# Does angular momentum depend on coordinate frame?

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.

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