4.2 Algebraic Relations

Proposition 12 Basic identities for \(L\), \(A\), and \(H\)

✓

Away from collision, the following four identities hold:

\(L \cdot A = 0\) (\(L\) and \(A\) are perpendicular)
\(A \in \mathrm{span}\{ q, p\} \) (\(A\) lies in the orbital plane)
\(A \cdot q = \| L\| ^2 - m\mu r\)
\(\| A\| ^2 = (m\mu )^2 + 2mH\| L\| ^2\).

Proof ▶

(i). Since \(L = q \times p\), the vector \(L\) is perpendicular to both \(q\) and \(p\). Therefore:

\[ L \cdot A = L \cdot (p \times L) - \frac{m\mu }{r}(L \cdot q) = 0 - 0 = 0. \]

(The scalar triple product \(L \cdot (p \times L) = (L \times p) \cdot L\) vanishes because a cross product is perpendicular to both factors.)

(ii). Apply the BAC-CAB identity to \(p \times L = p \times (q \times p)\):

\[ p \times (q \times p) = q\| p\| ^2 - p(q \cdot p). \]

Both terms are scalar multiples of \(q\) and \(p\), so \(A = p \times L - (m\mu /r)q \in \mathrm{span}\{ q, p\} \).

(iii). Use the cyclic property of the scalar triple product:

\[ (p \times L) \cdot q = L \cdot (q \times p) = L \cdot L = \| L\| ^2. \]

Therefore:

\[ A \cdot q = (p \times L) \cdot q - \frac{m\mu }{r}\| q\| ^2 = \| L\| ^2 - m\mu r. \]

(iv). Since \(p \perp L\), we have \(\| p \times L\| ^2 = \| p\| ^2\| L\| ^2\). Using the result of (iii) for the cross term:

\begin{align*} \| A\| ^2 & = \| p \times L\| ^2 - \frac{2m\mu }{r}(p \times L) \cdot q + (m\mu )^2 \\ & = \| p\| ^2\| L\| ^2 - \frac{2m\mu }{r}\| L\| ^2 + (m\mu )^2 \\ & = \left(\| p\| ^2 - \frac{2m\mu }{r}\right)\| L\| ^2 + (m\mu )^2 \\ & = 2mH\| L\| ^2 + (m\mu )^2. \end{align*}