Kepler's Two-Body Problem Choice of Sign

Question

I have been going over the solution to Kepler's problem and there is a subtlety I am missing. The notes I am following (my own notes from a while ago in fact...) get to the following expression $$ \frac{d\phi}{dr}=\pm\frac{L}{r^{2}\sqrt{2\mu(E-U_{eff}(r))}} $$ with $U_{eff}(r)=\frac{L^{2}}{2\mu r^{2}}-\frac{\alpha}{r}$ and in Kepler's case $\alpha=GMm$

Before going on to integrate this to get $\phi(r)$, there is the statement "Choose $+$ sign in order to start describing the motion from $r_{min}$", with $r_{min}$ being a turning point (satisfying $E-U_{eff}=0$). Taking only the positive square root, the analysis continues to arrive finally at the (correct) answer $$ r=\frac{r_{0}}{1+\epsilon\cos(\phi)}\text{ , }r_{0}=\frac{L^{2}}{\mu\alpha} $$ I don't see how this selection of only the $+$ square root is valid. It suggests that $\frac{d\phi}{dr}$ can only ever be positive, which is not the case (it should have, for a given $r$, two values one the negative of the other for the two places on the ellipse (/conic section) with that $r$). By doing this it is like we are only describing one half of the solution. I have carried on the calculation keeping the $\pm$ the whole way through and get the same answer (up to a choice of constant), and Fetter and Walecka don't seem to neglect the minus sign either (though their treatment of Kepler's problem is quite brief). Other sources seem to implicitly take just the $+$ sign without any comment.

It seems that the neglected portion of the solution comes back due to the multi-valued nature of inverse trigonometric functions which are then turned into (not-inverse) trig functions, but the fact that it all works out in the end anyway doesn't really satisfy me.

What is the reason that allows us to neglect the $-$ sign in $\frac{d\phi}{dr}$?

Answer 1

From the angular momentum conservation you have $ L=m r^2 \dot{\phi} \rightarrow L \, dt =mr^2 \, d \phi $ so, from $dt>0$, you have that $L>0$ implies $d\phi>0$. Now, from your first equation you have $ d \phi= \pm \frac{L}{r^2 \sqrt{2\mu(E-U_{eff}(r)}} \, dr $ so if you have a minus sign you start from the afelium because $dr<0$ and this implies that in your extremum of integration you must set $r_{\max}$. Note that in your solution you start from $r_{\min}$ because you have a "+" like sign. Another way can be seeing your solution in $\phi=0$ and check that $r_{\min}=r_0/(1+e)$.