- #1

vancouver_water

- 77

- 10

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

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!