Free shell in inertial time has the square-root Hamiltonian

In affine Minkowski space with $x^0=ct$, let $m\ge 0$, $p_i$ be spatial momentum components and $p_i\mathrm{\ne }0$ for $m=0$. On the future shell, $p_0=-E\mathrm{/}c$, where

$E(\mathbf{p})=\sqrt{m^2{c}^4+c^2\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2}\mathrm{.}$

On any compact interval $J=[t_1,t_2]$ of inertial time with $t_1<t_2$, the gauge $x^0=ct$ reduces the first-order shell action to

$I_J={\int }_{t_1}^{t_2}(p_i{\dot{q}}^i-E(\mathbf{p}))dt,q^i=x^i\mathrm{.}$

For this variational statement, take $q^i$ and $p_i$ smooth on a neighborhood of $J$ and restrict to variations that are smooth one-parameter families, fix $q$ at both endpoints, and vary $p$ freely; in the massless case the paths and sufficiently small variations stay in $\mathbf{p}\mathrm{\ne }0$. The resulting Hamilton equations hold on the interior of $J$.

The gauge pair $C_m=0$, $\psi =x^0-ct=0$ is second class because $\{C_m,\psi \}=-P^0\mathrm{\ne }0$ on the future shell; its Dirac reduction gives the canonical spatial bracket. The reduced Hamilton equations are ${\dot{q}}^i=c^2{p}_i\mathrm{/}E$, ${\dot{p}}_i=0$, agreeing with the unreduced characteristic trajectories. On this full flat future shell every characteristic line meets each inertial-time slice once; this global assertion does not extend by this proof to a curved temporal chart.