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Physics
Classical Physics
Mechanics
2 Body Radial Equation, Effective Potential derivation
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[QUOTE="vancouver_water, post: 5449293, member: 538179"] According to my textbook, in the derivation for the effective potential [itex]U_{eff}[/itex], starting with the Lagrangian [itex]L = \frac{1}{2}\mu(\dot r^2 +r^2\dot\phi^2) -V(r)[/itex], substituting into Lagrange's 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 substitution [itex]\dot\phi = \frac{l}{r^2\mu}[/itex] is made after substituting into Lagrange's equation and the effective potential is [itex]V(r) + \frac{l^2}{2\mu r^2}[/itex] and [itex]l[/itex] is the angular momentum. 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! [/QUOTE]
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2 Body Radial Equation, Effective Potential derivation
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