I was told that the angular velocity vector does not always have to point in the same direction as the angular momentum vector. This is due to the fact that they are related by the equation $L=I \omega$. But in general, $I$ is a tensor and so the result might not be in the same direction as $\omega$. Because the equations for linear and angular motion are very symmetrical that leads me to ask -- does the velocity vector always point in the same direction as the momentum vector?
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$\begingroup$ The answer is, in general, no. See my comment in the answer below. $\endgroup$– yuggibCommented Mar 28, 2015 at 10:40
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Because the equations for linear and angular motion are very symmetrical
In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object.
In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor.
So, the 'symmetry' isn't really there. Linear momentum and angular momentum are different kinds of objects.
does the velocity vector always point in the same direction as the momentum vector
As far as I know, linear kinetic momentum and linear velocity are parallel in both classical (three-vector) mechanics
$$\mathbf p = m \mathbf v$$
and relativistic (four-vector) mechanics.
$$\mathbf P = m \mathbf U $$
However, as yuggib points out in a comment, the canonical momentum
$$\mathbf P = \frac{\partial \mathcal L}{\partial \dot {\mathbf q}}$$
is not generally parallel to $\dot {\mathbf q}$. For example, the canonical momentum of a non-relativistic charged particle is
$$\mathbf P = m\mathbf v + q\mathbf A = \mathbf p + q\mathbf A$$
where $\mathbf A$ is the magnetic vector potential.
answered Mar 28, 2015 at 2:03
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$\begingroup$ The momentum in classical mechanics is usually defined as the derivative of the lagrangian $\mathscr{L}$ with respect to the velocity: $p_i=\frac{\partial\mathscr{L}}{\partial \dot{x}_i}$; and given a Lagrangian like that of a particle in an electromagnetic field $\mathscr{L}=\frac{1}{2}m\dot{x}^2+e\dot{x}A(x)-e\phi$, you get $p= m\dot{x}+ A(x)$ that does not have the same direction of velocity. $\endgroup$– yuggibCommented Mar 28, 2015 at 10:37
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$\begingroup$ @yuggib, good point about canonical momentum. $\endgroup$ Commented Mar 28, 2015 at 11:11
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$\begingroup$ Well, actually I realize now that with the Lagrangian point of view, also the angular momentum is a canonical momentum :-) $\endgroup$– yuggibCommented Mar 28, 2015 at 11:18
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$\begingroup$ @yuggib, I've edited my answer to address your comment. $\endgroup$ Commented Mar 28, 2015 at 11:59