Kepler’s Laws from Comoment Maps

7.1 Geometric Parameters

Definition 23 Eccentricity and semi-axes

For a phase point \(z = (q, p)\) away from collision, define:

\[ e = \frac{\| A\| }{m\mu }, \qquad \ell = \frac{\| L\| ^2}{m\mu }, \qquad a = -\frac{\mu }{2H}, \qquad b = a\sqrt{1-e^2}. \]

Here \(e\) is the eccentricity. \(\ell \) is the semi-latus rectum. \(a\) is the semi-major axis, meaning half the length of the long axis of the ellipse. \(b\) is the semi-minor axis, meaning half the length of the short axis of the ellipse. Since \(L\), \(A\), and \(H\) are conserved, all four quantities are constant along a solution. Geometric data. The center of the ellipse is not at the origin. The origin is a focus. The distance from the center to the focus is \(ae\). The relation \(b^2 = a\ell \) will be used later.

Assume \(H {\lt} 0\) and \(L \neq 0\). Then:

  1. Planarity. The position \(q(t)\) and momentum \(p(t)\) stay in the fixed plane \(\Pi = \{ y \in \mathbb {R}^3 : L \cdot y = 0\} \). The Lenz vector \(A\) also lies in \(\Pi \).

  2. \(0 \le e {\lt} 1\),  \(a {\gt} 0\),  \(b {\gt} 0\).

  3. \(b^2 = a\ell \).

  4. \(b = a\sqrt{1-e^2}\).

  5. \(\sqrt{a^2 - b^2} = ae\)  (focal distance formula).

Proof

(a). Since \(L = q \times p\), the vector \(L\) is perpendicular to both \(q\) and \(p\) at every instant. Because \(L\) is constant, the plane \(\Pi \) is fixed, and \(q(t)\) stays in it for all time. The vector \(A\) lies in \(\Pi \) by identity (i): \(L \cdot A = 0\).

(b). From identity (iv):

\[ e^2 = \frac{\| A\| ^2}{(m\mu )^2} = 1 + \frac{2H\| L\| ^2}{m\mu ^2}. \]

Since \(H {\lt} 0\) and \(\| L\| {\gt} 0\), we get \(e^2 {\lt} 1\), so \(0 \le e {\lt} 1\). Since \(H {\lt} 0\), \(a = -\mu /(2H) {\gt} 0\). Also \(1-e^2 {\gt} 0\), so \(\sqrt{1-e^2} {\gt} 0\). Hence \(b = a\sqrt{1-e^2} {\gt} 0\).

(c). Substitute the definitions:

\[ a\ell = \frac{-\mu }{2H} \cdot \frac{\| L\| ^2}{m\mu } = \frac{-\| L\| ^2}{2mH} = b^2. \]

(d). From (b) and (c): \(e^2 = 1 - \ell /a = 1 - b^2/a^2\), so \(b^2 = a^2(1-e^2)\), giving \(b = a\sqrt{1-e^2}\).

(e). From (d): \(a^2 - b^2 = a^2 - a^2(1-e^2) = a^2 e^2\), so \(\sqrt{a^2-b^2} = ae\).