PHYS216 Practical Astrophysics Lecture 3 – Coordinate Systems 2 Takes just over an hour… (break may not be needed) Module Leader: Dr Matt Darnley Course Lecturer: Dr Chris Davis

Calendars & Julian Date Gregorian calendar - used universally for civil purposes Julian calendar - its predecessor in the western world The two 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. The Julian Date (JD) - a continuous count of days and fractions of days since Greenwich noon on 1st January, 4713 BCE. Modified Julian Date (MJD) – a more convenient form of JD. MJD = JD – 2400000.5. Swedish version of the Gregorian Calendar, c. 1712.

Calculating Julian Date The Julian Day Number 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 x Y) D = rounddown(30.6001 x (m + 1)) JD= B + C + D + d + 1720994.5 This is the Julian Day Number for the beginning of the date in question at 0:00 hours (Greenwich time). Note half day extra because the Julian Day begins at noon. e.g. JD on 2014 July 31 – A = rounddown(2014/100) = 20 B = 2 – 20 + roundown(20/4) = -13 C = rounddown(365.25 x 2014) = 735613 D = rounddown(30.6001 x (7+1)) = 244 JD = -13 + 735613 + 244 + 31 + 1720994.5 = 2456869.5000 (at 0:00 hrs)

Calculating Julian Date 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. 3) Dropping the fractional part of all results of all calculations (except JD), let A = integ(y/100) B = 2 – A + integ(A/4) C = integ(365.25 . y) D = integ(30.6001 . (m + 1)) JD= B + C + D + d + 1720994.5 JD is the Julian Day Number for the beginning of the date in question at 0 hours (Greenwich time). e.g. JD on 2014 July 31 - at 00:00 hrs JD = 2456869.5000 at 06:30 hrs JD = 2456869.7708 Correspond to seconds!

Universal Time Remember: UT (or UTC) = GMT 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. Versions of UT: UT1: The mean solar time at 0° longitude (Greenwich). The Sun transits at noon, UT1. Derived from observations of distant quasars as they transit the Greenwich meridian. UTC – Coordinated Universal Time: Time given by broadcast time signals since 1972 Derived from atomic clocks. UTC is kept to within 1 second of UT1 by adding or deleting a leap second Remember: UT (or UTC) = GMT

Solar vs Sidereal Time Solar Day: Time between successive transits of the sun (noon to noon) Sidereal Day: Time between successive transits of distance stars and galaxies … 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.

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: LST is the Hour Angle of the Vernal Equinox, g, (shown later) The RA of a star = its Hour Angle relative to g. At meridian transit of any star, LST = RA LST tells us which RA is currently going through transit LST - RA of an object = Hour Angle of the object

When is an object observable? On March 21st, the Sun is at the Vernal Equinox, i.e. on March 21st the RA of the Sun = 00h On March 21st at noon, LST is exactly 12 hrs ahead of the local time (synchronise watches!) 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 (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.

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… (would probably still need a spray tan) On March 21st, as the earth rotates, the HA of g 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 g LST = 12 hrs Local time = 10 pm g To Vernal Equinox, g LST = 0 hrs Local time = noon LST = 12 hrs Local time = midnite

When is an object observable? LST = 18 hrs Local time = 2 am LST = 15 hrs Local time = 11 pm LST = 12 hrs Local time = 8 pm g g To Vernal Equinox, g LST = 12 hrs Local time = midnite

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 (why 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.

What is the approximate Local Sidereal Time Example: Its 22:05 PDT on 1st June 2014 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 March 22-March 31 = 9 days April 1 – Apiril 30 = 30 days May 1 – May 31 = 31 day 1 June = 1 day 2 June = 4.75 hrs = 0.198 days Total = 71.198 days

How does this relate to GMT (UT)? Earth viewed from above… June 1st 2014 Time in California PDT = 10.05 pm LST ≈ 14 hrs UT (time in Greenwich) Is 05.05am LST ≈ 22 hrs March 22-March 31 = 9 days April 1 – Apiril 30 = 30 days May 1 – May 31 = 31 day 1 June = 1 day 2 June = 4.75 hrs = 0.198 days Total = 71.198 days RA ~ 18 hrs) June 21st (toward g March 21st (toward RA ~ 12 hrs)

Calculating Local Sidereal Time To the nearest hour (good enough for a small telescope) Convert local time at the Observatory to UT/GMT UT = tloc + Dt tloc is the local time in decimal hours; Dt is the time difference between local and GMT/UT. Calculate the LST at the Observatory LST ~ (UT - 12) + Dd . (4/60) – l . (4/60) LST = GMST – l . (4/60) Where ’12’ corrects for the 12 hr time difference between LST and UT on 21 March. Dd is the number of days AFTER the Vernal Equinox (noon on 21 March, when LST = 0 hrs) l is the longitude WEST, in decimal degrees. The factors 4/60 convert both Dd and l to decimal hours. Your answer will therefore be in decimal hours. Try this example: What is the LST at the Armagh Observatory, l = 6.6500o W, at 19.00 BST on 28 March? UT =19.00 – 1.0 = 18.00 Delta-d = 7 days 18 hrs = 7.75 days GMST = 6.5167 hrs LST = 6.5167 - 6.65x(4/60) = 6.0734 = 6 hrs 04 min

Calculating Local Sidereal Time more precisely The precise formula for calculating LST must take into account the Earth’s 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 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 +/- l . (4/60)

Calculating Local Sidereal Time An example: LST at 20.00 UT on 31st July, 2014 in Armagh (6.65o W) 1. Convert UT Date and time to a precise Julian date JD = 2456870.3333 2. Calculate the number of days, D, since 1 January, 2000 at 12 hrs UT. D = (2456870.3333 – 2451545) = 5325.3333 3. Calculate GMST GMST = 18.697374558 + (24.06570982 . 5325.3333) = 128176.6241 hrs (- 5340 . 24) = 16.6241 hrs 4. Correct for Longitude of observatory (add if E of Greenwich, subtract if W) LST = 16.6241 – 0.4433 = 16.1807 hrs, or 16:10:51 (Calculation done with Excel. You may get slightly different numbers depending on how your calculator handles these very big numbers; but you should get to within a minute of time.)

Messier objects at 14 hrs RA?

Messier objects at 14 hrs RA?

Calculating Alt-Az from RA, Dec, and Sidereal Time So how do I point my telescope at M3 ? Need to know: RA (a) and Dec (d) Latitude of the observatory, f Local Sidereal Time, LST Remember: to calculate Alt and Az, you ONLY need HA, d , and f. 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: Target is M3, RA: 13h 42m 11.6s Dec: +28° 22’ 38.2″ (current epoch) LST is 14 hrs 03 min on Mount Laguna (latitude, f = 32.8400o) Now, calculate (i) HA from the RA and LST and (ii) the Altitude and Azimuth HA = LST – RA : 14.050 hrs - 13.703 hr HA = 0.347 hrs (HA is +ve, target is setting; 0 < HA < 12 hrs) HA = 5.205 deg Dec: 28.37728 deg. HA = 5.2050 deg. , Latitude = 32.8400o Altitude, a = 83.6786 deg = 83o 40’ 43” Azimuth, A = 226.4629 deg = 226o 27’ 47”

Other Things Which Affect Sky Positions – 1 1. Nutation 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

Other Things Which Affect Sky Positions – 2 2. Refraction 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. (only accurate for z << 90o, because tan90 = ∞ )

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! 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.

Other Things Which Affect Sky Positions – 3 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, q ’, above the observed horizon = q + a where displacement, a, in arcmins is given by: a = 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?

Mauna Kea Observatory Big island, Hawaii The summit of Mauna Kea in Hawaii is 4200 m above sea-level, h = 4200 m; therefore, a = 115 arcmin – that’s almost 2 degrees!

Other Things Which Affect Sky Positions – 4 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 rain and the speed at which the person is running: sin q = 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 q = vearth / c vearth = 30 km/s and the speed of light, c = 300,000 km/s. Therefore, q = 0.0057 deg = 20 arcsec

Angular Separations and converging lines of RA Stars 1 & 2: RA: 10h and 12h Dec: 0o Stars 3 & 4: Dec: +60o Stars 1 & 2 are 2 hrs apart in RA Stars 3 & 4 are 2 hrs apart in RA But 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 ★4 ★2 ★1

Angular Separations and converging lines of RA Stars 1 & 2: RA: 10h and 12h Dec: 0o Stars 3 & 4: Dec: +60o Angular sep of Stars 1 & 2 in degrees: 2 hrs = 360o x 2/24 hrs x cos d = 360o x 2/24 hrs x cos 0 = 30o = 360o x 2/24 hrs x cos 60 = 15o ★3 ★4 ★2 ★1

Small Angular Separations How to calculate the angular separation, q, of 2 objects on the sky For two objects, A and B, with coordinates (RA and Dec) aA , dA and aB , dB Dd = dA – dB Da = (aA – aB) cos dmean Where dmean is mean declination of both objects, in degrees. Angular separation, q : q = √ (Da2 + Dd2) *** This is only valid if q < 1o ***

Small Angular Separations (an example) Star A: 18h 29m 49.6s +20o17’05” Star B: 18h 29m 46.0s +20o16’25” Dd = 40” dmean = +20o16’45” = +20o16.67’ = +20.28o Da = 3.6 seconds of time Key: 1 sec of time = 15”.cosdmean Therefore: 3.6 sec of time = 3.6 × 15” × cos20.28o Da = 51” Angular separation, q , is given by: q = √ (Da2 + Dd2) = √ (40×40 + 51×51) = 65 arcsec A 20:17:00 q Dd B Da 20:16:00 18:29:50 48 46

Large Angular Separations To calculate the angular separation, q , of 2 objects with a large separation (q > 1o) or in the general case, the following formula can be used:

Tangential or “Proper” motions Stars move with respect to “stationary” background galaxies. The brightest star in the sky, Sirius, has the following position and proper motion, m: The get the precise, current epoch coordinates of the star you need to: (a) precess the coordinates AND (b) correct for the star’s proper motion. Correct the RA and Dec coordinates separately NOTE: These are SPEEDS !!

Proper motions Typical proper motions of nearby stars ≈ 0.1 arcsec/year Star with highest proper motion is Barnard’s star; PM = 10.25 arcsec/year Barnard’s star Evolution of the Great Bear: The changing appearance of the Big Dipper (Ursa Major) between 100,000 BC and 100,000 AD.

Asteroids and Comets 18h 29m 49.6s +20o17’05” Asteroids and comets can have very high proper motions (arcseconds per second!) Ephemeris – a table of coordinates over a range of dates EXAMPLE Near-Earth Asteroid NEO 2012-DA On 1 July at 12.00 UT its coordinates are: 18h 29m 49.6s +20o17’05” Its Proper motions is ma cos d = 1.0 arcsec/minute md = 2.0 arcsec/minute Q. What are its coordinates on 2 July at 12.00 UT? 2 July Dd 1 July Da

And finally…. A bit of astrology! (Sorry, couldn’t resist) RA~3hr At mid-day on March 21st, the sun (when viewed from the Earth) is at RA = 0 hrs, midway between the constellations of Aquarius and Pisces (according to the IAU)… What’s the star sign of someone born on March 21st? And why is it “wrong”? RA~1hr RA~5hr E March 21st - star sign is Aries (March 21st – April 20th) RA~13hr RA~17hr RA~15hr