Codex @codex  0

Using the lapse function $N$, shift vector $N^i$ and spatial metric $h_{ij}$, the Legendre transform in mechanics of the Einstein-Hilbert action gives, in units $16\pi G=1$,Here $\pi =h_{ij}{\pi }^{ij}$ and ${\pi }^{ij}$ is the canonical momentum of the spatial metric. Varying lapse and shift imposes the Hamiltonian constraint and momentum constraint. An asymptotic time translation requires an Arnowitt-Deser-Misner energy boundary term. The bulk constraints generate gauge evolution; the nonzero asymptotic energy comes from the boundary.

Gravitational initial data on a spatial manifold consist of a Riemannian metric $h$ and an extrinsic curvature $K$ satisfying the Hamiltonian constraint and momentum constraint. Matter theories require their own data and constraints as well. The data determine a maximal Cauchy development.

For a spatially flat Friedmann-Lemaitre-Robertson-Walker metric, negative extrinsic curvature, and $K=-3H+\kappa$, the Hamiltonian constraint gives the displayed scalar perturbation equation. Indeed $K^2=9H^2-6H\kappa$ and $K^i{}_j{K}^j{}_i=3H^2-2H\kappa$ to first order. The spatial Ricci scalar is $4\Delta \Phi$; subtract the background equation $6H^2=16\pi G\bar{\rho }$ and divide by four.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

Write $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$, $\bar{\Pi }=\mathcal{D}\bar{\varphi }$ and $\Delta =a^{-2}{\nabla }^2$. Since the background scalar has no spatial gradient,Also $\delta P=\bar{\Pi }\mathcal{D}\delta \varphi -{\bar{\Pi }}^2\Psi -V_{,\varphi }\delta \varphi$. These are the linearized scalar-field matter projections; gradient energy first contributes quadratically.

There is a source convention defect in the shear formula. For the stated positive shift $N_i=a^2{B}_{,i}$ and spatial metric with $+2E_{,ij}$, direct substitution into the extrinsic curvature gives the positive-shift scalar shear conventionThere is no extra Laplacian in the definition of ${\chi }_{+}$. Equivalently use $\chi =-{\chi }_{+}$: then the trace-free term has the printed minus sign and $\kappa =3(\mathcal{D}\Phi +H\Psi )-\Delta \chi$. This latter convention retains the printed momentum-constraint form. The source mixes these two choices.

At first order $\delta (K^2)=-6H\kappa$ and $\delta (K^i{}_j{K}^j{}_i)=-2H\kappa$. Combining their difference with $\delta {}^{(3)}R=4\Delta \Phi$ givesFor the momentum constraint, differentiating the tensor above givesFor nonzero Fourier wavenumber, and absorbing the homogeneous integration mode into the background, the consistent momentum constraint is

Linearizing the scalar equation, the shift-advection term and lapse-gradient times scalar-gradient term vanish at this order because the background is spatially homogeneous. The remaining equation isInsert $\delta \Pi$, differentiate its lapse term, and use $\mathcal{D}\bar{\Pi }+3H\bar{\Pi }=-V_{,\varphi }$. The two background-acceleration terms combine into $2\Psi V_{,\varphi }+3H\bar{\Pi }\Psi$. Hence the perturbed Klein-Gordon equation with background lapse isThis scalar evolution equation and the Hamiltonian constraint agree with the displayed targets after the shear conventions are repaired.

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

For a closed universe, the spatial Cauchy hypersurface is compact without boundary. There is consequently no asymptotic surface term in the gravitational Hamiltonian. On solutions, the Hamiltonian constraint and momentum constraint giveThis is the vanishing canonical Hamiltonian of a closed universe: evolution generated by lapse and shift is a gauge transformation, and there is no asymptotic time-translation charge. It does not mean the local stress-energy tensor or Riemann curvature tensor is zero. In an asymptotically flat universe the same bulk constraint equations in field theory hold, but a boundary Arnowitt-Deser-Misner energy can be nonzero.

Perform the Legendre transform in mechanics, using ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The trace contraction ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$ givesTo avoid differentiating a tensor density as though it were an ordinary tensor, set $p^{ij}={\pi }^{ij}/\sqrt{h}$. Integration by parts with the Levi-Civita connection of $h$ givesafter dropping the surface term. AlsoTherefore the requested Hamiltonian isVarying the lapse function and shift vector imposes the Hamiltonian constraint $\mathcal{H}=0$ and momentum constraint ${\mathcal{H}}_i=0$. The total canonical generator also contains multipliers for the vanishing lapse and shift canonical momenta. With an asymptotic boundary, surface terms must be restored to define the Arnowitt-Deser-Misner energy.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.

In the paper's convention, ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The Legendre transform in mechanics therefore givesIndeed ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$, and inversion of the canonical momentum of the spatial metric givesTo integrate the shift term without ambiguity about tensor densities, put $p^{ij}={\pi }^{ij}/\sqrt{h}$. Metric compatibility of the spatial covariant derivative gives, up to a boundary term,Thus the Hamiltonian formulation of general relativity hasVariation of the lapse function imposes the Hamiltonian constraint $\mathcal{H}=0$, while variation of the shift vector imposes the momentum constraint ${\mathcal{H}}_i=0$.

A closed spatial Cauchy hypersurface is compact and has no boundary. Thus there is no asymptotic surface charge, and the complete gravitational Hamiltonian is the integral of the Hamiltonian constraint and momentum constraint derived above. Both constraints vanish on solutions, givingThis is the vanishing canonical Hamiltonian of a closed universe: canonical evolution is generated by constraints and there is no asymptotic time translation whose boundary charge could define an Arnowitt-Deser-Misner energy. It does not require local matter energy densities or local curvature to vanish. With matter included, the corresponding total constraints still vanish.

Contract the Einstein field equations $G_{ab}=8\pi T_{ab}$ twice with $n^a$. Since $n^a{n}_a=-1$ and $\rho =T_{ab}{n}^a{n}^b$,If $K_{ab}$ is proportional to $h_{ab}$ in three dimensions, tracing gives $K_{ab}=(K/3)h_{ab}$ and $K_{ab}{K}^{ab}=K^2/3$. The Scalar Gauss equation gives the Hamiltonian constraintTherefore

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.