Galactic Coordinates

In the essay on spherical trigonometry I demonstrated how to convert Equatorial coordinates, the astronomers= right ascension and declination, into the equivalent Ecliptic coordinates. Now I want to demonstrate the somewhat harder task of converting the Equatorial coordinates of a star into the equivalent galactic coordinates.

Astronomers conceive the system of galactic coordinates as a heliocentric grid projected onto the celestial sphere and anchored to the Milky Way. Galactic latitude coordinates with the location of the galactic north pole (whose Equatorial coordinates J2000 stand at right ascension 12 hr 51 min 26.282 sec (192.8595˚) and declination 27˚ 07' 42.01" (27.1283˚)). Galactic longitude coordinates with the location of the galactic center (whose Equatorial coordinates J2000 stand at right ascension 17 hr 45 min 37.224 sec (266.4051˚) and declination -28˚ 56' 10.23" (-28.9362˚)). Astronomers established this system before they knew precisely where the center of our galaxy lies on the sky, so the coordinate center of the galaxy actually lies a small distance northeast of the real center of the galaxy, Sagittarius A*, which UCLA's Professor Andrea Ghez and her team have gone and studied for over a decade: the five-million-solar-masses compact object that occupies the real center lies at the galactic coordinates of -0˚ 2' 46.2" latitude and 359˚ 56' 39" longitude.

Again we get an exercise in spherical trigonometry. In this case we solve three spherical triangles, one to get the galactic latitude of a star, one to get an intermediate angle, and the third to get the galactic longitude of the star. To aid us in the effort to work out the conversion I will carry out a specific conversion as we go; specifically, I will convert the Equatorial coordinates of Alpha Canis Majoris (Sirius) [right ascension J2000, ρ=6 hr 45 min 8.871 sec = 101.2869625˚, declination J2000, δ=-16˚ 42' 57.99" = -16.7161˚].

For our first triangle we make the celestial north pole, the projection of Earth's north pole on the sky, the vertex A, the point on the sky occupied by the galactic north pole the vertex B, and the point occupied by the star we have under study the vertex C. We have the angular length of the line AB as c=90˚-27.1283˚=62.8717˚, the angular distance that the galactic north pole lies due south of the celestial north pole. We also have the angular length of the line AC as b=90˚-(-16.7161˚)=106.7161˚, the distance that Sirius lies due south of the celestial north pole. And then we have the vertex angle A equal to the difference between the right ascensions of the galactic north pole and of our chosen star, A=192.8595˚-101.2870˚=91.5725˚.

That last statement needs some justification. If we draw two different lines from the north celestial pole to the celestial equator, they each meet the celestial equator at a right angle. Each of those lines, the side opposite the vertex where the other meets the celestial equator, has an angular length of ninety degrees. Those two lines form a spherical triangle with the segment of the celestial equator between them, so the sine rule necessitates that the angular length of the equatorial segment equals the width of the angle at which those two lines meet each other at the north celestial pole. Thus, in any spherical triangle in which we make the north celestial pole one of the vertices, the vertex angle at that pole equals the difference between the right ascensions of the other two vertices, even if they don't lie on the celestial equator.

Now we want to solve the first triangle for the side a. We use the cosine rule,

(Eq'n 1)

and get a=98.8904˚, the angular distance that Sirius lies south of the galactic north pole. That result gives us the galactic latitude directly by way of

(Eq'n 2)

which yields, in this case, ψ=-8.8904˚.

For our second, intermediate, triangle we define the celestial north pole as the vertex D, the center of the galaxy as the vertex E, and the star's location as the vertex F. We want to solve this triangle for the angular distance on the sky between the star and the center of the galaxy; that is, we want to find the length of the arc EF=d. We already know the length of the arc DF=e=106.7161˚ and the arc DE has length f=118.9362˚ (90˚ + the galactic center's declination). The vertex angle D equals the difference between the right ascensions of the star and the center of the galaxy; D=165.1181˚ in this case. Again we apply the cosine rule,

(Eq'n 3)

and get d=132.1363˚.

Finally, we establish a triangle by defining its vertices as G at the north pole of the galaxy, H at the center of the galaxy, and K at the location of the star. That triangle has sides GH=k=90˚, HK=g=132.1363˚, and GK=h=98.8904˚. We now want to solve that triangle for the vertex angle G, which gives us the galactic longitude of the star directly. We use the cosine rule in the form

(Eq'n 4)

In this case we calculate G=132.7698˚. Naively we assume that

(Eq’n 5)

because that gives us the vertex angle between the arcs extending from the galactic north pole to the center of the galaxy and to our target star. But we have merely calculated the smaller of the two angles between those two lines; we have no indication as to whether that smaller angle puts the star galactic east of the galactic center or galactic west of the galactic center. In the later case we would have to calculate

(Eq’n 6)

because we measure galactic longitude eastward from the galactic center.

To resolve that dilemma we must repeat steps two and three above, but replace the center of the galaxy (vertex E) with the ascending node of the galactic equator (vertex E’), the point where the galactic plane crosses the celestial equator with galactic longitude increasing from south to north. That point has the Equatorial coordinates J2000 of right ascension 18 hrs 51 min 26.275 sec (282.8595E) and declination zero and it has the galactic coordinates J2000 of latitude zero and longitude 32.9319E. We thus have the arc DE’=f’=90E and the vertex angle D’=282.8595E-D, which gives us D’=181.5725E in the present example. We thus have Equation 3 in the form

(Eq’n 7)

We then have Equation 4 in the form

(Eq’n 8)

In this case we made the substitution sinh=sin(90E-R)=cosR. We then subtract G from G’. If the subtraction yields a positive 32.9319 degrees, the we have calculated the longitudinal displacement of the star westward from the galactic center and we must use Equation 6 to calculate the correct galactic longitude. If the subtraction yields a negative 32.9319 degrees or a number smaller than 32.9319 degrees, then we have calculated the longitudinal displacement of the star eastward from the galactic center and, thus, Equation 5 gives us the correct galactic longitude. We notice from the right ascensions that the shortest distance across the sky from the galactic center to Sirius runs more or less westward, so we have actually calculated the galactic longitude as a westward displacement, which confirms our calculation of a positive 32.9319 degrees in our subtraction (G’-G=165.7017E-132.7698E=+32.9319E). We measure galactic longitude eastward, so we calculate the actual galactic longitude of Sirius by subtracting G from 360E and get 8=227.2302E.

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