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Where a celestial body is in your sky depends on the rotation angle of the planet at your location, relative to the stars. That latter angle is expressed by the sidereal time $$θ$$ (theta). The sidereal time is the right ascension that is on the celestial meridian at that moment. If the sidereal time is again the same (at the same location), then the stars are again in the same directions in the sky. We measure the sidereal time in degrees here. It can also be measured in hours. You can transform between degrees and hours using 1 hour = 15°.

$$∆J$$

How much time has passed since 1 January 2000, 00:00 UTC (Julian Day number 2451544.5), measured in (whole and fractional) days.

$$L$$

The heliocentric mean longitude of the Earth, measured in degrees.

$$l_\text{w}$$

The west longitude of the observation location, measured in degrees. West longitude is equal to negative east longitude.

$$t$$

The clock time, in the local time zone, measured in hours.

$$t_\text{z}$$

UTC minus the clock time, measured in hours. In other words: how many hours you have to add to the clock time to get UTC.

$$θ$$

The sidereal time, measured in degrees. Divide by 15 to get sidereal time measured in hours.

## 1. From ClockTime to Sidereal Time

The mean sidereal time (relative to the mean equinox of the date) at a particular moment as seen from a particular location is equal to

\begin{align} θ \| ≡ L_0 + L_1 ∆J + L_2 ∆J^2 + L_3 ∆J^3 - l_\text{w} \pmod{360°} \label{eq:basis} \\ L_0 \| = 99.967794687° \\ L_1 \| = 360.98564736628603° \\ L_2 \| = 2.907879×10^{−13}° \\ L_3 \| = −5.302×10^{−22}° \end{align}

The $$\pmod{360°}$$ (modulus 360°) part means that you can add (or subtract) multiples of 360° to the result; usually we do that until the result ends up between 0 and 360°. This also means that a given sidereal time keeps coming back (just like a given clock time).

$$∆J$$ must take into account not just the date but also the time in UTC (Universal Time). If $$∆J_d$$ is number of whole days since 0:00 local time at the beginning of 1 January 2000, then we have

$$∆J ≡ ∆J_d + \frac{t + t_\text{z}}{24} \label{eq:fulljd}$$

If you combine equations \ref{eq:basis} and \ref{eq:fulljd} then you get

$$θ ≡ θ_0 + θ_1 t \pmod{360°} \label{eq:sterrentijd}$$

where

\begin{align} θ_0 \| ≡ L_0 + L_1 ∆J_d + θ_p \pmod{360°} \label{eq:θ_0} \\ θ_p \| ≡ L_2 ∆J_d^2 + L_3 ∆J_d^3 - l_\text{w} + θ_1 t_z \\ θ_1 \| ≡ M_0 + M_1 ∆J_d + M_2 ∆J_d^2 \label{eq:θ_1} \\ M_0 \| = 15.04106864026192° \\ M_1 \| = 2.423233×10^{−14}° \\ M_2 \| = −6.628×10^{−23}° \end{align}

Because $$∆J_d$$ is always a whole number, and because we calculate $$θ_p$$ modulus 360°, we can use $$L_1 \bmod 360°$$, i.e., 0.98564736628603° instead of 360.98564736628603°. This keeps the numbers in the calculations smaller.

$$θ_p$$ and $$θ_1$$ change with time only very slowly. In practice, you need to calculate them only once for a given problem, except if you need very high accuracy or if you want to calculate sidereal times for a very long period of time (for example, of at least a century).

What is the sidereal time at 23:00 hours CET on 2006 December 1st, as seen from 5° east longitude (roughly the longitude of the Netherlands and Belgium)? Then $$t = 23$$, $$t_\text{z} = −1$$, $$l_\text{w} = −5$$, and $$∆J_d = 2526$$ (see the Calculation Page about the Julian Day Number). Equation \ref{eq:fulljd} then yields

$∆J = 2526 + \frac{23 - 1}{24} = 2526.91667$

and equation \ref{eq:basis} yields

\begin{align*} θ \| = 99.967794687° + 360.98564736628603° × 2526.91667 \\ \| + 2.907879×10^{−13}° × 2526.91667^2 \\ \| - 5.302×10^{−22}° × 2526.91667^3 - (−5°) \\ \| = 912285.61655° = 45.61655° \pmod{360°} \end{align*}

We can also use formulas \ref{eq:sterrentijd}ff, and then find

\begin{align*} θ_1 \| = 15.04106864026192° + 2.423233×10^{−14}° × 2526 - 6.628×10^{−23}° × 2526^2 \\ \| = 15.0410686403231° \\ θ_0 \| = 99.967794687° + 0.98564736628603° × 2526 \\ \| + 2.907879×10^{−13}° × 2526^2 \\ \| - 5.302×10^{−22}° × 2526^3 \\ \| - (−5°) + 15.0410686403231° × −1 \\ \| = 2579.67198° = 59.67198° \pmod{360°} \\ θ \| = 59.67198° + 15.0410686403231° × 23 = 45.61655° \pmod{360°} \end{align*}

Both methods yield the answer 45.61655° which is equivalent to 45.61655°/15° = 3.0411 hours = 03:02 hours, so the sidereal time at the chosen date, time, and location is 03:02 hours.

## 2. From SiderealTime to Clock Time

In the opposite direction, you find the local time $$t$$ from

\begin{align} t \| = \frac{θ - θ_0}{θ_1} = \frac{θ - θ_p - L_0 - L_1 ∆J_d}{θ_1} \pmod{t_\text{s}} \label{eq:lokaal} \\ t_\text{s} \| ≡ \frac{360°}{θ_1} \end{align}

$$t_\text{s}$$ says how many clock-hours correspond to 24 sidereal hours. $$θ_p$$ and $$θ_1$$ depend a bit on the date ($$∆J_d$$), so $$t_\text{s}$$ depends (a little) on the date, too. For the most accurate calculations you should calculate $$θ_p$$ and $$t_\text{s}$$ anew for each day. For quicker calculations, you can calculate $$θ_p$$ and $$t_\text{s}$$ for a date in the middle of the desired interval, use those to calculate $$t$$ (from formula \ref{eq:lokaal}) between 0 and $$t_\text{s}$$ for that middle date, and then add multiples of $$t_\text{s}$$ to it to find estimates for nearby clock times. Those estimates will be pretty good, as long as the interval is not too long (e.g., not longer than a few years).

As seen from 5° east longitude, at what clock times (in the Central European Time zone) is it 03:00 sidereal time? Then $$θ = 3 × 15 = 45°$$, $$l_\text{w} = −5$$, and $$t_\text{z} = −1$$. We found earlier for 1 December 2006 that $$θ_1 = 15.0410686403231°$$ and $$θ_0 = 59.67198°$$, and then equations \ref{eq:lokaal}ff yield

\begin{align*} t_\text{s} \| = \frac{360°}{15.0410686403231°} = 23.93446959 \\ t \| = \frac{45 - 59.67198°}{15.0410686403231°} = −0.97546 = 22.95901 = \text{22:58} \pmod{23.93446959} \end{align*}

so it is 03:00 sidereal time at 22:58 CET on 1 December 2006, and also every 23.93446959 hours = 23:56:04 earlier or later (but not infinitely far).

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Last updated: 2017-12-28