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 1


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.

  • $\begingroup$ But should it always be $m\dot{q}$, where $q$ is the generalized coordinate? $\endgroup$
    – roi_saumon
    Commented Mar 19, 2019 at 15:13
  • $\begingroup$ @roi_saumon Please see the edited answer and let me know whether I should clarify anything more. $\endgroup$
    – SRS
    Commented Mar 19, 2019 at 15:29

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