- #1
gitano
- 11
- 0
Hi,
I have a question that has been bothering me for a while now. For Kepler's problem we know that angular momentum [tex]M_{z}[/tex] is conserved and that the angular velocity [tex]\frac{d\phi}{dt}[/tex] is equal to [tex]\frac{M_{z}}{mr^{2}}[/tex]. When we substitute for [tex]\frac{d\phi}{dt}[/tex] in the expression for energy, we get an effective potential
[tex]U_{eff}(r) = U(r) + \frac{M_{z}^{2}}{2mr^{2}}[/tex],
which is correct.
However, when we substitute this into the Lagrangian, one of the signs changes and we arrive at an erroneous effective potential
[tex]U_{eff}(r) = U(r) -\frac{M_{z}^{2}}{2mr^{2}}[/tex]
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.
I have a question that has been bothering me for a while now. For Kepler's problem we know that angular momentum [tex]M_{z}[/tex] is conserved and that the angular velocity [tex]\frac{d\phi}{dt}[/tex] is equal to [tex]\frac{M_{z}}{mr^{2}}[/tex]. When we substitute for [tex]\frac{d\phi}{dt}[/tex] in the expression for energy, we get an effective potential
[tex]U_{eff}(r) = U(r) + \frac{M_{z}^{2}}{2mr^{2}}[/tex],
which is correct.
However, when we substitute this into the Lagrangian, one of the signs changes and we arrive at an erroneous effective potential
[tex]U_{eff}(r) = U(r) -\frac{M_{z}^{2}}{2mr^{2}}[/tex]
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.