Cyclic Coordinates in Hamiltonian Mechanics I was reading up on Hamiltonian Mechanics and came across the following: 

If a generalized coordinate $q_j$ doesn't explicitly occur in the
  Hamiltonian, then $p_j$ is a constant of motion (meaning, a constant, independent
  of time for a true dynamical motion). $q_j$ then becomes a linear function of time.
  Such a coordinate $q_j$ is called a cyclic coordinate.

The above quote is taken from p. 4 in Ref. 1.
What I don't understand is why $q_j$ is a linear function of time if $p_j$ is constant in time. In other words, why does $p_j$ constant in time imply partial $\frac{\partial H}{\partial p_j}$ is a constant? (In particular, $\frac{\partial H}{\partial p_j}$ could depend on any of the other coordinates or momenta.)
Reference:


*

*Patrick Van Esch, Hamilton-Jacobi Theory in Classical Mechanics, Lecture notes. The pdf file is available from the author's homepage here.

 A: It's an old question and already has a fine answer, I just wanted to quickly add a more physical example.
Take a 2D particle in a central potential in polar coordinates, the Lagrange function is
$$ L(r,\dot r,\theta,\dot\theta) = \frac 1 2 \left( \dot r^2 + r^2 \dot\theta^2 \right) - V(r) \;. $$
Now the angle $\theta$ is a cyclic variable, which gives us the conservation law
$$ \frac{\partial L}{\partial \dot\theta} = r^2 \dot\theta = \text{const} \equiv c \;. $$
However,
$$ \theta(t) = \theta_0 + \int_{t_0}^t \frac{c\, \mathrm dx}{r^2(x)}$$
is not linear in time in general.
A: OP is right. The text has an error. A cyclic coordinate $q_j$ does not have to be an linear function of $t$.
Example: Consider two canonical pairs $(q,p)$ and $(Q,P)$ with Hamiltonian $H= p Q +P$.
Then $q$ is cyclic, and therefore $p$ is a constant of motion. 
$\dot{Q} =\frac{\partial H}{\partial P}=1$, so $Q$ is a linear function of time.
$\dot{q}= \frac{\partial H}{\partial p} = Q$, and hence $q$ is a quadratic function of time.
