Hello! I am considering the motion of particle in a central field and appear to derive inconsistent equations of motion when the Euler Lagrange equations are applied directly to (1) the Lagrangian $L(r,\dot r, \dot \phi)$ and (2) when conservation of angular momentum is used eliminate $\dot \phi$ to write $L(r,\dot r)$:
Method 1: The Lagrangian for a particle in a central field is $$ L = \frac{1}{2}m\left(\dot r^2 + r^2 \dot \phi^2\right) - U(r)$$ Langrange's equation, ${\rm d}(\partial L/\partial \dot q)/{\rm d}t = \partial L/\partial q$, for $q=\phi$ yields conservation of ang. momentum:
\begin{align} &\frac{{\rm d}}{{\rm d}t}(mr^2\dot \phi) = 0 \quad \Rightarrow \quad mr^2\dot \phi = \text{constant} = M (\text{say})\\ \end{align} Langrange's equation for $q=r$ becomes \begin{align} m\ddot r = mr\dot \phi^2 - {\rm d}U/{\rm d}r = \frac{M^2}{mr^3} - {\rm d}U/{\rm d}r \end{align} where the second equality holds from conservation of ang. momentum.
Method 2: Now if we use conservation of ang. momentum $\dot \phi = M/(mr^2)$ to eliminate $\dot \phi$ from our original Lagrangian we have $$ L = \frac{1}{2}m\dot r^2 + \frac{M^2}{2mr^2} -U(r)$$ Lagrange's equation for $r$, ${\rm d}(\partial L/\partial \dot r)/{\rm d}t = \partial L/\partial r$, gives $$m\ddot r = -\frac{M^2}{mr^3} - {\rm d} U/{\rm d} r$$ Compare this to the radial momentum equation of method 1. Note the difference in sign of the term $M^2/(mr^3)$!!
Any ideas of how to resolve the disparity?
Resolution of Disparity. Ref: How can you solve this "paradox"? Central potential in particular the answer by @Craig J Copi. Essentially when computing $\partial L/\partial r$ in method 2 one must remember that although $M$ is a constant of the motion ($\dot M = 0$) that $\partial M/\partial r = 2mr\dot \phi = 2M/r \neq 0$
