<p>The equation of the equinoxes only applies to the origin, so it is assumed <span class="math-container">$ \delta = 0 $</span>. So <span class="math-container">$\tan \delta = 0, \sin \delta = 0 $</span>, which means your second equation simplifies to <span class="math-container">$\Delta\psi \cos \epsilon $</span>.</p>

CC BY-SA 4.0
2022-10-14T12:32:56.910Z
2022-10-14T12:32:56.910Z
Greg Miller
<p>I am studying about apparent and mean positions of Sun using quite old books: <em>Astronomical algorithms</em> (1991) and <em>Explanatory supplement to the astronomical almanac</em> (1992).</p>
<p>The question is about the mismatch between the equation of the equinoxes and the correction for a right ascension. It seems like these two are the same concept. However, their equations do not seem to be matched.</p>
<p>According to the p.116 of the second book, the <em>equation of the equinoxes</em> refers to the nutation in right ascension. It is defined as follows:
<span class="math-container">$$ \Delta \psi \cos \epsilon $$</span>
where <span class="math-container">$\Delta \psi$</span> is the <em>nutation in longitude</em> and <span class="math-container">$\epsilon$</span> is the <em>true obliquity of the ecliptic</em> (i.e., correction about the nutation is applied).</p>
<p>However, according to the p.120 of the same book, the first order correction <span class="math-container">$\Delta \alpha$</span> to the right ascension <span class="math-container">$\alpha$</span> is as follows:
<span class="math-container">$$\Delta \alpha = \left ( \cos \epsilon + \sin \epsilon \sin \alpha \sin \delta \right ) \Delta \psi - \cos \alpha \tan \delta \Delta \epsilon$$</span>
where <span class="math-container">$\delta$</span> is the declination.</p>
<p>Because <span class="math-container">$\epsilon$</span> is about 23.4 degrees and others are arbitrary values, <span class="math-container">$\Delta \alpha$</span> cannot be approximated as <span class="math-container">$ \Delta \psi \cos \epsilon $</span>.</p>
<p>It seems like there is some assumptions that I am missing or there is a problem in my understanding. I would like to ask help for this problem.</p>

CC BY-SA 4.0
2022-10-13T13:44:07.970Z
This may be helpful: https://en.wikipedia.org/wiki/Astronomical_nutation#Effect_on_position_of_astronomical_objects
PM 2Ring
<coordinate><positional-astronomy>
Mismatch between the equation of the equinoxes and the correction for a right ascension
J. Choi
# Mismatch between the equation of the equinoxes and the correction for a right ascension

(Asked by: username_0 on 2022-10-13)

I am studying about apparent and mean positions of Sun using quite old books: *Astronomical algorithms* (1991) and *Explanatory supplement to the astronomical almanac* (1992).

The question is about the mismatch between the equation of the equinoxes and the correction for a right ascension. It seems like these two are the same concept. However, their equations do not seem to be matched.

According to the p.116 of the second book, the *equation of the equinoxes* refers to the nutation in right ascension. It is defined as follows:
$$ \Delta \psi \cos \epsilon $$
where $\Delta \psi$ is the *nutation in longitude* and $\epsilon$ is the *true obliquity of the ecliptic* (i.e., correction about the nutation is applied).

However, according to the p.120 of the same book, the first order correction $\Delta \alpha$ to the right ascension $\alpha$ is as follows:
$$\Delta \alpha = \left ( \cos \epsilon + \sin \epsilon \sin \alpha \sin \delta \right ) \Delta \psi - \cos \alpha \tan \delta \Delta \epsilon$$
where $\delta$ is the declination.

Because $\epsilon$ is about 23.4 degrees and others are arbitrary values, $\Delta \alpha$ cannot be approximated as $ \Delta \psi \cos \epsilon $.

It seems like there is some assumptions that I am missing or there is a problem in my understanding. I would like to ask help for this problem.

**Tags:** coordinate, positional-astronomy

> **Comments:**
> **username_1**: This may be helpful: https://en.wikipedia.org/wiki/Astronomical_nutation#Effect_on_position_of_astronomical_objects  

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## 1 answers

### Answer by username_2 on 2022-10-14. Score: 3 | Accepted answer

The equation of the equinoxes only applies to the origin, so it is assumed $ \delta = 0 $. So $\tan \delta = 0, \sin \delta = 0 $, which means your second equation simplifies to $\Delta\psi \cos \epsilon $.


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