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

Consider the Newtonian gravitational potential at a distance of Sun:

$$\varphi \left ( r \right )~=~-\frac{GM}{r}.$$

I write the classical Lagrangian in spherical coordinates for a planet with mass $m$:

$$L ~=~ \frac{1}{2}m (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{GM}{r},$$

and find that the canonical momentum $p_{\phi }$ is a constant of motion, because:

$$\dot{p_{\phi }}~=~ \frac{\partial L}{\partial \phi} ~=~ 0.$$

  1. What is the physical interpretation of the canonical momentum?

  2. How can we from the Lagrangian see that it is a constant of motion?

share|cite|improve this question
The Lagrangian for the classical two body problem is: $$ L = \frac{M m}{M\!+\!m} \frac{1}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{GMm}{r} $$ the Lagrangian per unit mass is: $$ L =\frac{1}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{G(M\!+\!m)}{r} $$ For motion about a fixed mass M, the Lagrangian is: $$ L = \frac{m}{2} (\dot{r}^{2} + r^{2}\dot{\theta ^{2}} + r^{2}\dot{\phi ^{2}}\sin^{2}\theta ) + \frac{G M m}{r} $$ – Nick May 12 '13 at 22:24
up vote 2 down vote accepted

The azimuthal momentum

$$p_{\phi}~:=~\frac{\partial L}{\partial \dot{\phi}}$$

is the (polar) $z$-component of the angular momentum $L_z$ of the point mass $m$ relative to the heliocentric reference frame. It is a constant of motion because the azimuthal angle $\phi$ is a cyclic coordinate.

share|cite|improve this answer
Just in case some readers don't understand "cyclic coordinate" let me just note that it is a variable NOT present in the Lagrangian. Its derivative is, but that's not the same thing. – Paul J. Gans Nov 3 '12 at 20:20

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.