# Angular momentum definition? [closed]

The definition of linear momentum is this: Momentum is a vector quantity defined as the product of an object's mass, $m$, and its velocity, $\vec v$.

So According to that definition,The definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis".

But how can we define angular momentum as this: "a pseudovector $\vec r \times \vec p$, the cross product of the particle's position vector $\vec r$ (relative to some origin) and its momentum vector $\vec p = m\vec v$"?

## closed as unclear what you're asking by Mike, ACuriousMind♦, AccidentalFourierTransform, CuriousOne, user36790 Apr 26 '16 at 17:46

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So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis".

That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and (ii) it's not even correct in the general case.

To be more specific, the 'general case' means a rigid body with an arbitrary shape, in which case the moment of inertia is a full 3×3 matrix $I$, which relates the angular momentum $\vec L$ and the angular velocity $\vec \omega$ via a linear transformation, $$\vec L=I\vec \omega\tag1$$ which in component language reads $$L_k=\sum_{k=1}^3I_{kj}\omega_j.\tag 2$$ In certain specific frames (e.g. with axes along the axes of symmetry if those exist) the matrix $I_{kj}$ can be diagonal, but in general it's not.

Note, in particular, that having nonzero off-diagonal elements of $I_{kj}$ means that $\vec \omega$ and $\vec L$ need not be parallel. That is, the angular momentum is not always directed along the axis of rotation. (For a rotating free body of arbitrary shape, since $\vec L$ is conserved, this means that the instantaneous axis of rotation changes over time, which is again in line with how things are.)

But how can we define angular momentum as this: "a pseudovector $\vec r \times \vec p$, the cross product of the particle's position vector $\vec r$ (relative to some origin) and its momentum vector $\vec p = m\vec v$"?

As always with questions of the form "why do we define X as Y?", the answer is "because we can". More specifically, we can define anything we want, and the definition only 'sticks' if it's useful. The definition of $\vec L=\vec r\times\vec p$ for a free particle sticks because it is useful: it is still conserved for a system that's isolated or under the influence of a central force, and it can be used in broader contexts than just a body in pure rotational motion.

In addition, it reduces to the alternative definition in terms of a moment of inertia and an angular velocity $\omega$ when the latter exists (i.e. when the particle is in a rotational motion, defined by $\vec v=\vec \omega\times \vec r$, which requires that $\vec v\cdot\vec r=0$, which is not always the case), via standard manipulations which are found in any suitable classical mechanics textbook (e.g. Goldstein).

More generally, though, one should have a thorough understanding of the existing definitions before claiming that they 'should' be something else, particularly if that something else is incorrect.