# Correct Definition of Angular Momentum of a Charged Particle in an Electromagnetic Field? (Classical Mechanics) [duplicate]

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What is the more correct definition of angular momentum $\vec{\mathbf{M}}$ in three dimensions? (I.e. classically/Lagrangian/Hamiltonian, not necessarily quantum or relativistic)

$$\vec{\mathbf{M}}=m\cdot \vec{\mathbf{r}}\times \vec{\mathbf{v}}?$$

or

$$\vec{\mathbf{M}}=\vec{\mathbf{r}} \times \frac{\partial L}{\partial \vec{\mathbf{v}}}?$$

Obviously these two expressions are usually the same, but not always. (I think, I'm not sure, that is why I am asking -- I might be confusing "actual" momentum with a generalized momentum. I would have thought "canonical momentum" would refer to the former and not the latter).

If I am reading them correctly, for a single particle in an electromagnetic field, we have (classically) that

$$\frac{\partial L}{\partial \vec{\mathbf{v}}} = m\vec{\mathbf{v}}+q\vec{\mathbf{A}}\not=m\vec{\mathbf{v}}$$

Hence, if I want to calculate the angular momentum for such a particle, I need to know better what the proper definition of angular momentum is, since there are at least two choices available, both which seem appealing.

(Also the reason why $\frac{\partial L}{\partial \vec{\mathbf{v}}}\not=m\vec{\mathbf{v}}$ here is because we're dealing with a non-conservative system, right? So that force only equals the time derivative of momentum for conservative systems?)

## marked as duplicate by Community♦May 28 '16 at 23:52

In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular momentum". The canonical momentum depends on your choice of generalized coordinates to describe the system, the kinetic momentum does not.