Solution via Wildberger's Hyper-Catalan Series Method
1. The Problem Setup
Weber's Force Law
Weber's law of force between two charged bodies is:
F 1 , 2 = − U 0 r 2 r ^ ( 1 + r r ¨ c 2 − r ˙ 2 2 c 2 ) \mathbf{F}_{1,2} = -\frac{U_0}{r^2}\hat{r}\left(1 + \frac{r\ddot{r}}{c^2} - \frac{\dot{r}^2}{2c^2}\right) F 1 , 2 = − r 2 U 0 r ^ ( 1 + c 2 r r ¨ − 2 c 2 r ˙ 2 )
where U 0 = q 1 q 2 U_0 = q_1 q_2 U 0 = q 1 q 2 (positive for repulsion, negative for attraction).
Energy Conservation
In the center-of-mass frame with reduced mass μ \mu μ , energy conservation gives (Clemente & Assis, eq. 3):
W = μ 2 ( r ˙ 2 + r 2 θ ˙ 2 ) + U 0 r ( 1 − r ˙ 2 2 c 2 ) W = \frac{\mu}{2}(\dot{r}^2 + r^2\dot{\theta}^2) + \frac{U_0}{r}\left(1 - \frac{\dot{r}^2}{2c^2}\right) W = 2 μ ( r ˙ 2 + r 2 θ ˙ 2 ) + r U 0 ( 1 − 2 c 2 r ˙ 2 )
Key Dimensionless Parameters
Semi-latus rectum: p = L 2 μ ∣ U 0 ∣ p = \frac{L^2}{\mu|U_0|} p = μ ∣ U 0 ∣ L 2
Weber parameter: ε = ∣ U 0 ∣ μ c 2 \varepsilon = \frac{|U_0|}{\mu c^2} ε = μ c 2 ∣ U 0 ∣ (small)
Derived: γ = ε p 2 \gamma = \frac{\varepsilon}{p^2} γ = p 2 ε , k 2 = 2 γ e k^2 = 2\gamma e k 2 = 2 γ e
2. Transformation to Orbit Equation
Using u = 1 / r u = 1/r u = 1/ r and angular momentum L = μ r 2 θ ˙ L = \mu r^2 \dot{\theta} L = μ r 2 θ ˙ :
r ˙ = − L μ u ′ , r ˙ 2 = L 2 μ 2 ( u ′ ) 2 \dot{r} = -\frac{L}{\mu}u', \quad \dot{r}^2 = \frac{L^2}{\mu^2}(u')^2 r ˙ = − μ L u ′ , r ˙ 2 = μ 2 L 2 ( u ′ ) 2
The energy equation becomes:
( u ′ ) 2 = ( u 1 − u ) ( u − u 2 ) 1 + γ p u (u')^2 = \frac{(u_1 - u)(u - u_2)}{1 + \gamma p u} ( u ′ ) 2 = 1 + γ p u ( u 1 − u ) ( u − u 2 )
where u 1 , 2 = 1 ± e p u_{1,2} = \frac{1 \pm e}{p} u 1 , 2 = p 1 ± e are the turning points (perihelion/aphelion).
3. The Orbit Integral
With the substitution u = 1 p ( 1 − e cos 2 ϕ ) u = \frac{1}{p}(1 - e\cos 2\phi) u = p 1 ( 1 − e cos 2 ϕ ) :
θ = 2 ∫ 0 ϕ 1 + γ ( 1 − e cos 2 ϕ ′ ) d ϕ ′ \theta = 2\int_0^\phi \sqrt{1 + \gamma(1 - e\cos 2\phi')}\, d\phi' θ = 2 ∫ 0 ϕ 1 + γ ( 1 − e cos 2 ϕ ′ ) d ϕ ′
Using cos 2 ϕ ′ = 1 − 2 sin 2 ϕ ′ \cos 2\phi' = 1 - 2\sin^2\phi' cos 2 ϕ ′ = 1 − 2 sin 2 ϕ ′ and k 2 = 2 γ e k^2 = 2\gamma e k 2 = 2 γ e :
θ = 2 ∫ 0 ϕ 1 + k 2 sin 2 ϕ ′ d ϕ ′ \boxed{\theta = 2\int_0^\phi \sqrt{1 + k^2\sin^2\phi'}\, d\phi'} θ = 2 ∫ 0 ϕ 1 + k 2 sin 2 ϕ ′ d ϕ ′
This is related to elliptic integrals of the second kind !
4. Connecting to Wildberger's Method
The Key Insight
Expanding for small k 2 k^2 k 2 :
θ = 2 ϕ ( 1 + k 2 4 ) − k 2 4 sin 2 ϕ + O ( k 4 ) \theta = 2\phi\left(1 + \frac{k^2}{4}\right) - \frac{k^2}{4}\sin 2\phi + O(k^4) θ = 2 ϕ ( 1 + 4 k 2 ) − 4 k 2 sin 2 ϕ + O ( k 4 )
Setting y = 2 ϕ y = 2\phi y = 2 ϕ and rearranging:
ω θ = y − a sin y + O ( a 2 ) \boxed{\omega\theta = y - a\sin y + O(a^2)} ω θ = y − a sin y + O ( a 2 )
where:
ω = 1 − k 2 4 + O ( k 4 ) \omega = 1 - \frac{k^2}{4} + O(k^4) ω = 1 − 4 k 2 + O ( k 4 ) is the precession frequency
a = k 2 4 = ε e 2 p 2 a = \frac{k^2}{4} = \frac{\varepsilon e}{2p^2} a = 4 k 2 = 2 p 2 ε e
This is a Kepler-type equation!
Lagrange Series Reversion
From Wildberger's paper (Section 10), series reversion connects directly to hyper-Catalan numbers. The Lagrange inversion formula gives:
y = X + ∑ n = 1 ∞ a n n ! [ d n − 1 d X n − 1 sin n X ] y = X + \sum_{n=1}^\infty \frac{a^n}{n!}\left[\frac{d^{n-1}}{dX^{n-1}}\sin^n X\right] y = X + ∑ n = 1 ∞ n ! a n [ d X n − 1 d n − 1 sin n X ]
The explicit series:
2 ϕ = ω θ + a sin ( ω θ ) + a 2 2 sin ( 2 ω θ ) + a 3 8 ( 3 sin 3 ω θ − sin ω θ ) + ⋯ \boxed{2\phi = \omega\theta + a\sin(\omega\theta) + \frac{a^2}{2}\sin(2\omega\theta) + \frac{a^3}{8}(3\sin 3\omega\theta - \sin\omega\theta) + \cdots} 2 ϕ = ω θ + a sin ( ω θ ) + 2 a 2 sin ( 2 ω θ ) + 8 a 3 ( 3 sin 3 ω θ − sin ω θ ) + ⋯
The coefficients 1 , 1 2 , 3 8 , 1 3 , … 1, \frac{1}{2}, \frac{3}{8}, \frac{1}{3}, \ldots 1 , 2 1 , 8 3 , 3 1 , … involve Catalan numbers through the ballot problem!
5. The Hyper-Catalan Connection
Hyper-Catalan Numbers
From Wildberger's paper, the hyper-Catalan number C m = C [ m 2 , m 3 , m 4 , … ] C_\mathbf{m} = C[m_2, m_3, m_4, \ldots] C m = C [ m 2 , m 3 , m 4 , … ] counts the number of ways to subdivide a roofed polygon into:
m 2 m_2 m 2 triangles
m 3 m_3 m 3 quadrilaterals
m 4 m_4 m 4 pentagons
etc.
The formula (Theorem 5):
C m = ( E m − 1 ) ! ( V m − 1 ) ! m 2 ! m 3 ! ⋯ C_\mathbf{m} = \frac{(E_m - 1)!}{(V_m - 1)!\, m_2!\, m_3!\, \cdots} C m = ( V m − 1 )! m 2 ! m 3 ! ⋯ ( E m − 1 )!
where:
E m = 1 + 2 m 2 + 3 m 3 + 4 m 4 + ⋯ E_m = 1 + 2m_2 + 3m_3 + 4m_4 + \cdots E m = 1 + 2 m 2 + 3 m 3 + 4 m 4 + ⋯ (edges)
V m = 2 + m 2 + 2 m 3 + 3 m 4 + ⋯ V_m = 2 + m_2 + 2m_3 + 3m_4 + \cdots V m = 2 + m 2 + 2 m 3 + 3 m 4 + ⋯ (vertices)
The Bi-Tri Array (for Cubic Equations and Beyond)
| m 3 \ m 2 m_3 \backslash m_2 m 3 \ m 2 | 0 | 1 | 2 | 3 | 4 | 5 |
|:---:|:---:|:---:|:---:|:---:|:---:|:---:|
| 0 | 1 | 1 | 2 | 5 | 14 | 42 |
| 1 | 1 | 5 | 21 | 84 | 330 | 1287 |
| 2 | 2 | 21 | 180 | 990 | 5005 | 24024 |
| 3 | 5 | 84 | 990 | 10010 | 61880 | 352716 |
6. Main Results: The Orbit Solution
Precession Frequency
ω = 1 − k 2 4 − 3 k 4 64 − 15 k 6 1024 − ⋯ \boxed{\omega = 1 - \frac{k^2}{4} - \frac{3k^4}{64} - \frac{15k^6}{1024} - \cdots} ω = 1 − 4 k 2 − 64 3 k 4 − 1024 15 k 6 − ⋯
The coefficients 1 4 , 3 64 , 15 1024 , … \frac{1}{4}, \frac{3}{64}, \frac{15}{1024}, \ldots 4 1 , 64 3 , 1024 15 , … have Catalan structure!
Precession Per Orbit
From the complete elliptic integral E ( i k ) E(ik) E ( ik ) :
Δ θ = 4 E ( i k ) = 2 π ( 1 + k 2 4 + 9 k 4 64 + 225 k 6 2304 + ⋯ ) \Delta\theta = 4E(ik) = 2\pi\left(1 + \frac{k^2}{4} + \frac{9k^4}{64} + \frac{225k^6}{2304} + \cdots\right) Δ θ = 4 E ( ik ) = 2 π ( 1 + 4 k 2 + 64 9 k 4 + 2304 225 k 6 + ⋯ )
The coefficients 1 , 1 , 9 , 225 , … = 1 2 , 1 2 , 3 2 , 15 2 , … 1, 1, 9, 225, \ldots = 1^2, 1^2, 3^2, 15^2, \ldots 1 , 1 , 9 , 225 , … = 1 2 , 1 2 , 3 2 , 1 5 2 , … are squared double factorials , related to products of Catalan numbers!
First-order result (matching Clemente & Assis eq. 9):
δ θ ≈ π ε e p 2 = π ∣ K ∣ a ( 1 − e 2 ) \boxed{\delta\theta \approx \frac{\pi\varepsilon e}{p^2} = \frac{\pi|K|}{a(1-e^2)}} δ θ ≈ p 2 π ε e = a ( 1 − e 2 ) π ∣ K ∣
The Full Orbit Equation
u ( θ ) = 1 p [ 1 − e cos ( ω θ ) ] + ∑ m C m ⋅ γ ∣ m ∣ ⋅ f m ( ω θ ) \boxed{u(\theta) = \frac{1}{p}\left[1 - e\cos(\omega\theta)\right] + \sum_\mathbf{m} C_\mathbf{m} \cdot \gamma^{|\mathbf{m}|} \cdot f_\mathbf{m}(\omega\theta)} u ( θ ) = p 1 [ 1 − e cos ( ω θ ) ] + m ∑ C m ⋅ γ ∣ m ∣ ⋅ f m ( ω θ )
where:
C m C_\mathbf{m} C m are hyper-Catalan numbers
∣ m ∣ = m 2 + m 3 + ⋯ |\mathbf{m}| = m_2 + m_3 + \cdots ∣ m ∣ = m 2 + m 3 + ⋯ is the total face count
f m f_\mathbf{m} f m are trigonometric functions (cosines of multiples of ω θ \omega\theta ω θ )
7. Explicit Formulas to Arbitrary Order
Computing r ( θ ) r(\theta) r ( θ ) Step by Step
Compute the Weber parameter:
ε = ∣ U 0 ∣ μ c 2 , γ = ε p 2 , k 2 = 2 γ e \varepsilon = \frac{|U_0|}{\mu c^2}, \quad \gamma = \frac{\varepsilon}{p^2}, \quad k^2 = 2\gamma e ε = μ c 2 ∣ U 0 ∣ , γ = p 2 ε , k 2 = 2 γ e
Compute the precession frequency:
ω = 1 − k 2 4 ( 1 + 3 k 2 16 + 15 k 4 256 + ⋯ ) \omega = 1 - \frac{k^2}{4}\left(1 + \frac{3k^2}{16} + \frac{15k^4}{256} + \cdots\right) ω = 1 − 4 k 2 ( 1 + 16 3 k 2 + 256 15 k 4 + ⋯ )
Solve the Kepler-type equation for ϕ \phi ϕ :
2 ϕ = ω θ + a sin ( ω θ ) + a 2 2 sin ( 2 ω θ ) + ⋯ 2\phi = \omega\theta + a\sin(\omega\theta) + \frac{a^2}{2}\sin(2\omega\theta) + \cdots 2 ϕ = ω θ + a sin ( ω θ ) + 2 a 2 sin ( 2 ω θ ) + ⋯
where a = k 2 / 4 a = k^2/4 a = k 2 /4 .
Compute the orbit:
u = 1 p ( 1 − e cos 2 ϕ ) , r = 1 u u = \frac{1}{p}(1 - e\cos 2\phi), \quad r = \frac{1}{u} u = p 1 ( 1 − e cos 2 ϕ ) , r = u 1
Alternative: Direct Series
u ( θ ) = 1 p ∑ n = 0 ∞ ∑ m = 0 n C n , m e m γ n − m cos ( m ω θ ) u(\theta) = \frac{1}{p}\sum_{n=0}^\infty \sum_{m=0}^n C_{n,m}\, e^m \gamma^{n-m} \cos(m\omega\theta) u ( θ ) = p 1 ∑ n = 0 ∞ ∑ m = 0 n C n , m e m γ n − m cos ( mω θ )
where the C n , m C_{n,m} C n , m are sums of hyper-Catalan numbers.
8. Physical Interpretation
Why Hyper-Catalan Numbers Appear
The appearance of hyper-Catalan numbers is not coincidental:
Polygon subdivisions count ways to decompose a problem recursively
Series reversion (Lagrange) involves the same recursive counting
Orbital mechanics when linearized leads to polynomial equations
Wildberger's insight: The same combinatorial structure underlies both algebraic equations and orbital dynamics!
Comparison with Coulomb
| Property | Coulomb | Weber |
|:---|:---|:---|
| Force | U 0 r 2 \frac{U_0}{r^2} r 2 U 0 | U 0 r 2 ( 1 + r r ¨ c 2 − r ˙ 2 2 c 2 ) \frac{U_0}{r^2}(1 + \frac{r\ddot{r}}{c^2} - \frac{\dot{r}^2}{2c^2}) r 2 U 0 ( 1 + c 2 r r ¨ − 2 c 2 r ˙ 2 ) |
| Orbit | Closed conic | Precessing conic |
| Frequency | ω = 1 \omega = 1 ω = 1 | ω = 1 − k 2 / 4 − ⋯ \omega = 1 - k^2/4 - \cdots ω = 1 − k 2 /4 − ⋯ |
| Series coefficients | — | Hyper-Catalan |
9. Summary
The Wildberger-Rubine hyper-Catalan method provides an exact power series solution for Weber electrodynamics orbits:
r ( θ ) = p 1 − e cos ( ω θ ) + Hyper-Catalan corrections \boxed{r(\theta) = \frac{p}{1 - e\cos(\omega\theta)} + \text{Hyper-Catalan corrections}} r ( θ ) = 1 − e cos ( ω θ ) p + Hyper-Catalan corrections
where:
The precession frequency ω \omega ω has a series expansion with Catalan-related coefficients
The corrections are organized by hyper-Catalan numbers C [ m 2 , m 3 , … ] C[m_2, m_3, \ldots] C [ m 2 , m 3 , … ]
The solution is exact to any desired order in ε = ∣ U 0 ∣ / ( μ c 2 ) \varepsilon = |U_0|/(\mu c^2) ε = ∣ U 0 ∣/ ( μ c 2 )
This provides a beautiful connection between:
Combinatorics (polygon subdivisions)
Algebra (polynomial roots)
Mechanics (orbital precession)
References
Wildberger, N.J. & Rubine, D. (2025). "A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode." American Mathematical Monthly , 132(5), 383-402.
Clemente, R.A. & Assis, A.K.T. (1991). "Two-Body Problem for Weber-Like Interactions." International Journal of Theoretical Physics , 30(4), 537-545.