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I'm having trouble understanding precisely what a stellar day is. Neither the USNO nor the IERS sites provide a definition. And Wikipedia's description as the "rotation period relative to the fixed stars" as "the span of time it takes for the Earth to make one entire rotation with respect to the celestial background or a distant star" is confusing, since, relative to fixed stars, the earth is both rotating and precessing, so that the rotation period relative to fixed stars as such, is not a single value, but will be different for stars with different equatorial coordinates.

I assume that the stellar day is simply the earth's rotation period on its axis, but I'm not sure that's right, or how to state it formally (e.g. with respect to inertial frames).

I understand why the stellar day is distinct from and longer than the sidereal day, since the coordinate system that defines a sidereal day is rotating slowly against the rotation of the earth.

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According to the Wikipedia article you reference, "stellar day" is supposedly a new name for a planet's sidereal rotation period. However, I cannot find any documentation of this new name anywhere, and that includes my copy of Explanatory Supplement to the Astronomical Almanac, 3rd edition, edited by Urban and Seidelmann (University Science Books, 2012) which was just published within the month. This source is definitive and the term doesn't appear therein (okay at least not in the index). However, further digging found reference to it here

but I've yet to find an actual statement of the change in terminology from "sidereal rotation period" to "stellar day" anywhere in the IERS conventions. Anywhere, the distinction is that the sidereal day is measured relative to the moving vernal equinox, which accounts for precession, whereas the sidereal rotation period (stellar day) is relative to the fixed inertial frame of background stars.

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So that last point is the important subtlety (and the one missed in the Wikipedia definition): it's not about "a rotation with respect to" the celestial sphere, which suggests returning to the same point (a time that varies over the celestial sphere); it's about the rotation with respect to the inertial frame defined by the celestial sphere. – raxacoricofallapatorius Nov 1 '12 at 1:29

The sidereal day is the (mean) time between two transits of the R.A. origin, i.e. the vernal equinox. You are right, and this is strictly not equal to the time it takes the Earth to do a rotation in relation to the fixed stars, because the Vernal Equinox itself is precessing (and nutating).

The amount of time it takes the Vernal Equinox to do a full rotation is suposed to be well known (26000 yr) so you can simply make the correction: since the Vernal Equinox moves towards west, that period you call "stellar day" must be slightly larger, and you can do the correction very easily (I don think there is a more formal definition based upon any reference star, but please post it here if you find it).

Just for curiosity, in which context are you using that "stellar day"? I had never heard about it. Astronomers use the sidereal day as a synonym. The difference is surely much smaller than the "imprecisions" of the Earth movement itself, so it should have had no sense in the past (nor today with atomic clocks).

(Not to be confused with the Solar Day, i.e. the time between two transits of the Mean Sun, which is the "normal" 24-hour day, about 4 minutes larger than the Sidereal Day)

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The term "stellar day" isn't one I'd come across before either. I first saw it on Wikipedia, actually, where it appears in several articles. It is also referred to here. – raxacoricofallapatorius Nov 1 '12 at 1:17
Another complication (at least if I'm visualizing this right): The process you describe above for making the correction between the length of a stellar and sidereal day in an average one, since, even though $\Delta\lambda$ varies only slightly, due to the eccentricity of the earth's orbit, and may be taken as constant, the corresponding $\Delta\alpha$, which determines the difference between stellar and sidereal day length, varies with the $\delta$ of the point being tracked along the ecliptic. – raxacoricofallapatorius Nov 1 '12 at 1:23
That is similar to the question that leads the Solar Day to be defined by the Mean Sun (ideal point at constant speed in the celestial equator instead of the eclyptic), since the eclyptic plane is tilted with respect to the equator (defined as normal to the rotation axis). Only average ang speeds are used in these definitions. In the sidereal day too. – Eduardo Guerras Valera Nov 1 '12 at 7:00
Even through the eclyptic, the real Sun has no constant speed (thank Kepler for that), not to mention the variations in the orbital parameters of the Earth, a fraction being totally unpredictable... So, take only mean ideal speeds in the equator for definitions such as your stellar day. – Eduardo Guerras Valera Nov 1 '12 at 7:11
My comment above isn't quite right. Even assuming an idealized constant stellar day, and an idealized constant rate of precession, it seems to be that the duration of the sidereal day must vary systematically. But that's totally different question. – raxacoricofallapatorius Nov 1 '12 at 21:48

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