Generalized momenta

Consider the motion of a single particle moving in one dimension. The kinetic energy is

$$K = \frac{1}{2}\,m\,\skew{3}\dot{x}^{\,2},$$ (7.33)

where $m$ is the mass of the particle and $x$ its displacement. The particle's linear momentum is $p=m\,\skew{3}\dot{x}$. However, this can also be written

$$p = \frac{\partial K}{\partial \skew{3}\dot{x}}= \frac{\partial {\cal L}}{\partial\skew{3}\dot{x}},$$ (7.34)

because ${\cal L}=K-U$ and the potential energy $U$ is independent of $\skew{3}\dot{x}$.

Consider a dynamical system described by ${\cal F}$ generalized coordinates $q_i$, for $i=1,{\cal F}$. By analogy with the preceding expression, we can define generalized momenta of the form

$$p_i = \frac{\partial {\cal L}}{\partial\skew{3}\dot{q}_i},$$ (7.35)

for $i=1,{\cal F}$. Here, $p_i$ is sometimes called the momentum conjugate to the coordinate $q_i$. Hence, Lagrange's equation (7.22) can be written

$$\frac{d p_i}{dt} = \frac{\partial {\cal L}}{\partial q_i},$$ (7.36)

for $i=1,{\cal F}$. Note that a generalized momentum does not necessarily have the dimensions of linear momentum.

Suppose that the Lagrangian ${\cal L}$ does not depend explicitly on some coordinate $q_k$. It follows from Equation (7.36) that

$$\frac{d p_k}{dt} = \frac{\partial {\cal L}}{\partial q_k}=0.$$ (7.37)

Hence,

$$p_k = {\rm const.}$$ (7.38)

The coordinate $q_k$ is said to be ignorable in this case. Thus, we conclude that the generalized momentum associated with an ignorable coordinate is a constant of the motion.

For example, the Lagrangian [Equation (7.24)] for a particle moving in a central potential is independent of the angular coordinate $\theta$. Thus, $\theta$ is an ignorable coordinate, and

$$p_\theta = \frac{\partial {\cal L}}{\partial\skew{5}\dot{\theta}} = m\,r^{\,2}\,\skew{5}\dot{\theta}$$ (7.39)

is a constant of the motion. Of course, $p_\theta$ is the angular momentum about the origin; this is conserved because a central force exerts no torque about the origin.