6.1 Fixed Negative-Energy Surface

Definition 19 Negative-energy shell and rescaled observables

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Fix $h {\lt} 0$. The negative-energy shell $M_h$ is the set of all phase points $(q, p)$ with $H(q, p) = h$ and $r \neq 0$. Equivalently, it is the fixed-energy surface cut out by the equation $H = h$ inside the non-collision phase space, so it has dimension five.

On $M_h$, define the rescaled Lenz vector:

\[ K = \frac{A}{\sqrt{-2mh}}, \qquad K_u = u \cdot K. \]

This is well-defined because $h$ is a fixed negative number, so $\sqrt{-2mh}$ is an honest real constant on the whole surface. The rescaling is chosen so that $\{ K_u, K_v\} = L_{u \times v}$ on $M_h$ (proved in Theorem 21).

Proposition 20 The model algebra is $\mathfrak {so}(4)$

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On $\mathbb {R}^3 \oplus \mathbb {R}^3$, define the bracket:

\[ [(u,v),(u',v')] = \bigl(u \times u' + v \times v',\, u \times v' + v \times u'\bigr). \]

This is isomorphic to $\mathfrak {so}(4)$ via the linear bijection $\Phi \colon \mathbb {R}^3 \oplus \mathbb {R}^3 \to \mathfrak {so}(4)$:

\[ \Phi (u,v) = \begin{pmatrix} 0 & -u_3 & u_2 & v_1 \\ u_3 & 0 & -u_1 & v_2 \\ -u_2 & u_1 & 0 & v_3 \\ -v_1 & -v_2 & -v_3 & 0 \end{pmatrix}. \]

Proof ▶

The matrix $\Phi (u, v)$ is antisymmetric, so it lies in $\mathfrak {so}(4)$. The map $\Phi $ is linear and bijective (both spaces are six-dimensional). A direct matrix multiplication verifies the bracket identity $[\Phi (u,v), \Phi (u',v')] = \Phi ([(u,v),(u',v')])$, so $\Phi $ is a Lie algebra isomorphism.

Theorem 21 Hidden $\mathfrak {so}(4)$ comoment map

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On the shell $H = h {\lt} 0$, the map

\[ \mu _h(u, v) = L_u + K_v \]

is an $\mathfrak {so}(4)$-comoment map for the bracket of Proposition 20.

Proof ▶

We check directly that $\{ \mu _h(u,v), \mu _h(u',v')\} = \mu _h([(u,v),(u',v')])$. The only subtlety is that the formula

\[ \{ A_u, A_v\} = -2mH\, L_{u \times v} \]

holds on the full phase space, where $H$ is still a function. We may replace $H$ by the constant $h$ only after restricting to the surface $M_h$.

From Theorem 18 and the definition of $K_u$:

\[ \{ K_u, K_v\} = \frac{1}{-2mh}\{ A_u, A_v\} = \frac{-2mH}{-2mh}L_{u \times v} = L_{u \times v} \quad \text{(since $H = h$ on $M_h$)}. \]

Also $\{ L_u, K_v\} = K_{u \times v}$ because $L_u$ rotates $A$ (and hence $K$) by the same formula as for $L$. Therefore:

\begin{align*} \{ L_u + K_v,\, L_{u'} + K_{v'}\} & = \{ L_u, L_{u'}\} + \{ L_u, K_{v'}\} + \{ K_v, L_{u'}\} + \{ K_v, K_{v'}\} \\ & = L_{u \times u'} + K_{u \times v'} + K_{v \times u'} + L_{v \times v'} \\ & = L_{u \times u' + v \times v'} + K_{u \times v' + v \times u'} \\ & = \mu _h\bigl([(u,v),(u’,v’)]\bigr). \end{align*}

Corollary 22 Two commuting copies of $\mathfrak {so}(3)$

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On the shell $M_h$, define:

\[ J^+_u = \tfrac {1}{2}(L_u + K_u), \qquad J^-_u = \tfrac {1}{2}(L_u - K_u). \]

Then $J^+$ and $J^-$ each satisfy the $\mathfrak {so}(3)$ bracket, and they commute with each other:

\[ \{ J^+_u, J^+_v\} = J^+_{u \times v}, \qquad \{ J^-_u, J^-_v\} = J^-_{u \times v}, \qquad \{ J^+_u, J^-_v\} = 0. \]

So $\mathfrak {so}(4) \cong \mathfrak {so}(3) \oplus \mathfrak {so}(3)$ via the splitting $(L, K) \mapsto (J^+, J^-)$.

Proof ▶

For the commutator, expand using the brackets from Theorem 21:

\begin{align*} \{ J^+_u, J^-_v\} & = \tfrac {1}{4}\bigl(\{ L_u, L_v\} - \{ L_u, K_v\} + \{ K_u, L_v\} - \{ K_u, K_v\} \bigr) \\ & = \tfrac {1}{4}\bigl(L_{u \times v} - K_{u \times v} + K_{u \times v} - L_{u \times v}\bigr) = 0. \end{align*}

The self-brackets $\{ J^+_u, J^+_v\} = J^+_{u \times v}$ and $\{ J^-_u, J^-_v\} = J^-_{u \times v}$ follow similarly.