# How Does A Precessing Sphere Precess?

In particular, consider a perfectly spherical object or radius $r$, with a certain axis $L$, and this axis titled relative to a vertical axis by some amount $\theta$.

Say that it's "wobbling" in a stable, circular manner -- the axis $L$ is revolving around the vertical axis, but comes back to where it started after a revolution.

In other words, the sphere looks like this one.

My question is, how can I describe the motion of any point on that sphere (assuming it's a perfect sphere)?

First, notice that if you look at the bottom-most point of Africa, it rotates in an ellipse counterclockwise. But, the North & South Poles as viewed from the top are rotating clockwise!

But if you look at areas around the equator (the green areas of Africa), they seem to have an 8-shaped motion with the smaller circle being formed a the bottom.

How can this be? I'm totally mystified. How can different points of the sphere rotate in different directions (and opposite directions, too) in that way?

I also imagine that some points from behind the Earth in that video will be rotating clockwise at the same time as Africa is rotating counterclockwise. But then again, it seems like the motions change from upper to lower hemispheres, and the directions seem to be the same in the same hemisphere if you view it overhead.

Is there a way for me to know how a certain point is moving? Or is this animation misleading me?

This seems to me like a purely mathematical problem if we assume the ideal case.

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Not an answer but "precession" will give better search results than "wobbling." – daniel Jun 22 '13 at 15:24
Thank you, I'll edit accordingly :) – markovchain Jun 22 '13 at 15:32
Although, because the sphere isn't spinning, can it still be considered precession? – markovchain Jun 22 '13 at 15:36
I think the answer is yes, if precession is a "change in the orientation of the axis of rotation." The absence of rotation doesn't mean there isn't an axis of rotation--in the abstract we could choose one arbitrarily. Of course if there is no spin the motion of a point on the surface of the sphere will be different. – daniel Jun 22 '13 at 15:42

# Summary

A perfect sphere will not precess.

• What the video shows is not true precession
• The earth's axis does precess, but that's exactly because the earth is not perfectly spherical

You may have noticed that while the earth goes round the sun, the direction of the depicted orbit of the earth does not change. However it is quite obvious from looking on an image of the complete orbit that the direction does change (e.g. if drawn on a vertical plane, on the left of the sun it's vertical, while below the sun it's horizontal). Also, in the video, the sun is always on the right.

So why does the direction not change in the video? Well, it's because the video is in a rotating frame of reference. That is, the "camera" is rotating around the (simulated) earth, in a way that the sun is always on the right of the earth. That's why there's an apparent precession: What you are seeing is the apparent rotation of the complete universe due to the movement of the camera. If background stars had been depicted, you would have seen the stars "rotating" as well.

The best way to see that the earth's axis does not do a full rotation during a single year is to notice that the earth axis does point (approximately) to the polar star both in the summer and in the winter. It wouldn't if the earth's axis would actually process a noticeable amount per year.

Now the video does not show how the earth really rotates today, but how it would rotate in a tidal lock. However the only difference is that the earth would rotate along its axis once a year. Indeed, you'd get the very same movie with our fast-rotating year if you'd make a picture once every 24 hours (well, with some correction for the intervals because of the excentricity of the earth's orbit, but let's ignore those complications).

OK, so how to mathematically describe the movement? Well, if we describe the (active) rotation of the earth with a rotation matrix $R_a(t)$ and the (passive) rotation of the frame of reference with $R_p(t)$, then the rotation is described by the combined rotation $R_p(t)R_a(t)$. Now the passive rotation is in an axis perpendicular to the ecliptic; in the video that's mapped to the $z$ axis. The earth's axis it tilted by an angle $\alpha$ from that direction. So if $R_z(\phi)$ is the rotation matrix around the $z$ axis, $R_y(\phi)$ is the same for the $y$ axis, $\omega_p$ is the angular speed of the earth orbit ($1/\text{year}$) and $\omega_a$ is the angular speed of the earth rotation (in the tidal lock case, $\omega := \omega_a = -\omega_p$), and we put the time $t=0$ at the point when the earth's axis in the rotating frame is tilted in the $y$ direction, the complete rotation matrix for time $t$ for the movie is $$R(t) = R_z(-\omega t)R_y(-\alpha)R_z(\omega t)R_y(\alpha)$$

Now $$R_z(\phi) = \begin{pmatrix} \cos\phi & \sin\phi & 0\\ -\sin\phi & \cos\phi & 0\\ 0 & 0 & 1 \end{pmatrix}$$ and $$R_y(\phi) = \begin{pmatrix} \cos\phi & 0 & \sin\phi\\ 0 & 1 & 0\\ -\sin\phi & 0 & \cos\phi \end{pmatrix}$$

So the total depicted rotation is described by the rotation matrix $$\left( \begin{array}{ccc} \cos (\omega t) \sin ^2(\alpha )+\cos (\alpha ) \left(\cos (\alpha ) \cos ^2(\omega t)+\sin ^2(\omega t)\right) & (\cos (\alpha )-1) \cos (\omega t) \sin (\omega t) & \sin (\alpha ) \left(\sin ^2(\omega t)+\cos (\alpha ) (\cos (\omega t)-1) \cos (\omega t)\right) \\ \left(\cos (\omega t) \cos ^2(\alpha )-\cos (\omega t) \cos (\alpha )+\sin ^2(\alpha )\right) \sin (\omega t) & \cos ^2(\omega t)+\cos (\alpha ) \sin ^2(\omega t) & (\cos (\alpha ) (\cos (\omega t)-1)-\cos (\omega t)) \sin (\alpha ) \sin (\omega t) \\ \cos (\alpha ) (\cos (\omega t)-1) \sin (\alpha ) & \sin (\alpha ) \sin (\omega t) & \cos ^2(\alpha )+\cos (\omega t) \sin ^2(\alpha ) \end{array} \right)$$ For $\alpha=0$ you'd get the identity matrix. That's because in that case, the passive rotation just cancels out the active rotation.

However note again, that this is not the actual rotation, but the apparent rotation due to the rotating reference frame. For the actual rotation, omit $R_z(-\omega t)$. Then you get a simple rotation around a fixed axis tilted against the $z$ axis by the angle $\alpha$.

Since the axis of rotation rotates around the $z$ axis, a rotation along the $x$ axis (or a combination of $x$ and $y$ axis rotations) would just amount to shifting the start time. So basically it would not create anything qualitatively new. – celtschk Jun 23 '13 at 11:55