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)

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


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

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).

See here: http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec5.pdf
and here: http://insti.physics.sunysb.edu/itp/lectures/01-Fall/PHY505/09c/notes09c.pdf

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?)


1 Answer 1


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.

If you are told to "compute angular momentum", usually, the kinetic angular momentum will be what is meant. If you want to be absolutely sure that what you are computing is truly "angular momentum", you have but one choice: Determine the Noether charge for the rotations, since, like energy is the Noether charge of time translations and momentum is the Noether charge of spatial translations, angular momentum is the Noether charge of rotations.

  • $\begingroup$ Just some follow-ups: so the canonical momentum can still not equal the kinetic momentum even when the Lagrangian is with respect to the actual position and velocity? That seemed to be the conclusion I got from the links I mentioned. Is this discrepancy because the system is not conservative? (I.e. the canonical (angular) momentum with respect to the actual position and actual velocity is equal to the kinetic (angular) momentum if and only if the system is conservative?) $\endgroup$ May 28, 2016 at 21:34
  • $\begingroup$ Also in reference to Noether charges, these are always the "kinetic" varieties? Or only when the system has the requisite symmetries to allow the quantities to be conserved? $\endgroup$ May 28, 2016 at 21:36