I 2 Body Radial Equation, Effective Potential derivation

1. Apr 20, 2016

vancouver_water

According to my textbook, in the derivation for the effective potential $U_{eff}$, starting with the Lagrangian $L = \frac{1}{2}\mu(\dot r^2 +r^2\dot\phi^2) -V(r)$, substituting into Lagrange's equation gives $\mu\ddot r = -\frac{\partial V}{\partial r} + \frac{l^2}{\mu r^3} = -\frac{\partial}{\partial r}(V(r) + \frac{l^2}{2\mu r^2})$, where the substitution $\dot\phi = \frac{l}{r^2\mu}$ is made after substituting into Lagrange's equation and the effective potential is $V(r) + \frac{l^2}{2\mu r^2}$ and $l$ is the angular momentum.

However, when I do the derivation by first substituting the angular momentum, it goes like this: Starting with $L = \frac{1}{2}\mu(\dot r^2 +\frac{l^2}{\mu^2 r^2}) -V(r)$, then substituting into lagranges equation gives $\mu\ddot r = -\frac{\partial V}{\partial r} - \frac{l^2}{\mu r^3} = -\frac{\partial}{\partial r}(V(r) - \frac{l^2}{2\mu r^2})$, where the effective potential is $V(r) - \frac{l^2}{2\mu r^2}$.

This is the wrong result, but I can't see why substituting angular momentum before or after taking the derivatives should make any difference.

Thanks for any help!

2. Apr 20, 2016

drvrm

actually we use r,theta description.
here theta is replaced by phi....
no matter but r, dr/dt, phi, dphi/dt are the generalised coordinates and velocities ;
naturally one has two Lagranges equations - and if the result of second Lagranges equation is put in the definition of L itself to describe r-ewuation ; you have problem - the two are independent equations in r and phi.
so the action principle must be getting affected.
r^2. phi(dot) is a constant -where r and phi(dot) both can vary such that the product is a constant.
the error lies in the r-equation as this condition which implicitly connects r and phi(dot) is no longer an independent equation of motion in r.