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Generalized Momenta
Consider the motion of a single particle moving in one dimension. The
kinetic energy is
 |
(649) |
where
is the mass of the particle, and
its displacement.
Now, the particle's linear momentum is
. However,
this can also be written
 |
(650) |
since
, and the potential energy
is independent of
.
Consider a dynamical system described by
generalized coordinates
, for
. By analogy with the above expression, we can
define generalized momenta of the form
 |
(651) |
for
. Here,
is sometimes called the momentum conjugate to the coordinate
. Hence, Lagrange's equation (613) can be written
 |
(652) |
for
. Note that a generalized momentum does not necessarily have
the dimensions of linear momentum.
Suppose that the Lagrangian
does not depend explicitly on some coordinate
. It follows from Equation (652) that
 |
(653) |
Hence,
 |
(654) |
The coordinate
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
. Thus,
is an ignorable coordinate,
and
 |
(655) |
is a constant of the motion. Of course,
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
. It follows that
is an ignorable coordinate,
and
 |
(656) |
is a constant of the motion. Of course,
is the total linear momentum in the
-direction. This is conserved because there is no external force acting on
the system in the
-direction.
Next: Spherical Pendulum
Up: Lagrangian Dynamics
Previous: Sliding down a Sliding
Richard Fitzpatrick
2011-03-31