Orbital and spin angular velocity?

• etotheipi

etotheipi

The Wikipedia page for angular velocity makes a big fuss over "spin" and "orbital" angular velocities, but I have checked through Gregory and Morin's textbooks on classical mechanics and haven't found any reference to them at all. They just work with a single quantity, the angular velocity.

Alarmingly, Wikipedia says that the "orbital" angular velocity is dependent on the choice of origin. That immediately seemed odd, since those other two sources both defined ##\vec{\omega}## w.r.t. the instantaneous axis of rotation of the particle/rigid body, which should be invariant under translations (unlike the angular momentum, which is evidently dependent on origin). Furthermore, this notion of "orbital angular velocity" doesn't seem to satisfy any of the theorems given for angular velocities in those texts (e.g. addition of angular velocities in different reference frames, etc.)

I wondered if Gregory and Morin are always referring to spin angular velocities? And if so, is there ever any need for the "orbital" variety of angular velocities? Thanks!

Last edited by a moderator:
I make a distinction between orbital angular momentum, ##\vec L_{orb}=\vec r \times \vec p## and (rigid body) spin angular momentum ##\vec L_{spin}=I_{cm}\vec \omega##. The former is position-dependent while the latter is about the CM and does not depend on a position vector. That much is clear. I am not familiar with the textbooks you mentioned, so I cannot express an opinion. Can you be more specific and explain your statement "Furthermore, this notion of "orbital angular velocity" doesn't seem to satisfy any of the theorems given for angular velocities in those texts." Which notion doesn't seem to satisfy which theorems and how did you come to this conclusion?

As for the need for the "orbital" variety of angular velocity, how would describe the ages-old observation that the Moon always shows the same side to the Earth? I would say "The orbital angular velocity and the spin angular velocity are equal." What would you say?

etotheipi and Motore
That immediately seemed odd
The moment of inertia depends upon the choice of calculation axis. Surely the angular momentum cannot! Therefore...

etotheipi
I make a distinction between orbital angular momentum, ##\vec L_{orb}=\vec r \times \vec p## and (rigid body) spin angular momentum ##\vec L_{spin}=I_{cm}\vec \omega##. The former is position-dependent while the latter is about the CM and does not depend on a position vector. That much is clear. I am not familiar with the textbooks you mentioned, so I cannot express an opinion. Can you be more specific and explain your statement "Furthermore, this notion of "orbital angular velocity" doesn't seem to satisfy any of the theorems given for angular velocities in those texts." Which notion doesn't seem to satisfy which theorems and how did you come to this conclusion?

The textbooks I am using only refer to "angular velocity", and do not use the terms "orbital" and "spin" as in the Wikipedia article. For angular momenta I can understand the distinction between orbital and spin, but not so much for angular velocities. For instance, the rate of change of a vector ##\vec{a}## in a frame ##\mathcal{F}## is related to that in ##\mathcal{F}'## via$$\left( \frac{d\vec{a}}{dt} \right)_{\mathcal{F}'} = \left( \frac{d\vec{a}}{dt} \right)_{\mathcal{F}} + \vec{\Omega} \times \vec{a}$$here ##\vec{\Omega}## is independent of any chosen origin, it only depends on the relative orientations of the two frames. If the angular velocity of frame ##\mathcal{F}## w.r.t. ##\mathcal{F'}## is ##\vec{\Omega}_1## and the angular velocity of frame ##\mathcal{F'}## w.r.t. ##\mathcal{F''}## is ##\vec{\Omega}_2##, then the angular velocity of frame ##\mathcal{F}## w.r.t. ##\mathcal{F''}## is ##\vec{\Omega}_1 + \vec{\Omega}_2##. Here the ##\vec{\Omega}##s are only dependent on relative orientation, not any choice of origin.

I would write the angular velocity of a particle moving as a circle as invariant of the chosen origin, since it is the instantaneous axis of rotation which is important. This is also contrary to "orbital angular velocities".

Last edited by a moderator:
As for the need for the "orbital" variety of angular velocity, how would describe the ages-old observation that the Moon always shows the same side to the Earth? I would say "The orbital angular velocity and the spin angular velocity are equal." What would you say?

For this example I would just express the angular velocity of the moon w.r.t. the Earth as a single ##\vec{\Omega}##, which could either be decomposed into the body fixed moon frame (0) plus that of the body fixed frame w.r.t. the Earth frame (##\vec{\Omega}##), or decomposed into a moon fixed frame with axes always aligned with the Earth (##\vec{\Omega}##), plus that w.r.t. the Earth frame (0).

Last edited by a moderator:
kuruman
For this example I would just express the angular velocity of the moon w.r.t. the Earth as a single ##\vec{\Omega}##, which could either be decomposed into the body fixed moon frame (0) plus that of the body fixed frame w.r.t. the Earth frame (##\vec{\Omega}##), or decomposed into an inertial, moon fixed Earth frame (##\vec{\Omega}##) plus that w.r.t. the Earth frame (0).
Great. Now tell me how you would explain to an eight year old why the same side of the Moon faces the Earth at all times. Sometimes the decomposition of a resultant into components has more explanatory power than the resultant itself. Another example is the decomposition of the single contact force exerted by a surface on a mass into orthogonal components, also known as the "normal force" and the "force of friction".

Last edited:
vanhees71 and etotheipi
I think I figured it out, so I'll make a note of it just in case anyone else has a similar question in the future.

Orbital Angular Velocities
Let the position vector ##\vec{r}## and the velocity vector ##\vec{v}## define an instantaneous plane, with normal vector ##\vec{u}##. Then the orbital angular velocity is defined by $$\vec{\omega} = \omega \vec{u} = \frac{d\phi}{dt} \vec{u}$$such that$$\vec{v}_{\bot} = \vec{\omega} \times \vec{r}$$The orbital angular velocity is defined w.r.t. a chosen coordinate system, and is not invariant under constant translations and rotations. Orbital angular velocities generally do not add between frames, since the perpendicular components of velocity differ under time independent translations.

Spin Angular Velocities
The spin angular velocity is identical to the orbital angular velocity about the instantaneous centre of rotation of the body in the frame ##\mathcal{F}##. That is to say that ##\vec{\omega} = \omega \hat{n}##, if ##\hat{n}## is the instantaneous axis of rotation in the frame ##\mathcal{F}##. As such, they refer to the rates of change of orientation of a frame w.r.t. another frame and are invariant under time independent translations. About the centre of rotation, ##\vec{v} = \vec{v}_{\bot}##.

The spin angular velocity is related to the time derivatives of the Euler angles as$$\vec{\omega} = \dot{\alpha}\vec{u}_1 + \dot{\beta} \vec{u}_2 + \dot{\gamma}\vec{u}_3$$the components of which could be converted into a body fixed frame.

Spin angular velocities do add between frames, i.e. ##\vec{\omega}= \vec{\omega}_1 + \vec{\omega}_2##.

Summary
I think I am convinced that Gregory and Morin both refer to the spin angular velocity whenever they mention ##\vec{\omega}##, and I don't think we ever need to use orbital angular velocities.

Last edited by a moderator:
For this example I would just express the angular velocity of the moon w.r.t. the Earth as a single ##\vec{\Omega}##, which could either be decomposed into the body fixed moon frame (0) plus that of the body fixed frame w.r.t. the Earth frame (##\vec{\Omega}##), or decomposed into an inertial, moon fixed Earth frame (##\vec{\Omega}##) plus that w.r.t. the Earth frame (0).

How would you explain that we do not always see exactly the same side of the Moon?

I would say that the spin angular velocity remains constant while the orbital angular velocity varies (approximately according to Kepler 2) and has another direction.

If a particle rotates in a circle...
Do you mean translate along a circle? I prefer to reserve the word "rotation" for changing the orientation of the body-fixed axes w.r.t some coordinate axes. This also means, that angular velocity/momentum w.r.t. some point, doesn't imply "rotation".

etotheipi
Do you mean translate along a circle? I prefer to reserve the word "rotation" for changing the orientation of the body-fixed axes w.r.t some coordinate axes. This also means, that angular velocity/momentum w.r.t. some point, doesn't imply "rotation".

Yes, that's better wording - thanks for catching it! I updated post #7 to reflect what I understand so far. I believe the "spin" variety refers to changing orientation, i.e. relative rotation of two frames (and think this is the more common usage for angular velocity).

Whilst an object traveling in a straight line might have some orbital angular velocity, since the radius of curvature of its trajectory is zero its spin angular velocity will be zero.

Last edited by a moderator:
Yes, that's better wording - thanks for catching it! I updated post #7 to reflect what I understand so far. I believe the "spin" variety refers to changing orientation, i.e. relative rotation of two frames (and think this is the more common usage for angular velocity).
Yes, if I read "angular velocity", without further qualifiers (reference point), I assume the changing orientation is meant. If I read "angular momentum" without reference point definition, I assume it's taken around the center of mass.

Whilst an object traveling in a straight line might have some orbital angular velocity, since the radius of curvature of its trajectory is zero its spin angular velocity will be zero.
For a straight line the radius of curvature is infinite, while the curvature is zero.

etotheipi
Yes, if I read "angular velocity", without further qualifiers (reference point), I assume the changing orientation is meant. If I read "angular momentum" without reference point definition, I assume it's taken around the center of mass.

Thanks, that's very reassuring ☺. I just found it quite unnerving that all mentions of angular velocity in my textbook seemed to refer to the spin angular velocity, whilst Wikipedia spent so much time explaining the other. I think it's clear now!

For a straight line the radius of curvature is infinite, while the curvature is zero.

Sorry, you're of course right

I was hoping there wouldn't be need for this, but here it is anyway. Consider mass element ##m_i## on a rigid, moving and rotating body. Its angular momentum relative to an arbitrary origin is$$\vec L_{i } = m_i\vec r_{i} \times \vec v_{i}$$The position of the element can be written as $$\vec r_i=\vec R+\vec r_i'$$where ##\vec R## is the position of the CM and ##\vec r_i'## is the position vector of the element relative to the CM. Then,$$\vec L_i=m_i(\vec R+\vec r_i')\times\left(\frac{d\vec R}{dt}+\frac{d\vec r_i'}{dt}\right)~~~~~(1)$$ Note that $$M \vec R=\sum_i m_i \vec r_i=\sum_i m_i \vec R+\sum_i m_i \vec r'_i=M \vec R +\sum_i m_i \vec r'_i~~\rightarrow ~~\sum_i m_i \vec r'_i=0.$$ Now sum equation (1) over ##i##, remove the parentheses and rearrange to get,\begin{align} \vec L_{tot.}=\sum_i \vec L_i=M\vec R\times \frac{d\vec R}{dt}+\sum_i m_i\left( \vec r'_i\times\frac{d\vec r_i'}{dt}\right) \nonumber \\ + \vec R\times \frac{d}{dt}\sum_i m_i\vec r_i' +\left(\sum_i m_i\vec r_i'\right)\times \frac{d\vec R}{dt}.\nonumber \end{align}The last two terms are zero and we are left with $$\vec L_{tot.}=M\vec R\times \frac{d\vec R}{dt}+\sum_i m_i\left( \vec r'_i\times\frac{d\vec r_i'}{dt}\right).$$ Define orbital and spin angular velocity vectors through $$\vec v_{orb.}=\frac{d\vec R}{dt}=\vec{\omega}_{orb.}\times \vec R~;~~\vec v_{spin}=\frac{d\vec r_i'}{dt}=\vec{\omega}_{spin}\times \vec r_i'.$$Note that ##\vec{\omega}_{spin}## is the same for all mass elements on a rigid body. Then, $$\vec L_{tot.}=M\vec R\times(\vec{\omega}_{orb.}\times \vec R)+\sum_i m_i \vec r'_i\times (\vec{\omega}_{spin}\times \vec r'_i).$$Finally, expand the triple products to get, $$\vec L_{tot.}=MR^2\vec{\omega}_{orb.}+\sum_i (m_i r_i^2) \vec{\omega}_{spin}=MR^2\vec{\omega}_{orb.}+I_{cm}\vec{\omega}_{spin}=\vec L_{orb.}+\vec L_{spin}.$$This says that the total angular momentum can be decomposed into the orbital angular momentum of the center of mass (with the rigid body being treated as a point mass), plus the spin angular momentum about the center of mass. This decomposition is entirely analogous to the total kinetic energy split between the energy of the center of mass (##=\frac{1}{2}MV_{cm}^2##) and the kinetic energy about the center of mass (##=\frac{1}{2}I_{cm}\omega_{spin}^2##).

Last edited:
vanhees71, Leo Liu and etotheipi
@kuruman thank you for writing this out, you are right, this is the König theorem for angular momenta and it is a very useful tool!

My only gripe is that sometimes we will not be able to express ##\frac{d\vec{R}}{dt}## in the form ##\vec{\omega}_{orb} \times \vec{R}## because it is not so often that the reference point on the rigid body is also moving in a circular velocity field. More generally it looks like we would only have ##\vec{v}_{orb, \bot} = \vec{\omega}_{orb} \times \vec{r}##.

Nonetheless, it does work fine in the case of something like the moon orbiting the Earth, which is what I assume you were considering for the calculation. Thanks, I really appreciate your insight!

Last edited by a moderator:
@kuruman
My only gripe is that sometimes we will not be able to express ##\frac{d\vec{R}}{dt}## in the form ##\vec{\omega}_{orb} \times \vec{R}## because it is not so often that the reference point on the rigid body is also moving in a circular velocity field. More generally it looks like we would only have ##\vec{v}_{orb, \bot} = \vec{\omega}_{orb} \times \vec{r}##.
I am not sure I understand your gripe. In cases where there is a radial ##\dot R## component to the velocity of the CM, it will not contribute to the angular momentum about the origin and the decomposition is still valid. Did I miss something?

etotheipi
I am not sure I understand your gripe. In cases where there is a radial ##\dot R## component to the velocity of the CM, it will not contribute to the angular momentum about the origin and the decomposition is still valid. Did I miss something?

Yes sorry I did not explain myself too well... I agree with your final result, but noted that the intermediate step writing ##\frac{d\vec{R}}{dt} = \vec{\omega}_{orb} \times \vec{R}## only holds if the centre of mass velocity describes a circular field. More generally, we must wait a few more steps before introducing the orbital angular velocity$$\vec{L}_{orb} = M\vec{R} \times \frac{d\vec{R}}{dt} = M\vec{R} \times \vec{V}_{\bot} = M\vec{R} \times (\vec{\omega}_{orb} \times \vec{R}) = MR^2 \vec{\omega}_{orb}$$where I have used that ##\vec{R} \times \frac{d\vec{R}}{dt} = \vec{R} \times (\vec{V}_{\bot} + \vec{V}_{\parallel}) = \vec{R} \times \vec{V}_{\bot}## since ##\vec{R} \times \vec{V}_{\parallel} = \vec{0}##

But really, a very small quibble ☺

Last edited by a moderator:
Yes, of course. I jumped ahead a step or two in the LaTeX composition. Thanks for providing the missing steps.

etotheipi
It get's even more involved if you don't take the CM as the origin of the body-fixed frame. Then you get in addition a term you can call "spin-orbit coupling" ;-), taking into account the rotation of the center of mass around the arbitrary body-fixed point.

It's too hot here in Germany right now to type all this lengthy derivation into the forum. Unfortunately I've worked this out only in a German manuscript. Perhaps you can nevertheless read it, because of the high "density of formulas" ;-)):

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

Sect. 4.2.

kuruman and etotheipi