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I'm new to the field so I'm sure this is a basic misunderstanding. As I currently understand it, J2000 equatorial coordinates describe the position of objects by referencing a specific point in time (12:00 Jan 1st 2000) to account for the effect of the precessional cycle in the reference point for RA. Presumably this means that the same object measured on different dates should still have the same J2000 coordinates?

In Stellarium, if I look at the RA/Dec (J2000) of a star on one date, then compare to the previous year the J2000 coordinates don't match:

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I also notice that if I set the current date/time to 12:00 1/1/2000 the RA/Dec (J2000) and RA/Dec (on date) values are different when I would have expected them to be the same:

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Where am I going wrong here?

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Apparently, Stellarium's "J2000.0" reports the coordinates of stars in J2000 frame, but at the epoch of the date you specify, instead of reporting the coordinates of the star at epoch 2000.0, which can be misleading to say the least. So the difference in J2000 coordinates that you noticed corresponds to the proper motion of the star. And the star you picked, 61 Cygni, actually has one of the largest proper motion in all of the catalogued stars, with about 4.1"/year in RA and 3.2"/year in declination. Looking at the differences that Stellarium reports in your example, we can notice that it matches perfectly with this proper motion : 31.3" - 28.2" = 3.1" ± 0.1" in declination, for RA: (55.27s - 54.92s) x 15 x cos(declination) = 4.1" ± 0.1".

If you pick a star with a smaller proper motion in Stellarium, like Deneb (only 2 mas/year for each axis), the J2000 coordinates change much more slowly.

As for the difference between J2000 RA/Dec and apparent (on date) RA/Dec, this is caused by the fact that the J2000 frame's equator/equinox was fixed at the mean position of the equator/equinox in 2000 (mean position means that the small periodic effects of nutation are averaged out). So the apparent coordinates on Jan 1 2000 are slightly different because apparent coordinates take into account nutation N of the rotational axis R around the mean P (precession):

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  • $\begingroup$ Very much appreciated the detail and calculations in your answer! Thank you. $\endgroup$
    – John
    Oct 20, 2019 at 17:11

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