1
$\begingroup$

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)?

$\endgroup$
1
  • $\begingroup$ Where have you seen these conflicting definitions? Do you have any references? $\endgroup$
    – Philip
    Jul 7 at 3:04
2
$\begingroup$

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 $$ {\boldsymbol\omega}_{spin}=\left(\begin{array}{c}0\\ 0 \\1\end{array}\right)\,. $$

$\endgroup$
9
  • $\begingroup$ 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 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 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 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 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 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.