Assume we have a position vector $\mathbf r$ dictating the position of a particle in a rotating frame. Then, for a time dependent rotation matrix $\mathbf R(t)\in SO(3)$ that describes the motion of the particle, we can say that for the inertial frame, the particle's position $\mathbf r^*$ is $$\mathbf r^* = \mathbf R(t) \,\mathbf r$$ Likewise, it should follow pretty obviously that for particle's velocity $\mathbf v$ and acceleration $\mathbf a$ in its rotating frame, $$\mathbf v^* = \mathbf R(t)\, \mathbf v,\qquad \mathbf a^* = \mathbf R(t)\, \mathbf a$$

Again to clarify, $\mathbf x^*$ is the inertial frame, and $\mathbf x$ is the rotating frame.

Anyhow, I want to find the expression for the angular velocity tensor. For linear velocity, we can see that $$\mathbf v^* = \frac{d\mathbf r^*}{dt} = \mathbf R(t)\, \dot{\mathbf r} + \dot{\mathbf R}(t) \, \mathbf r$$

Since we know that for a rotation matrix, $\mathbf R^\intercal \mathbf R=I$, we can factor out a $\mathbf R(t)$ from above to get $$\mathbf v^* = \mathbf R(t)(\dot{\mathbf r} + \mathbf R^\intercal(t)\dot{\mathbf R}(t) \, \mathbf r)$$

This manipulation seems arbitrary, but then, apparently we somehow define the angular velocity tensor to be $$\Omega = \mathbf R^\intercal(t)\dot{\mathbf R}(t) \iff \mathbf v^* = \mathbf R(t)(\dot{\mathbf r} + \Omega \, \mathbf r) = \mathbf R(t)\, \mathbf v$$

This seems even more arbitrary, and honestly magical even that $\Omega$ even returns anything physically meaningful.

My question is basically, how does the definition $\Omega = \mathbf R^\intercal(t)\dot{\mathbf R}(t) $ "work" in a sense. What is the motivation for the definition, and why does it give us the angular velocity tensor? Intuitively, I want to think that it would instead be $\Omega = \dot{\mathbf R}(t)$, but obviously this isn't the case.

In addition, the angular velocity vector can be interpreted as a bivector or 2-vector because bivectors are isomorphic to skew-symmetric matrices and the angular velocity vector is skew-symmetric (which can be shown by differentiating $\mathbf R^\intercal \mathbf R=I$).

We know from the expression of $\mathbf v^*$ that $$\dot{\mathbf r} + \Omega \, \mathbf r = \mathbf v$$

If we look at the second term on the LHS, this is unusual because normally, we would have expressions like $$\tau_\text{bivector} = \mathbf r \wedge \mathbf F \tag{1}$$ where we wedge two vectors to return a bivector. If we take the Hodge dual, we can see the isomorphism between the bivector and vector cross product $$\star (\tau_\text{bivector}) = \star (\mathbf r \wedge \mathbf F) \iff \tau_{\text{axial vector}} = \mathbf r \times \mathbf F$$

Yet here, we somehow have an expression where we are mutiplying a bivector with a vector, and this multiplication operation

$$\Omega_{\text{bivector}} \,???\, \mathbf r_\text{vector}$$

which gives us a vector and also is isomorphic to the cross product (since the cross product formula is $\omega \times \mathbf r = \mathbf v$)

What is the interpretation for the multiplication of the bivector with a vector? Are we "wedging" a bivector with a vector (and if so, how does it work?)? Or, in general, what is this operation and how is it isomorphic to a cross product that somehow returns a regular vector?

The wikipedia article on the angualr velocity tensor has a section on this but I have never seen the idea of a musical isomorphism so I didn't quite understand it.


1 Answer 1


It’s clearer if you distinguish the spaces: the lab frame (inertial) and the moving frame. $R$ sends the latter to the former. This is true for $\dot R$ as well.

Physically, you want define angular velocity within the same frame and not as a hybrid object. Mathematically, this means you want it to be an endomorphism, ie it sends within one of the two spaces within itself. This is why you apply the inverse. $\Omega=R^{-1}\dot R$ is a valid endomorphism of the moving frame.

Note that an equally valid choice is to take $\omega=\dot RR^{-1}$. It is also skew symmetric as well and is this time an endomorphism of the lab frame. This is why it is interpreted as the angular velocity in the lab frame.

Mathematically, your group $SO(3)$ is not flat. You proposition, $\dot R$ lies in the tangent space of $R$ and can be seen as the velocity of the rotation in $SO(3)$. However, it is hard to compare it with another velocity at a different orientation, since there is no canonical identification of the tangent spaces. This is the same as the previous argument, $\dot R$ depends on the two frames so it depends on their relative orientation as well and not on their intuitive velocity.

The trick is to use the group structure and send the tangent plane to the tangent plane at unity (the Lie algebra). You can do this by right multiplication or left multiplication which gives you the angular velocity in the lab frame and the moving frame respectively.

This approach is explained in more detail in Arnold’s Mathematical Methods of Classical Mechanics.

For the cross product, the discussion of bibecvtors is needlessly complicated. It is simpler to look directly at the link between cross product and skew symmetric operators which is what is relevant at the end of the day. The natural isomorphism between vectors and skew symmetric endomorphisms given by: $$ a\to A = (x\to a\times x) $$ i.e. for a vector $a$, you can associate the endomorphism $A$ which sends $x$ to $a\times x$. Indeed, $A$ is a valid endomorphism by the linearity of the cross product (second factor). Again by the linearity of the cross product (first factor), it is a vector space morphism. To check that it is surjective and injective, the easiest way is to look at it in an orthonormal basis: $$ \begin{align} (Ax)_i &= A_{ij}x_j &= \epsilon_{ijk}a_jx_k \end{align} $$ with $\epsilon$ the Levi-Civita symbol. From $A$ you can uniquely determine $a$ using: $$ a_k = \frac{1}{2}\epsilon_{ijk}A_{ij} $$ which directly gives you subjectivity and injectivity. Thus you did define an isomorphism. The angular velocity vector is the vector you obtain from the angular velocity skew symmetric operator via this isomorphism.

Hope this helps.

  • $\begingroup$ Thank you for the detailed response :D (+1). Just to clarify on a very general/abstract sense for the first question, taking a time derivative of $R$ results in a matrix that is angular velocity, but converts between two frames, and we use left or right multiplication with an inverse to keep it within a given frame? Also, I've edited my second question to make it more clear, so if youre able to look at that, and perhaps clarrify what you meant by the notation here $\omega\to (r\to \omega\times r)$ that would be much appreciated! $\endgroup$
    – Max0815
    Sep 15, 2023 at 4:20
  • $\begingroup$ Yes, you by the additional multiplication you "reset" the configuration so that the relative orientation does not matter in the angular velocity. I've added a bit more clarification for the second part. $\endgroup$
    – LPZ
    Sep 15, 2023 at 13:25
  • $\begingroup$ Thanks again! I will give this some thorough thought to understand it fully xd $\endgroup$
    – Max0815
    Sep 15, 2023 at 15:01

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