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I am working through a problem and getting what seems to be different answers in different reference frames. The angular momentum of a point mass with respect to an axis that doesn't go through the mass is parallel to the axis in some frames, but not in others.

I am trying to understand the definition of the inertia tensor that I have found in several places online (for instance on Wikipedia). The angular momentum of a point mass revolving around an axis is given by:

$\vec{L}=\mathbf{I}\vec{\omega}$

where the inertia tensor $\mathbf{I}$ is given by:

$\begin{eqnarray*} \mathbf{I}&=&\begin{bmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{zy} & I_{zz}\\ \end{bmatrix}\\ I_{xx}&=&m(y^2+z^2)\\ I_{yy}&=&m(x^2+z^2)\\ I_{zz}&=&m(x^2+y^2)\\ I_{xy}=I_{yx}&=&-mxy\\ I_{yz}=I_{zy}&=&-myz\\ I_{xz}=I_{zx}&=&-mxz \end{eqnarray*}$

So, I took a unit point mass at $x=1$ and figured its inertia tensor and got:

$\mathbf{I}=\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

If I choose as the rotation velocity $\vec{\omega}=\vec{z}=\begin{bmatrix}0 & 0 & 1\end{bmatrix}^T$ I get $\vec{L}$ happens to equal $\vec{\omega}$ because it's a unit point pass a unit distance away from the axis rotating at unit velocity. This makes sense to me. If I change any of the unit values to something else I will get a different magnitude of $\vec{L}$ but it will always be parallel to $\vec{\omega}$. This makes sense to me.

But, I stuck an arbitrary rotation velocity $\vec{\omega}=\begin{bmatrix}1 & 2 & 3\end{bmatrix}^T$ and ended up with an angular momentum of $\vec{L}=\begin{bmatrix}0 & 2 & 3\end{bmatrix}^T$. This is not even parallel to $\vec{\omega}$ any more!

I know that there are cases when the angular momentum of an extended object is not parallel to its angular velocity. That's why we have the off-diagonal terms in the inertia tensor in the first place.

What I don't understand is how that can be the case with a point mass. It's easy to imagine another frame where the $z'$ axis is parallel to $\vec{\omega}$ and in that frame, my intuition suggests that the angular momentum should be parallel to the angular velocity.

That suggests that the angular momentum depends on the frame. I didn't expect that since vectors are supposed to be geometric objects with lives independent of any frame that might be used to give them coordinates. How can it be that if I calculate a vector property in one frame, it is parallel to another vector, but if I calculate the same vector property just in a different frame, it isn't parallel any more?

Is my intuition on angular momentum wrong? Am I applying the math wrong? I don't doubt the math itself, as I have looked it up in several sources and have worked through the algebra myself and convinced myself that it is correct.

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OK, I think I figured it out -- it's not the rotation that is important, it is the fact that the "center of rotation" isn't at the origin any more.

Rather than think about the $\begin{bmatrix}1 & 2 & 3\end{bmatrix}^T$ that I had before as angular velocity, let's think about an easier case. Suppose we are rotating my point mass on the X axis around an axis rotated $45^\circ$ from the X axis. Because it will make the numbers come out easier, let's say that the point mass is at $x=\sqrt{2}$.

Now if we change our point of view so that our reference frame is parallel to the rotation axis, the point is no longer at $\begin{bmatrix} \sqrt{2} & 0 & 0\end{bmatrix}^T$, but is now at $\begin{bmatrix} 1 & 1 & 0 \end{bmatrix}^T$ in the new frame. In this frame, the inertia tensor will have one nonzero product of inertia:

$\mathbf{I}'=\begin{bmatrix}1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}$

Now that it has a nonzero product of inertia, it makes sense that the angular momentum isn't parallel to the rotation any more.

I would have guessed that just sliding along the rotation axis wouldn't change the inertia tensor, but further consideration (and math) shows that it does. The inertia about the rotation axis we are sliding along doesn't change, but the inertia about other axes can, and therefore the inertia tensor will no longer be diagonal and the angular momentum will no longer be parallel to the rotation axis.

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