What is sidereal time?
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Imagine the equator and poles of the Earth projected onto the
Celestial Sphere. Each instant, stars move from East to West
across the sky. Most are on small circles parallel to the
Celestial Equator. For instance, Mintaka in Orion's belt happens
to be almost exactly on the equator, and if you follow the
Hunter's progress tonight (I'm writing this in January) this star
will trace out the Celestial Equator. The Earth's axis is tilted
at about 23.5 degrees to the plane of the Earth's orbit around
the Sun - and the path of the Sun in the sky is called the
Ecliptic. The Equator and Ecliptic are 'great circles' - the
planes that contain them pass through the centre of the Celestial
sphere. The two circles intersect at two points on the Celestial
Sphere, and the Sun passes through one of the intersection points
around March 21st each year. At this vernal equinox, the Sun has
a declination of exactly zero, and the Sun will be above the
horizon for half the day everywhere on Earth between the arctic

The star charts will show Mintaka as having a declination of
about zero and a right ascension of 5h 32m roughly. This means
that Mintaka is roughy 83 degrees (5 and 32/60 times 15) around
the sky from an imaginary point on the celestial equator called
the First Point of Aries. 

A third great circle on the Celestial Sphere is your Meridian.
This circle passes through the celestial poles, your zenith and
nadir, and crosses your horizon due South and North. This great
circle is moving across the face of the celestial sphere - at
about 15 degrees per hour, or one complete revolution per day.
Your zenith passes through the stars of declination equal to your
latitude in turn each sidereal day.

The Hour Angle (often measured in hours and minutes) of (say)
Mintaka at any given time is the angle round the sky from your
southern meridian to the star. This Hour Angle tracks the changes
from minute to minute as the star rises, transits and then sets.
When the star transits, the hour angle is taken as exactly zero.

Your local sidereal time connects the hour angle (that changes
from minute to minute as the Earth turns on its axis) with the
Right Ascension (essentially constant except for precession - a
slow process over time). The formula is

HA = LST - RA (all in degrees)

So, to find the HA of Mintaka, you just find your LST and
subtract the RA of the star. All three quantities must be
measured in the same units, either degrees or decimal hours. As
there are 24 hours and 360 degrees in a complete circle, you can
convert from decimal hours to degrees by multiplying by 15. From
the above formula, when the LST is the same as the RA, then the
HA must be zero - i.e the star is transiting. Does that mean we
can just look at our watch, see its 05:32 in the morning, look
out the door to see Mintaka crossing the meridian each night?

No. sidereal time is different to UT. It goes faster. Suppose you
start a stopwatch the moment you see the star Mintaka in Orion's
belt cross your meridian, and you stop the stopwatch when the
star transits again the next night. The elapsed time will be 23
hours 56 minutes 4.10 seconds if you have very good reactions and
a decent stopwatch! So Mintaka will transit 4 minutes earlier
each day, and two hours earlier each month. Today, Mintaka will
transit about 9pm civil time. At the end of February, it will
transit about 7pm civil time. An horizon star map of the kind you
find in the astronomy magazines is usually calculated for 9pm. If
you have the April map, you can use it at 1 am in February.

Why should the UT day and the star day be different? Because the
UT day is based on an average of the time between two successive
transits of the Sun. The Earth moves in its orbit around the Sun
a bit less than 1 degree per day, so the Sun takes that little
bit longer to cross your meridian a second time. Lets see, 1
degree extra means 1/15 hours extra or... 4 minutes! Aha!

Because the duration of sidereal time is shorter, the sidereal
time corresponding to a given UT will shift through a cycle of
roughly a year. You can find tables giving rough hours and
minutes of sidereal time for midnight each day, and these will
remain correct within minutes from year to year. This article
looks at calculations accurate enough for celestial navigation
(but please don't trust anything here until you have tested it
completely yourself).

Mean and Apparent positions - nutation and precession

Apparent means 'what you could measure tonight if you had access
to the USNO observatory and a lot of patience and skill'. Mean
means 'referred to averaged out coordinate systems'.

The position of a body on the Celestial Sphere can be referred to
the First Point of Aries using the RA and Dec coordinates.
However, the First Point of Aries is shifting around. Some of the
shift is due to the slow precession of the equinoxes - the
Earth's axis of rotation is wobbling like a toy top, and
completes a circle around the pole of the ecliptic every 25,800
years. The Equinox shifts by 360 / 25800 of a degree each year,
about 50 arcseconds - or about 3% of the width of the Moon. Over
70 years this adds up to a whole degree!

In order to save having to print new star charts yearly,
astronomers refer star positions to standard Epochs. The current
one is J2000.0 - star charts will have their epoch printed on the
title page. It is quite normal to use a precession formula to
translate coordinates in a standard epoch to those of the mean
equinox of date. These are 'mean coordinates'

If that isn't bad enough, the Earth is strongly coupled to the
Moon by gravity, and the Moon pulls the Earth's axes slighly with
a period of 18.6 years and an amplitude of 17.5 arcseconds. 

This causes the Earth to nutate (nod slightly up and down), so
the stars appear to be going round in little tiny cicles of size
17.5 arcseconds on the sky every 18.6 years, even after you have
allowed for precession. Mean coordinates that have been corrected
for nutation are called 'apparent coordinates'. They are what you
could in principle measure on the sky.

Further reading

If you found the above hard going, then you might find


useful. Print it out and read it on the bus / train. The diagrams
will help if nothing else.

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Keith Burnett (kburnett@btinternet.com) - 28 Jan 2001