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?

  • $\begingroup$ 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 $\endgroup$
    – Triatticus
    Sep 14, 2019 at 3:42

1 Answer 1


Angular momentum and angular velocity are only parallel when the rotation is around one of the three principal axes found by diagonalization. What we gained was knowing what those special axes are where the dynamics are simple. The general rotational dynamics around an arbitrary axis are still complicated.

For an irregularly shaped object, like, say this comet

enter image description here

it should seem rather surprising that there is any axis around which the rotational dynamics are simple, much less three of them, in orthogonal directions!

  • $\begingroup$ Why can't I see that mathematically though? Where is the equation that says $L$ has the same direction as $ \omega$? $\endgroup$
    – khaled014z
    Sep 14, 2019 at 3:31
  • $\begingroup$ 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. $\endgroup$
    – G. Smith
    Sep 14, 2019 at 3:34
  • $\begingroup$ Oh that cleared things up, thank you. $\endgroup$
    – khaled014z
    Sep 14, 2019 at 3:46

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