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In an online book

Geodätische Astronomie: Grundlagen und Konzepte

by Albert Schödlbauer I saw a picture of precession and nutation of Earth's axis. enter image description here

In German but I suppose it is understandable.
The precession is: $$ 360 ^\circ / (25.728 \text{ years}) = 50.2 '' \text{ per year} $$

In the book the autor states that the true rotation axis moves along a small cone wrt the average rotation axis.
This is the small elliptical cone with axes a/b in the picture.
Then he goes on and states that the true axis moves along the great precesion cone with an angular velocity of $$ 50.2'' \cdot \sin{(23.5)} = 20.1'' \text{ per year}.$$

Can someone explain why this is so, I don't get it by myself.

In another book the same is stated in the following way.

The earth's rotation vector is inclined $23.5^\circ$ to the pole of it's orbital plane, the ecliptic. The period of the resulting precession is about $26,000$ years, corresponding to a motion of the rotation vector of $20$ arcsec per year [$2\pi \sin{(23.5^\circ) / 26,000}$ radians per year].

Why ??

While: The line of intersection of the ecliptic and celestial equator precesses at a rate of $50$ arcsecs per year

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2 Answers 2

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This corresponds roughly to the angle between the rotation axis in year $n$ and the rotation axis in year $n+1$. It simply has to do with the way to account for angles in spherical coordinates. This is because if you look at the circle drawn by the tip of a unit vector precessing around $z$ with a constant angle $\theta$, the radius of this circle will not be $1$ but rather $\sin(\theta)$ (this is also why the elementary solid angle in 3D is $d\theta \sin(\theta) d\phi$ and not simply $d\theta d\phi$).

If you look at the angle formed between a unit vector of angle $(\theta, \phi)$ in spherical coordinates, with a vector of angle $(\theta, \phi + d\phi)$, it will be $\sin(\theta) d\phi$. Indeed, if $\theta = \pi/2$ for instance, the vector is moving along the equator, which is a great circle on the sphere, so you can directly convert $d \phi$ into an angle. In the other extreme case, if $\theta = 0$, then no matter the value of $\phi$, the vector will be exactly vertical so that all measured angles will be $0$. In between, you get a $\sin(\theta)$ factor.

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    $\begingroup$ Although I knew this, I could not connect it with the astronomical picture of equatorial and ecliptical planes. These planes intersect in a great circles and their line of intersection rotates $50''$ p.y. What in this picture is now moving at $20''$ p.y.? Now I see it is the North star or Polaris, the star at which the axis of rotation of the earth points to. This point projected on the celestial sphere (not the star itself of course) rotates not in a great circle but in a circle with diameter multiplied with $\sin{23.5}$, so this point of projection moves at $20''$ p.y. Do you agree? $\endgroup$
    – Gerard
    May 4, 2020 at 8:53
  • $\begingroup$ Yes, that is probably an even better way to phrase it. The line of intersection between the equatorial and ecliptical planes rotate by $50''$ a year (and go a full $360^{\circ}$ turn in $\sim 26000$ years), whereas the projection of the Earth rotation axis on the celestial sphere does not describe a great circle so the angle between two successive projections is smaller by a factor $\sin(23.5)$, which gives $50'' \sin(23.5) \simeq 20''$ per year. $\endgroup$
    – QuantumApple
    May 4, 2020 at 11:34
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    $\begingroup$ Well only this. In 26,000 : 2 yrs the great circle goes a half $180^{\circ}$ turn but it's hard to imagine the inclined Earth rotation axis to turn only $180^{\circ} \sin{23.5} = 71.8^{\circ}$. Not equal to $2\cdot 23.5=47^{\circ}$ probably bc. the $71.8^{\circ}$ is an integrated value and $47^{\circ}$ just a difference. $\endgroup$
    – Gerard
    May 4, 2020 at 14:10
  • $\begingroup$ Yes, that's also true. The "distance" that is "travelled" by the Earth rotation axis vector on the unit sphere during $26000/2$ years is $\pi \sin(23.5) = 1.25$ "rad" ($= 71.8^{\circ}$). But the actual angle $\theta$ between the two vectors is such that $\sin(\theta/2) = r$, where $r = \sin(23.5)$ is the radius of the small circle which is drawn by the vector on the unit sphere. It is because you're adding vectorial quantities together, so you need to take into account the fact that the infinitesimal vector quantity $d \phi e_{\phi}$ is rotating during the integration. $\endgroup$
    – QuantumApple
    May 4, 2020 at 16:52
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So like remarked by QuantumApple it's just spherical coordinates. The angle that is meant in the referenced texts is the angle between two positions of Earth's rotation axis. When this angle is small e.g. $\phi$ and the angle in the equatorial plane - or any other horizontal plane - is also small $\theta$, then the cosine rule gives:

$$ r^2 + r^2 - 2 \cdot r\cdot r \cos{\phi} = (r\sin{(23.5)} \cdot \theta)^2 $$

and with $\cos{\phi} = 1 - 0.5 \phi^2$ this gives $\phi = \theta \sin{(23.5)}$.

So for small angles $d\phi = 20''$, $d\theta = 50''$.
I didn't realize that during a full $360^\circ$ rotation in a horizontal plane the $23.5^\circ$ tilted axis of rotation does not turn $360^\circ$.

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