Generalized Momenta

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

$$K = \frac{1}{2} m \dot{x}^{ 2},$$ (B.23)

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

$$p = \frac{\partial K}{\partial \dot{x}}= \frac{\partial L}{\partial\dot{x}},$$ (B.24)

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

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

$$p_i = \frac{\partial L}{\partial\dot{q}_i},$$ (B.25)

for $i=1,{\cal F}$ . Here, $p_i$ is sometimes called the momentum conjugate to the coordinate $q_i$ . Hence, Lagrange's equation, (B.22), can be written

$$\frac{d p_i}{dt} = \frac{\partial L}{\partial q_i},$$ (B.26)

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

Suppose that the Lagrangian, $L$ , does not depend explicitly on some coordinate $q_k$ . It follows from Equation (B.26) that

$$\frac{d p_k}{dt} = \frac{\partial L}{\partial q_k}=0.$$ (B.27)

Hence,

$$p_k = {\rm constant}.$$ (B.28)

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.