Moment of inertia tensor and symmetry of the object

Question

What information does the moment of inertia tensor give on the structure of an item.

I was told that its eigenvectors give the principal axes of the object.

Do you know more about this?

Answer 1

The inertia tensor is a bit more descriptive in the spherical tensor basis (so instead of having nine basis dyads made from combinations like $\hat x \hat y$, you have 9 basis tensors that transform like the nine $Y_l^m$ for $l \in \{0,1,2\}$).

Since $I_{ij}$ is symmetric, all $l=1$ spherical tensors are zero. The $l=0$ portion is:

$$ I^{(0,0)} = \frac 1 3 {\rm Tr}(I)\delta_{ij}$$

and that is the spherically symmetric part of the object.

Removing the spherically symmetric part leaves a "natural" (read: symmetric, trace-free) rank-2 tensor:

$$ S_{ij} = I_{ij} -I^{(0,0)}$$

The spherical components are:

$$ S^{(2,0)} = \sqrt{\frac 3 2}S_{zz} $$

This tells you if your object is prolate or oblate.

$$ S^{(2,\pm 2)} = \frac 1 2 [S_{xx}-S_{yy}\pm 2iS{xy}]$$

You will find that $S^{(2,+2)} = (S^{(2,-2)})^*$, and that if you are in diagonal coordinates, they are real and equal. If the value is 0, then the object is cylindrically symmetric.

$$ S^{(2,\pm 1)} = \frac 1 2 [S_{zx}\pm iS_{zy}]$$

Here: $S^{(2,+1)} = -(S^{(2,-1)})^*$, and the term is zero in diagonal coordinates.

Answer 2

Using the inertia tensor $I_{jk}$ of an object, you can construct an ellipsoid satisfying the equation $$1=x_{j}I_{jk}x_{k} .$$ As you said, the eigenvectors of $I_{jk}$ are the principal axis $\vec{r}_{j}$ of your object with the respective moment of inertia $\theta_{j}$ as eigenvalue. Thus, changing coordinates to the principal axis of the object diagonalises the inertia tensor and the equation for the ellipsoid is then $$1=\theta_{1}x_{1}^{2}+\theta_{2}x_{2}^{2}+\theta_{3}x_{3}^{2} .$$ The semi-axles of this ellipsoid are parallel to the principal axles of your object and their length is the inverse square root of the moment of inertia in that direction $a_{j}=\frac{1}{\sqrt{\theta_{j}}}$.

Answer 3

For a rigid body, the inertia tensor together with the angular velocity provide the angular momentum of the body about a selected point. Specifically, $\vec L = \bf I \cdot \vec \omega$ where $\vec L$ is the angular momentum, $\bf I$ the inertia tensor, and $\vec \omega$ the angular velocity. In general, the diagonal elements of $\bf I$ are the moments of inertia and the non-diagonal elements are the products of inertia. The three eigenvalues of $\bf I$ give the directions of the three principal axis, and the three eigenvectors give the moments of inertia with respect to each of these axis. Using eigenvalues, $\bf I$ is in diagonal form; the products of inertia are all zero. In practice, the body axes fixed in the rigid body are chosen to be the principal axes for describing three-dimensional rotation of a rigid body.

See an advanced physics mechanics textbook such as Classical Mechanics by Goldstein, or Mechanics by Symon.