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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

$$\frac{d\omega_{12}}{dt}=\frac{a_1^2-a_2^2}{a_1^2+a_2^2}\big(\omega_{13}\omega_{23}+\omega_{14}\omega_{24}\big)$$

$$\frac{d\omega_{13}}{dt}=\frac{a_1^2-a_3^2}{a_1^2+a_3^2}\big(\omega_{12}\omega_{32}+\omega_{14}\omega_{34}\big)$$

$$\frac{d\omega_{23}}{dt}=\frac{a_2^2-a_3^2}{a_2^2+a_3^2}\big(\omega_{21}\omega_{31}+\omega_{24}\omega_{34}\big)$$

$$\frac{d\omega_{14}}{dt}=\frac{a_1^2-a_4^2}{a_1^2+a_4^2}\big(\omega_{12}\omega_{42}+\omega_{13}\omega_{43}\big)$$

$$\frac{d\omega_{24}}{dt}=\frac{a_2^2-a_4^2}{a_2^2+a_4^2}\big(\omega_{21}\omega_{41}+\omega_{23}\omega_{43}\big)$$

$$\frac{d\omega_{34}}{dt}=\frac{a_3^2-a_4^2}{a_3^2+a_4^2}\big(\omega_{31}\omega_{41}+\omega_{32}\omega_{42}\big)$$

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

$$\text{sn}'=\text{cn}\cdot\text{dn}$$

$$\text{cn}'=-\text{sn}\cdot\text{dn}$$

$$\text{dn}'=-m\cdot\text{sn}\cdot\text{cn}.$$

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,

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

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

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

$$\theta=(B-A)\omega_{12}t$$

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 $\omega_{12}=0$.)


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  • $\begingroup$ @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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