The following question regards the moment of inertia tensor of classical mechanics, given by

$$\mathbf{I} = \begin{bmatrix} I_{xx}&I_{xy}&I_{xz}\\ I_{yx}&I_{yy}&I_{yz}\\ I_{zx}&I_{zy}&I_{zz} \end{bmatrix} $$

Where are $I_{ii}$ are moments of inertia and all $I_{ij}$ are products of inertia. We also define the angular momentum $\vec{L}$ by $\vec{L} = \mathbf{I}\vec{\omega}$, where $\vec{\omega}$ is the angular velocity of the rotating body.

We say that if a choice of axes $x, y, z$ yields a angular momentum vector that is parallel to the angular velocity vector ($\vec{L} = \mathbf{I}\vec{\omega} = \beta\vec{\omega}$, $\beta$ = const.), then those axes $x, y, z$ are the principal axes of the body.


Okay, so, consider a cube of side length $\alpha$ rotating about its main diagonal (the axis of rotation passes through $(0,0)$ and $(\alpha, \alpha)$). Skipping the integrals and just presenting the result, the angular momentum $\vec{L}$ of such a body is

$$\vec{L} = \mathbf{I}\vec{\omega} = \dfrac{M\alpha^2}{12}\begin{bmatrix} 8&-3&-3\\-3&8&-3\\-3&-3&8\end{bmatrix}\omega\hat{\omega}=\dfrac{M\alpha^2}{6}\vec{\omega}$$

EDIT: clarification: The $x, y, z$ axes are parallel to the cube edges, with the origin at a lower corner. Since $\vec{\omega}$ is acting through the main diagonal of the cube, we have

$$\mathbf{I}\vec{\omega} = \mathbf{I}\dfrac{\omega}{\sqrt{3}}\begin{bmatrix}1\\1\\1\end{bmatrix},$$

since the direction of $\vec{\omega}$ is from the origin at one corner at $(0,0,0)$ to the opposite corner at $(1,1,1)$ (and I have normalized this direction with a factor of $1/\sqrt{3}$). Then, you can see that using the inertia tensor given above, we obtain

$$\textbf{I}\vec{\omega} = \dfrac{M\alpha^2}{12}\dfrac{\omega}{\sqrt{3}}\begin{bmatrix}2\\2\\2\end{bmatrix} = \dfrac{M\alpha^2}{12}2\vec{\omega} = \dfrac{M\alpha^2}{6}\vec{\omega}$$


Next, consider the same cube rotating about any axis through it's center (a more general case of the above), with the same axes as defined above. The angular momentum here is

$$\vec{L} = \mathbf{I}\vec{\omega} = \dfrac{M\alpha^2}{6}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\vec{\omega} = \dfrac{M\alpha^2}{6}\vec{\omega}$$


And, finally, consider a cone rotating about an axis passing through it's point and through the center of it's base. The angular momentum here is

$$\vec{L} = \mathbf{I}\vec{\omega} = \dfrac{3}{20}M\begin{bmatrix}R^2+4h^2&0&0\\0&R^2+4h^2&0\\0&0&2R^2\end{bmatrix}\vec{\omega} = \begin{bmatrix}\lambda_1\omega_x\\ \lambda_2\omega_y \\ \lambda_3\omega_z\end{bmatrix}$$

where $h$ is the height of the cone and $R$ is the radius of its base.

In the first case, the inertia tensor is diagonal and hermitian. In the second case, these are still true, but with the added quality that the inertia tensor is a multiple of the identity matrix. In the last case, the inertia tensor is still hermitian, and is the identity matrix multiplied by a vector rather than a scalar.

My question is; how do I properly discuss the differences in "flavors" of principal axes? These all are principal axes, correct? Since The angular velocity is pointing in the same direction as the angular momentum in all cases.

Edit: I was mistaken stating that the inertia tensor of the first case is diagonal. My textbook states that an inertia tensor expressed in the basis of principal axes must be diagonal, having zero for all off-diagonal elements. But, this is not fully consistent with the definition that principal axes are those that, when evaluating the inertia tensor on that basis, the angular momentum is parallel with the angular velocity.

  • $\begingroup$ There seem to be a few missing details here. Can you explain how you got the first equation -- in particular, what coordinate system you did it in? How did you conclude that $I \omega$ is equal to a multiple of $\omega$, when $I$ isn't diagonal? $\endgroup$
    – knzhou
    Commented Feb 14, 2017 at 5:35
  • $\begingroup$ @knzhou In that case, the cube is rotating about it's main diagonal, with the $x, y, z$ axes being parallel to the cube's edges, with the origin at a lower corner. I will add a more detailed derivation above. $\endgroup$
    – user97626
    Commented Feb 14, 2017 at 6:07
  • $\begingroup$ @knzhou see my edit please $\endgroup$
    – user97626
    Commented Feb 14, 2017 at 6:18

1 Answer 1


The correct statements are the following:

  • If $\vec{\omega}$ is directed along a principal axis, then $I \vec{\omega}$ is proportional to $\vec{\omega}$.
  • If $I$ is expanded in components, where the coordinate axes are principal axes, then the components $I_{xy}$ are a diagonal matrix.

Your three examples work like this:

  • In your first example, you choose $\vec{\omega}$ to be a principal axis, but you didn't work in principal axis coordinates, so the matrix $I_{xy}$ wasn't diagonal.
  • In your second example, everything works alright. Actually, it turns out that in this case, every axis is a principal axis, which is why you got the identity matrix. You'd get the same result in any coordinate system. This only happens in cases of very high symmetry.
  • Your third example is a more typical case, where the principal axes have different moments of inertia. This is what happens for generic objects.

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