As an outsider to the field, I am surprised to learn that astronomers commonly use the equatorial coordinate system, in which fixed stars are not actually fixed.

Isn't that making life harder than it needs to be? Why would one not define a coordinate system in terms of fixed stars?

For example like this: Fixed star A is defined as (0°, 0°), fixed star B is defined as (0°, X). That would constrain the 3 degrees of freedom of a spherical coordinate system.

Just to be clear, by fixed stars I mean stars that do not change their apparent relative position over long time scales, say a million years. So stars in the Milky Way would likely be excluded.

When making a new observation, these two reference stars may not be visible, which would make calibration difficult. But there could be a reference catalog of fixed stars with precisely known coordinates. Any new observation could be made relative to a visible subset of that catalog. This would be similar in principle to the International Temperature Scale of 1990, which defines temperature relative to a set of reference materials and states.

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    $\begingroup$ The standard equatorial system is defined in terms of “fixed” extragalactic sources: en.m.wikipedia.org/wiki/… $\endgroup$ – Peter Erwin Jul 17 '19 at 7:59
  • $\begingroup$ I don't understand your question, all this exactly happens in astronomy? Distant quasars provide your calibration, and during one night they, as well as all other stars rotate with the celestial sphere, leaving their coordinates on this rotating coordinate system constant. On long timescales stars then show their proper motion, which is measured again relative to the distant 'non-moving' quasars. $\endgroup$ – AtmosphericPrisonEscape Jul 17 '19 at 9:52
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    $\begingroup$ I think this question is about axial precession, which causes the coordinates to change, necessitating the use of epochs etc. $\endgroup$ – James K Jul 17 '19 at 10:56
  • $\begingroup$ @PeterErwin that seems to be the answer. $\endgroup$ – uhoh Jul 18 '19 at 2:08
  • $\begingroup$ @JamesK I agree. $\endgroup$ – uhoh Jul 18 '19 at 2:08

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