# How can the 'spin' angular momentum vector NOT always be parallel to the spin angular velocity?

From Wikipedia:

The orbital angular momentum vector of a point particle is always parallel and directly proportional to the orbital angular velocity vector $$\omega$$ of the particle, where the constant of proportionality depends on both the mass of the particle and its distance from origin. The spin angular momentum vector of a rigid body is proportional but not always parallel to the spin angular velocity vector $$\Omega$$, making the constant of proportionality a second-rank tensor rather than a scalar.

So, in a rigid body, the vector(s)/tensor(s) might NOT be pointing in the same direction?

Huh?

• The angular momentum vector and the angular velocity vector are not necessarily in the same direction. Commented Jan 8, 2021 at 1:40
• @G.Smith How is that possible? I thought angular momentum was angular velocity times mass. How can factoring the mass change the direction? Commented Aug 9, 2022 at 7:33

The spin angular momentum, that is the angular momentum with respect to the center of mass, is conserved if there are no torques applied on the object.

The spin angular momentum is:

$$\mathbf L = \int_v \mathbf r \times d\mathbf p = \int_v \mathbf r \times \frac{d\mathbf r}{dt} \rho dv$$

Where $$\mathbf r$$ is the position vector of the points of the object from the COM. Because the object is a rigid body, the distance between points is constant, including distance from the COM. So the only possible movement of all points relative to the COM is a common instantaneous rotation.

But nothing requires that this rotation must be constant in time! For each point, the infinitesimal rotation means a displacement $$\Delta \mathbf r$$

$$\Delta \mathbf r = R\mathbf r - \mathbf r = (R - I)\mathbf r \implies \frac{d \mathbf r}{dt} = \Omega \mathbf r$$

Where $$\mathbf r$$ are the position vectors relative to the selected origin in the body, $$I$$ is the identity matrix, $$R$$ is the infinitesimal rotation matrix:

$$\begin{Bmatrix} 1 & -\theta_3 & \theta_2\\ \theta_3 & 1 & -\theta_1 \\ -\theta_2 & \theta_1 & 1 \end{Bmatrix}$$

and $$\Omega$$ is the matrix:

$$\begin{Bmatrix} 0 & -\omega_3 & \omega_2\\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{Bmatrix}$$

The $$\omega$$'s are the instantaneous angular velocities relative to the coordinates axis. The cross product in the integral of the angular momentum becomes:

$$\mathbf r \times \frac{d\mathbf r}{dt} = \mathbf r \times \Omega \mathbf r$$

Expanding the cross product, the angular momentum at any given time, relative to the point in the body, can be expressed as: $$\mathbf L = (\int_v \rho \mathbf M dv) \omega = \mathbf I\omega$$

where $$\mathbf I$$ is the inertia matrix, $$M$$ is the square matrix:

$$\begin{Bmatrix} (y^2 + z^2) & -xy & -xz \\ –yx & (z^2 + x^2) & -yz \\ -zx & –zy & (x^2 + y^2) \end{Bmatrix}$$

and $$\omega$$ is the column matrix:

$$\begin{Bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{Bmatrix}$$

As can be seen, $$\mathbf I$$ and $$\omega$$ can change with time, keeping $$\mathbf L$$ constant. There are however 3 axis in the object for that $$\mathbf L$$ is indeed parallel to $$\omega$$, corresponding to the principal moments of inertia.

• Way above my head. How about something for the layman? Commented Aug 9, 2022 at 7:35
• The angular momentum L is a vector, the same for the angular velocity. But they are related by a matrix called inertia matrix. When a vector (in the case $\omega$) is multiplied by a matrix, it changes not only magnitude but also direction. So L is not in general parallel to $\omega$. Commented Aug 9, 2022 at 13:53

For general 3D rotation of a rigid body about a fixed point or about its center of mass, $$\vec L = {\bf I} \vec \omega$$ where $$\vec L$$ is the angular momentum, $$\bf I$$ is the inertia tensor, and $$\vec \omega$$ is the angular velocity. If the body rotates about one of its principal axes, then $$\vec L$$ is parallel to $$\vec \omega$$.

The angular momentum vector $$\vec{L}$$ is parallel to the angular velocity vector $$\vec{\omega}$$ only for very symmetric bodies (for example a sphere or a cube).

For general bodies $$\vec{L}$$ is not necessarily parallel to $$\vec{\omega}$$. As an example let us consider a thin circular ring (like a hula hoop ring) of mass $$m$$ and radius $$r$$.

(image from Lists of moments of inertia)

Around the $$z$$-axis it has a moment of inertia $$I_z=mr^2$$. And around the $$x$$ and $$y$$ axes it has a moment of inertia $$I_x=\frac{1}{2}mr^2$$ and $$I_y=\frac{1}{2}mr^2$$. The important feature of this example is: the moments of inertia are not all the same. We can write these moments of inertia as a tensor $$I=\begin{pmatrix} \frac{1}{2}mr^2 & 0 & 0 \\ 0 & \frac{1}{2}mr^2 & 0 \\ 0 & 0 & mr^2 \end{pmatrix}$$

Then, in the general case, the angular momentum vector is given by $$\vec{L}=I\vec{\omega}$$

In our example (the hula hoop ring) this simplifies to \begin{align} L_x&=\frac{1}{2}mr^2 \omega_x \\ L_y&=\frac{1}{2}mr^2 \omega_y \\ L_z&=mr^2 \omega_z \end{align}

Now suppose the ring rotates around a tilted axis, with the angular velocity vector $$\vec{\omega}=\begin{pmatrix} 0 \\ \Omega \\ \Omega \end{pmatrix}$$

You get $$\vec{L}=\begin{pmatrix} 0 \\ \frac{1}{2}mr^2 \Omega \\ mr^2 \Omega \end{pmatrix}$$

Now you can see, this $$\vec{L}$$ is not parallel to $$\vec{\omega}$$.