# What does diagonalizing inertia tensor do

In the normal case with moment of inertia, the angular momentum is parallel to the angular velocity

$$\vec{L} = I\,\vec{\omega}\tag{1}$$

But when the object is rotating instantaneously about a point, the moment of inertia becomes a second rank tensor and the angular momentum is no longer always parallel to the angular velocity

So I'm new to tensors stuff and what is typically done to the inertia tensor is diagonalizing it to find the fixed axes about which the object rotates about

But how is it possible, if we have $$\vec{L} = I\,\vec{\omega}\tag{1}$$ where $$I$$ is now a tensor which is diagonalized, the equation now says

$$L_x = I_{xx} \omega_{x}$$

$$L_y = I_{yy} \omega_{y}$$

$$L_z = I_{zz} \omega_{z}$$

So writing $$L$$ as a vector

$$\vec L = I_{xx} \omega_{x} \hat e_1 + I_{y} \omega_{y} \hat e_2 + I_{zz} \omega_{z} \hat e_3$$

However $$\vec \omega = \omega_{x} \hat e_1 + \omega_{y} \hat e_2 + \omega_{z} \hat e_3$$

Since $$I_{xx} , I_{yy} , I_{zz}$$ are not the same, $$\vec L$$ is not alligned always with $$\vec \omega$$ in these eigenvector axes, so what did we gain?

• Interestingly the process of bringing the angular momentum vector into alignment with a principle axis of an object is the entire basis of balancing the wheels of a car Sep 14, 2019 at 3:42

• Why can't I see that mathematically though? Where is the equation that says $L$ has the same direction as $\omega$? Sep 14, 2019 at 3:31
• Your $\hat{e}_1$, etc. are along the principal axes. So, for example, when it is rotating around $\hat{e}_1$, $\omega_x$ is nonzero but $\omega_y$ and $\omega_z$ are zero. Sep 14, 2019 at 3:34