# Difference between kinematic momentum and conjugated momentum in purely mechanical setup

I don't know much about physics, but I wanted to understand what was the difference between the "kinematic momentum" and the conjugated momentum. As I understand it, kinematic momentum is mass times speed and the conjugated momentum should be the one that verifies the Lagrangian equation (somehow I read it is $$\frac{\partial L}{\partial \dot{q}}$$). I was looking at this example to see when they could be different. But is there any simple purely mechanical examples, without charged particles?

## 1 Answer

The conjugate momentum corresponding to an angular coordinate is not a linear momentum but an angular momentum. For example, consider the Lagrangian of a particle of mass $$m$$ moving in a central force field described by the potential $$V(r)$$ and written in plane polar coordinates: $$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r).$$ You can easily verify that the quantity $$\frac{\partial L}{\partial\dot{r}}=m\dot{r}$$ is a linear momentum while $$\frac{\partial L}{\partial\dot{\theta}}=mr^2\dot{\theta}=mr^2\omega$$ is the angular momentum about the origin.

Addendum If $$q$$ is a generalized coordinate of linear dimension (e.g., the radial coordinate $$r$$, or cartesian coordinates $$x,y,z$$ etc), and if the potential is independent of the generalized velocities $$\dot{q}$$, the generalized velocity corresponding to $$q$$, defined as $$p_q:=\frac{\partial L}{\partial\dot{q}}$$ will always be of the form $$m\dot{q}$$, simply from dimensional analysis.

• But should it always be $m\dot{q}$, where $q$ is the generalized coordinate? – roi_saumon Mar 19 '19 at 15:13
• @roi_saumon Please see the edited answer and let me know whether I should clarify anything more. – SRS Mar 19 '19 at 15:29