Selected Questions - Answers From All Experts Astronomy Forum (Sidereal time)

Question:  Can you explain what is sidereal time and how one can find it?  - Marge M., Montreal

Answer:

"Sidereal time" basically means time by the stars, i.e. using star positions to obtain the time at one's location.   The diagram below is useful as a reference:

Here, the observer ('Obs') at longitude L with respect to Greenwich (G, or the Greenwich meridian at 0 degrees). The L.S.T. or local sidereal time is 6h 00m as denoted by the fact that 6h Right Ascension (RA) is on the observer's meridian. Then the hour angle (HA) for the star will be as shown. (Note: the view shown is a POLAR one, i.e. the Earth as seen from above Earth's north pole).

The crucial thing to note is how the sidereal time is related to RA as well as hour angle. So to find it one usually needs to know - or find- these measures as well.

The Right Ascension (RA) is measured in units (hours, minutes, seconds) associated with time, and the Right Ascension of the star is clearly equal to the local sidereal time (L.S.T.) plus the hour angle (which in this case is negative). Thus, we can write:

HA = RA of observer meridian - RA of object

Again, the hour angle is the angular distance in hours separating your local meridian and the RA of the object. It increases westwards, unlike RA - which increases eastwards. As may be seen, the RA of the observer's meridian (6 h in Figure 1) is none other than the local sidereal time. If you know what Right Ascension is on your meridian you know the L.S.T.

Another useful expression used often is:

Local time of transit (L.T.T.) = Star's RA - Sun's RA

Where L.T.T. is the local mean solar time of transit (or L.M.T.) of the star or other object and Sun's RA is the Sun's Right Ascension on the particular date. A simple interpolation method is adequate (for most purposes) to obtain the latter, given the base data for the specific dates of the solstices and equinoxes below:

Date: ----- March 21---------June 22-----------Sept. 23--------Dec. 22

Sun R.A. ----( 0 hr.)--------(6 hr.)-----------(12 hr.)----------(18 hr.)

March 21 is just the (approx.) date of the Vernal Equinox, June 22 is the summer solstice, Sept. 23 the autumnal equinox and Dec. 22 the winter solstice. Since the Earth complete one circuit of the Sun in 365 days, the Sun (as seen from the Earth) will be apparently displaced along the ecliptic (the plane of the Earth's orbit projected onto the celestial sphere) by about one degree per day or equivalent to a 4 minute time difference. (Since 24 hours of Right Ascension corresponds to a 360 deg angular difference).

Thus, using either of the four known RA dates for the Sun - and knowing the Sun's RA changes by 4 mins/day, the new solar RA on any day can be found.

Example (1)

Find the Right Ascension of the Sun on July 1st.

We note that 8 days separates July 1st from June 22 when the Sun's R.A. is known to be 6h 00m.

The time difference is: 8 days x (4 min/day) = 32 mins.

So the Sun's RA on July 1st = 6h 00m + 32 min = 6h 32 m

Example  (2):

Aldebaran (RA = 4h 34m) is found to be 45 degrees east of your meridian on a particular date. What is your L.S.T.?

Solution

First, find the hour angle (HA) of Aldebaran. We have 45 deg = 45 deg/(15 deg/h) = 3 h. But since the HA is measured east of the meridian then one must use the negative value (-3 h) so:

24 h + (- 3h) = 21 h 00m

We know:

HA = RA of the meridian - RA (object)

and RA of the meridian = L.S.T.

Therefore:

L..S.T.= HA + RA of object

L.S.T. = 21h 00m + 4 h 34 m = 25h 34m = 1h 34m (25h 34m - 24h 00m)

A nice additional problem for practice, to see if you understand sidereal time, is given below:

The star Canopus (RA = 6h 20m) is observed to have a local hour angle = 45 deg on Feb. 10th for a given location.

What is the local sidereal time ?

Math Drives Astronomy (Pt. 2)

By the time the aspiring Astronomy major reaches his sophomore year, he will be facing even more math in courses such as Spherical Astronomy.  The emphasis here is on astronomical time (sidereal time and position - the latter based on different coordinate systems used (equatorial, horizontal, ecliptic).

Each such coordinate system is defined by a different set of poles and equator. For example, the coordinate system depicted above - the equatorial system - is based on the projection of the Earth's N. and S. poles into the sky, as well as its equator. The poles then become the North and South Celestial poles, and the equator becomes the celestial equator. If these poles are defined respectively at +90 degrees (NCP) and -90 degrees (SCP) and the celestial equator at 0 degrees, then a consistent system of celestial latitude and longitude  can be defined in a consistent system, to locate any celestial object. We call the longitude coordinate (Θ) the Right Ascension (R.A.) while the celestial latitude coordinate is called declination and is measured in degrees north or south of the celestial equator. (In the diagram, the complementary angle of the declination is shown, φ, which we call the zenith distance.)

One of the first things the astronomy sophomore learns is how to find directions around the celestial sphere, including how to relate the R.A. to time and time keeping.  He must also do exercises showing he can find the hour angle - and distinguish it from the Right Ascension. This starts with identifying the R.A. of his local meridian (the imaginary celestial longitude passing through his zenith or highest point.)  The Right Ascension of the star is clearly equal to the local sidereal time (L.S.T.) plus the hour angle. Thus, we can write:  HA = RA of observer meridian - RA of object
Of course, before he gets very far, he will be adept at sketching any number of diagrams to solve time and position problems, such as the diagram below - with perspective looking down onto the North Celestial pole:

For example, in the diagram shown (Fig. 2) if the star's R.A. is 7h 00m and the observer is at a local sidereal time of 6h 00m, then the hour angle becomes: HA = 6h - 7h = -1h or -15 degrees. (Since every hour of longitude corresponds to 15 degrees angle, i.e. Earth turns through 15 degrees every hour, 360 degrees in 24 hours.)

The budding astronomy student will also have to know how to sketch a three dimensional diagram to enable conversion between coordinates, say from the horizontal (observer -based) to the celestial sphere. This will always include what is called the fundamental "astronomical triangle" (shaded region of Fig. 3) from which spherical trig relationships can be obtained and conversions can be made to different coordinate systems.

Using spherical trig, the student can write the law of sines and law of cosines for spherical triangles (such as shown in Fig. 3)which are analogs of the law of sines and cosines for triangles in plane trig.

We have for the law of sines:

Sin A/ sin a = sin B/ sin b = sin C/ sin c

where A, B, C denote ANGLES and a,b,c denote measured arcs. (Note: we could also have written these by flipping the numerators and denominators).

We have for the law of cosines:

cos a = cos b cos c + sin b sin c cos A

Where a, b, c have the same meanings, and of course, we could write the same relationship out for any included angle.

Now, we use Fig. 3, for a celestial sphere application, in which we use the spherical trig relations to obtain an astronomical measurement.

Using the angles shown in Fig. 3 each of the angles for the law of cosines (given above) can be found. They are as follows:

cos a = cos (90 deg - decl.)

where decl. = declination

cos b = cos (90 deg - Lat)

where 'Lat' denotes the latitude. (Recall from Fig. 1 if φ is polar distance (which can also be zenith distance) then φ = (90 - Lat))

cos c = cos z

where z here is the zenith distance.

sin b = sin (90 deg - Lat)

sin c = sin z

and finally,

cos A = cos A

where A is azimuth.

Let's say we want to find the declination of the star if the observer's latitude is 45 degrees N, and the azimuth of the star is measured to be 60 degrees, with its zenith distance z = 30 degrees. Then one would solve for cos a:

cos a = cos (90 deg - decl.)=

cos (90 deg - Lat) cos z + sin (90 deg - Lat) sin z cos (A)

cos (90 deg - decl.) =

 

cos (90 - 45) cos 30 + sin (90 - 45) sin 30 cos 60

 

And:

cos (90 deg - decl.)= cos (45) cos 30 + sin (45) sin 30 cos 60

We know, or can use tables or calculator to find:

cos 45 = [2]^½/ 2

cos 30 = [3]^½/ 2

sin 45 = [2]^½/ 2

sin 30 = ½

cos 60 = ½

Then:

cos (90 deg - decl.)= {([2]^½/ 2 )([3]^½/ 2)} + {[2]^½/ 2} (½) (½)

cos (90 deg - decl.)= [6]^½/ 4 + [2]^½/ 8

 

= {2[6]^½ + [2]^½}/ 8

cos (90 deg - decl.)= 0.789

arc cos (90 deg - decl.)= 37.9 deg

Then:

decl. = 90 deg - 37.9 deg = 52.1 deg

Or,  decl. (star) = + 52.1 degrees

 

The student will also become familiar with matrix methods for converting between the different coordinate systems.

 

The basic principle involves relating the Cartesian coordinates (rectilinear) of a point on the celestial sphere (diagram) to the curvilinear coordinates measured in the primary and secondary reference planes. One has then, for example:

(x)
(y)
(z) u,v =


(cos v .....cos u)
(cos v .....sin u)
(sin v..............)

After conversion the curvilinear coordinates may be calculated according to:

u = arctan (y/x) and v = arcsin (z)

Now, consider conventional orthogonal matrices of 3 x 3 dimensions, given as functions: R1(Θ), R2(Θ) and R3(Θ), to rotate the general system by the angle Θ about axes x, y and z, respectively. Thus we obtain:

R1(Θ) =

(1..........0................0)
(0.....cos(Θ)..... sin(Θ))
(0......-sin(Θ)....cos(Θ))


R2(Θ) =

(cos (Θ)......0........- sin(Θ))
(0................1...............0.. )
(sin(Θ)........0.......cos(Θ) )


R3(Θ) =


(cos(Θ)..........sin(Θ)..........0)
(-sin (Θ)......cos(Θ)...........0)
(0 ..................0..................1)


To fix ideas, say you wish to obtain the horizontal coordinates (A, a) for some object in the sky and you know its R.A. and decl. from an almanac.  Then the procedure is fairly straightforward, and entails writing:

 

R3(Θ) = R3(-180 deg)

R2(Θ) = R2(90 - lat.)

so that:

(x)
(y)
(z) A,a = R3(-180 deg) R2(90 - lat.) (XYZ(h, decl.))

 

where : (XYZ(h, decl.)) =

(x)
(y)
(z) h,decl.

 

Bear in mind: R3(-180 deg) =

(cos 180........-sin 180................0)
(sin 180........cos 180................0)
(0......................0......................1)

And:R3(Θ) =


(cos(Θ)..........sin(Θ)..........0)
(-sin (Θ)......cos(Θ)...........0)
(0 ..................0..................1)

Therefore: R3(-180) =

(-1.....0.......0)
(0.......-1.....0)
(0.......0.......1)

Obtaining R2(Θ) = R2(90 - lat.) is just as easy, if one recalls the basic trig identity:

cos (90 - φ) = sin (φ)

 

Not surprisingly, it is the sophomore year at which point most wannabe Astronomy majors change their minds and either drop out entirely or change their major - say to a less mathematically demanding subject. At the Univ. of South Florida, nearly 60% had dropped out by the end of the quarter which featured spherical astronomy.