I was recently introduced to Euler Angles in a Dynamics course, but I am confused on the difference between precession and spin angles. Both precession and spin consist in rotating a coordinate system about the $z$-axis, which means that they have the same transformation matrices.

# [Physics] the difference between precession and spin angles

precessionreference framesrigid-body-dynamicsrotational-dynamics

#### Related Solutions

Why does not the precession rate of the angular velocity vector give exactly the frequency of wobbling? In other words, how come the precession of $\vec\omega$ is different to the line of nodes rotation?

Short answer: Because the precession rate $\Omega$ corresponds to the precession of $\vec\omega$ in the body frame, not in the Earth frame. The variables in the Euler equations, $I_i$, $\omega_i$ and $\tau_i$, are in the body frame.

The angular velocity of the disk has two contributions, the component $\vec\omega'$ due to the spin and the component $\vec\omega''$ due to a rotation of the inclined disc about a vertical axis. The latter corresponds to the wobbling as viewed by someone in the Earth frame.

As we can see from the figure, the resultant angular velocity $\vec\omega$ is always off the symmetry axis of the disk, which means it has a non-vanishing projection in the plane of the disk (dashed line). At same time, the axes fixed in the plane of the disk, $\rho_1$ and $\rho_2$ rotates with spin $s$ (viewed by someone in the Earth frame). Hence, someone in the disk frame will see the axes fixed and the projection of $\omega$ in this plane rotating with rate $s$. That is why the precession rate equals the spin.

On the other hand, the wobbling effect is due to the non-vanishing $\vec\omega''$. A $2\pi$ rotation about the vertical line corresponds to a complete oscillation (wobbling) therefore the frequency of the wobbling is actually equal to the magnitude of $\vec\omega''$.

I think that you are confusing more classical arguments about magnetism. Maybe you already know what I am going to write, but I think this is the point.

First, precession is typical of the classical model of diamagnetism (here and here you may find something relevant). Consider an hydrogen atom in the Bohr model; you can demonstrate that the orbiting electron generates an orbital angular momentum. When we apply a magnetic field, on that momentum will act a couple that can be written as: $$\vec{τ} = \vec{μ} \times \vec{B} $$ with $B$ magnetic field and $μ$ magnetic moment of the system (in this case, due to the orbiting). If you call $L$ the angular orbital momentum of the electron, you can write the Euler's equation for the dynamic of a rigid body: $$ \vec{τ} =\frac{d\vec{L} }{dt}. $$ By reasoning on the derivative of this vector and on the direction of the torque, you can see that this will cause a precession of $L$ around $B$. The projection of $L$ in the direction of $B$ is constant.

The second part, about alignment, make me think about the classical model of paramagnetism (I can't actually find a really satisfying link to it, but I think it will be explained in all the books about magnetism from a classical point of view). In this theory, you assume that your atom has a certain magnetic dipole moment (cause by the angular orbital momentum or by the spin, with a semiclassical approach) that in a magnetic field will have an energy: $$ U = - \vec{μ} \cdot \vec{B} = - μB \cos{θ} $$ where $θ$ is the angle between the two vectors. Using Boltzmann's statistics, you can evaluate the mean value of $\cos{θ} $ and then explain the properties of paramagnetic materials. As you can see, the energy of this interaction between the magnetic dipole $μ$ and the magnetic field $B$ will the smallest possible if the two vectors are aligned.

In these classical or semiclassical models, the spin is usually added "artificially" as another magnetic dipole that can precede or can interact with the magnetic field. Even if they lead to a correct result, these classical models aren't right; other models, based on quantum mechanics, have been developed to discuss magnetic properties of materials.

I would like to add some references or sources, but I don't have any book written in English, only my teacher's notes. I will add them in the future, eventually.

## Best Answer

Yes, but in between the two there is a rotation about $x$. (Note: Some use $y$ for the middle rotation). This makes all the difference in the world for describing the orientation of a world with respect to the ecliptic, the original usage of Euler rotation sequences.

A bit overly simplified, the initial rotation about the original $z$-axis is the precession angle. Precession for a planet is very slow. The second rotation about the once-rotated $x$-axis is the nutation angle. For the Earth, this closely corresponds to the nearly constant obliquity. The final rotation about the twice-rotated $z$ axis is the daily rotation angle.

With two exceptions, given any orientation, the Euler sequence that rotates the initial $x$, $y$, and $z$ axes to the $X$, $Y$, and $Z$ axes that correspond to the desired orientation is unique with the constraint that the rotations about $z$ are between 0° (inclusive) and 360° degrees (exclusive) and the intermediate rotation about the once-rotated$x$-axis is between 0° and 180° (exclusive of both).

The two exceptions occur when the intermediate rotation about the once-rotated $x$-axis is 0° or 180°. These situations are called "gimbal lock." Here the distinction between precession and rotation

iscompletely arbitrary; the standard approach is to arbitrarily set one of the two to zero.