# Solving Euler's equation for rigid-body rotational velocity in 4D

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 $$\omega_{12}=0$$.)

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