# Can any physical rigid body be represented by an ellipsoid with the same angular dynamics?

According to wikipedia, the inertia tensor of an ellipsoid with semi-axes $a,b,c$ and mass $m$ is

$$\left[\begin{array}{ccc} \frac{m}{5}(b^2+c^2)&0&0\\ 0&\frac{m}{5}(a^2+c^2)&0\\ 0&0&\frac{m}{5}(a^2+b^2)\\ \end{array}\right].$$

If you create an arbitrary 3x3 positive diagonal matrix and try to solve for the $a,b,c$, it's very easy to wind up with imaginary dimensions. If I try to place separate point masses, I seem to run into the same problem.

Does that mean that the tensor doesn't represent a physically possible distribution of mass, or just not a uniform density solid? Intuitively, at least, it seems that it must be impossible for an inertia tensor to a have a single large value and two small values since a single point mass with a non-zero radius will always affect two dimensions equally and an ring of infinitesimal height still leaves the two minor dimensions with half the momentum of the large principal axis.

• Another way to express the triangle inequality is $\operatorname{Tr}(I) \ge 2\lambda_i(I) \ge 0$: The trace of the inertia tensor must be at least twice as large as each of the eigenvalues of the inertia tensor, each of which must be non-negative. One consequence is that an inertia tensor cannot have one large eigenvalue and two small ones. Jul 24 '18 at 9:03

1. Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body, and wrt. to an arbitrary choice of Cartesian coordinates $x$, $y$, and $z$), then the diagonal elements $I_{xx}$, $I_{yy}$, and $I_{zz}$ of the inertia tensor satisfy the triangle inequality, $$I_{xx} +I_{yy} ~\geq~ I_{zz}, \qquad I_{yy} +I_{zz} ~\geq~ I_{xx}, \qquad I_{zz} +I_{xx} ~\geq~ I_{yy}. \tag{1}$$ Sketched proof: Write down the definition of moment of inertia.$\Box$
2. Observation: It follows from the triangle inequality (1) alone that $$I_{xx}, I_{yy},I_{zz}~\geq~0 \tag{2}$$ are non-negative. (The ineq. (2) of course also follows from the definition of moment of inertia.)
3. Corollary of Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body), then the three moments of inertia $I_x$, $I_y$, and $I_z$, around the three principal axes (which we will call $x$, $y$, and $z$) satisfy the triangle inequality, $$I_x +I_y ~\geq~ I_z, \qquad I_y +I_z ~\geq~ I_x, \qquad I_z +I_x ~\geq~ I_y. \tag{3}$$
4. In other words, if a semi-positive definite symmetric real $3\times 3$ matrix with non-negative eigenvalues $I_x$, $I_y$, and $I_z$ does not satisfy the triangle inequality (3), it doesn't represent a physically possible distribution of mass.
5. Conversely, one may show that given three eigenvalues $I_x$, $I_y$, and $I_z$ that satisfy (3), they may be reproduced by a solid ellipsoid with a unique choice of non-negative semi-axes $a$, $b$, and $c$ (unique up to the scaling of the total mass $m$). $$\frac{2}{5}m a^2~=~I_y +I_z -I_x~\geq~0,$$ $$\frac{2}{5}m b^2~=~I_z +I_x -I_y~\geq~0,$$ $$\frac{2}{5}m c^2~=~I_x +I_y -I_z~\geq~0.\tag{4}$$