PHYS216 Practical Astrophysics Lecture 2 – Coordinate Systems 1 *** TAKES JUST OVER AN HOUR – PROBABLY DON’T NEED A BREAK. Module Leader: Dr Matt Darnley Course Lecturer: Dr Chris Davis

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

Altitude-Azimuth Mounts Jodrell Bank (radio) & Liverpool telescope (optical)

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

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.

Equatorial Coordinates - 1 HA and Declination – NOT a fixed coordinate system 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!

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 (in hrs mins secs), 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

Origin of the Equatorial System Right Ascension reference - the First Point of Aries (FPoA), g The zero-point of RA - the great circle 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).

Origin of the Equatorial System First Point of Aries: RA: 0h 0m 0s Dec: 0o 0’ 0. Not a specific object in space. 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 Equatorial plane crosses the Ecliptic plane (see next slide)

Equatorial vs Ecliptic Planes 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.

First Point of Aries and the Vernal Equinox When the Sun reaches the First Point of Aries, as it does once each year, an equinox occurs. In the northern hemisphere, this is the Vernal Equinox, before which the north pole is tipped away from the Sun, and days are shorter than nights. After the equinox, the north pole is angled toward the Sun, so that days start to become increasingly longer than nights, and moving the northern hemisphere into spring and summer. For the southern hemisphere, these effects are reversed: the Autumnal Equinox occurs, and days start to become shorter than nights. The First Point of Aries is not a fixed point in space: it moves along the Ecliptic at a rate of roughly one degree every seventy years. When the Equinox was first observed, thousands of years ago, the First Point actually lay in the constellation of Aries. Due to the effect of precession, the First Point of Aries crossed into the neighbouring constellation of Pisces in about 70 BCE. It has taken about 2,000 years to cross Pisces, and it will cross into the next zodiacal constellation, Aquarius, in about the year 2600. Following its journey along the Ecliptic, it will return to Aries once again in about 23,000 years. Because the First Point of Aries is the zero-point for calculating coordinates on the Celestial Sphere, its own coordinates are always fixed, regardless of its motion. They are, of course, zero hours Right Ascension and zero degrees Declination. FPoA: one of the two points on the Celestial Sphere where the Ecliptic Plane and 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!

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.

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

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: 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” N E ------ W S Targets with an angular distance > 90o from the latitude of the observer can’t be reached. Which of the above three targets CAN be observed from La Palma – Latitude 28.7o N)?

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

Converting to Alt-Az Coordinates 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 HA N 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!

Converting to Alt-Az Coordinates Given: the latitude of the observer, f, 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. By the sine rule: sin(360°-A)/sin(90°-δ) = sin(HA)/sin(90°-a) sin(A)/cos(δ) = sin(HA)/cos(a) then: sin(A) = - sin(HA) cos(δ) / cos(a) … This gives us the azimuth A (IF d > 0). HA

Converting to Alt-Az Coordinates Given: the latitude of the observer, f, 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)

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)

Converting to Alt-Az Coordinates (an example…) Planetary Nebula, M76, equatorial coordinates: Convert to degrees. d = 51o 19’ 30”  51o 19.5’  51.325o We will observe at 3am when the target has transitted: 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) a = 56.624o = 56o 37’ 29” 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 = 321o 46’ 50” Observatory in La Palma, latitude f = 28.7624 (NB: a –ve angle, or very large +ve angle, means target is West of North)

Making sense of the numbers… M76, equatorial coordinates: a = 1h 39m 10s = 51o 19’ 30” HA ~ 2.2 hr 56o When we want to observe it: Hour Angle, HA = 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 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

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 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, f. THIS IS NOT TRUE FOR THE OTHER STARS!

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

An example to try for yourself! Convert the Equatorial Coordinates of the Crab Nebula, M1, to Alt-Az coords. Coordinates of the Target: a = 1h 39m 10s , d = 51o 19’ 30” Coordinates of the Observatory: 19.8207o N, 155.4681o W HA: -2.0 hrs (or +22.0 hrs) Dec = 51.325 deg ; HA = -30.0 deg ; latitude = 19.8207 deg sin(a) = sin(51.325).sin(19.8207) + cos(51.325).cos(19.8207).cos(-30) a = 50.7000 deg = 50o 42’ 0” HA > 12hrs (its 22hrs!) , therefore cos(A) = [ sin(51.325) – sin(19.8207).sin(50.7) ] / cos(19.8207).cos(50.7000) A = +29.5544 deg = 29o 33’ 29” (+ve, still rising) Crab Nebula, observed by the LT with IO:O on 30 Oct 2013.

Precession The Earth's rotation axis precesses slowly, 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.

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 dE are 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)

A worked example, M76, Equinox 1950

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”

A worked example, M76, Equinox 2000

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”

Converting current Epoch to Alt-Az Convert Epoch 2013.5 declination to decimal 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 = 28.760o (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(51.645). sin(28.760) + cos(51.645).cos(28.760).cos(-15) Altitude, a = 64.522o 12 < HA 24 hrs: cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) ] = [ sin(51.645) – sin(28.760).sin(64.522) ] / cos (28.760).cos(64.522) Azimuth, A = 21.923o

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 = 28.760 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(HA) Azimuth, A: cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a) RA: 5.8744 hrs = 88.116 deg Dec: 7.3994 deg Precess to 2014.5 RA(2014.5) = 88.116 + [ 0.014 . 64.5 (cos23.5 + sin23.5.sin88.116.tan7.3994) ] = 88.116 + [ 0.014 . 64.5 (0.9171 + 0.05176) ] = 88.991 deg (5h 55min 58.0sec) Dec(2014.5) = 7.3994 + [ 0.014 . 64.5 (0.9171 + 0.05176) ] = 7.3994 + 0.01184 = 7.4112 (7deg 24 arcmin 41 arcsec Convert to Alt-Az: Sin(a) = sin(7.4112).sin(28.76) + cos(7.4112).cos(28.76).cos(0) a = 68.650 deg cos(A) = [ sin(7.4112) – sin(28.76) sin(68.65) ] / cos(28.76) cos(68.65) A = 0 deg

Coordinates of Solar System Objects But bodies within the Solar System move a lot within the equatorial coordinate system. 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.

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 Annual parallax – the maximum displacement of a star from its mean position A star with an annual parallax of 1 arcsec is at a distance of 1 parsec Distance, D = 1/q E.g.Proxima Centauri, Annual parallax = 0.772 arcsec, D = 1/0.772 = 1.30 pc

The Gaia Space Telescope