$\def\va{\vec{\alpha}} \def\vw{\vec{\omega}} \def\vn{\vec{n}}$Let $\va(t)$ be a rotation vector such that its direction is the rotational axis and its length $\alpha=|\va|$ is the angle describing the rotation. In Is there a formula for the rotation vector in terms of the angular velocity vector? the formula $$ \vec{\omega}= \dot{\vec{\alpha}} + \frac{1 - \cos \alpha}{\alpha^2} \left(\vec{\alpha} \times \dot{\vec{\alpha}}\right) + \frac{\alpha - \sin \alpha}{\alpha^3} \left(\vec{\alpha} \times \left(\vec{\alpha} \times \dot{\vec{\alpha}}\right)\right)\, $$ is given which relates angular velocity $\vw$ to the rotation vector $\va$ and its time derivative.

If I multiply the formula with $\va$, the two elaborate terms on the right disappear, because both contain a cross product of $\va$ such that their dot product with $\va$ is zero. I get:

$$\vw\cdot\va = \dot\va\cdot\va \tag{1}$$

Because $\vw$ and $\va$ are parallel, we also have $$\omega \alpha = \dot\va\cdot\va \tag{1}$$

Now let $\va(t) = \alpha(t)\cdot \vn(t)$ for unit vector $\vn(t)$. Then we get $$ \dot\va(t) = \dot\alpha(t) \cdot\vn(t) + \alpha(t)\cdot\dot\vn(t)\,.$$

In the case $\alpha(t)=const$, and leaving out the $(t)$ for better readability, this simplifies to $\dot\va = \alpha\cdot\dot\vn$ and inserting into (1) results in

\begin{align} \omega\alpha = \vw\cdot\va &= \alpha \va \cdot\dot\vn\\ &= \alpha^2 \vn\cdot\dot\vn \end{align} Dividing by $\alpha$ we get

$$ \omega = \alpha\vn\cdot\dot\vn\,.$$

Since $\vn(t)$ is a unit vector for every $t$, any change of $\vn(t)$ must always only change its direction, never its length, which means that $\vn\cdot\dot\vn=0$ and therefore

$$ \omega = 0$$

in the case $\dot\alpha=0$, even for $\dot\vn\neq0$.

Question: How can this be that the axis of rotation changes its direction but $\omega$ and thereby angular momentum is zero? One of my assumptions of how $\va$ works is probably wrong. Yet all I have assumed is that $\va$ is just three numbers that change over time and that it can be decomposed into $\alpha\vn$. OK and that this is in line with the formula cited for $\omega$. Where is the mistake? Or can I have a change in rotation axis without having angular momentum?


1 Answer 1


Early on you make the assumption that $\vec{\omega}$ and $\vec{\alpha}$ are parallel. This is not in general true.

This may have arisen from a basic misconception about the meaning of $\vec{\alpha}(t)$. The direction of $\vec{\alpha}(t)$ is not the instantaneous axis of rotation. The axis-angle variables give you the rotation necessary to obtain the current orientation of a body (for example, at time $t$) relative to a reference orientation (for example, at $t=0$). It is true that any orientation can be expressed in this way: a rotation $\alpha(t)$ about a unit vector $\vec{n}(t)$, which we can put together as a combined vector $\vec{\alpha}(t)\equiv \alpha(t)\vec{n}(t)$. But this rotation depends on the entire history of the trajectory up to time $t$, and as was made clear in the page you referenced, Is there a formula for the rotation vector in terms of the angular velocity vector?, the relation is quite complicated. There is no particular reason why the axis $\vec{n}(t)$ describing the current orientation should have any relation to the direction of the current angular velocity $\vec{\omega}(t)$.

One can conceive of a special case where this is true: it is the simple one where $\vec{n}$ has been constant throughout the trajectory, and both $\vec{\omega}$ and $\vec{\alpha}$ have been parallel to $\vec{n}$ for the whole time. So $$ \vec{\omega}(t)=\omega(t)\vec{n} \qquad\text{and}\qquad \vec{\alpha}(t)=\alpha(t)\vec{n} $$ Then the angle of rotation $\alpha(t)$ is just the time integral of the magnitude of the angular velocity $\omega(t)$. In this case, though, things are less interesting. The second and third terms of your first equation vanish identically, so $$\vec{\omega}=\dot{\vec{\alpha}}\qquad\text{and}\qquad \omega=\dot{\alpha} $$ Your derivation is correct until we get to the special case $\alpha(t)=$ constant, which of course correctly implies $\omega(t)=0$. But that is as one would expect, for this special case, it doesn't illustrate anything awry.


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