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Generalized Momenta

Consider the motion of a single particle moving in one dimension. The kinetic energy is
\begin{displaymath}
K = \frac{1}{2}\,m\,\dot{x}^{\,2},
\end{displaymath} (649)

where $m$ is the mass of the particle, and $x$ its displacement. Now, the particle's linear momentum is $p=m\,\dot{x}$. However, this can also be written
\begin{displaymath}
p = \frac{\partial K}{\partial \dot{x}}= \frac{\partial L}{\partial\dot{x}},
\end{displaymath} (650)

since $L=K-U$, 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 above expression, we can define generalized momenta of the form

\begin{displaymath}
p_i = \frac{\partial L}{\partial\dot{q}_i},
\end{displaymath} (651)

for $i=1,{\cal F}$. Here, $p_i$ is sometimes called the momentum conjugate to the coordinate $q_i$. Hence, Lagrange's equation (613) can be written
\begin{displaymath}
\frac{d p_i}{dt} = \frac{\partial L}{\partial q_i},
\end{displaymath} (652)

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 (652) that

\begin{displaymath}
\frac{d p_k}{dt} = \frac{\partial L}{\partial q_k}=0.
\end{displaymath} (653)

Hence,
\begin{displaymath}
p_k = {\rm const.}
\end{displaymath} (654)

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, in Section 9.5, the Lagrangian (615) for a particle moving in a central potential is independent of the angular coordinate $\theta $. Thus, $\theta $ is an ignorable coordinate, and

\begin{displaymath}
p_\theta = \frac{\partial L}{\partial\dot{\theta}} = m\,r^2\,\dot{\theta}
\end{displaymath} (655)

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.

Again, in Section 9.7, the Lagrangian (642) for a mass sliding down a sliding slope is independent of the Cartesian coordinate $x$. It follows that $x$ is an ignorable coordinate, and

\begin{displaymath}
p_x = \frac{\partial L}{\partial \dot{x}} = M\,\dot{x} + m\,(\dot{x}+\dot{x}'\,\cos\theta)
\end{displaymath} (656)

is a constant of the motion. Of course, $p_x$ is the total linear momentum in the $x$-direction. This is conserved because there is no external force acting on the system in the $x$-direction.


next up previous
Next: Spherical Pendulum Up: Lagrangian Dynamics Previous: Sliding down a Sliding
Richard Fitzpatrick 2011-03-31