Central-Force Reduction and Effective Potentials

For nonzero angular momentum, central-force motion becomes a one-dimensional radial problem because the conserved angular-momentum vector fixes the orbital plane and its magnitude replaces the angular velocity by a constant. The radial coordinate then moves with conserved energy in an effective potential containing both the physical interaction and a centrifugal barrier. Turning points, circular orbits, and radial stability become properties of one function; zero angular momentum is the degenerate radial case.

Required background. Classical and Variational Mechanics provides action principles, canonical variables, and mechanical conservation laws; Newtonian Gravity provides the gravitational potential, Poisson equation, and inverse-square force law. Helpful background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes; Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Planar reduction of a central force

Let $V(r)$ depend only on $r=|\mathbf r|$. The torque vanishes,

$$\dot{\mathbf L}=\mathbf r\times(-\nabla V)=0,$$

so $\mathbf L$ is constant. When $\mathbf L\ne0$, the trajectory remains in the unique plane perpendicular to it. When $\mathbf L=0$, position and velocity are collinear and the motion is radial along a fixed line, so there is no unique orbital plane selected by angular momentum. In polar coordinates for the nondegenerate planar case,

$$L=\frac m2(\dot r^2+r^2\dot\phi^2)-V(r).$$

The angle is cyclic, hence

$$\ell=mr^2\dot\phi$$

is conserved. Substituting $\dot\phi=\ell/(mr^2)$ into the energy gives

$$E=\frac m2\dot r^2+V_{\rm eff}(r), \qquad V_{\rm eff}(r)=V(r)+\frac{\ell^2}{2mr^2}.$$

This is a first integral of the radial equation,

$$m\ddot r=-\frac{dV_{\rm eff}}{dr}.$$

The term $\ell^2/(2mr^2)$ is not a new interaction; it is angular kinetic energy expressed using the conserved angular momentum. Substitute the conserved quantity into the energy, as above, rather than directly into the original Lagrangian before varying: the latter shortcut reverses the centrifugal-force sign. Eliminating a cyclic velocity at the action level requires the Routhian $L-\ell\dot\phi$, which gives $m\dot r^2/2-V_{\rm eff}$. The planar reduction, effective potential, orbit classification, and circular-orbit test are given in Tong 2013, Chapter 4 introduction and §§4.1–4.2.2, pp. 48–54, PDF.

Turning points and circular orbits

Allowed radii obey $E\ge V_{\rm eff}(r)$. A simple root of $E-V_{\rm eff}$ is a turning point where $\dot r=0$ and the radial motion reverses. A circular orbit at $r=r_0$ requires both

$$E=V_{\rm eff}(r_0), \qquad V_{\rm eff}'(r_0)=0.$$

Set $r=r_0+\delta r$. Linearization gives

$$m\delta\ddot r=-V_{\rm eff}''(r_0)\delta r.$$

Thus $V_{\rm eff}''(r_0)\gt0$ produces oscillations with $\omega_r^2=V_{\rm eff}''(r_0)/m$ and radial linear stability; a negative second derivative produces exponential departure. This tests perturbations within the reduced central-force model. It says nothing about additional degrees of freedom suppressed by the assumptions.

For $V=-GMm/r$,

$$V_{\rm eff}=-\frac{GMm}{r}+\frac{\ell^2}{2mr^2}.$$

The circularity condition yields

$$r_0=\frac{\ell^2}{GMm^2}, \qquad E_0=-\frac{GMm}{2r_0},$$

and $V_{\rm eff}''(r_0)=GMm/r_0^3\gt0$. The independent force check $mv^2/r_0=GMm/r_0^2$, together with $\ell=mvr_0$, reproduces the same radius and energy.

What the reduction retains and loses

The one-dimensional equation determines $r(t)$, but reconstructing the orbit also requires

$$\dot\phi=\frac{\ell}{mr^2}.$$

Different signs of $\ell$ give the same radial motion with opposite orientations. Setting $\ell=0$ removes the barrier and allows radial collision for a singular attractive potential. For $\ell\ne0$, $V_{\rm eff}\to+\infty$ as $r\to0$ in the Kepler problem, preventing the reduced orbit from reaching the origin.

The energy equation also gives useful quadratures. On any interval where the sign of $\dot r$ is fixed,

$$t-t_0=\pm\int_{r_0}^{r} \frac{\sqrt{m/2}\,dr'}{\sqrt{E-V_{\rm eff}(r')}},$$

and division by $\dot\phi$ gives

$$\phi-\phi_0 =\pm\int_{r_0}^{r} \frac{\ell\,dr'}{r'^2\sqrt{2m[E-V_{\rm eff}(r')]}}.$$

These formulas separate radial accessibility from orbital shape. Near a simple turning point the denominator behaves as $\sqrt{|r-r_t|}$, so the time to reach it is finite. Near an unstable circular orbit the root is double and the integral can diverge logarithmically, reflecting prolonged motion close to the unstable radius.

For the Kepler potential, $E\lt0$ with two positive turning points describes a bound radial oscillation, $E=0$ is the parabolic boundary, and $E\gt0$ allows an unbound branch extending to infinity. Those energy labels are not universal for arbitrary $V$: they use the isolated normalization $V(\infty)=0$. What is universal is comparison of $E$ with the chosen $V_{\rm eff}$ and its asymptotic values.

At a turning point the radial velocity vanishes, not the full velocity: unless $\ell=0$, the particle still has the tangential speed $|\ell|/(mr)$. Confusing these statements incorrectly treats periapsis or apoapsis as a moment of rest.

The construction assumes an inertial frame, a time-independent central potential, a point-particle description, and constant mass. Drag destroys energy conservation; a noncentral force changes $\ell$; explicit time dependence prevents a fixed effective-potential diagram. In relativistic orbital dynamics, a similar-looking potential must be derived from the spacetime metric and normalization condition rather than imported from Newtonian mechanics.

Exercise: power-law circular stability

Let $V(r)=ar^n$ and suppose a nonzero-angular-momentum circular orbit exists at $r_0$. Determine its radial stability. Explain separately what happens for $n=-2$.

Solution

Circularity gives $anr_0^{n-1}=\ell^2/(mr_0^3)$, or $\ell^2/m=anr_0^{n+2}$. For $n\ne-2$, existence therefore requires $an\gt0$ and fixes a positive $r_0$. Then

$$V_{\rm eff}''(r_0) =an(n-1)r_0^{n-2}+\frac{3\ell^2}{mr_0^4} =an(n+2)r_0^{n-2}.$$

Because $an\gt0$, the orbit is stable for $n\gt-2$ and unstable for $n\lt-2$. At $n=-2$, circularity does not select a radius: it requires the tuned value $\ell^2/m=an=-2a$, for which

$$V_{\rm eff}(r) =\frac{a+\ell^2/(2m)}{r^2}=0$$

at every radius. The resulting continuum of constant-radius solutions has no restoring force: the linearized equation is $\delta\ddot r=0$, so an arbitrarily small radial velocity produces secular drift. It is not bounded linear stability. Without the angular-momentum tuning, no circular orbit exists. Kepler gravity has $n=-1$ and is radially stable.

Kepler Orbits and Orbital Elements integrates the inverse-square radial and angular equations into conic sections.

References

• Tong, David. Dynamics and Relativity. University of Cambridge lecture notes, Lent 2013. Chapter 4 introduction and §§4.1–4.2.2, pp. 48–54, PDF.