$j_a = - T_{0a}$ and $S_{ab} = T_{ab}$. Define $$\begin{aligned}
\label{hamiltonian1}
C & = & 2 G_{00} - 2 T_{0 0} \\
& = & -{k^a}_b \, {k^b}_a + ( {\mbox{tr}}\, k )^2 + R - 2 \rho. \nonumber\end{aligned}$$ $C=0$ is the Hamiltonian constraint. Similarly, $C_a = 0$ is the momentum constraint, where $$\begin{aligned}
\label{momentum1}
C_a & = & -G_{0a} + T_{0a} \\
& = & \nabla_b {k^b}_a - \nabla_a ({\mbox{tr}}\, k) - j_a. \nonumber\end{aligned}$$ In Gaussian coordinates, the relation between the metric and the extrinsic curvature is $$\label{dtg}
\partial_t g_{ab} = -2 k_{ab}.$$ The evolution equation for the extrinsic curvature is obtained by setting ${E^a}_b=0$, with $$\begin{aligned}
\label{eveq1}
{E^a}_b & = & ^{(D)}{R^a}_b - {T^a}_b + {1 \over(D- 2)} T \, {\delta^a}_b \\
\label{eveq}
\Rightarrow \; \partial_t {k^a}_b & = & 
 {R^a}_b + ({\mbox{tr}}\, k) \, {k^a}_b - {M^a}_b.\end{aligned}$$ Here $${M^a}_b = {S^a}_b - {1 \over D-2} (({\mbox{tr}}\, S) - \rho) {\delta^a}_b.$$

Scalar and Maxwell fields in four dimensions
============================================

\[scaMax4D\]

Equations of motion
-------------------

As said above, let us start by considering in detail, as archetypal system, the system defined by the action (\[action1\]), [*i.e.,*]{} the spacetime dimension is $D=4$ and the matter fields are a massless scalar 