for .

The directions of the three mutually orthogonal unit vectors
define the three so-called *principal axes
of rotation* of the rigid body under investigation. These axes are special because when the body rotates about
one of them (*i.e.*, when is parallel to one of them) the angular momentum vector
becomes *parallel* to the angular velocity vector .
This can be seen from a comparison of Equation (466) and Equation (487).

Suppose that we reorient our Cartesian coordinate
axes so the they coincide with the mutually orthogonal principal axes of rotation. In this new reference frame, the eigenvectors of are the unit vectors,
, , and , and the eigenvalues
are the moments of inertia about these axes, , , and , respectively. These latter quantities are referred to as the
*principal moments of inertia*.
Note that the products of inertia are all *zero* in the new
reference frame. Hence, in this frame, the moment
of inertia tensor takes the form of a *diagonal* matrix: *i.e.*,

(488) |

When expressed in our new coordinate system, Equation (466)
yields

(489) |

(490) |

In conclusion, there are many great simplifications to be had by choosing a coordinate system whose axes coincide with the principal axes of rotation of the rigid body under investigation. But how do we determine the directions of the principal axes in practice?

Well, in general, we have to solve the eigenvalue equation

(491) |

where , and . Here, is the angle the unit eigenvector subtends with the -axis, the angle it subtends with the -axis, and the angle it subtends with the -axis. Unfortunately, the analytic solution of the above matrix equation is generally quite difficult.

Fortunately, however, in many instances the rigid body under investigation possesses some kind of symmetry, so that at least one principal axis can be found by inspection. In this case, the other two principal axes can be determined as follows.

Suppose that the -axis is known to be a principal axes (at the origin)
in some coordinate system. It follows that the two products of inertia
and are zero [otherwise, would not be an eigenvector in Equation (492)]. The other two principal axes must lie in the
- plane: *i.e.*, . It then follows that
, since
.
The first two rows in the matrix equation (492) thus
reduce to

Eliminating between the above two equations, we obtain

But, . Hence, Equation (495) yields

There are two values of , lying between and , which satisfy the above equation. These specify the angles, , that the two mutually orthogonal principal axes lying in the - plane make with the -axis. Hence, we have now determined the directions of all three principal axes. Incidentally, once we have determined the orientation angle, , of a principal axis, we can then substitute back into Equation (493) to obtain the corresponding principal moment of inertia, .

As an example, consider a uniform rectangular lamina of mass and sides and which lies
in the - plane, as shown in Figure 29. Suppose that the axis of
rotation passes through the origin (*i.e.*, through a corner of the lamina).
Since throughout the lamina, it follows from Equations (464) and
(465) that
. Hence, the -axis
is a principal axis. After some straightforward integration, Equations (460)-(463) yield

(497) | |||

(498) | |||

(499) |

Thus, it follows from Equation (496) that

(500) |