γ

As was mentioned in an earlier article, the backbone of space science is the ability to transform from one convenient coordinate system to another. This can be done in one of two ways: spherical trigonometry and vector analysis. The two methods do arrive at the same conclusions; however, the ease with which they arrive at the conclusions is vastly different.

A simple comparison between the two methods can be achieved by challenging both to a simple task: find the angle between two stars; each having specific right ascension (RA) and declination (dec) coordinates. Of course, if the two stars have identical RAs or decs, then the angle between them is simply the dec difference or RA difference, respectively. However, this ideal case is not usually the general reality, therefore a more robust method is required. Which method is more practical when determining this angle? Let us begin with spherical trigonometry.

SPHERICAL TRIGONOMETRIC ANGULAR SEPARATION

I had originally derived a number of spherical trigonometric equations beginning in late 1993 and ending in 1999. They all had different uses, but the first one determined the angle between two stars (and any other celestial objects) with the RAs (α1 and α2) and decs (δ1 and δ2) of both. The equation took me several days to derive. When completed, it was tested with a number of known stars with known equatorial coordinates (RA and dec) in order to determine its reliability and effectiveness. I found that the final equation was trustworthy and could be used for more serious work.

The spherical trigonometric equation to determine angular separation, otherwise known as the "angular separation equation", began with the two general objects of interest: two stars with some RAs and decs, as shown in Fig. 1. The angles in Fig. 1 depict the RAs, decs the differences between them, and the direct angular separation between the stars. The RA and dec. differences (deltas) are defined in Eq. 1.

Fig. 1: Two stars with coordinates α1, δ1 and α2, δ2 separated by angles Δα, Δδ, and γ

 Δα = α2 - α1 and Δδ = δ2 - δ1 (1)

My challenge in late 1993 was to determine the angular separation angle (that I called γ) with only the RA and dec information of both objects of interest. I imagined that the stars were on a surface of an immense sphere, centered at the Earth, with an abstract radius "R". Therefore, lines can be drawn between the stars along this spherical surface, as shown in Fig. 1. A "spherical trapezoid" will therefore be formed. A chord can be drawn between each line corresponding to Δα, Δδ, and γ, so that a 2-D trapezoid is formed, as shown in Fig. 2. Each chord line has an abstract distance formed by the angles and the abstract radii.

Fig. 2: Spherical and abstract trapezoid formed by the abstract radii and the angles Δα, Δδ, and γ

The distance dγ can be determined from both triangles within the trapezoid of Fig. 2 with the law of cosines, as shown in Eq. 2 and Eq. 3. The angle "ν" was required to complete the information required to use the cosine law. The "ν" variable was chosen because it rhymed with "new", because the angle was unknown at the time, therefore the "new (nu) angle". A cosine identity that relates the cosine of an angle with its supplementary angle is shown in Eq. 4. The cosine of the supplementary angle in Eq. 3 can be replaced with the equivalent cosine shown in Eq. 4 and results in Eq. 5. Since Eq. 2 and Eq. 5 are equal to each other, the right hand sides of both equations can be equated, as shown in Eq. 6. Collecting terms and isolating the cosine results in Eq. 7.

 (dγ)2 = (dα1)2 + (dδ)2 - 2 (dα1) (dδ) cos(ν) (2)
 (dγ)2 = (dα2)2 + (dδ)2 - 2 (dα2) (dδ) cos(180° - ν) (3)
 cos(180° - ν) = -cos (ν) (4)
 (dγ)2 = (dα2)2 + (dδ)2 + 2 (dα2) (dδ) cos(ν) (5)
 (dα1)2 + (dδ)2 - 2 (dα1) (dδ) cos(ν) = (dα2)2 + (dδ)2 + 2 (dα2) (dδ) cos(ν) (6)
 2 (dδ) [ dα1 + dα2 ]  cos(ν) = (dα1)2 - (dα2)2 (7)

The solution of the "cosine nu" term is finally shown in Eq. 8. This can be substituted back into Eq. 2, such that Eq. 9 results.

 cos(ν) = ( dα1 - dα2 ) / 2 (dδ) (8)
 (dγ)2 = (dδ)2 + (dα1) (dα2) (9)

The abstract distances dγ, dδ, dα1, and dα2 are determined from their corresponding angles and the abstract radius "R" with the 2-D trigonometric relationships shown in Eq. 10, Eq. 11, Eq. 12, and Eq. 13. Substituting these into Eq. 9 results in Eq. 14. Note that the 4R2 terms will all cancel and therefore so will the abstract radius "R", which is good, since we never knew what it was anyway. Therefore, Eq. 15 results when the abstract radius is cancelled out.

 dγ = 2R sin(γ/2) (10)
 dδ = 2R sin(Δδ/2) (11)
 dα1 = 2Rcosδ1 sin(Δα/2) (12)
 dα2 = 2Rcosδ2 sin(Δα/2) (13)
 [ 2R sin(γ/2) ]2 = [ 2R sin(Δδ/2) ]2 + [ 2Rcosδ1 sin(Δα/2) ] [ 2Rcosδ2 sin(Δα/2) ] (14)
 sin2(γ/2) = sin2 (Δδ/2) + cosδ1 cosδ2 sin2 (Δα/2) (15)

Equation 15 was one of the most exciting equations that I had derived because it represented an entirely new realm of trigonometry that was based solely on angles rather than on 2-D Euclidean trigonometry. I had originally called the equation "The Pythagorean Theorem in Three Dimensions" because it looked like an angular representation of the Pythagorean Theorem. In fact, of all of the equations I had derived, this one was the most used. I had used the equation in the form shown in Eq. 15 for several years before I noticed that it could be simplified further by using the simple trigonometric identity shown in Eq. 16. This resulted in Eq. 17, which could be simplified even more. After several more cancellations, Eq. 18 results. Expanding the terms and rearranging results in Eq. 19. Eq. 1 can now be substituted into Eq. 19 so that Eq. 20 results. After cleaning up the terms, Eq. 21 becomes the final expression. Eq. 21 was simpler to use than Eq. 15, but they both produced the same results, as they should.

 sin2(γ/2) = ( 1 - cosγ ) / 2 (16)
 ( 1 - cosγ ) / 2 = [ 1 - cos(Δδ) ] / 2 + cosδ1 cosδ2 [ 1 - cos(Δα) ] / 2 (17)
 cosγ = -cos(Δδ) + cosδ1 cosδ2 [ 1 - cos(Δα) ] (18)
 cosγ = cos(Δδ) + cosδ1 cosδ2 cos(Δα) - cosδ1 cosδ2 (19)
 cosγ = cosδ1 cosδ2 + sinδ1 sinδ2 + cosδ1 cosδ2 cos(Δα) - cosδ1 cosδ2 (20)
 cosγ = sinδ1 sinδ2 + cosδ1 cosδ2 cos(Δα) (21)

UNIT VECTOR ANGULAR SEPARATION

In what seems to be a stark contrast to spherical trigonometry, vector analysis begins with two unit vectors pointing toward the two example stars, as shown in Fig. 3. The angular separation (γ) is the angle that is subtended by the two unit vectors. Unit vectors have a length of 1 unit and a direction that can be expressed in any number of coordinate systems, such as Cartesian, cylindrical, or spherical. For simplicity, we will first use Cartesian coordinates and then transform these into their equivalent spherical coordinates.

Cartesian coordinates contain three axes, normally denoted by î, ĵ, and , as shown in Fig. 4. Each unit vector can be split into its Cartesian components, as shown in Eq. 22. The x, y, and z components are all less than one unit, but when the sum of their squares is calculated, the final answer is one (the original unit vector).

Fig. 3: Unit vectors, each pointing in the direction of an example star

Fig. 4: The Cartesian coordinate system

 = x î + y ĵ + z (22)

Figure 4 also features the RA and dec. angles of the unit vector (still pointing to one of the stars). The x, y, and z Cartesian coordinates can also be expressed as these spherical angles, with the "1" of the unit vector being the radial component, as demonstrated in Eq. 23, Eq. 24, and Eq. 25. Substituting Eq. 23, Eq. 24, and Eq. 25 into the x, y, and z terms of Eq. 22, Eq. 26 and Eq. 27 result for the two unit vectors pointing at the two stars.

 x = (1) cosδ cosα (23)
 y = (1) cosδ sinα (24)
 z = (1) sinδ (25)
 1 = (cosδ1 cosα1) î + (cosδ1 sinα1) ĵ + (sinδ1) (26)
 2 = (cosδ2 cosα2) î + (cosδ2 sinα2) ĵ + (sinδ2) (27)

The angular separation angle (γ) is best determined from the two unit vectors by using the dot-product, as defined in Eq. 28. The cross-product can also be used, but it is more difficult to simplify the result. Equation 23, Eq. 24, Eq. 25, Eq. 26, and Eq. 27 can be substituted into Eq. 28 and Eq. 29 will result. Simplification of Eq. 29 occurs in Eq. 30 and Eq. 31. Finally, Eq. 32 is the final answer.

 cos (γ) = 1 · 2 = x1x2 + y1y2 + z1z2 (28)
 cosγ = cosδ1 cosα1 cosδ2 cosα2 + cosδ1 sinα1 cosδ2 sinα2 + sinδ1 sinδ2 (29)
 cosγ = cosδ1 cosδ2 ( cosα1 cosα2 + sinα1 sinα2 ) + sinδ1 sinδ2 (30)
 cosγ = cosδ1 cosδ2 cos ( α1 - α2 ) + sinδ1 sinδ2 (31)
 cosγ = sinδ1 sinδ2 + cosδ1 cosδ2 cos(Δα) (32)

Note that Eq. 32 is identical to the spherical trigonometric solution in Eq. 21; however, Eq. 32 was derived more quickly and with much less labour than Eq. 21. Therefore, in this one case, vector analysis is more straightforward than spherical trigonometry. An added bonus of vector analysis is that directions are also included, especially when considering the cross-product. Spherical trigonometry does not identify directions, just angles, although directions can be assigned after the fact (but then vectors would be created).

Once the equation has been derived (whatever method is used), it can be used again and again to determine the angle between two objects in the night sky or the angle between two vectors whether they are pointing to celestial objects or not. The equation is really incredibly powerful and occurs many times over in many disciplines of science.

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