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### PHYS216 Practical AstrophysicsLecture 2 – Coordinate Systems 1

Module Leader:

Dr Matt Darnley

Course Lecturer:

Dr Chris Davis

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

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.

Alt-Az coordinates of a star are specific to the timeand the observer's location.

a

S

E

A

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 ‘em, 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.

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

(remember: HA increases to the west)

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!

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 (g). RA is analogous to longitude on Earth; lines of R.A. are great circles which pass through the poles, and whose planes pass through the centre of the celestial sphere (and Earth!).

g

Ecliptic

Equatorial

Lines of

equal RA

Origin of the Equatorial System

Right Ascension is referenced to a fixed point on the celestial sphere, called the First Point of Aries (FPoA), g

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.

- R.A. has units in hrs/mins/sec: 24 hr = 360o, so 1 hr is equivalent to 15o (same as HA).

Origin of the Equatorial System

The RA and Dec of the FPoA are 0,0.

What does this point in space look like!? Pretty boring, unfortunately… that’s because its not a specific object in the sky; it was defined, thousands of years ago, around the passage of the seasons (see later).

Because of precession, the FPoA is now in the constellation of Pisces (also explained later)!

Optical Image from the Digital Sky Survey

- The First Point of Aries is defined by the point on the Celestial sphere where the crosses the Equatorial plane crosses the Ecliptic plane (see next slide)

Equatorial vs EclipticPlanes

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

g

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.

First Point of Aries and the Vernal Equinox

- The FPoA represents one of the two points on the Celestial Sphere where the Ecliptic Planeand the Equatorial Plane cross one another.
- The First Point of Ariesisthe 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)

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
- Example coordinates:
- Supernova SN1987A:
- 05h 35m28.0s,-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 that is > 90o from the latitude of the observer can’t be reached

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

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.

Note:f is the latitude of the observer

(remember, angular height of Polaris above the horizon is roughly equal to f)

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)
- 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 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 –vedeclination)

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 these 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
- 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

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, of the observer.

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-Azcoords.

- 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 hrs)

Crab Nebula, observed by the LT with IO:O on, 30 Oct 2013.

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.

Precession

- Converting coordinates between two Eqinoxes, or updating to current Epoch:
- And:
- Where:
- aT and dT are the RA and Dec (both 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”

*** NOTE ***

Equinox defines a standardised coordinate system (1950, 2000)

Epoch can be any time (its usually when you want to observe)

Converting between Equinox 1950 and Equinox 2000

Convert Equinox 1950 coords to degrees:

a1950 = 1h 39m 10s1h 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”

Converting Equinox 2000 to the current Epoch

Convert Equinox 2000 coords to degrees:

a2000 = 1h 42m 20s1h 42.33m 1.706h25.583o

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 11s25.797o

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= 28.76
- 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(HA)

Azimuth, A: cos(A) = [ sin(δ) - sin(φ) sin(a) ] / cos(φ) cos(a)

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 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.

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 – 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/q

E.g.Proxima Centauri,

Annual parallax = 0.772 arcsec,

D = 1/0.772 = 1.30 pc

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