The integrals of the motion take the form From (2.5), (2.6) and (2.7) we find We point out that the “+” and “ − ” signs correspond to the two possible orientations of the angular momentum. We choose, without loss of generality, the sign “+” and J > 0 throughout this work. Now, from Eq. (2.8) we find that the trajectories should satisfy where ω = c R = κ $^{2}$R p 0 = c √ 3 κ 2 2 E . So in this case we have a uniform motion in a circle with the linear speed | v | = ωR = c . This solution can also be obtained from Eq. (2.2) by demanding x$_{·}$p = 0. Indeed, it can be easily checked that (2.2) and (2.3) imply the system From (2.11) it follows easily that the orthogonality of x and p refers to the motion of a particle in a circle with radius | x | = R = √ 2 E 3 κ 2 . 2) For r ( E − κ 2 2 r 2 ) > Jc , the path lies entirely within the annulus bounded by the circles r = r$_{min}$ and r = r$_{max}$ , that is we have r$_{min}$ ≤ r ≤ r$_{max}$, (2.12) d dt x$_{·}$p = E − 3 2 κ $^{2}$x$^{2}$, d dt p 2 = − 2 κ $^{2}$x$_{·}$ p , d dt x 2 = 2 c | p | x$_{·}$p . (2.11) 1) For r ( E − κ 2 2 r 2 ) = Jc , we get r = R = √ 2 E 3 κ $_{2}$, ϕ = ωt + ϕ$_{0}$, p = p$_{0}$ = 2 E 3 c , θ = ωt + ϕ$_{0}$ + π 2 , (2.10) r ( E − κ 2 2 r 2 ) ≥ Jc, (2.9) and we can classify the types of motion as folows. The first two possibilities refer to J = 0. Namely, r √ r 2 ( E − κ 2 2 r 2 ) 2 − ( Jc ) 2 dϕ = ±| J | cdr. (2.8) E = cp + 1 2 κ $^{2}$r$^{2}$, (2.6) J ≡ J$_{3}$ = rp sin( θ − ϕ ) . (2.7) 4