Effective Potential and Radial Reduction

Derivation

Effective Potential and Radial Reduction

D-130 Home PU-201 Threads force · energy Depends on Cyclic Coordinates and Conserved Conjugate Momenta, Recovery of Newton's Second Law from L = T - V

Statement

For a single particle of mass m moving in a central potential V(r), the planar two-degree-of-freedom problem in (r, φ) reduces to a single one-dimensional equation for the radial coordinate, m r̈ = −dV_eff/dr, in which the conserved angular momentum ℓ is folded into an effective potential V_eff(r) = V(r) + ℓ²/(2m r²). The angular motion is eliminated, not ignored: it survives entirely inside the centrifugal term.

Why it matters

Every bound-orbit, scattering, and stability question in a central field — planetary orbits, the Kepler problem, Rutherford scattering, the radial wavefunctions of hydrogen — is decided by the shape of one function of one variable. Reducing two coupled equations to a single V_eff(r) turns orbital mechanics into the far simpler study of a particle sliding in a 1D potential well.

The reduction is also the cleanest illustration of what a cyclic coordinate buys you: because φ does not appear in the Lagrangian, its momentum is conserved, and that constant of motion can be used to remove a whole degree of freedom before any orbit is computed.

Assumptions

The potential is central, V depends only on r = |r|. If V depended on direction, the torque would not vanish, angular momentum would not be conserved, and ℓ could not be treated as a constant to be substituted away.

The force is conservative, so a scalar V(r) with F = −dV/dr exists. Otherwise the Lagrangian L = T − V is not defined and the whole variational route is unavailable.

Motion is confined to a plane, guaranteed because L = r × p is a fixed vector, so r stays perpendicular to it. If ℓ = 0 exactly the argument degenerates to purely radial (one-dimensional) motion instead.

The mass is the (reduced) inertial mass of the radial motion, for a genuine two-body problem m must be replaced by the reduced mass μ = m₁m₂/(m₁+m₂) and r by the separation. Dropping this silently gives orbital periods wrong by the mass ratio.

Non-relativistic kinematics, T = ½m v². At speeds near c the kinetic term and the relation between ℓ and φ̇ change, and the centrifugal term ℓ²/(2mr²) is no longer the correct barrier.

Derivation

L = ½m(ṙ² + r²φ̇²) − V(r)

Lagrangian L = T − V with the plane-polar kinetic energy; v² = ṙ² + r²φ̇². A

∂L/∂φ = 0 ⇒ d/dt(∂L/∂φ̇) = 0

φ does not appear in L, so it is a cyclic coordinate; its Euler–Lagrange equation says the conjugate momentum is constant. A

p_φ = ∂L/∂φ̇ = m r²φ̇ ≡ ℓ = const

Evaluate the conjugate momentum; this is the conserved angular momentum (cyclic-coordinates → conserved momenta). A

φ̇ = ℓ/(m r²)

Solve the constraint for the angular velocity so it can be removed from the radial dynamics. A

d/dt(∂L/∂ṙ) − ∂L/∂r = 0

Euler–Lagrange equation for the remaining coordinate r. A

∂L/∂ṙ = m ṙ , ∂L/∂r = m r φ̇² − dV/dr

Differentiate L term by term; the rφ̇² piece is the centrifugal contribution from the kinetic energy. A

m r̈ − m r φ̇² + dV/dr = 0

Assemble the radial Euler–Lagrange equation. It still contains φ̇. B

m r̈ − m r · ℓ²/(m²r⁴) + dV/dr = 0

Substitute φ̇ = ℓ/(mr²) from step 4. This is the key move: the angular degree of freedom is now gone, replaced by the constant ℓ. Legal because ℓ is conserved along the actual trajectory. B

m r̈ = −dV/dr + ℓ²/(m r³)

Simplify m·ℓ²/(m²r⁴) · r = ℓ²/(mr³) and rearrange. B

ℓ²/(m r³) = −d/dr[ ℓ²/(2m r²) ]

Recognise the centrifugal term as minus the derivative of a potential-like function of r alone, so it can be absorbed into a gradient. B

m r̈ = −d/dr[ V(r) + ℓ²/(2m r²) ] ≡ −dV_eff/dr

Combine both r-derivatives under one gradient and define V_eff. This is the one-dimensional radial equation. C

m r̈ = −dV_eff/dr , V_eff(r) = V(r) + ℓ²/(2m r²)

Reading. The radius obeys Newton's second law for a fictitious particle of mass m in a one-dimensional potential V_eff. The extra term ℓ²/(2m r²) is the centrifugal barrier: it is the rotational kinetic energy re-expressed through the conserved ℓ, and being positive and steeply rising as r → 0, it repels the particle from the origin whenever ℓ ≠ 0. Turning points are where V_eff(r) = E; a minimum of V_eff is a stable circular orbit.

Units check. ℓ has units kg·m²·s⁻¹, so ℓ²/(2m r²) has (kg²m⁴s⁻²)/(kg·m²) = kg·m²·s⁻² = J, the same as V(r). And m r̈ is kg·m·s⁻² = N, matching −dV_eff/dr (J/m = N). Consistent.

Limiting cases

ℓ = 0: the barrier vanishes, V_eff = V, and the motion is a straight radial fall or climb — a genuine 1D problem with no orbit.
Circular orbit: at r₀ where V_eff′(r₀) = 0, r̈ = 0 and r stays constant; V′(r₀) = ℓ²/(m r₀³), i.e. the attractive force supplies exactly the centripetal requirement.
Kepler, V = −k/r: V_eff = −k/r + ℓ²/(2m r²) has a single minimum → stable bound ellipses; E < 0 bound, E ≥ 0 unbound.
Large r, attractive: V_eff → V(r) → 0⁻ so the barrier is negligible far out; the centrifugal term dominates only at small r.
Small oscillations about r₀: V_eff ≈ V_eff(r₀) + ½V_eff″(r₀)(r−r₀)² gives radial oscillation frequency ω_r = √(V_eff″/m).

Breaks when

Non-central forces (e.g. a magnetic Lorentz force, tidal or triaxial fields, or velocity-dependent forces): torque no longer vanishes, ℓ is not conserved, and step 4 — solving for φ̇ in terms of a constant — is invalid, so no single V_eff exists.
Relativistic speeds: T ≠ ½mv² and p_φ = γmr²φ̇, so the barrier is not ℓ²/(2mr²). In GR the effective potential acquires an extra −G M ℓ²/(m c² r³) term that destroys the barrier at small r and produces perihelion precession and photon capture.
Dissipation (drag, gravitational-wave radiation): energy and ℓ are not constant, the orbit is not a fixed curve in V_eff, and the 1D conservative picture only holds adiabatically.
ℓ itself time-varying because of an external torque or a slowly changing potential: V_eff becomes time-dependent and r̈ = −∂V_eff/∂r no longer captures the full dynamics.

Failure modes

Sign error on the barrier: writing V_eff = V − ℓ²/(2mr²). The centrifugal term is positive (repulsive); a wrong sign predicts collapse to the origin.
Confusing r̈ = 0 with equilibrium: a turning point (r̈ = 0 but V_eff ≠ E′ slope nonzero) is instantaneous; only V_eff′ = 0 gives a sustained circular orbit.
Forgetting r²φ̇² in T: using v² = ṙ² alone drops the centrifugal term entirely and yields purely radial (wrong) motion.
Substituting ℓ before deriving the E–L equation for φ: inserting the constant into L and then varying φ loses the conservation law — the substitution is only legal in the radial equation, after φ has been eliminated.
Using m instead of the reduced mass μ in a two-body problem, giving periods and barrier heights off by the mass-ratio factor.
Treating V_eff″ < 0 as still "bound": a maximum of V_eff is an unstable circular orbit; small perturbations grow.

Discussion

The centrifugal term is not a new force added by hand; it is bookkeeping. The particle's total kinetic energy is ½m ṙ² + ½m r²φ̇², and once φ̇ = ℓ/(mr²) is imposed, the angular piece becomes ℓ²/(2mr²) — a function of r alone. It sits on the same footing as V(r) in the energy budget, so it is natural to move it there. What looks like a barrier in a 1D picture is simply the cost, in radial energy, of conserving angular momentum as r shrinks: to keep ℓ = mr²φ̇ fixed while r falls, φ̇ must rise, and the rotational energy climbs like 1/r².

This is why the same E − V_eff = ½m ṙ² ≥ 0 condition classifies every orbit. Bound motion is oscillation of r between two turning points where V_eff = E; a single tangency gives a circular orbit; and if E exceeds the barrier the particle escapes. The whole qualitative zoo of central-force orbits — ellipses, hyperbolae, rosettes, capture — is read straight off the graph of one function.

The construction connects directly to quantum mechanics. Separating the hydrogen Schrödinger equation in spherical coordinates produces exactly V_eff(r) = V(r) + ℎ²ℓ(ℓ+1)/(2m r²), with the classical ℓ² replaced by ℎ²ℓ(ℓ+1). The centrifugal barrier is why wavefunctions with ℓ > 0 vanish at the origin and why higher-angular-momentum states feel a shielded nucleus. The classical reduction is the skeleton of the radial equation students later solve for atomic orbitals.

At the level of Noether's theorem, the reduction is the statement that the SO(2) rotational symmetry of a central Lagrangian yields a conserved charge (ℓ) which can be used to reduce the phase space by two dimensions — one for the cyclic coordinate φ and one for its momentum, now a fixed parameter. In the Hamiltonian picture this is symplectic reduction: the ℓ = const level set quotiented by the φ-flow gives a reduced 2D phase space (r, p_r) governed by H_red = p_r²/(2m) + V_eff(r). Bertrand's theorem then singles out only V ∝ −1/r and V ∝ r² as the potentials whose bound orbits close for every ℓ and E — a fact invisible until the problem is cast in this one-dimensional form.

Common misconceptions. The centrifugal barrier is often called "the centrifugal force" and mistaken for a real interaction in the rotating frame. Here it is nothing of the kind: we work in the inertial frame, and the barrier is a term in the energy that appears purely because we chose to eliminate φ. It is real in its consequences (it sets the orbit's inner turning point) but it is not a force acting on the particle in the lab frame.

Worked examples

Circular orbit radius for the gravitational Kepler potential. V(r) = −GMm/r, given ℓ.

Set up: V_eff = −GMm/r + ℓ²/(2mr²); circular orbit at V_eff′(r₀) = 0. A

V_eff′(r) = GMm/r² − ℓ²/(m r³) = 0 ⇒ r₀ = ℓ²/(GMm²)

Differentiate and solve for r₀; symbols before numbers. A

m = 5.97×10²⁴ kg (Earth), M = 1.989×10³⁰ kg (Sun), G = 6.674×10⁻¹¹ N·m²/kg², ℓ = 2.66×10⁴⁰ kg·m²/s

Insert Earth's orbital angular momentum. A

r₀ = (2.66×10⁴⁰)² / [(6.674×10⁻¹¹)(1.989×10³⁰)(5.97×10²⁴)²]

Numerator = 7.08×10⁸⁰; denominator = (6.674×10⁻¹¹)(1.989×10³⁰)(3.564×10⁴⁹) = 4.73×10⁶⁹. B

r₀ ≈ 1.50×10¹¹ m ≈ 1.0 AU

Reading. The V_eff minimum reproduces the Earth–Sun distance to the expected accuracy, confirming that the effective-potential reduction encodes the real orbit.

Units check. (kg·m²/s)² / [(N·m²/kg²)(kg)(kg²)] = (kg²m⁴s⁻²)/(kg·m·s⁻²·m²) = m. Correct.

Radial oscillation frequency about a circular orbit. For V = −k/r, find ω_r at r₀ and compare to the orbital frequency ω_φ.

Set up: ω_r² = V_eff″(r₀)/m from small-oscillation expansion. B

V_eff″(r) = −2k/r³ + 3ℓ²/(m r⁴) , with ℓ² = m k r₀ at the circular orbit

Second derivative; use V_eff′(r₀)=0 ⇒ k/r₀² = ℓ²/(mr₀³), i.e. ℓ² = mkr₀. B

V_eff″(r₀) = −2k/r₀³ + 3k/r₀³ = k/r₀³ ⇒ ω_r = √(k/(m r₀³))

Substitute ℓ² = mkr₀ and simplify; symbols only. B

ω_φ = φ̇ = ℓ/(m r₀²) = √(k/(m r₀³)) = ω_r

Compute the angular frequency at r₀; the two frequencies are equal for the 1/r potential. C

Numbers: k = GMm = (6.674×10⁻¹¹)(1.989×10³⁰)(5.97×10²⁴) = 7.92×10⁴⁴ J·m; r₀ = 1.50×10¹¹ m; m = 5.97×10²⁴ kg

ω = √[7.92×10⁴⁴ / (5.97×10²⁴ · 3.375×10³³)]. A

ω_r = ω_φ ≈ 1.99×10⁻⁷ s⁻¹ (T ≈ 3.16×10⁷ s ≈ 1 yr)

Reading. Because ω_r = ω_φ the orbit closes after one radial cycle — the ellipse does not precess. This equality is special to the 1/r potential (Bertrand's theorem) and is why Kepler orbits are fixed closed ellipses.

Units check. √[(J·m)/(kg·m³)] = √[(kg·m²s⁻²·m)/(kg·m³)] = √(s⁻²) = s⁻¹. Correct.

Problems

Show that for V(r) = ½mω²r² (the isotropic 2D oscillator) the effective potential has a single minimum, and find r₀ in terms of ℓ, m, ω.
Solution
V_eff = ½mω²r² + ℓ²/(2mr²). Then V_eff′ = mω²r − ℓ²/(mr³) = 0 ⇒ r⁴ = ℓ²/(m²ω²) ⇒ r₀ = √(ℓ/(mω)). Since V_eff→+∞ at both r→0 and r→∞ and there is one stationary point, it is a minimum — a single stable circular orbit.
For the Kepler potential V = −k/r, compute the height of the effective-potential minimum and hence the energy of the circular orbit.
Solution
r₀ = ℓ²/(mk). V_eff(r₀) = −k/r₀ + ℓ²/(2mr₀²) = −k·mk/ℓ² + ℓ²/(2m)·(mk/ℓ²)² = −mk²/ℓ² + mk²/(2ℓ²) = −mk²/(2ℓ²). The circular orbit has E = V_eff,min (since ṙ = 0), giving E_circ = −mk²/(2ℓ²), the most bound orbit for that ℓ.
A particle has ℓ = 4.0×10⁻³⁴ kg·m²/s (order ℎ), m = 9.11×10⁻³¹ kg, in V = −e²/(4πε₀r) with k = e²/(4πε₀) = 2.31×10⁻²⁸ J·m. Find the classical circular-orbit radius.
Solution
r₀ = ℓ²/(mk) = (4.0×10⁻³⁴)² / [(9.11×10⁻³¹)(2.31×10⁻²⁸)] = 1.60×10⁻⁶⁷ / 2.104×10⁻⁵⁸ = 7.6×10⁻¹⁰ m. This is ~1.4 Bohr radii, confirming the effective potential reproduces atomic-scale orbits when ℓ ~ ℎ. (Exact a₀ uses ℓ = ℎ = 1.055×10⁻³⁴, giving 5.3×10⁻¹¹ m.)
Classify the circular orbit stability for a power-law potential V = −α/rⁿ (α > 0). For which n are circular orbits stable?
Solution
V_eff = −α/rⁿ + ℓ²/(2mr²). V_eff′ = nα/rⁿ⁺¹ − ℓ²/(mr³) = 0 fixes r₀. Stability requires V_eff″(r₀) > 0. V_eff″ = −n(n+1)α/rⁿ⁺² + 3ℓ²/(mr⁴). Using ℓ²/(mr₀²) = nα/r₀ⁿ from V_eff′=0: V_eff″(r₀) = [−n(n+1) + 3n]α/r₀ⁿ⁺² = n(2−n)α/r₀ⁿ⁺². This is positive iff n < 2. So attractive power laws with n < 2 (including Kepler n=1) give stable circular orbits; n = 2 is marginal, n > 2 unstable — the classic result that an inverse-cube or steeper attraction cannot hold a stable orbit.
Starting from energy conservation E = ½m ṙ² + V_eff(r), derive the orbit equation dφ/dr and set up the integral for Δφ between turning points.
Solution
From E = ½mṙ² + V_eff: ṙ = ±√[(2/m)(E − V_eff)]. Also φ̇ = ℓ/(mr²). Divide: dφ/dr = φ̇/ṙ = [ℓ/(mr²)] / √[(2/m)(E−V_eff)] = ℓ/(r²√[2m(E−V_eff)]). Integrating between the turning points r_min, r_max (where E = V_eff): Δφ = ∫_{r_min}^r_max ℓ dr / [r²√(2m(E−V_eff(r)))]. The orbit closes iff 2Δφ is a rational multiple of 2π; for Kepler 2Δφ = 2π exactly.