A rigid object, with no torques applied to it, rotates with constant angular momentum. But its angular velocity $\omega$ is not constant in general; it follows the differential equations







where $\omega_{ij}=-\omega_{ji}$ is a component of $\omega$ with respect to the object's principal axes, and $a_i^2$ is proportional to a second moment of mass. (A box or ellipsoid has $a_i$ proportional to the length of its $i$'th axis.) We may choose the axes such that $a_1\geq a_2\geq a_3\geq a_4\geq0$. If the object has symmetry, then some $a_i-a_j=0$ and the equations simplify.

If $a_1=a_2=a_3=a_4$, then $\omega$ is constant.

If $a_1=a_2=a_3\neq a_4$, then $\omega_{12},\omega_{13},\omega_{23}$ are constant, and $\omega_{14},\omega_{24},\omega_{34}$ satisfy a linear differential equation whose solution is a circle.

If $a_1=a_2\neq a_3=a_4$, then $\omega_{12},\omega_{34}$ are constant; and changing to isoclinic coordinates $(\omega_{12}\pm\omega_{34}),(\omega_{13}\mp\omega_{24}),(\omega_{14}\pm\omega_{23})$ shows that the solution is a pair of circles (with different frequencies, in the ratio of $(\omega_{12}+\omega_{34})/(\omega_{12}-\omega_{34})$).

If $a_1=a_2\neq a_3\neq a_4\neq a_1$, then $\omega_{12}$ is constant. The other five are still non-linearly related. There are easily found particular solutions with some other $\omega_{ij}$ being constant or $0$.

In general, if we assume $\omega_{14}=\omega_{24}=\omega_{34}=0$, then this reduces to the classical problem in 3D, which can be solved with Jacobi's elliptic functions (or geometrically by intersecting ellipsoids). This is because




But can the full 4D problem (with or without $a_1=a_2$) be solved with elliptic functions?

Of course I'm also interested in the $n$-dimensional generalization.

...It appears that the differential equations are not Lipschitz-continuous, so it's not clear that they have solutions for all $t$. (Compare with $dy/dt=y^2$.) But this is helped by conservation of energy $(a_1^2+a_2^2)\omega_{12}^2+(a_1^2+a_3^2)\omega_{13}^2+\cdots$ ; the equations are Lipschitz-continuous on this restricted domain (an ellipsoid), so we do have existence and uniqueness of solutions for all $t$.

Still working on the $a_1=a_2$ case, I tried variation of parameters, based on the linear part (with $\omega_{12}$ as a factor). With the following definitions,

$$A=\frac{a_1^2-a_3^2}{a_1^2+a_3^2},\quad B=\frac{a_1^2-a_4^2}{a_1^2+a_4^2},\quad C=\frac{a_3^2-a_4^2}{a_3^2+a_4^2}$$

$$M = \begin{bmatrix} 0 & -A\omega_{12} & 0 & 0 & 0 \\ A\omega_{12} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -B\omega_{12} & 0 \\ 0 & 0 & B\omega_{12} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} \omega_{13} \\ \omega_{23} \\ \omega_{14} \\ \omega_{24} \\ \omega_{34} \end{bmatrix} = \exp(tM) \begin{bmatrix} u_{13} \\ u_{23} \\ u_{14} \\ u_{24} \\ u_{34} \end{bmatrix}$$


the equation becomes

$$\frac{d}{dt} \begin{bmatrix} u_{13} \\ u_{23} \\ u_{14} \\ u_{24} \\ u_{34} \end{bmatrix} = \cos\theta \begin{bmatrix} Au_{14}u_{34} \\ Au_{24}u_{34} \\ -Bu_{13}u_{34} \\ -Bu_{23}u_{34} \\ C(u_{13}u_{14}+u_{23}u_{24}) \end{bmatrix} + \sin\theta \begin{bmatrix} -Au_{24}u_{34} \\ Au_{14}u_{34} \\ -Bu_{23}u_{34} \\ Bu_{13}u_{34} \\ C(u_{23}u_{14}-u_{13}u_{24}) \end{bmatrix}.$$

(This is pointless if $\omega_{12}=0$.)

  • $\begingroup$ @YuiTo Cheng -- Why the edit? Surely there should be some tag for 4D geometry, or higher dimensions in general. $\endgroup$ – mr_e_man Jul 7 at 17:54
  • $\begingroup$ Is this from some reference? $\endgroup$ – Qmechanic Jul 8 at 11:32
  • $\begingroup$ @Qmechanic -- No. $\endgroup$ – mr_e_man Jul 9 at 14:01

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