Flat shifted chart with negative coordinate Hamiltonian

Use the inherited smooth bundle structure under [A1] and the temporal-chart data [F1]. The coordinate transformation is invertible with determinant one and pulls back the Minkowski line element $-(d{X}^0{)}^2+(d{X}^1{)}^2+(d{X}^2{)}^2+(d{X}^3{)}^2$ to $g$, proving flatness. The lapse is positive and $h=\delta$ is positive definite, and its unit normal $n={\mathrm{\partial }}_0-2{\mathrm{\partial }}_1=T$ is future by the chosen orientation, so the chart satisfies [F1]; ${\mathrm{\partial }}_0$ is spacelike since $g({\mathrm{\partial }}_0,{\mathrm{\partial }}_0)=-1+4=3>0$, while $d{x}^0$ is temporal.

Here $\kappa =\sqrt{m^2{c}^2+p_1^2}$, so [F2] gives $H=c(\kappa -2p_1)$. The inequality $p_1>mc\mathrm{/}\sqrt{3}$ implies $m^2{c}^2<3p_1^2$, hence ${\kappa }^2=m^2{c}^2+p_1^2<4p_1^2$, so $\kappa <2p_1$ and $H<0$.

Here $n={\mathrm{\partial }}_0-2{\mathrm{\partial }}_1$, so the shell covector has $p(n)=p_0-2p_1=-\kappa$; [F3] gives $E_n=c\kappa >0$. Put $r:=p_1\mathrm{/}\kappa$; $m>0$ gives $0<r<1$. Differentiating the Hamiltonian of [F2] gives ${\dot{q}}^1=c(r-2)$, ${\dot{q}}^2={\dot{q}}^3=0$ and ${\dot{p}}_i=0$. Thus, for any $q_0\in {\mathbb{R}}^3$, $x(t)=(ct,q_0+ct(r-2,0,0))$, $t\in \mathbb{R}$ in seconds, is the smooth lifted Hamiltonian trajectory. Its tangent is $V:=dx\mathrm{/}dt=c(n+r{\mathrm{\partial }}_1)$, with $g(V,n)=-c$ and $g(V,V)=-c^2(1-r^2)<0$, so it is future timelike. Define its dimensionless spatial velocity relative to $n$ by $w:=V\mathrm{/}[-g(V,n)]-n\in n^{\perp }$, and its physical spatial velocity by $v_n:=cw$. Indeed, the observer time increment is $d{t}_n=-g(V,n)dt\mathrm{/}c$, and the projected displacement is $(V+g(V,n)n)dt=v_{n}d{t}_n$. Here $w=r{\mathrm{\partial }}_1$, so $\mathrm{\mid }{v}_n{\mathrm{\mid }}_h=c\sqrt{h_{ij}{w}^i{w}^j}=cr<c$. Meanwhile $\mathrm{\mid }\dot{q}{\mathrm{\mid }}_{\delta }=c(2-r)>c$. These are respectively the normal-observer speed and the coordinate speed magnitude in this chart.