Dedekind ellipsoid

Example

Consider a homogeneous, incompressible fluid of density $\rho$ that attracts itself by Newtonian gravity. A Dedekind ellipsoid is a triaxial ellipsoid $x^2/a^2 + y^2/b^2 + z^2/c^2 \le 1$, with $a > b$, that is a steady equilibrium of this fluid of the following kind. The boundary does not move at all in an inertial frame. The fluid inside moves with a velocity field of uniform vorticity about the $z$-axis, tangent to the boundary. Check: the velocity field $v = \lambda \left( -\frac{a}{b} y, \frac{b}{a} x, 0 \right)$, for a constant $\lambda$, has zero divergence and is tangent to every ellipse $x^2/a^2 + y^2/b^2 = \text{const}$, so the boundary stays fixed. Its vorticity is $\left(0, 0, \lambda \left( \frac{b}{a} + \frac{a}{b} \right) \right)$, which is uniform. Let $P$ be the pressure and $\Phi$ the gravitational potential, which is quadratic inside the ellipsoid. Since $(v \cdot \nabla) v = -\lambda^2 (x, y, 0)$, the steady Euler equations reduce to $P + \rho \Phi - \frac{1}{2} \rho \lambda^2 (x^2 + y^2) = \text{const}$. Requiring $P = 0$ on the boundary gives conditions on $a$, $b$, $c$ and $\lambda$. These conditions are met exactly by the ellipsoids with the same axes as the Jacobi ellipsoids. This is an instance of Dedekind's theorem, which exchanges the roles of the rotation of the figure and of the internal vorticity.