How can angular velocity or angular momentum be a vector?

Rotations in 3 dimensions are not commutative; however they are in the plane. In classical mechanics, are we allowed to say that angular momentum is a vector because particles only rotate along a single axis? Or do we make some kind of argument that you can assign angular velocity to an object because you can approximate it locally as being along a single rotation axis and so its "locally commutative" and hence appropriate to call it a vector? If that is the case what happens if we have a particle rotating in a way that isn't differentiable? Is there some physical reason that particles can't rotate in that sort of way?

I'm really confused by this and haven't taken physics since high school, any help would be appreciated. :)

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This is more physics than math, but angular velocity and momentum are actually pseudovectors: en.wikipedia.org/wiki/Pseudovector – J. M. Aug 28 '10 at 12:45
I tagged it physics first, but it rearranges the tags in alphabetical order (I think). Thank you for helping, but I'm still confused. I'm still don't understand how angular velocity can be defined when angular displacements aren't commutative. – WWright Aug 28 '10 at 13:32
To me, the notion is only a convenience. It's convenient to put an imaginary vector along the axis of rotation because that's what you get when you cross-product $r$ and your force/velocity (which is perpendicular to $r$). It gets more convenient because forces in opposite directions have opposite-facing vectors, and when those add up, they cancel out. The idea isn't that there is an actual vector there; it's only a very convenient thing to do for the maths. – Justin L. Aug 28 '10 at 19:01
I find the notion of pseudovector quite artificial. The most natural way to look at angular velocity and angular momentum are as bivectors: en.wikipedia.org/wiki/Bivector – Hans Lundmark Sep 30 '10 at 20:11

There are a couple of things at play here, which coincidentally(?) mirror the difference between mathematicians who say "vector" to mean anything can be added commutatively and multiplied by a scalar, and physicists who say "vector" to mean a 3- (or 4-)dimensional quantity that transforms properly under change of coordinates.

1. While arbitrary rotations in 3 dimensions do not commute, infinitesimal rotations do. (In fact any "infinitesimal" transformations commute, as you can see by multiplying $I + \epsilon A$ and $I + \epsilon B$ and ignoring second-order terms in $\epsilon$.) Since angular velocity can be thought of as "infinitesimal rotation per infinitesimal time", it ends up being a vector (in the mathematical sense) even though rotation itself isn't. Similarly, angular momentum is the derivative of kinetic energy with respect to angular velocity, so it is a gradient, which is a (dual) vector (to angular velocity).

2. There's an interesting mapping between these infinitesimal rotation matrices and vectors (in the physical sense) which only works in 3 dimensions. If you think of an infinitesimal rotation as $I + \epsilon A$, one can show that $A$ must be antisymmetric. This means its diagonal entries are 0, leaving only 3 degrees of freedom in the off-diagonal entries. Such a matrix can be associated with a vector using the cross product:

$$\begin{bmatrix}0 & -\omega_z & \omega_y \\\ \omega_z & 0 & -\omega_x \\\ -\omega_y & \omega_x & 0\end{bmatrix} \mathbf{r} = \mathbf{\omega} \times \mathbf{r}$$

So angular velocity isn't "naturally" a vector in physical space, but rather lives in a different 3-dimensional vector space. That's why once you interpret it as a physical vector, it turns out to transform slightly differently, as J.M. has pointed out in the comments.

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Heh, "mirror". :D That's the key to looking at pseudovectors! – J. M. Aug 28 '10 at 14:05
Thanks. I think I see what's going on now. – WWright Aug 28 '10 at 14:12
There's probably something that can be said here about tangent spaces of manifolds, or the Lie algebra of the rotation group, but I wish I had any expertise in those areas. – Rahul Aug 28 '10 at 14:13
One has the Lie group O(n) of all distance preserving linear transformations (matrices) of R^n. These are just the n x n matrices that satisfy A^tA = Id. The Lie algebra of this Lie group is o(n), which is just all the antisymmetric matrices. To see this, one picks a path of matrices A(s) with A(0) = Id and just differentiate the equation A^t(s)A(s) = Id at s = 0. The right hand side gives 0 since Id is constant, and the left hand side, after using the fact that differentiation and transpose commute, gives A'(0) + A'^t(0) = 0, which is just the condition that A is antisymmetric. – Jason DeVito Aug 28 '10 at 16:06