It is well known that rigid body inertia tensors are 3 by 3 positive semidefinite matrices, which is the same as saying that their eigenvalues are all non-negative.

A little less known is the fact that those eigenvalues also satisfy the triangle inequalities, which means that the sum of any two eigenvalues is always greater or equal than the remaining one. This arises from the definition of the tensor itself which involves integrals of an always non-negative distribution of mass. Per https://physics.stackexchange.com/a/48273/116038 :

In other words, if a semi-positive definite symmetric real $3×3$ matrix with non-negative eigenvalues [...] does not satisfy the triangle inequality (1), it doesn't represent a physically possible distribution of mass.

When estimating inertia tensors using, for example, regression techniques, one can obtain matrices that do not represent physically possible rigid bodies (if we do not constrain the regression solution to physically consistent values).

Using a non-positive semidefinite $3×3$ symmetric matrix in places where an inertia tensor is expected -- e.g., in some rigid-body formulations, in simulation, in control schemes, etc. -- would entail major problems. For example, a non-positive semidefinite matrix used as inertia tensor would give rise to negative kinetic energies, $ E_k = \frac{1}{2} \omega^\top I \omega $.

Now, my question is: What problems (in formulations, simulation, control, etc.), if any, can arise if we use as inertia tensor some matrix that, while satisfying the positive semidefinite condition, do not satisfy the triangle inequality conditions on the eigenvalues?

Update: I've checked that "inertia tensors" that are positive semidefinite but do not satisfy the triangle inequalities still verify the energy conservation law with respect to kinetic energy and work.


A rigid body's principal moments of inertia are obtained from these equations :

$$I_1=\int_V\,(x_2^2+x_3^2)\,\rho\,dV$$ $$I_2=\int_V\,(x_3^2+x_1^2)\,\rho\,dV$$ $$I_3=\int_V\,(x_1^2+x_2^2)\,\rho\,dV$$

where $x=x_1~,y=x_2~,z=x_3$ and the inertia tensor is:

$$I= \left[ \begin {array}{ccc} I_{{1}}&0&0\\ 0&I_{{2}}&0 \\ 0&0&I_{{3}}\end {array} \right] $$

with $$i_\alpha=\int_V\,x_\alpha^2\,\rho\,dV> 0~,\alpha=1,2,3$$ thus: $$I_1=i_2+i_3$$ $$I_2=i_3+i_1$$ $$I_3=i_1+i_2$$

and $$I_1+I_2=i_1+i_2+2\,i_3=I_3+2\,i_3 > I_3$$ $$I_2+I_3=2\,i_1+i_2+i_3=I_1+2\,i_1 > I_1$$ $$I_3+I_1=i_1+2\,i_2+i_3=I_2+2\,i_3 > I_2$$

thus the triangle inequality is a physical feature of a rigid body inertia tensor. If the rigid body is symmetric then the symmetry axes are principal axes and the principal moment of inertia must obey the triangle inequality, otherwise you don't describe the rigid body that you want to describe.


The triangle inequality for the moment of inertia tensor ultimately stems from the positivity of inertial mass (as shown in Eli's answer.) A body whose inertia tensor did not satisfy the triangle inequalities would necessarily have some region where $\rho < 0$.

We could then consider this body as composed of two sub-bodies, one with strictly positive inertial mass $m_1 > 0$ and one with strictly negative inertial mass $m_2 < 0$. Any internal forces between these bodies would result in these two sub-bodies accelerating in the same direction spontaneously: the internal forces on $m_1$ and $m_2$ would act in opposite directions, according to Newton's Third Law, but their acceleration vectors would point in the same direction due to their differing signs of mass.

Such "runaway solutions" are generally considered a Bad Thing. The only loopholes I see are (a) abandoning Newton's Third Law, but that would make Emmy Noether sad; or (b) requiring that these sub-bodies never exert any net force on each other, which would be problematic if we want the body to be rigid.


It is useful to look at inertia tensors in the spherical basis--in which the basis-tensors are eigentensors of rotations about the z-axis. So instead of considering 3 x 3 = 9 $T_{ij}$ for i, j $\in (x, y, z)$, on has 1 + 3 + 5 = 9 basis states as follows:

$T^{(0)} = \frac{1}{3}T_{ii}\ {\bf I}$

is spherically symmetric and proportional to the identity. As an inertial tensor, it corresponds to a sphere.

The antisymmetric part of a tensor:

$T^{(1)} = A_{ij} \equiv \frac{1}{2}(T_{ij}-T_{ji})$

transforms like a vector (much like a cross product), and has NO PLACE in the inertia tensors. There are 3 components.

Finally, the trace-free symmetric part has 5 components:

$T^{(2)} = S_{ij} \equiv \frac{1}{2}(T_{ij}+T_{ji})) - T^{(0)}$

that transform like spherical harmonics: $Y_{l=2}^m\ \ (\theta, \phi)$ for $m \in\ (-2, -1, 0, 1, 2) $.

The $|m|=1$ components:

$T^{(2, m=\pm 1)}\ \ \ \ = S_{xz} \pm i S_{yz} $

can be diagonalized away by a suitable coordinate rotation leaving:

$l=2, m=0$:

$T^{(2, 0)} \ \ = \sqrt{\frac{1}{2}} S_{zz} $

This part corresponds to the oblate/prolate-ness of the object; so for an ellipsoid, that would be with the eccentricity with the axis of symmetry aligned with the z-axis.

$l=2, |m|=2$:

$T^{(2, m=\pm 2)} \ \ \ \ \ = \frac{1}{2}(S_{xx} - S_{yy} \pm 2iS_{xy})$

For inertia tensors, the m=2 and m=-2 are the same, so that the Cartesian components are real. This part of the tensors corresponds to any cylindrical asymmetry--so it is zero for an ellipsoid of revolution.

  • 1
    $\begingroup$ Thanks for this other way to see inertia tensors, but I could not get any further insights about my question. $\endgroup$ – Cristóvão D. Sousa Aug 29 '17 at 18:29

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