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.)


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

2 Answers 2


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.


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.


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