Kovalevskaya top

Example

The Kovalevskaya top is a heavy rigid body turning about a fixed point $O$ in a uniform gravitational field, with the following mass distribution. Its principal moments of inertia at $O$ are $I_1 = I_2 = 2I_3$. Its centre of mass lies in the plane of the first two principal axes, at $\ell e_1$ with $\ell \ne 0$. Let $m$ be its mass and $g$ the gravitational acceleration, and put $c = m g \ell / I_3$. Let $\omega$ be the angular velocity and $\gamma$ the upward unit vertical in the body frame. The Euler–Poisson equations read $$2\dot\omega_1 = \omega_2\omega_3, \quad 2\dot\omega_2 = -\omega_1\omega_3 + c\gamma_3, \quad \dot\omega_3 = -c\gamma_2, \qquad \dot\gamma = \gamma \times \omega.$$

Besides the energy and the vertical component of the angular momentum, it has the Kovalevskaya integral $$K = (\omega_1^2 - \omega_2^2 - c\gamma_1)^2 + (2\omega_1\omega_2 - c\gamma_2)^2.$$ The check: put $z = \omega_1 + i\omega_2$ and $\xi = \gamma_1 + i\gamma_2$, so that $K = \abs{z^2 - c\xi}^2$. The equations give $\dot z = -\tfrac{i}{2}\omega_3 z + \tfrac{i}{2}c\gamma_3$ and $\dot\xi = -i\omega_3\xi + i\gamma_3 z$. Hence $\frac{d}{dt}(z^2 - c\xi) = -i\omega_3 (z^2 - c\xi)$, and the modulus of $z^2 - c\xi$ is constant. With three integrals in involution it is an integrable system.