Reduced-order Landau–Lifshitz local ODE and mass shell

Antisymmetry gives $\eta (U,f_0)=\alpha U^a{F}_{ab}{U}^b=0$. On the unit shell, $U_a{\mathrm{\Pi }}_U{}^a{}_b=U_b+(\eta (U,U)\mathrm{/}{c}^2)U_b=0$, so $\eta (U,a_{\mathrm{L}\mathrm{L}})=0$. The coordinate right-hand side $G=(U,a_{\mathrm{L}\mathrm{L}}(x,U))$ is $C^1$ on the open state domain, identified by the inertial chart with an open subset of ${\mathbb{R}}^8$: $F$ is $C^2$, its partials [F3] are $C^1$, and the remaining factors are polynomial in $U$, using [F4]. In any closed rectangular box in this domain, [F6] bounds all first coordinate partial derivatives of the eight components of $G$ by a common constant $D$. For two states $z,z^{\mathrm{\prime }}$ in the box, change their coordinates one at a time; each intervening coordinate segment stays in the box, so [F7] gives $\mathrm{\mid }{G}_i(z^{\mathrm{\prime }})-G_i(z)\mathrm{\mid }\le D{\sum }_{j=1}^8\mathrm{\mid }{z}_j^{\mathrm{\prime }}-z_j\mathrm{\mid }\le \sqrt{8}D\mathrm{\parallel }{z}^{\mathrm{\prime }}-z{\mathrm{\parallel }}_2$ for each component $i$. Therefore $\mathrm{\parallel }G(z^{\mathrm{\prime }})-G(z){\mathrm{\parallel }}_2\le 8D\mathrm{\parallel }{z}^{\mathrm{\prime }}-z{\mathrm{\parallel }}_2$, proving that $G$ is locally Lipschitz. Thus [F5] gives a unique local full-state solution through $(x_0,U_0)$.

For arbitrary $U$, antisymmetry gives $\eta (U,f_0)=0$ and the projector formula gives $\eta (U,{\mathrm{\Pi }}_{U}B)=(1+\eta (U,U)\mathrm{/}{c}^2)\eta (U,B)$. Along the full solution put $h(\tau )=\eta (U,U)+c^2$; then $h^{\mathrm{\prime }}(\tau )=2\eta (U,a_{\mathrm{L}\mathrm{L}})=\frac{2{\tau }_e}{c^2}\eta (U,B)h(\tau )\mathrm{.}$ The coefficient is continuous. The zero function solves this scalar linear IVP with $h(0)=0$, and the scalar vector field is locally Lipschitz in $h$ because it is linear in that variable, so uniqueness [F5] gives $h=0$ on the full common interval. Thus the proper-time normalization $\eta (U,U)=-c^2$ is preserved.

In the inertial frame of [F2], $e_0$ has coordinates $(1,0,0,0)$. For any future timelike $W$, convexity of $C^{+}$ makes $(1-s)e_0+sW$ timelike for $0\le s\le 1$. A timelike vector cannot have zeroth component zero, since then its squared norm would be the nonnegative spatial norm square. If $W^0\le 0$, the continuous zeroth component along this segment takes the value zero by [F8], a contradiction; hence every future timelike vector has $W^0>0$. Conversely, if $W$ is timelike and $W^0>0$, then $W^0>\mathrm{\mid }\mathbf{W}\mathrm{\mid }$ in this frame, and $(1-s)e_0+sW$ has time component $1-s+s{W}^0>s\mathrm{\mid }\mathbf{W}\mathrm{\mid }$ for every $0\le s\le 1$. It is therefore timelike throughout and joins $e_0$ to $W$, so $W$ belongs to the same timelike component $C^{+}$. Thus a timelike vector is future exactly when its zeroth component is positive. Along the solution, the preserved shell makes $U$ timelike. Its initial vector is future, so $U^0(0)>0$; if $U^0$ became nonpositive, [F8] on the interval between that time and $0$ would force a zero component, again impossible. Hence $U$ remains future directed throughout the local interval. No continuation beyond this interval is asserted.