Kepler's Problem as Geodesic Flow on S³

Jul 28, 2026 · updated Jul 29, 2026

The Kepler problem is usually approached through its first integrals. Energy and angular momentum reduce the motion to a plane, where the orbit can be calculated in polar coordinates. This is an effective way to solve the problem, but it is not the only geometry hidden in it.

For negative energy, the angular momentum and the Runge–Lenz vector reveal an $SO (4)$ symmetry. We will not derive their complete algebra here. Instead, we use this symmetry as a clue: $SO (4)$ is the rotation group of $ℝ^{4}$ and the orientation-preserving isometry group of the round three-sphere $𝕊^{3}$ . This suggests a natural question: can the negative-energy Kepler problem be understood as free motion on $𝕊^{3}$ ?

Following Moser’s regularization, we will construct this correspondence directly. This construction identifies the regularized Kepler flow with the geodesic flow on $𝕊^{3}$ and lets us recover Kepler ellipses from great circles.

Stereographic Projection of the Three-Sphere

Write a point of $ℝ^{4}$ as $(𝜉_{0}, 𝜉_{⟂}) \in ℝ \times ℝ^{3}$ , and let

𝕊^{3} = {(𝜉_{0}, 𝜉_{⟂}) \in ℝ \times ℝ^{3} | 𝜉_{0}^{2} + {‖ 𝜉_{⟂} ‖}^{2} = 1} .

We denote the north pole by $𝑁 = (1, 0)$ . Stereographic projection from $𝑁$ gives a coordinate chart

𝜎_{𝑁} : 𝕊^{3} - {𝑁} \to ℝ^{3}, 𝑥 = 𝜎_{𝑁} (𝜉_{0}, 𝜉_{⟂}) = \frac{𝜉_{⟂}}{1 - 𝜉_{0}} .

Its inverse is

𝜉_{0} = \frac{{‖ 𝑥 ‖}^{2} - 1}{1 + {‖ 𝑥 ‖}^{2}}, 𝜉_{⟂} = \frac{2 𝑥}{1 + {‖ 𝑥 ‖}^{2}} .

Thus, $𝕊^{3}$ is the one-point compactification of $ℝ^{3}$ : the omitted north pole appears at $‖ 𝑥 ‖ \to \infty$ .

The round metric pulls back to a conformal multiple of the Euclidean metric,

𝑔_{rd} = \frac{4}{{(1 + {‖ 𝑥 ‖}^{2})}^{2}} \sum_{𝑖 = 1}^{3} {(d 𝑥_{𝑖})}^{2},

and its co-metric is

𝑔_{rd}^{- 1} = \frac{{(1 + {‖ 𝑥 ‖}^{2})}^{2}}{4} \sum_{𝑖 = 1}^{3} 𝜕_{𝑥_{𝑖}}^{2} .

For a covector $𝑦 \in 𝑇_{𝑥}^{*} ℝ^{3}$ , the free-particle Hamiltonian of the round sphere is therefore

𝐹 (𝑥, 𝑦) ≔ \frac{1}{2} 𝑔_{rd}^{- 1} (𝑦, 𝑦) = \frac{{(1 + {‖ 𝑥 ‖}^{2})}^{2}}{8} {‖ 𝑦 ‖}^{2} .

The unit-speed geodesic flow lives on the level set $𝐹 = \frac{1}{2}$ .

The base projection alone is not yet a map of dynamical systems: the Kepler problem lives in phase space. We need the cotangent lift, whose local coordinates are $(𝑥, 𝑦)$ . The unexpected part of the construction will be that $𝑥$ corresponds to the Kepler momentum, while the Kepler position becomes the conjugate covector $𝑦$ .

The Negative-Energy Kepler Surface

After reduction by the center-of-mass motion, the three-dimensional Kepler problem has phase space

𝒫︀ = 𝑇^{*} (ℝ^{3} - {0})

with position $𝑞 \in ℝ^{3} - {0}$ , momentum $𝑝 \in ℝ^{3}$ , and Hamiltonian

𝐻 (𝑞, 𝑝) = \frac{1}{2} {‖ 𝑝 ‖}^{2} - \frac{1}{‖ 𝑞 ‖} .

The collision locus $𝑞 = 0$ is excluded, and the Hamiltonian vector field becomes singular as a trajectory approaches it.

After rescaling the energy, it is enough to study the normalized surface $Σ_{- 1 / 2}$ . Its energy equation is

- \frac{1}{2} = \frac{1}{2} {‖ 𝑝 ‖}^{2} - \frac{1}{‖ 𝑞 ‖},

or equivalently

\frac{{(1 + {‖ 𝑝 ‖}^{2})}^{2}}{4} {‖ 𝑞 ‖}^{2} = 1 .

Now introduce the canonical transformation

Φ (𝑞, 𝑝) = (- 𝑝, 𝑞) ≕ (𝑥, 𝑦) .

The normalized energy condition becomes

\frac{{(1 + {‖ 𝑥 ‖}^{2})}^{2}}{4} {‖ 𝑦 ‖}^{2} = 1 .

Comparing this equation with the spherical Hamiltonian $𝐹$ , we immediately obtain

Φ (Σ_{- 1 / 2}) = {𝐹 = \frac{1}{2}} = 𝑆^{*} (𝕊^{3} - {𝑁}) .

Thus, the normalized Kepler energy surface is exactly the unit-speed constraint for the round sphere in stereographic cotangent coordinates.

Matching the Flows

The equality of the energy hypersurfaces does not yet identify their parameterized flows. Let

𝐾 ≔ 𝐻 \circ Φ^{- 1} = \frac{1}{2} {‖ 𝑥 ‖}^{2} - \frac{1}{‖ 𝑦 ‖} .

Since $Φ$ is canonical, the pushforward of the Kepler vector field is the Hamiltonian vector field of $𝐾$ :

Φ_{*} 𝑋_{𝐻} = \sum_{𝑖 = 1}^{3} [\frac{𝑦_{𝑖}}{{‖ 𝑦 ‖}^{3}} 𝜕_{𝑥_{𝑖}} - 𝑥_{𝑖} 𝜕_{𝑦_{𝑖}}] .

For the spherical free-particle Hamiltonian, using

d 𝐹 = - 𝜄_{𝑋_{𝐹}} 𝜔, 𝜔 = \sum_{𝑖 = 1}^{3} d 𝑦_{𝑖} \land d 𝑥_{𝑖},

we obtain

𝑋_{𝐹} = \sum_{𝑖 = 1}^{3} [\frac{{(1 + {‖ 𝑥 ‖}^{2})}^{2}}{4} 𝑦_{𝑖} 𝜕_{𝑥_{𝑖}} - \frac{1 + {‖ 𝑥 ‖}^{2}}{2} {‖ 𝑦 ‖}^{2} 𝑥_{𝑖} 𝜕_{𝑦_{𝑖}}] .

On the common energy hypersurface

Σ = {𝐹 = \frac{1}{2}} = {𝐾 = - \frac{1}{2}},

the constraint $(1 + {‖ 𝑥 ‖}^{2}) ‖ 𝑦 ‖ = 2$ reduces this vector field to

𝑋_{𝐹} |_{Σ} = \sum_{𝑖 = 1}^{3} [\frac{𝑦_{𝑖}}{{‖ 𝑦 ‖}^{2}} 𝜕_{𝑥_{𝑖}} - ‖ 𝑦 ‖ 𝑥_{𝑖} 𝜕_{𝑦_{𝑖}}] = ‖ 𝑦 ‖ Φ_{*} 𝑋_{𝐻} .

Therefore,

Φ_{*} 𝑋_{𝐻} = \frac{1}{‖ 𝑦 ‖} 𝑋_{𝐹} on Σ .

If $𝑡$ denotes Kepler time and $𝑠$ denotes the arc-length parameter of the spherical geodesic flow, then

d 𝑡 = ‖ 𝑦 ‖ d 𝑠 = ‖ 𝑞 ‖ d 𝑠, d 𝑠 = \frac{d 𝑡}{‖ 𝑞 ‖} .

This is the Sundman time reparametrization. It turns the normalized Kepler flow into the unit-speed geodesic flow on the punctured sphere.

Collision Regularization

So far, the construction gives

Σ_{- 1 / 2} ≅ 𝑆^{*} (𝕊^{3} - {𝑁}) .

At fixed energy, approaching collision means

𝑞 \to 0, ‖ 𝑝 ‖ \to \infty .

Under $𝑥 = - 𝑝$ and $𝑦 = 𝑞$ , this becomes $‖ 𝑥 ‖ \to \infty$ , precisely the missing north pole of the stereographic chart.

Adding the north pole to the base is not enough by itself. At collision, the remaining datum is the direction of the regularized trajectory. The natural completion adds the unit covector sphere

𝑆_{𝑁}^{*} 𝕊^{3} ≅ 𝕊^{2}

over $𝑁$ . The completed energy surface is therefore

{\hat{Σ}}_{- 1 / 2} ≅ 𝑆^{*} 𝕊^{3} ≅ 𝑇_{1} 𝕊^{3} .

The geodesic flow is smooth on this completed space. Great circles through $𝑁$ give the regularized radial collision–ejection trajectories. At fixed negative energy, these circles are the $𝐿 = 0$ degenerate ellipses.

From Great Circles to Kepler Orbits

We can now read the Kepler orbit directly from a great circle. Let $𝜓$ be its unit-speed angle, shifted so that $𝜓 = 0$ corresponds to periapsis, and choose orthonormal vectors $𝑒_{1}, 𝑒_{2} \in ℝ^{3}$ in the orbital plane. Applying the cotangent stereographic map and using $𝑞 = 𝑦$ gives

𝑞 (𝜓) = (cos 𝜓 - 𝑒) 𝑒_{1} + \sqrt{1 - 𝑒^{2}} sin 𝜓 𝑒_{2}, ‖ 𝑞 (𝜓) ‖ = 1 - 𝑒 cos 𝜓 .

The parameter $𝑒$ measures how close the great circle comes to the north pole. Eliminating $𝜓$ gives

{(𝑞 \cdot 𝑒_{1} + 𝑒)}^{2} + \frac{{(𝑞 \cdot 𝑒_{2})}^{2}}{1 - 𝑒^{2}} = 1,

so the cotangent projection of the great circle is a Kepler ellipse of eccentricity $𝑒$ .

Because the great circle has unit speed, $d 𝜓 = d 𝑠$ . The Sundman relation therefore becomes

d 𝑡 = ‖ 𝑞 (𝜓) ‖ d 𝜓 = (1 - 𝑒 cos 𝜓) d 𝜓,

and hence

𝑡 - 𝜏 = 𝜓 - 𝑒 sin 𝜓 .

Thus, the great-circle angle, the regularized time, and the eccentric anomaly are the same variable up to an additive constant.

From geodesics on the three-sphere, we have therefore reproduced and solved the negative-energy branch of Kepler’s problem.

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