Is the notion of angular velocity $\boldsymbol{\omega}$ defined for arbitrary particle trajectories (e.g. a crazy one like $\textbf x(t)=(t^3\ln(t),\sqrt{t}\operatorname{erf}(2t),\cot^{42}(t))$) or only when the position and velocity vectors of the particle are constantly orthogonal (i.e. $\textbf x(t)\cdot\dot{\textbf x}(t)=0$ for all $t\geq 0$, which implies that the particle is constrained to move on a sphere)? Or even only when the particle engages in circular motion (i.e. it is constrained to travel on a great circle of the aforementioned sphere)? I have seen two conflicting definitions; for example, one can attempt to define angular velocity directly via the formula $$\boldsymbol{\omega}:=\frac{\textbf x\times{\dot{\textbf x}}}{\textbf x\cdot\textbf x}$$ which has the advantage that (as long as the particle never comes to the origin), the angular velocity at each moment in time $t\geq 0$ is completely well-defined, regardless of how sophisticated the trajectory $\textbf x$ is. This definition also has the property that at every moment $t\geq 0$ in time, $\boldsymbol \omega(t)\in(\operatorname{span}\{\textbf x(t),\dot{\textbf x}(t)\})^{\perp}$ is orthogonal to the instantaneous plane of the particle's rotation and does so in a right-handed way. One can check that its magnitude $||\boldsymbol \omega||$ also behaves in a reasonable manner. On the other hand, sometimes people seem to define the angular velocity $\boldsymbol \omega$ indirectly as the unique vector satisfying: $$\dot{\textbf x}=\boldsymbol \omega\times{\textbf x}$$ which also lies along "the axis of rotation". These definitions are incompatible with each other for a number of reasons, the obvious one being that in the latter we have already assumed a priori that $\textbf x\cdot\dot{\textbf x}=0$ (indeed, one can prove the following general result about $3$-dimensional Euclidean vector spaces: if $\textbf a\in{\textbf R^3}$ and $\textbf c\in{\textbf R^3\setminus{\{\textbf 0}}\}$ are two $3$-dimensional Euclidean vectors with $\textbf c$ not equal to the zero vector $\textbf 0$, then there exists a vector $\textbf b\in \textbf R^3$ such that $\textbf a=\textbf b\times\textbf c$ if and only if $\textbf a\cdot\textbf c=0$). So how should one conceptualize angular velocity (for a single point particle)?

  • $\begingroup$ Where have you seen these conflicting definitions? Do you have any references? $\endgroup$
    – Philip
    Jul 7, 2021 at 3:04

1 Answer 1


The second equation $\dot{\boldsymbol x}=\boldsymbol\omega\times \boldsymbol x$ is the general definition of spin angular velocity, $\boldsymbol\omega\,,$ in space for a rotating rigid body (see Arnold's book on Classical Mechanics).

The first equation $\boldsymbol\omega=\boldsymbol x\times\dot{\boldsymbol x}/|{\boldsymbol x}|^2$ is the correct definition for orbital angular velocity for a single particle with trajectory $\boldsymbol{x}(t)\,.$

Because every point in a rotating rigid body is a single particle it is a natural question when both concepts agree.

Consider the single particle that moves such that $\boldsymbol x$ and $\dot{\boldsymbol x}$ are always orthogonal. Using the Grassmann identity $a\times(b\times c)=(a\cdot c)\,b-(a\cdot b)\,c$ it is easy to see that the orbital angular velocity satisfies the equation for spin angular velocity. However, the fully correct definition of spin angular velocity requires that the $\boldsymbol\omega$ must be the unique vector that satisfies $\dot{\boldsymbol x}=\boldsymbol\omega\times \boldsymbol x$ for all $\boldsymbol x$ in the rigid body.

For example consider the particles $$ {\boldsymbol x}(t)=\left(\begin{array}{c}\cos t\\ \sin t\\1\end{array}\right),~~ \dot {\boldsymbol x}(t)=\left(\begin{array}{c}-\sin t\\ \cos t\\0\end{array}\right),~~ {\boldsymbol\omega}_{\text orbital}(t)=\frac{1}{2}\left(\begin{array}{c}-\cos t\\ -\sin t\\1\end{array}\right)\,, $$

$$ {\boldsymbol y}(t)=\left(\begin{array}{c}-\sin t\\ \cos t\\1\end{array}\right),~~ \dot {\boldsymbol y}(t)=\left(\begin{array}{c}-\cos t\\ -\sin t\\0\end{array}\right),~~ {\boldsymbol\omega}_{\text orbital}(t)=\frac{1}{2}\left(\begin{array}{c}\sin t\\ -\cos t\\1\end{array}\right)\, $$ and assume they are part of the same rigid body. It is easy to see that both their orbital angular velocities satisfy $\dot{\boldsymbol x}=\boldsymbol\omega\times \boldsymbol x$, resp. $\dot{\boldsymbol y}=\boldsymbol\omega\times \boldsymbol y\,.$ The uniqe spin angular velocity that satisfies both these equations is however $$ {\boldsymbol\omega}_{spin}=\left(\begin{array}{c}0\\ 0 \\1\end{array}\right)\,. $$

  • $\begingroup$ I have another question, just to make sure I definitely understand: if a particle's trajectory goes like: $\textbf {x}(t)=(\cos(t),\sin(t),1)$ (i.e. the particle sweeps $1$ radian of angle per unit of time "counter-clockwise" around a circle centered about the $z$-axis and which is also one unit above the $xy$ plane), what is the correct formula for its angular velocity $\boldsymbol \omega$ as a function of time? $\endgroup$ Jul 7, 2021 at 11:09
  • $\begingroup$ Is it $\boldsymbol \omega(t)=2^{-1/2}(-\cos(t),-\sin(t),1)$ which is obtained through the first formula that you seemed to say was correct. Or is it just the constant function $\boldsymbol \omega(t)=\hat{\textbf{k}}$? $\endgroup$ Jul 7, 2021 at 11:09
  • $\begingroup$ Actually, I think you're saying that the second formula is the more appropriate one...but then how can we have the second formula while simultaneously defining the angular velocity for general trajectories? I mentioned in my post that it assumes ahead of time that the position $\textbf x$ and velocity $\dot{\textbf x}$ are orthogonal so that there even exists a vector $\boldsymbol \omega$ which relates the two via the cross product... so how can $\boldsymbol \omega$ be defined then for general trajectories if one adopts the second definition? $\endgroup$ Jul 7, 2021 at 11:18
  • $\begingroup$ Seems your $\boldsymbol \omega(t)$ should be $(-\cos(t),+\sin(t),1)/\sqrt{2}\,$. How $\boldsymbol\omega$ is defined exactly for general trajectories you can learn from Arnold's book. $\endgroup$
    – Kurt G.
    Jul 7, 2021 at 12:38
  • 1
    $\begingroup$ Your understanding is correct. Thanks for being so keen to get to the bottom of this. I will add some updates to the answer. $\endgroup$
    – Kurt G.
    Jul 8, 2021 at 6:13

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