Cotangent phase data, observables and action dimensions — Examples

The periodic quotient atlas makes $T^2$ a smooth manifold. By [F1], its cotangent form is $\omega =dx\wedge d{p}_x+dy\wedge d{p}_y$. The differential $dC=d{p}_x+ad{p}_y$ is nowhere zero, so $C^{-1}(0)$ is a regular embedded hypersurface. The canonical bracket is skew, hence $\{C,C\}=0$ and the single constraint is first class.

Contraction with $dx\wedge d{p}_x+dy\wedge d{p}_y$ gives $X_C={\mathrm{\partial }}_x+a{\mathrm{\partial }}_y$. Integrating this constant vector field yields the stated flow and leaves momenta fixed.

Fix a starting point $(x_0,y_0)$ and any target first coordinate $u\in [0,1)$. For each $n\in \mathbb{Z}$, the flow parameter $s=n-x_0+u$ makes the first coordinate equal to $u$ modulo one and the second equal to $y_0+a(n-x_0+u)$ modulo one. The residues $na<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1$ are dense: for any integer $N\ge 2$, divide $[0,1)$ into $N$ half-open intervals of length $1\mathrm{/}N$. Two of the $N+1$ residues $0,a,\dots ,Na<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1$ lie in one interval, so for some nonzero integer $k$ and integer $r$ we have $0<\delta :=\mathrm{\mid }ka-r\mathrm{\mid }<1\mathrm{/}N$; the strict lower bound follows from irrationality. Replacing $k$ by $-k$ if needed, $\delta$ is an integer multiple of $a$ modulo one. The residues $j\delta$ for $j=0,1,\dots ,\lfloor 1\mathrm{/}\delta \rfloor$ have gaps at most $\delta$ around the unit interval, so they form a $1\mathrm{/}N$-net. Since $N$ is arbitrary, the multiples of $a$ are dense modulo one. Therefore the orbit is dense in the second coordinate for the chosen $u$; as $u$ was arbitrary, the flow orbit is dense in $T^2$.

The orbits through $(0,0,0,0)$ and $(0,1\mathrm{/}2,0,0)$ are distinct: equality would require integers $n,k$ with $an+k=1\mathrm{/}2$; if $n\mathrm{\ne }0$ this would make $a$ rational, and if $n=0$ it would assert that an integer equals $1\mathrm{/}2$. Both orbits are dense by step 3.1. Let $U$ be a saturated open subset of the constraint surface that meets the fixed-momentum torus. Its intersection with that torus is nonempty, open, and saturated. Its complement is closed and saturated; if nonempty, it contains a whole dense orbit and hence the whole torus. Therefore the intersection is the whole torus. Any quotient neighborhoods of the two distinct orbit classes have such preimages and therefore intersect, so the orbit space is non-Hausdorff. This does not contradict the local symplectic leaf spaces in [F3].

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.