Jacobi ellipsoid

Example

Consider a homogeneous, incompressible fluid of density $\rho$ that attracts itself by Newtonian gravity and rotates rigidly with angular velocity $\Omega$ about the $z$-axis. A Jacobi ellipsoid is a triaxial ellipsoid $x^2/a^2 + y^2/b^2 + z^2/c^2 \le 1$, with $a > b > c > 0$, that is a figure of equilibrium of this fluid. This means that in the rotating frame the fluid is in hydrostatic equilibrium and the pressure vanishes on the boundary. Write $\Delta(u) = \sqrt{(a^2 + u)(b^2 + u)(c^2 + u)}$ and let $G$ be the gravitational constant. Equilibrium holds exactly when the axes satisfy $$a^2 b^2 \int_0^\infty \frac{du}{(a^2 + u)(b^2 + u) \Delta(u)} = c^2 \int_0^\infty \frac{du}{(c^2 + u) \Delta(u)},$$ and then the angular velocity is given by $$\frac{\Omega^2}{\pi G \rho} = 2 a b c \int_0^\infty \frac{u \, du}{(a^2 + u)(b^2 + u) \Delta(u)}.$$ Check: the gravitational potential inside a homogeneous ellipsoid is a quadratic form in $x, y, z$ plus a constant, and so is the centrifugal potential $-\frac{1}{2} \Omega^2 (x^2 + y^2)$. The total potential is constant on the boundary exactly when its three coefficients are proportional to $1/a^2$, $1/b^2$, $1/c^2$. These are two conditions, and they give the two displayed equations. For each ratio $b/a$ in a certain range, the first equation fixes $c$, so the Jacobi ellipsoids form a one-parameter family up to scale.