Transforming from Geographic to Celestial Coordinates Michael McEllin 1 Introduction The simplest astronomical observation of all is that the stars appear to move around the Earth (which, of course is the result of the Earth rotating with respect to the fixed 1 stars). We therefore catalogue the position of astronomical bodies against a Celestial coordinate system that is fixed with respect to the stars, but moves around with respect to Earth based coordinates. Imagine the Earth s equator projected on the sky: instead of latitude and longitude we use Right Ascension to measure angle around the equator and Declination to measure the angle above or below the celestial equator. While declination, often denoted by δ, is measured in degrees, right ascension, α or RA, is traditionally measured in hours, minutes and seconds, which was convenient for early sky mapping because astronomers simply had to note the time on the clock when a star passed through the meridian (the line on the sky going north to south). Four coordinate conversion purposes, of course, we need to deal in angles measured in degrees or radians and convert to the RA convention. Looking down from the North Pole, RA angles are measured anti-clockwise around the celestial equator. It is important to remember this conventional right-handed orientation of coordinates, because it is opposite to the convention used for compass bearings (going clockwise from north) and we will not get our signs right if we do not take this into account. 1 The stars do also have their own proper motions with respect to each other, and there is an apparent parallax motion due to the Earth orbiting the Sun, but these are small and only apparent with sophisticated measurements. One of the reasons we need a good Celestial coordinate system is so we can detect such motions. For this reason the current celestial coordinate system is tied to the positions of very distant quasars which given current technology really do not appear move. 1
From the Earth s surface it is, of course, easiest to log the position of something we see in the sky (or detect with other instruments) using measurements such as compass bearing (which astronomers call azimuth - the angle from the direction North) and altitude (the angle above the horizon) - see Figure 1. How do we convert our observations in alt-azimuth form to celestial coordinates? Or how do we convert known position of an object in the sky in celestial coordinates into instructions for identifying the position in alt-azimuth coordinates? What we need is a machine that takes two numbers (such as altitude and azimuth) and spits out Figure 1: Altitude-Azimuth Coordinates right ascension and declination. (Or conversely a machine that takes right ascension and declination and spits out altitude and azimuth.) This is a very common problem in astronomy and there are well known solutions. You can look them up. You can also find simple programs, library subroutines and spreadsheets that perform this function. (See, for example, the astropy Python library). This note is about explaining how it works, and providing formulae than for example could be used in a spreadsheet or python program. Once you know the trick, the geometry and basic trigonometry involved is not difficult, but the conventional way used to describe the algorithm usually involves representing the simulates equations as a matrix equation. This is a technique normally learned at A-level. Matrix methods are enormously important for anyone wishing to program numerical calculations because many problems (such as weather forecasting, for example) usually turn into very large systems of simultaneous equations, often with tens or hundreds of thousands of variables. Using the concepts of matrix algebra is the only way to keep track of all the manipulations that need to be performed. Our coordinate transformation problem is simple enough that it could be written without a matrix equation. Although it would be a bit harder to follow the algorithm it would be just about manageable. If you are up for it, see if you can follow the maths in the next section. It 2
is long rather than difficult, and we have to introduce the advanced maths concepts of vectors and matrices in order to keep track of the algebra (otherwise it just gets too difficult to see where you are going without drawing lots of complicated 3D geometry diagrams). The important result, however, is given at the end and you do not need to understand how it comes about to use it. 2 The Algorithm Our initial coordinates and final coordinates are in polar form (that is using only angles. Two numbers at the start and two numbers at the end. In spite of this, the easiest way to work out the conversion is to go through a couple of three dimensional coordinate systems on the way. We should not be surprised at this, in fact, because time is a hidden third coordinate in our polar coordinates, because the Earth is rotating and any conversion must include a time parameter. The plan will involve four stages: 1. Convert to a cartesian ( xyz ) coordinate system, with x pointing north, y pointing west, and z to the sky. (Y to the west gives us a right-handed coordinate system, which is the same as the celestial coordinate system.) 2. Rotate the axes of our coordinate system around the east-west direction so that the z axis is now parallel with the rotation axis of the Earth. 3. Rotate the axes around the new z direction (we will call this z ) until the x axis points towards the vernal equinox, which is the point on the sky where in its annual progress around the ecliptic plane the Sun crosses the celestial equator moving northwards (usually about 20th of March). 4. Convert back from cartesian to polar coordinates to get the celestial values. In practice, steps 2 and 3 will end up being combined into a single 3D rotation. This is where matrix algebra is most useful: it greatly reduces the possibility of making mistakes in the algebraic manipulations by helping us keep track of which lengths get multiplied by which trig function value. 3
2.1 Alt-Azimuth to XYZ Imagine that we point to the object on the sky in which we are interested with a meter rule. (Obviously this pointing direction is only valid at one instant - we will need to note the time - before the Earth s rotation moves the sky around. We will now need to work out the x,y,z coordinates of the far end of the meter rule against the x direction running from south to north, the y direction running west to east and the z direction running from the ground vertically upwards. If the Sun where directly overhead the meter rule would leave a shadow on the ground of length cos(a), where A is the altitude angle. The z coordinate - height of the end above the ground - would just be sin(a). Using simple trig. we can find the x-coordinate as cos(a)cos(z), where Z is the Azimuth angle, and the y coordinate as cos(a)sin(z). (Note the - sign when getting the Y value: this is because Azimuth angle comes round clockwise from North (i.e. through East first) but we need Y pointed to the West, 2.2 Coordinate Rotation Examine Figure 2. Here we have left the meter rule in the same position and rotated the rules marking the X and Z axes anticlockwise. (If we are in the northern hemisphere and the Y axis is going out of paper.) This is the right-handed rotation required to move the Z axis to be parallel with the Earths rotation axis. Assuming we are at a latitude, L, then the rotation angle would be θ = ( π 2 L). The thing to keep in mind is that the meter rule is a vector - a real object in space which continues to exist however we choose to lay out other meter rules (at right angles to each other) in order to define a rectangular coordinate system. We can move these rules that mark the axes around as we wish as long as they stay orthogonal, but the projections of the original meter rule on each of them to get coordinates will change as we move them around. It can be seen from the diagram and simple trigonometry that the new coordinate values, x, y and z are given by x = x.cos(θ) + z.sin(θ) (1) y = y (2) z = x.sin(θ) + z.cos(θ) (3) 4
Figure 2: Altitude-Azimuth Coordinates This is better expressed as a matrix equation: x cos(θ) 0 sin(θ) y = 0 1 0 z sin(θ) 0 cos(θ) x y z (4) This means exactly the same thing as the earlier set of equations, using the matrix multiplication rule, which says that we take the vector on the left and turn it horizontal, i.e: V = [x, y, z]. We then multiple the elements of this vector by the corresponding column values in each row to get back to the equations (1) to (3). The mathematician now thinks of this rotation matrix as a machine into which we put one vector of coordinates [x,y,z] and get out another coordinate vector V = [x, y, z ]. We can write this in matrix shorthand as V = R y (θ).v (5) 5
Furthermore, when we want to know how to write down the equations for rotating around the z axis by an angle φ (anti-clockwise looking down the z axis) we can immediately use the same pattern of trig. terms without a lot of detailed thought: x y z = cos(φ) sin(φ) 0 sin(φ) cos(φ) 0 0 0 1 We can write this in matrix shorthand as x y z (6) V = R z (φ).v (7) where the value φ, of course, will depend on the time at which we make the observation, since it has to take account of the Earth action orientation in space relative to the celestial coordinate system. The rules of matrix multiplication immediately now tell us that the complete rotation around both axes can be thought of as two rotation operation apply one after the other: V = R z.r y V (8) Which says that we put V into the machine R y to get V which is immediately fed to the machine R z to get V. Of course, to actually do the calculation we need to see this written in terms of the matrix components: x y z = cos(φ) sin(φ) 0 sin(φ) cos(φ) 0 0 0 1 cos(θ) 0 sin(θ) 0 1 0 sin(θ) 0 cos(θ) x y z (9) Now we just multiply out the two matrices using the standard rules, confident that all the trig. terms are in the right places: x cos(φ)cos(θ) sin(φ) cos(φ)sin(θ) x y = sin(φ)cos(θ) cos(φ) sin(φ)sin(θ) y (10) z sin(θ) 0 cos(θ) z If we want to we can multiply this out to get three separate equations. (However, software subroutine libraries generally like to keep things as vectors and matrices, represented as data arrays, because it makes the programming a lot easier.) Anyway, we would have: x = x.cos(φ)cos(θ) + y.sin(φ) + z.cos(φ)sin(θ) (11) 6
y = x.sin(φ)cos(θ) + y.cos(φ) + z.sin(φ)sin(θ) (12) z = x.sin(θ) + y.0 + z.cos(θ) (13) We can do a sanity check on these equation (though doing thing the matrix way is a pretty good guarantee we got it all right). If we assume that φ and θ are both zero, then the above equations reduce to x = x, y = y and z = z, as we would expect. Similarly, if we put φ = 0 we just revert to equation (4), and if we put θ = 0 we revert to equation (6). You might like to think about the other checks, where you put φ or θ equal to π/2 or π. 2.3 The Value of φ In order to complete the rotation calculation we need to know what value of φ we should feed into the rotation matrix. This involves using the time to work out how far we need to rotate to bring our x axis pointing to the vernal equinox. This is easy if we know the sidereal time, which runs slightly more than four minute a day faster than Solar time. At this point it is easiest to look up a standard formula for Greenwich Mean Sidereal Time: GM ST = (18.697374558 + 24.06570982441908D)modulus(24) (14) Where D is the number of days (including fractions of a day) since 12 Noon (Universal Time) on 1st January in the year 2000. The modulus reduces the GMST value to the range 0...24. (That is, we take out any whole multiples of 24 from the answer.) Some programming environments have a built time/date functions that allow you to get a calendar date to year-day count directly. With a language such as Python you can load the astropy library (but in that case you would be using it to do this whole coordinate conversion process anyway). An way to find the value D is to use an on-line Calendar date to Julian Day convertors, such as http://aa.usno.navy.mil/data/docs/juliandate.php, otherwise you have to program a slightly complex algorithm for converting date to days, properly taking account of leap years. (Fortunately, we no longer have to remember the millennium bug issue arising from the fact that year 2000 though divisible by 4 was a special case with no 29th February.) We then calculate φ as the angle by which the 0 of right ascension has rotated from the meridian, which at 0 of longitude would be: φ = GMST 2π 24 λ (15) 7
where λ is longitude east of the Greenwich meridian (all angles in radians). 2.4 Conversion back to Celestial Polar Coordinates In our new coordinate system, the ratio of the y and x coordinates must be the tangent of the hour angle, α - the distance around the celestial equator to the direction in which we are pointing, so the angle is obtained from the arc tangent. Because of the earlier care with the relative orientation of XYZ axes (right-handed) we can get the hour angle directly with: α = atan( y ) (16) x Note that in most computational environments there is a special function, often called ATAN2(y,x), which avoids the possibility of dividing by zero when x = 0 and ensures that the ambiguities about the sign of α are correctly dealt with. The declination, δ is the angle to the equatorial plane, will be: 3 Summary δ = asin(z ) (17) We got there in the end - though as you can see it involves some longdistance algebra and trigonometry (though none of it, line by line, is particularly difficult). Matrix algebra concepts help us to keep track of all the multiplications. In fact, the idea of a rotation matrix is something that comes up again and again in practical applications of maths not just in astronomy but also in physics (including high-energy physics), engineering and computer graphics. Matrix algebra and computers go together extremely well, because we find it easy to represent complex equations for numerical solution as large arrays, which are treated as matrices by widely used subroutine libraries. This allows us to use and re-use very reliable, well tested and very efficient software for handling matrices over and over again on many different problems that seem at first sight to have little relationship with each other. What we have explored is a relatively trivial application, though one that turns up again and again. If we were working in more dimensions, of course, (perhaps 4, 11 or even 10,000) we would simply not bother to worry about writing out all the separate equations explicitly: we just trust the matrix algebra to keep us straight. 8