Poincare semidirect group, dual pairing and moment components — Examples

Let $\mathcal{P}=(T^{\ast }\mathbb{R}{)}^2\cong {\mathbb{R}}^4$ have coordinates $(q_1,p_1,q_2,p_2)$, symplectic form $\omega ={\sum }_{a=1}^{2}d{q}_a\wedge d{p}_a$, and canonical bracket $\{F,G\}={\sum }_{a=1}^2(F_{q_a}{G}_{p_a}-F_{p_a}{G}_{q_a})$. Let $q_a$ be one-dimensional positions in metres, $p_a$ momenta in kg m/s, $t$ an external equal-time parameter in seconds, $m_a>0$ in kg, and $c>0$ in m/s. For $k>0$ in kg/s${}^2$ define $E_a(p_a)=\sqrt{m_a^2{c}^4+c^2{p}_a^2}$, $V(r)=k{r}^2\mathrm{/}2$ with $r=q_1-q_2$, $H=E_1+E_2+V(r)$, $P=p_1+p_2$, and $K=(q_1{E}_1+q_2{E}_2)\mathrm{/}{c}^2-tP$. The proposed assertion that this Hamiltonian and unchanged free boost satisfy $\{K,P\}=H\mathrm{/}{c}^2$ at every configuration is false.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ for the cited Poincare algebra input. Use the phase space, bracket, masses, and functions defined in the Statement, with $k$ having units kg/s${}^2$, so $V$ has energy units. The retained free total boost is

$K=\frac{q_1{E}_1+q_2{E}_2}{c^2}-t(p_1+p_2),P=p_1+p_2\mathrm{.}$

With canonical brackets, $\{K,P\}=(E_1+E_2)\mathrm{/}{c}^2$, while the Poincaré charge algebra requires $\{K,P\}=H\mathrm{/}{c}^2$. They disagree at every configuration with $q_1\mathrm{\ne }{q}_2$. The construction is therefore a counterexample to combining this instantaneous potential with the unchanged free boost. It is not a universal no-interaction theorem.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ for the cotangent/constraint supplier. Use the product cotangent domain $\mathcal{P}=T^{\ast }({\mathbb{R}}^4\times {\mathbb{R}}^4)$ with coordinates $(x_i^a,p_{ia})$ for $i=1,2$, Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$, canonical form $\omega ={\sum }_{i=1}^2{\sum }_{a=0}^{3}d{x}_i^a\wedge d{p}_{ia}$, and induced bracket $\{F,G\}={\sum }_{i=1}^2{\sum }_{a=0}^3(F_{x_i^a}{G}_{p_{ia}}-F_{p_{ia}}{G}_{x_i^a})$. Take $x_i^a$ in metres, $p_{ia}$ in kg m/s, $P_i^a:={\eta }^{ab}{p}_{ib}$, $p_i^2:={\eta }^{ab}{p}_{ia}{p}_{ib}$, $r^a:=x_1^a-x_2^a$, and $r^2:={\eta }_{ab}{r}^a{r}^b$. Use constraints

$C_1=\frac{1}{2}(p_1^2+m_1^2{c}^2)+V(r^2),C_2=\frac{1}{2}(p_2^2+m_2^2{c}^2)+V(r^2)\mathrm{.}$

Here $m_i>0$ in kg, $c>0$ in m/s, and $V:\mathbb{R}\to \mathbb{R}$ is smooth with units of momentum squared. Their canonical bracket is

$\{C_1,C_2\}=-2V^{\mathrm{\prime }}(r^2)r\cdot (P_1+P_2)\mathrm{.}$

For $V(z)=\lambda z$, with $\lambda >0$ in momentum squared per square metre, this bracket is nonzero at explicit points on the joint surface $C_1=C_2=0$ where $d{C}_1,d{C}_2$ are independent. Thus, by First-class closure for a regular constraint family, the pair fails first-class closure on the local regular surface around each such point, and arbitrary independent multipliers need not preserve both constraints there. Additional consistency conditions are required; no universal obstruction for all interacting relativistic models is claimed.

Let $\mathcal{P}=T^{\ast }\mathbb{R}$ have position coordinate $q\in \mathbb{R}$ in metres, momentum coordinate $p\in \mathbb{R}$ in kg m/s, and symplectic form $\omega =dq\wedge dp$. For fixed mass $m>0$ in kg and speed $c>0$ in m/s, take the free Hamiltonian $H(p)=E(p)=\sqrt{m^2{c}^4+c^2{p}^2}$, in joules, on $\mathcal{P}$; time $t$ is measured in seconds. The proposed assertion that every smooth initial action $S_0:\mathbb{R}\to \mathbb{R}$ produces a characteristic family extending as a smooth single-valued graph $p=S_q(t,q)$ with finite $S_{qq}$ through every finite future time $t\ge 0$ is false.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ for the cotangent-shell supplier. With the phase space and Hamiltonian from the Statement, let $a\in \mathbb{R}$ be the initial position in metres and $t$ seconds. Choose the smooth initial action $S_0(a)=-\kappa a^2\mathrm{/}2$, $\kappa >0$ in kg/s. Its characteristics have $p(a)=-\kappa a$ and

$q(t,a)=a+t\frac{c^{2}p(a)}{E(p(a))}\mathrm{.}$

For this characteristic family, call $(t,a)$ a caustic point when the projection $(t,a)\mapsto (t,q(t,a))$ is not locally invertible there.

The projection derivative is ${\mathrm{\partial }}_{a}q=1-t\kappa m^2{c}^6\mathrm{/}E(p{)}^3$. It stays positive for $0\le t<m\mathrm{/}\kappa$ and first vanishes at $(t,a)=(m\mathrm{/}\kappa ,0)$, where ${\mathrm{\partial }}_{a}p=-\kappa \mathrm{\ne }0$. Therefore this characteristic family cannot extend there as a smooth single-valued graph $p=S_q(t,q)$ with finite $S_{qq}$. The initial data and Hamiltonian are smooth; the failure is the projection singularity, not an initial-data defect.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ for the cotangent-shell supplier. Work with spatial position $q\in {\mathbb{R}}^3$ in metres and momentum parameter $\alpha \in {\mathbb{R}}^3$ in kg m/s, with the Euclidean dot product and norm $\mathrm{\mid }\alpha \mathrm{\mid }=\sqrt{\alpha \cdot \alpha }$; at fixed time the spatial phase space has canonical form $\omega ={\sum }_{i=1}^{3}d{q}^i\wedge d{p}_i$. Here $t\in \mathbb{R}$ is measured in seconds, $m\ge 0$ is in kg, and $c>0$ is in m/s. The parameter domain is all ${\mathbb{R}}_{\alpha }^3$ when $m>0$, and ${\mathbb{R}}_{\alpha }^3\setminus \{0\}$ when $m=0$. For a free massive particle with $m>0$, or a nonzero massless particle with $m=0$ and $\alpha \mathrm{\ne }0$, define

$S(t,q,\alpha )=q\cdot \alpha -tE(\alpha ),E(\alpha )=\sqrt{m^2{c}^4+c^2\mathrm{\mid }\alpha {\mathrm{\mid }}^2}\mathrm{.}$

Then $S_t+E(S_q)=0$, $S_{q\alpha }=I$, and its integration coordinates satisfy $\beta =S_{\alpha }=q-t{c}^2\alpha \mathrm{/}E$, equivalently $q=\beta +t{c}^2\alpha \mathrm{/}E$. The mixed Hessian is everywhere nonsingular on the stated domain, so this is a complete integral there. $[S]=\mathrm{J}\mathrm{s}$, $[\alpha ]=\mathrm{k}\mathrm{g}\mathrm{m}\mathrm{/}\mathrm{s}$, and $[\beta ]=\mathrm{m}$.