Angular momentum in Lagrangian Mechanics

1. Oct 15, 2012

TheDragon

In Newton's problem,and other central force problems in Classical Mechanics, you can get with decreasing the center of mass movement to the lagrangian:
$L=1/2m(r' ^2+r^2 \varphi'^2)-V(r)$
because $\varphi$ is cyclic, you can write:
$\frac{d}{dt}(mr^2 \varphi')=0$
or, defining the angular momentum:
$mr^2 \varphi'=l$

In order for you to understand my question we will go through two ways:

Way no.1) Now if you go ahead and use euler-lagrange equation to r you will get:
$\frac{d}{dt}(mr')=mr\varphi'^2-\frac{∂V(r)}{∂r}$
and exchanging for l you can get:
$\frac{d}{dt}(mr')=\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}$

Way no.2) But if you first excange phi and then use euler-lagrange to r you get:
$L=1/2m(r' ^2+\frac{l^2}{m^2r^2})-V(r)=1/2mr' ^2+\frac{l^2}{2mr^2}-V(r)$
and then euler lagrange for r:
$\frac{d}{dt}(mr')=\frac{∂}{∂r}(\frac{l^2}{2mr^2})-\frac{∂V(r)}{∂r}=-\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}$

The plus in way no.1 became a minus in way no.2. How can you explain it?
If you can, don't show me a solution using energy because I already know that one. I want to use only lagrangian formalism.

2. Oct 20, 2012

TheDragon

The answer to the question is that the angular momentum is a constant in relation to time, but the partial derivative in r is not zero. In other words:
$\frac{dl}{dt}=0$,

$\frac{∂l}{∂r}\neq0$.

I do think that:
$\frac{dl}{dr}=0$.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook