equinox, E. • aE and dEare the catalogue coordinates, RA and Dec (for equinox E). • q (the precession constant) = 50.4” per year (multiply by 25,600 yrs to get 360o!) • e is the angle between the equatorial and ecliptic planes, precisely 23o 27’ 8” Equinox defines a standardised coordinate system (1950, 2000) Epoch can be any time (its usually when you want to observe)

29. A worked example, M76, Equinox 1950

30. Converting between Equinox 1950 and Equinox 2000 Convert Equinox 1950 coords to degrees: a1950 = 1h 39m 10s1h 39.17m 1.653h (÷24 and ×360) 24.795o d1950 = 51o 19’ 30”  51o 19.5’ 51.325o Precess 1950 degrees to 2000 degrees and convert back to RA and Dec

31. Converting between Equinox 1950 and Equinox 2000 Convert Equinox 1950 coords to degrees: a1950 = 1h 39m 10s1h 39.17m 1.653h (÷24 and ×360) 24.795o d1950 = 51o 19’ 30”  51o 19.5’ 51.325o Precess 1950 degrees to 2000 degrees and convert back to RA and Dec a2000 = 24.795o + [0.014o. 50yrs . (cos.23.5o + sin23.5o.sin24.795o.tan51.325o)] = 25.583o = 1.7055h ≈ 1h 42m 20s Remember: Precession constant is 50.4” = 0.014o/ year. Angle between ecliptic and equatorial planes Is 23.5o) d2000 = 51.325o + [0.014o. 50yrs . (sin23.5o.cos24.795o) = 51.578 ≈ 51o 34’ 40”

32. A worked example, M76, Equinox 2000

33. Converting Equinox 2000 to the current Epoch Convert Equinox 2000 coords to degrees: a2000 = 1h 42m 20s1h 42.33m 1.706h25.583o d2000 = 51o 34’ 40”  51o 34.67’ 51.577o Time between 2000 baseline (1 Jan) and date we want to observe = 13.5 years. Therefore, precess 2000 coords forward by this period: a2013.5 = 25.583o + [0.014o. 13.5yrs . (cos23.5o + sin23.5o.sin25.583o.tan51.577o)] = 25.797o = 1.720h ≈ 1h 43m 11s d2013.5 = 51.577o + [0.014o. 13.5yrs . (sin23.5o.cos25.583o) = 51.645o = 51o 38’ 42”

34. Converting current Epoch to Alt-Az Convert Epoch 2013.5 coords to degrees: a2013.5 = 1h 43m 11s25.797o d2013.5 = 51o 38’ 42” 51.645o Finally, convert Epoch 2013.5 coords to Alt-Az system for La Palma, φ = 28.76o (assume target transits at 1.00 am; I get access to the telescope at midnight, so hour angle, H = -1 hrs, or -15o) sin(a) = sin(δ) sin(φ) + cos(δ) cos(φ) cos(H) = sin(51.645). sin(28.76) + cos(51.645).cos(28.76).cos(-15) Altitude, a = 64.48o sin(A) = - sin(H) cos(δ) / cos(a) = - sin(-15).cos(51.645) / cos(64.48) Or… cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) ] = [ sin(51.645) – sin(28.76).sin(64.48) ] / cos (28.78).cos(64.48) Azimuth, A = 21.89o

35. One for you to try…. • Step 1:precess Equinox 1950 coords to Epoch 2014.5 • Step 2: convert these new Equatorial (RA-Dec) coords to Alt-Az coordinates (assume HA = 0, i.e. object is transiting!) Object: Betelgeuse • Coords: a1950 = 5h 52m 28s , d1950 = +7o 23’ 58” • Obs latitude:f= 28.76 • Precession const: q= 0.014o,eclip/eq angle e = 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(H) Azimuth, A: sin(A) = - sin(H) cos(δ) / cos(a)

36. Coordinates of Solar System Objects But bodies within the Solar System move a lot within the equatorial coordinate system. • The SUN: • The Sun's declinationcan 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 Ascensioncan 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.

37. End.. See you next week….