Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On Goldstein's "Classical Mechanics" (first ed.), I have read that

if $q_j$ is a cyclic coordinate, its generalized momentum conjugate $p_j$ is costant.

He obtained that starting from Lagrange's equation: $\frac {d}{dt} \frac {\partial L}{\partial \dot q_j}- \frac {\partial L}{\partial q_j}=0$

But this Lagrange's equation refers to a conservative system. What would happen if I considered a system in which the potential is $U(q, \dot q)$?

share|cite|improve this question
up vote 1 down vote accepted

OP wrote(v1):

What would happen if I considered a system in which the potential is [velocity-dependent] $U(q, \dot q)$?

Well, if OP already knows that the generalized force$^1$

$$\tag{1} Q_j~=~\frac {d}{dt} \frac {\partial U}{\partial \dot q^j}- \frac {\partial U}{\partial q^j}$$

is given in terms of a velocity-dependent potential $U=U(q, \dot q, t)$, this means that Lagrange's equations

$$\tag{2}\frac {d}{dt} \frac {\partial T}{\partial \dot q^j}- \frac {\partial T}{\partial q^j}~=~Q_j,$$

can be written as

$$\tag{3}\frac {d}{dt} \frac {\partial L}{\partial \dot q^j}- \frac {\partial L}{\partial q^j}~=~0, \qquad L~=~T-U.$$

As long as one has Lagrange's equations (3), then it is still true that if $q^j$ is a cyclic coordinate $\frac {\partial L}{\partial q^j}=0$, then its generalized momentum conjugate $p_j:=\frac {\partial L}{\partial \dot q^j}$ is a constant of motion.

$^1$ Here we consider for simplicity a system with only one type of generalized force. In practice, there may be several types of forces (e.g. gravity force, Lorentz force, etc.). The generalization is straightforward.

share|cite|improve this answer
But if I consider a velocity-dependent potential, Lagrange's equation is $\frac {d}{dt} \frac {\partial L}{\partial \dot q_j}- \frac {\partial L}{\partial q_j}=Q_j$, isn't it? If $q_j$ is a cyclic coordinate have I always Lagrange's equation of the type: $\frac {d}{dt} \frac {\partial L}{\partial \dot q_j}- \frac {\partial L}{\partial q_j}=0$? – sunrise Nov 9 '12 at 13:20
I updated the answer. – Qmechanic Nov 9 '12 at 13:34
I got it! Thanks a lot! :) – sunrise Nov 9 '12 at 13:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.