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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?

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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

1 Answer 1

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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.

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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

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