13.7: Diagonalize the Inertia Tensor

Finding the three principal axes involves diagonalizing the inertia tensor, which is the classic eigenvalue problem discussed in appendix $19.1$. Solution of the eigenvalue problem for rigid-body motion corresponds to a rotation of the coordinate frame to the principal axes resulting in the matrix

$$\{\mathbf{I}\} \cdot \boldsymbol{\omega} = I\boldsymbol{\omega}$$

where $I$ comprises the three-valued eigenvalues, while the corresponding vector $\boldsymbol{\omega}$ is the eigenvector. Appendix $19.1$ gives the solution of the matrix relation

$$\{\mathbf{I}\} \cdot \boldsymbol{\omega} = I \{\mathbb{I}\} \boldsymbol{\omega} \label{13.26}$$

where $I$ are three-valued eigen values for the principal axis moments of inertia, and $\{\mathbb{I}\}$ is the unity tensor, equation $(A.2.4)$.

$$\{\mathbb{I}\} \equiv \begin{Bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{Bmatrix}$$

Rewriting $\ref{13.26}$ gives

$$(\{\mathbf{I}\} − I \{\mathbb{I}\}) \cdot \boldsymbol{\omega} = 0 \label{13.28}$$

This is a matrix equation of the form $\mathbf{A} \cdot \boldsymbol{\omega} =0$ where $\mathbf{A}$ is a $3 \times 3$ matrix and $\boldsymbol{\omega}$ is a vector with values $\omega_x , \omega_y, \omega_z$. The matrix equation $\mathbf{A} \cdot \boldsymbol{\omega} =0$ really corresponds to three simultaneous equations for the three numbers $\omega_x , \omega_y, \omega_z$. It is a well-known property of equations like $\ref{13.28}$ that they have a non-zero solution if, and only if, the determinant $\text{det}(\mathbf{A})$ is zero, that is

$$\text{det}(\mathbf{I}−I\mathbb{I})=0$$

This is called the characteristic equation, or secular equation for the matrix $\mathbf{I}$. The determinant involved is a cubic equation in the value of $I$ that gives the three principal moments of inertia. Inserting one of the three values of $I$ into equation $(13.4.6)$ gives the corresponding eigenvector $\omega$. Applying the above eigenvalue problem to rigid-body rotation corresponds to requiring that some arbitrary set of body-fixed axes be the principal axes of inertia. This is obtained by rotating the body-fixed axis system such that

$$\begin{align} L_1 & = & I_{11}\omega_1 + I_{12}\omega_2 + I_{13}\omega_3 = I\omega_1 \\ L_2 & = & I_{21}\omega_1 + I_{22}\omega_2 + I_{23}\omega_3 = I\omega_2 \notag \\ L_3 & = & I_{31}\omega_1 + I_{32}\omega_2 + I_{33}\omega_3 = I\omega_3 \notag \end{align}$$

or

$$\begin{align}(I_{11} − I) \omega_1 + I_{12}\omega_2 + I_{13}\omega_3 = 0 \\ I_{21}\omega_1 + (I_{22} − I) \omega_2 + I_{23}\omega_3 = 0 \notag \\ I_{31}\omega_1 + I_{32}\omega_2 + (I_{33} − I) \omega_3 = 0 \notag \end{align}$$

These equations have a non-trivial solution for the ratios $\omega_1 : \omega_2 : \omega_3$ since the determinant vanishes, that is

$$\begin{vmatrix} (I_{11} − I) & I_{12} & I_{13} \\ I_{21} & (I_{22} − I) & I_{23} \\ I_{31} & I_{32} & (I_{33} − I) \end{vmatrix} = 0$$

The expansion of this determinant leads to a cubic equation with three roots for $I$. This is the secular equation for $I$ whose eigenvalues are the principal moments of inertia.

The directions of the principal axes, that is the eigenvectors, can be found by substituting the corresponding solution for $I$ into the prior equation. Thus for eigensolution $I_1$ the eigenvector is given by solving

$$\begin{align}(I_{11} − I_1) \omega_{11} + I_{12}\omega_{21} + I_{13}\omega_{31} = 0 \\ I_{21}\omega_{11} + (I_{22} − I_1) \omega_{21} + I_{23}\omega_{31} = 0 \notag \\ I_{31}\omega_{11} + I_{32}\omega_{21} + (I_{33} − I_1) \omega_{31} = 0 \notag \end{align}$$

These equations are solved for the ratios $\omega_{11} : \omega_{21} : \omega_{31}$ which are the direction numbers of the principle axis system corresponding to solution $I_1$. This principal axis system is defined relative to the original coordinate system. This procedure is repeated to find the orientation of the other two mutually perpendicular principal axes.