$$\require{cancel}$$

# 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.