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I 2 Body Radial Equation, Effective Potential derivation

  1. Apr 20, 2016 #1
    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!
  2. jcsd
  3. Apr 20, 2016 #2
    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.
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