TD spin and momentum norms

The equations in [A1] act on the moment variables defined in [F2] and are exactly the pole-dipole system assumed in [F1]. On the region [A2], [F1] gives preservation of $S^{ab}{p}_b=0$ and constancy of $\mu$ and $S_{ab}{S}^{ab}\mathrm{/}2$ on the local solution. This is a conditional deduction within the adopted truncation.

Suppose also $S({\tau }_0)=0$. Step 1.1 preserves $C^a=S^{ab}{p}_b=0$ and the scalar $\frac{1}{2}{S}_{ab}{S}^{ab}$, while the admissible region keeps $p$ future timelike. At each $\tau$, set $\mu =\sqrt{-g(p,p)}>0$, $e_0=p\mathrm{/}\mu$, and $E:=e_0^{\perp }$; [F3] says $E$ is a three-dimensional positive-definite inner-product space. Take any basis $v_1,v_2,v_3$ of $E$ and form $e_1:=\frac{v_1}{\sqrt{g(v_1,v_1)}},w_2:=v_2-g(v_2,e_1)e_1,e_2:=\frac{w_2}{\sqrt{g(w_2,w_2)}},w_3:=v_3-g(v_3,e_1)e_1-g(v_3,e_2)e_2,e_3:=\frac{w_3}{\sqrt{g(w_3,w_3)}}\mathrm{.}$ The denominators are positive: positive definiteness gives $g(v_1,v_1)>0$, and linear independence of the $v_i$ makes $w_2,w_3$ nonzero. These formulas give an orthonormal basis of $E$; with $e_0$ they give an orthonormal basis of $T_{z(\tau )}M$. The TD constraint says $S^{ab}{p}_b=0$; since $p_b=(-\mu ,0,0,0)$ in this basis, it gives $S^{a0}=0$, and antisymmetry also gives $S^{0a}=0$. Therefore $\frac{1}{2}{S}_{ab}{S}^{ab}=\frac{1}{2}{\sum }_{i,j=1}^3(S^{ij}{)}^2\ge 0,$ with equality if and only if every component $S^{ij}$ vanishes, that is, $S=0$. Its conserved value is initially zero, so $S$ remains zero throughout the local solution. The zero-spin branch in [F1] now yields $p=(\mu \mathrm{/}c)U$ and $D_{\tau }p=0$. Since [F1] also gives constant $\mu >0$, covariant differentiation of $U=(c\mathrm{/}\mu )p$ gives $D_{\tau }U=0$. Therefore $z$ is a timelike geodesic in proper-time parametrization and $M=\mu \mathrm{/}c$ is constant. No continuation outside the admissible region is supplied.