In spherical potential, we consider the degrees of freedom of the system. In the Hamiltonian, we have six free parameters, that is, the generalized coordinates and momenta
(x,y,z,px,py,pz) or (r,θ,ϕ,pr,pθ,pϕ)
The vanished partial derivative of $H$ (with respect to $t$ and $\alpha$, see in last week's notes) conserves the energy and the angular momentum (three-dimensions), thus the DoF is reduced by four.
In general, to simplify our analysis, we would like to find out some integrals as functions of only the coordinates in the phase-space
I(q,p)=C
to help us reduce the DoF.
More generally we can discuss constants of motion, which are functions of both the coordinates in the phase-space and time, but is invariant in an orbit. Integrals are constants of motions, while the converse is not necessarily true.
Discuss the integrals in a spherical potential
Can we find a fifth integral in a spherical potential?
With the general form of how $u=1/r$ is related to $\psi$ in a plane
dψ2d2u+u=L2u21drdΦ(1/u)
we examine the potential
Φ(r)=−GM(r1+r2a)⇒∂rΦ(r)=GM(r21+r32a)
so that
dψ2d2u+(1−L22GMa)u=L2GM
The general solution is
u=Ccos(Kψ−ψ0)+L2GMK2,K=(1−L22GMa)−1/2
Hence
Kψ−ψ0=±arccos[C1(r1−L2GMK2)]+2πn,n∈N
which is a function of $r, E, L$
Non-isolating integrals: If $K$ is irrational, we can make $\psi-\psi_0$ modulo $2\pi$ approximately any given number as closely as we wish, so given $r, E, L$, an orbit that is known to have a given value of the integral $\psi_0$ can have an azimuthal angle as close as we please to any number between $0$ and $2\pi$.
As a result, even if we select a $\psi_0$ randomly, the phase-space distribution of an orbit will not be affected
Isolating integrals: If $K$ is rational, say, $K=k_1/k_2$ where $k_1,k_2\in N$
ψ−ψ0≡±arccos[C1(r1−L2GMK2)]+2πk2nk1(mod2π)
Since $nk_1$ modulo $k_2$ has no more than $k_2$ possible values, the number of $\psi$ is also limited
In this way, the 3-D motion can be reduced to a 2-D one in the $(R,z)$ plane (meridional plane) under the Hamiltonion
H=H=21(pR2+pz2)+Φeff(R,z)
The orbit is, obviously, restricted to the area in the meridional plane satisfying $E\ge \Phi_{eff}$
Consider the minimum of $\Phi_{eff}(R,z)$, where
0=∂R∂Φeff=∂R∂Φ−R3Lz2,0=∂z∂Φeff
The latter is satisfied everywhere on the $z=0$ plane, assuming the symmetry of $\Phi$ with respect to this plane. And when the former is also satisfied
(∂R∂Φ)(Rg,0)=Rg3Lz2=Rgϕ˙2
which is simply the condition for a circular orbit with angular speed $\dot\phi$. $R_g$ is known as the guiding-center radius, since $(R_g,0)$ is the center of a group of closed isopotential curves.
Surfaces of section
When it is difficult to visualize the 4-D motion ($R,z,p_R,p_z$), we can determine whether orbits in the $(R,z)$ plane admit an additional isolating integral.
Since the Hamiltonian is a constant, we could plot the motion in the reduced phase space, say $(R,z,p_R)$, and $|p_z|$ will be determined.
An even simpler solution for visualizing the 4-D motion indirectly is to plot the points where the star crosses some plane, say $z=0$. Such a plot is known as a surface of section.
Consequents: the points $(R,p_R)$ where a star crosses the plane $z=0$ (when $p_z>0$ to remove the sign ambiguity in $p_z$)
Zero-velocity curve: the curve bounding the possible orbits in the plane $z=0$, on which $E=\Phi_{eff}$
Shell orbit: the trajectory of the star restricted to a 2-D motion
Invariant curve: a smooth curve on which the consequents of an orbit lie, suggesting the existence of some isolating integral $I(R,pR){z=0}=Const$ - the third integral, no analytical form
Nearly circular orbits
Now we take a look at some nearly circular orbits. Since they are close to circles, $R$ should not be far from $R_g$ ($z=0$) and we can define $x=R-R_g$ to be the separation, with which we expand the effective potential
The first-order derivatives both vanish, as well as the second-order mixed partial derivative, because $\Phi_{eff}$ is assumed to be symmetric about $z=0$
Now we try to understand the shape of the epicycle, which, to the first order, should be an ellipse. The X-semi-axis is obviously $X$, while in the $y$ direction perpendicular to both $x$ and $z$
y=ΔϕRg=−γXsin(κt+α)⇒Y=γX
Thus
YX=2Ωκ
Usually
21<YX<1
while $X/Y=1/2$ for Kepler potential and $X/Y=1$ for harmonic oscillator potential (homogeneous sphere).