# Sidereal times occuring twice a day question

I'm reading Practical Astronomy With Your Calculator Or Spreadsheet by Duffett-Smith and Zwart. Discussing the conversion of GST to UT, the authors mention that because the sidereal day is slightly shorter than the solar day, for a given calendar date there is a small range of sidereal times that occur twice. (approximately if $$0h\leq\mathrm{UT}\leq03m\,56s$$ and $$23h\,56m\,04s\leq\mathrm{UT}<0h$$).

I can't find much online about this and, as far as I can see, it's not mentioned in the Wikipedia Sidereal Time article.

So my question is, in their day-to-day working lives is this ambiguity a practical problem for astronomers or is it only of theoretical interest?

• It's only (potentially) ambiguous if you try to specify an event by giving a sidereal time and a calendar date. So don't do that. ;) Sep 1, 2022 at 15:11
• @PM2Ring invoking Henny Youngman?
– uhoh
Sep 3, 2022 at 20:11

The algorithms for GMST avoids this ambiguity by using the number of Julian centuries of UT1 since J2000 (or some other epoch) rather than a time and date. It would only be a problem if someone gave you a date and sidereal time and for some reason wanted you to unambiguously determine the time, which I can't imagine a practical use for.

Here's one algorithm used for computing GMST, it's likely different than what's in Duffett-Smith's book, but reproduced here for others without access to to the book. Example implementation here: Greenwich Mean Sidereal Time

\begin{align*} \theta &= 0.7790572732640 + 0.00273781191135448D_u + frac(JD(UT1)) \\ T &=\frac{JD_{UT1} - 2451545.0}{36525} \\ GMST &= 2\pi\theta + (0.014506 + 4612.15739966T + 1.39667721T^2 -0.00009344T^3 + 0.00001882T^4)/60/60*\pi/180 \end{align*}

$$D_u$$= Number of days of UT1 since J2000 (JD(UT1) - 2451545.0)

$$\theta$$ = Earth Rotation Angle (in fraction of full revolutions)

T = Centuries of UT1 since J2000

GMST = Greenwich Mean Sidereal Time (in radians)

$$frac(JD(UT1))$$ = fraction of day

• Are you sure that formula for Earth Rotation Angle is correct? According to my Astronomical Almanac (my calculation, so it could be wrong) the ERA at J2000 (JD = 2451545.0) is 280.5 degrees. Your formula gives 0.7790572732640/3600 = 0.000216405 degrees. Sep 1, 2022 at 18:08
• It is equation 2.11 from USNO Circular 179. The units are actually full revolutions, so I need to change the description. So it matches the answer you got. Sep 1, 2022 at 19:31
• I don't think $D_{U}$ refers to whole days. For JD = 2451545.50, the Earth should have rotated about 180 degrees since JD = 2451545.0. You don't get this result by assuming $D_{U}=0$ in your equation. You do get it by letting $D_{U}=0.5$. Sep 2, 2022 at 7:51
• $D_u$ has to be a whole number, .5 is invalid. For that case you set $D_u$ to 0 and add .5. Sep 2, 2022 at 14:05
• Grovelling apologies if I've got this wrong, but for JD = 2451545.50, if you set $D_{U}=0$ and $\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ then $\theta=360^{\circ}\left(0.7790572732640+\left(0.00273781191135448\times0\right)+0.5\right)=460.460618^{\circ}.$ Less ERA at epoch ($280.460618^{\circ}$) gives $180.000000^{\circ}$ (wrong). Setting $D_{U}=\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ gives $460.953425^{\circ}$. Less $280.460618^{\circ}$ gives $180.492806^{\circ}$ (correct). Sep 2, 2022 at 15:09