Perturbation of central field potential

Question

i`d like to consider system with Coulomb potential: $U = -\frac{\alpha}{r}$ and constant magnetic field.It is easy to write Lagrangian function: $$ L = \frac{m}{2}(\dot{\rho}^2 + \rho^2\dot{\phi}^2) + \frac{eH}{2c}\rho^2 \dot{\phi} + \frac{\alpha}{r} $$ and get trajectory equation in cylindrical coordinate system. $$\phi = \sqrt{\frac{m}{2}}\int \frac{\frac{M}{mr^2} - \Omega}{\sqrt{E-U_{eff}(r)}}dr,$$ where $\Omega = \frac{eH}{2mc}$, $$U_{eff} = -\frac{\alpha}{r} + \frac{mr^2}{2}(\Omega - \frac{M}{mr^2})^2.$$

If we expand brackets, we can get such integral, that i want to analyze: $$\phi = \frac{M}{\sqrt{2m}}\int\frac{dr}{r^2\sqrt{E+M\Omega + \frac{\alpha}{r} - \frac{M^2}{2mr^2} - \frac{m\Omega^2r^2}{2}}} - \Omega\cdot t $$

So we can see, that effect of the magnetic field is reduced to the replacement of energy by the expression $E' = E + M\Omega$.

We have field: $$U = U_1 + U_2 = -\frac{\alpha}{r} + \frac{M^2}{2mr^2} + \frac{m\Omega^2r^2}{2},$$

Where $U_1 = -\frac{\alpha}{r} + \frac{M^2}{2mr^2}$ and $U_2 = \frac{m\Omega^2r^2}{2}$

i want to consider the case when $|U_2|<<|U_1|$ and $|U_1|<<|U_2|$.

First case:

we can Expand in taylor series denominator, but I expect divergences at the turning points. Also i know, that i can say "$|U_2|$ gives our system precession", and i can calculate it. But for the second case i can`t do the same thing, i think that there will be a drift of trajectory,but i'm not sure.

So i want to know, how we can expand denominator and get approximation of trajectory directly(including coordinates near the turning points).The problem of decomposition into a perturbation series interests me more than the solution of the problem itself.

Answer 1

Since you notation is not quite concise I will do some guesswork in order to understand the underlying problem. The basic setting is that of a test particle that moves in a central potential $V(r)$. It holds that the change in the polar angle $\phi$ is given by $$ \Delta \phi = 2 \int_{r_\text{min}}^{r_\text{max}} \frac{\left(\frac{l}{r^2}\right)\text{d}r }{\sqrt{2m(E-V(r)) - \frac{l^2}{r^2}}},\qquad(1) $$ where $E,l$ is the initial energy and angular momentum of the system and $r_\text{min}, r_\text{max}$ are the minimal and maximal radius reached of the test mass $m$.

If I understand correctly you want to study the potential $$ V(r) = -\frac{\alpha}{r} + \beta r^2 $$ in the regimes where the first term dominates the second and vice versa (the situation roughly translates into (for all times) small and (for all times) large distances from the origin). It seems that OP is already aware of the method for example explained in "Mechanics. Vol. 1" by L.Landau and E. Lifshitz (chapter 15 exercise 3). I think the misunderstand comes from the fact that it makes no sense to include the "centrifugal term" into the $1/r$-potential. The bookkeeping should be: $$ V(r) = U(r) + \delta U (r),\qquad(2) $$ where in case one $$ U(r)=-\frac{\alpha}{r},\quad\delta U (r) = \beta r^2,\qquad(2.1) $$ and in case two $$ U(r)=\beta r^2,\quad\delta U (r) = -\frac{\alpha}{r}.\qquad(2.2) $$ Using this bookkeeping the insertion of (2.1), (2.2) in (2) then in (1) and then expanding in orders of $\delta U$ should result in the desired corrections for the two cases.