Contradicting Effective Potentials for Kepler's Problem

[gitano]

Hi,

I have a question that has been bothering me for a while now. For Kepler's problem we know that angular momentum $$M_{z}$$ is conserved and that the angular velocity $$\frac{d\phi}{dt}$$ is equal to $$\frac{M_{z}}{mr^{2}}$$. When we substitute for $$\frac{d\phi}{dt}$$ in the expression for energy, we get an effective potential

$$U_{eff}(r) = U(r) + \frac{M_{z}^{2}}{2mr^{2}}$$,

which is correct.

However, when we substitute this into the Lagrangian, one of the signs changes and we arrive at an erroneous effective potential

$$U_{eff}(r) = U(r) -\frac{M_{z}^{2}}{2mr^{2}}$$

which is clearly wrong. There must be some subtlety which I am overlooking that explains why you can't substitute this expression into the Lagrangian and arrive at the correct effective potential, or for that matter the correct Lagrangian.

 

[gitano]

I am also curious that if in general you can express a generalized coordinate in terms of a constant of motion (the coordinate is cyclic) is it also wrong then to make such a substitution in the Lagrangian as is incorrect in the Kepler problem.

 

[dgOnPhys]

Hi,

have a look at this: http://www.aerostudents.com/files/dynamicsAndStability/lagrangianDynamics.pdf"

Section 3.4 has your answer: when suppressing a coordinate through a conserved momentum, the Lagrangian is not invariant, you have to use the Routhian (actually if you change its sign, things look better) to transform the Lagrangian so that its Euler-Lagrange equations are the equation of motion

Routhian will turn out to be R=-T+Ueff (the correct Ueff)

Hope this helps