Hamiltonian and Momentum Constraints

Einstein's field equations are usually written as one tensor statement, $G_{\mu\nu}=8\pi T_{\mu\nu}$ — ten equations bundled together. The moment you treat gravity as an initial-value problem — pick a "now," specify the field, let the equations tell you the future — that bundle comes apart along a seam it was hiding. Slice spacetime into a stack of spacelike three-surfaces $\Sigma_t$ , each carrying an induced metric $\gamma_{ij}$ and an extrinsic curvature $K_{ij}$ , and project the ten equations onto and across each slice. Six turn out to be genuine evolution equations, telling $\gamma_{ij}$ and $K_{ij}$ how to march forward in time. The other four contain no second time derivative at all: they are relations the data $(\gamma_{ij},K_{ij})$ must already satisfy on a single instant, before any evolution begins — the Hamiltonian constraint (one equation) and the momentum constraint (three). This is not an accident of bookkeeping; it is forced by the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ , which guarantees that the four components carrying a normal index drop one order in time. The parallel with electromagnetism is exact: Gauss's laws $\nabla\!\cdot\!\mathbf E=\rho$ and $\nabla\!\cdot\!\mathbf B=0$ constrain the fields on a slice, while the curl equations evolve them. Gravity simply has four such constraints instead of two. Geometrically they are the Gauss and Codazzi embedding conditions — the statement that not every abstract pair $(\gamma_{ij},K_{ij})$ can sit inside a spacetime obeying Einstein's equations. Practically they are where every numerical-relativity simulation begins: solve the four constraints once, then evolve. This article derives them, verifies them, and follows them from André Lichnerowicz's 1944 elliptic equation to the black-hole mergers LIGO hears.

Key takeaways

In the 3+1 (ADM) split, the ten Einstein equations separate into four constraints on the initial data $(\gamma_{ij},K_{ij})$ and six evolution equations — the constraints carry no second time derivative.
The Hamiltonian constraint is the normal-normal projection $G_{\mu\nu}n^\mu n^\nu = 8\pi\rho$ ; the momentum constraint is the three normal-tangential projections $\gamma^{i}{}_\mu G^{\mu}{}_{\nu}n^\nu = 8\pi S^i$ .
They are exactly the Gauss and Codazzi embedding relations: $\,{}^{(3)}R + K^2 - K_{ij}K^{ij}=16\pi\rho\,$ and $\,D_j(K^{ij}-\gamma^{ij}K)=8\pi S^i$ .
The Bianchi identities guarantee the constraints are preserved by the evolution — data satisfying them at $t=0$ satisfy them forever — which is why the Cauchy problem of general relativity is well posed.
Counting: 10 metric components − 4 gauge (lapse, shift) − 4 constraints = 2 physical degrees of freedom, the two gravitational-wave polarizations.
Solving them, usually by York's conformal method built on Lichnerowicz's equation, is the first step of every binary-black-hole and neutron-star simulation.
Key idea: the four constraints are the projections of $G_{\mu\nu}=8\pi T_{\mu\nu}$ along the slice normal $n^\mu$ ; they carry no second time derivative and so restrict the initial data on a single spacelike slice, while the six tangential projections evolve it.

The core insight

The four constraint equations are the projections of $G_{\mu\nu}=8\pi T_{\mu\nu}$ along the slice normal $n^\mu$ ; they contain no second time derivatives and so restrict the initial data $(\gamma_{ij},K_{ij})$ on a single spacelike slice, while the six tangential-tangential projections evolve them.

Where this lives today

The constraints are the first line of code in every gravitational-wave prediction. Before a binary-black-hole run takes a single step, a solver must produce data satisfying equations (1)–(2). Spectral solvers such as TwoPunctures (Ansorg–Brügmann–Tichy) and LORENE/FUKA solve the Lichnerowicz–York system to machine precision for boosted, spinning holes and neutron stars; the Bowen–York extrinsic curvature supplies momentum and spin by solving the momentum constraint analytically on a conformally flat maximal slice. That data feeds evolution codes — the Einstein Toolkit, SpEC, GRChombo — running constraint-damping formulations (BSSN, Z4c, generalized harmonic) precisely because discretization violates the constraints and something must suppress the residuals. Every template in the LIGO–Virgo–KAGRA banks that matched GW150914 or the binary neutron star GW170817 traces back to constraint-solved data. And the reach runs deeper: the ADM mass and momentum are surface integrals of the constraints at infinity, and in canonical quantum gravity the Hamiltonian constraint, promoted to an operator, becomes the Wheeler–DeWitt equation $\hat{\mathcal H}\Psi=0$ — the source of quantum gravity's notorious "problem of time."

The physical picture

Freeze the universe at one instant and you hold a curved three-dimensional surface $\Sigma$ sitting inside four-dimensional spacetime. Two tensors describe it. The induced metric $\gamma_{ij}$ is its intrinsic geometry — how you measure lengths and angles staying within the slice. The extrinsic curvature $K_{ij}$ is its embedding geometry — how the slice bends and tilts as you step off it along the unit normal $n^\mu$ , equivalently how fast $\gamma_{ij}$ is changing in time. Together $(\gamma_{ij},K_{ij})$ are the gravitational analogue of "position and velocity": the data you would hand to a computer to start a simulation. The deep fact of this article is that you are not free to choose them arbitrarily. Four equations bind them before the clock even starts.

Figure

The 3+1 foliation: spacetime is sliced into spacelike surfaces Σ_t carrying the intrinsic metric γ_ij and extrinsic curvature K_ij. The unit normal n^μ, lapse N and shift N^i connect neighbouring slices; the four constraints are relations that G_μν n^μ n^ν = 8πρ and γ^i_μ G^μ_ν n^ν = 8πS^i impose on a single slice

Why exactly four constraints, and why do they carry no acceleration? Because of the contracted Bianchi identity. The four components of $G^{\mu\nu}$ with one index along the normal, $G^{\mu\nu}n_\mu$ , are forced by $\nabla_\mu G^{\mu\nu}=0$ to be one time-derivative "lower" than the rest — the second time derivatives of $\gamma_{ij}$ live only in the purely spatial part $G^{ij}$ . So those four equations cannot be telling anything how to accelerate; they can only be relating quantities already present on the slice. This is the same structure that makes $\nabla\!\cdot\!\mathbf E=\rho$ a constraint on electromagnetic data rather than an equation of motion. The degree-of-freedom accounting falls out cleanly, and the second figure carries the one place in this subject where the constraint becomes an ordinary equation you can solve on paper.

Figure

Time-symmetric black-hole initial data: with K_ij = 0 the momentum constraint is automatic and the Hamiltonian constraint collapses to the flat Laplace equation ∇²ψ = 0, solved by ψ = 1 + M/(2r). The conformal factor diverges at the puncture r → 0, passes ψ = 2 at the throat r = M/2 (areal radius 2M, the horizon), and tends to 1 at infinity where the ADM mass reads off as M

The core equations

The two constraints are, at heart, the two contractions of the Gauss–Codazzi embedding relations with Einstein's equation substituted in. The first is a scalar balance on the slice: intrinsic curvature plus the "kinetic" combination of extrinsic curvature equals the local energy density. This is the Hamiltonian constraint.

{}^{(3)}R + K^2 - K_{ij}K^{ij} = 16\pi\rho \qquad (1)

where: ${}^{(3)}R$ — Ricci scalar of the intrinsic 3-metric $\gamma_{ij}$ ; $K_{ij}$ — extrinsic curvature of $\Sigma_t$ ; $K=\gamma^{ij}K_{ij}$ — its trace (mean curvature); $\rho=T_{\mu\nu}n^\mu n^\nu$ — energy density seen by the normal ("Eulerian") observer. Reading it: on a given slice you cannot dial $\gamma_{ij}$ and $K_{ij}$ independently — their curvatures must sum to the local energy. In vacuum the right side is zero and the slice's intrinsic and extrinsic curvatures must cancel. The second is a vector statement: the divergence of the trace-reversed extrinsic curvature is sourced by the local momentum density. This is the momentum constraint, three equations for $i=1,2,3$ .

D_j\!\left(K^{ij} - \gamma^{ij}K\right) = 8\pi S^i \qquad (2)

where: $D_j$ — covariant derivative of the spatial metric $\gamma_{ij}$ ; $K^{ij}$ — extrinsic curvature with indices raised by $\gamma^{ij}$ ; $S^i=-\gamma^{i}{}_\mu n_\nu T^{\mu\nu}$ — momentum density measured by the Eulerian observer. Reading it: shearing motion of the slice (the transverse part of $K_{ij}$ ) is tied to the flow of momentum through it, just as $\nabla\!\times\!\mathbf B$ is tied to current. Both descend from the same source — the four projections of the Einstein tensor that carry at least one normal index. Writing them side by side makes the origin explicit, and makes clear that the remaining six projections (two tangential indices) are what evolve the data.

G_{\mu\nu}n^\mu n^\nu = 8\pi\,T_{\mu\nu}n^\mu n^\nu,\qquad \gamma^{i}{}_\mu\,G^{\mu}{}_{\nu}\,n^\nu = 8\pi\,\gamma^{i}{}_\mu\,T^{\mu}{}_{\nu}\,n^\nu \qquad (3)

where: $n^\mu$ — future-pointing unit normal, $n_\mu n^\mu=-1$ ; $\gamma^{i}{}_\mu$ — projector onto the slice, from $\gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu$ ; $G_{\mu\nu}$ — Einstein tensor of the full spacetime. Reading it: one normal-normal projection gives the scalar Hamiltonian constraint, three normal-tangential projections give the vector momentum constraint — four constraints, exactly the count the Bianchi identity protects. The data do not sit still; the six evolution equations are packaged in ADM form as first-order equations for $\gamma_{ij}$ and $K_{ij}$ , driven by the freely chosen lapse and shift. The metric evolution equation is simply the definition of $K_{ij}$ rearranged.

\partial_t \gamma_{ij} = -2N K_{ij} + D_i N_j + D_j N_i \qquad (4)

where: $N$ — lapse (proper time per unit coordinate time along $n^\mu$ ); $N^i$ — shift (how spatial coordinates slide from slice to slice). Reading it: the extrinsic curvature literally is the time derivative of the spatial metric, up to the gauge freedom of lapse and shift. Feed $K_{ij}$ in and the slice moves. For actually constructing initial data one rescales the metric conformally, $\gamma_{ij}=\psi^4\tilde\gamma_{ij}$ , splitting off a single positive scalar $\psi$ . The Hamiltonian constraint then becomes a quasi-linear elliptic equation for $\psi$ — the Lichnerowicz–York equation, the workhorse of numerical relativity.

8\,\tilde D^2\psi - \psi\,{}^{(3)}\tilde R - \tfrac{2}{3}\psi^5 K^2 + \psi^{-7}\tilde A_{ij}\tilde A^{ij} = -16\pi\,\psi^5\rho \qquad (5)

where: $\psi$ — conformal factor; $\tilde\gamma_{ij}$ — chosen background metric; $\tilde D^2$ — its Laplacian; ${}^{(3)}\tilde R$ — its Ricci scalar; $\tilde A_{ij}$ — conformally rescaled trace-free part of $K_{ij}$ , with $A^{ij}=\psi^{-10}\tilde A^{ij}$ . Reading it: the messy nonlinear constraint has been organized into an elliptic PDE with a well-developed existence-and-uniqueness theory. The coefficient of the $\psi^5 K^2$ term is $\tfrac{2}{3}$ , from splitting the trace out of $K_{ij}$ (a factor sometimes dropped in careless treatments). On a maximal slice the trace $K=0$ , and the conformal momentum constraint loses its coupling to $\psi$ , becoming a linear elliptic equation you can solve first and independently.

\tilde D_j \tilde A^{ij} - \tfrac{2}{3}\,\psi^6\,\tilde D^i K = 8\pi\,\psi^{10}S^i \quad\xrightarrow{\ K=0\ }\quad \tilde D_j \tilde A^{ij}=8\pi\psi^{10}S^i \qquad (6)

where: $\tilde A^{ij}$ — conformal trace-free extrinsic curvature; the $K$ -term vanishes on maximal slices. Reading it: choosing $K=0$ decouples the four constraints into a linear momentum equation feeding a nonlinear Hamiltonian equation — the reason maximal and constant-mean-curvature slicings dominate practical initial-data construction.

Derivation AThe four constraints from the Gauss and Codazzi equations

The goal here is to obtain both constraints rigorously, not by analogy, by projecting the four-dimensional Einstein tensor onto and across the slice using the two embedding identities of a hypersurface. We start from the Gauss equation (which relates the intrinsic Riemann tensor to the ambient one plus extrinsic-curvature products) and the Codazzi equation (which relates a mixed projection of the ambient Riemann tensor to derivatives of $K_{ij}$ ), contract each appropriately, and substitute Einstein's equation at the very end. The payoff is that equations (1) and (2) fall out with their exact numerical coefficients and their source identifications $\rho$ and $S^i$ fixed unambiguously.

Step 1. Write the Gauss equation for the slice, ${}^{(3)}R_{ijkl} = \gamma^{\mu}{}_i\gamma^{\nu}{}_j\gamma^{\alpha}{}_k\gamma^{\beta}{}_l R_{\mu\nu\alpha\beta} + K_{ik}K_{jl} - K_{il}K_{jk}$ . Why: this is the fundamental embedding identity — it expresses the curvature you would measure inside $\Sigma$ in terms of the ambient curvature restricted to $\Sigma$ plus a purely extrinsic piece, and it is the only input needed for the scalar constraint.

Step 2. Contract twice with $\gamma^{ik}\gamma^{jl}$ and use $\gamma^{\mu}{}_\nu=\delta^\mu_\nu+n^\mu n_\nu$ . The extrinsic terms assemble into $K^2-K_{ij}K^{ij}$ and the projected Riemann terms reorganize into ${}^{(3)}R + K^2 - K_{ij}K^{ij}=R + 2R_{\mu\nu}n^\mu n^\nu$ . Why: the double contraction is what turns the full four-index Gauss relation into a single scalar equation, trading the ambient Riemann tensor for the ambient Ricci scalar and one normal Ricci projection.

Step 3. Convert the right side to the Einstein tensor. Since $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 g_{\mu\nu}R$ and $g_{\mu\nu}n^\mu n^\nu=-1$ , one finds $R+2R_{\mu\nu}n^\mu n^\nu = 2G_{\mu\nu}n^\mu n^\nu$ ; the bare $R$ cancels. Inserting $G_{\mu\nu}=8\pi T_{\mu\nu}$ and $\rho=T_{\mu\nu}n^\mu n^\nu$ gives ${}^{(3)}R + K^2 - K_{ij}K^{ij}=16\pi\rho$ . Why: only after this replacement does geometry become physics — the left side is pure embedding data, the right side the matter energy density, and the cancellation of $R$ is what guarantees no second time derivative survives.

Step 4. Now take the Codazzi equation, $\gamma^{\mu}{}_i\gamma^{\nu}{}_j\gamma^{\alpha}{}_k\,n^\beta R_{\mu\nu\alpha\beta} = D_j K_{ik} - D_i K_{jk}$ , which carries exactly one normal index. Why: this is the second embedding identity, and one normal index is precisely what a momentum (normal-tangential) projection needs, so it is the natural source of the vector constraint.

Step 5. Contract on $i$ and $k$ with $\gamma^{ik}$ . The left side collapses to the mixed Ricci projection $\gamma^{\nu}{}_j n^\beta R_{\nu\beta}$ and the right side to $D_i K^{i}{}_j - D_j K = D_i(K^{i}{}_j-\delta^i_j K)$ . Why: the single contraction converts the three-index Codazzi relation into the divergence of the trace-reversed extrinsic curvature, which is the exact combination the constraint contains.

Step 6. Trade Ricci for Einstein and insert the field equation. Because $\gamma^{i}{}_\mu n^\nu g_{\mu\nu}=\gamma^{i}{}_\mu n^\mu=0$ , the trace term $\tfrac12 g_{\mu\nu}R$ drops on projection, so $\gamma^{i}{}_\mu n^\nu R_{\mu\nu}=\gamma^{i}{}_\mu n^\nu G_{\mu\nu}=8\pi\gamma^{i}{}_\mu n^\nu T_{\mu\nu}=8\pi S^i$ , giving $D_j(K^{ij}-\gamma^{ij}K)=8\pi S^i$ . Why: the vanishing of the trace piece under the mixed projection is what makes the momentum constraint depend only on $K_{ij}$ and the momentum density, closing the derivation. *Sign convention used throughout: $K_{ij}=-\gamma^{\alpha}{}_i\gamma^{\beta}{}_j\nabla_\alpha n_\beta$ .*

Go deeper · Derivation B · 1 minWhy the constraints carry no second time derivative

Derivation BWhy the constraints carry no second time derivative

The goal is to make precise the claim that the four normal projections of the Einstein tensor are first order in time and therefore constraints rather than equations of motion. The argument is not about the particular form of $G_{\mu\nu}$ but about a structural fact — the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ — which forces the components $G^{\mu\nu}n_\mu$ to be one derivative "lower" than the spatial components. This is the general-relativistic version of the statement that Gauss's law in electromagnetism is a constraint while the curl equations are the dynamics, and it is what makes the initial-value formulation possible.

Step 1. Locate the second time derivatives. In ADM variables $\partial_t\gamma_{ij}$ defines $K_{ij}$ through equation (4), so $\partial_t^2\gamma_{ij}$ enters only through $\partial_t K_{ij}$ , and $\partial_t K_{ij}$ appears only in the spatial-spatial equations $G^{ij}=8\pi T^{ij}$ . Why: this pins the accelerations to the six tangential equations, leaving the four normal projections as candidates for constraints.

Step 2. Apply the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ , i.e. $\partial_t G^{0\nu} = -\partial_i G^{i\nu} - (\text{connection terms})$ . Why: this expresses the time derivative of the normal components $G^{0\nu}$ entirely through spatial derivatives of the other components, so $G^{0\nu}$ cannot itself contain the highest time derivative — that would make its own time derivative one order too high.

Step 3. Conclude that $G^{\mu\nu}n_\mu$ is at most first order in $\partial_t\gamma_{ij}$ . The four equations with a normal index therefore relate quantities on a single slice — one Hamiltonian and three momentum constraints — while $G^{ij}=8\pi T^{ij}$ carry the dynamics. The metric-component tally follows: $10 - 4\,(\text{gauge: lapse, shift}) - 4\,(\text{constraints}) = 2$ physical degrees of freedom. Why: this is the exact parallel of Maxwell theory, where the two Gauss laws are first order and the curl equations propagate the two photon polarizations; here four constraints and four gauge functions leave the two graviton polarizations.

Go deeper · Derivation C · 1 minThe Lichnerowicz–York equation from a conformal rescaling

Derivation CThe Lichnerowicz–York equation from a conformal rescaling

The aim is to show how the nonlinear Hamiltonian constraint (1) is tamed into the solvable elliptic equation (5) by writing the physical metric as a conformal factor times a freely chosen background, $\gamma_{ij}=\psi^4\tilde\gamma_{ij}$ . This is the single most consequential manipulation in initial-data construction: it isolates the "constrained" scalar $\psi$ from the freely specifiable conformal geometry, and it converts a nonlinear algebraic-differential entanglement into a Laplace-type problem with a mature existence theory. We use the three-dimensional conformal transformation of the Ricci scalar and the York rescaling of the trace-free extrinsic curvature.

Step 1. Split $K_{ij}$ into its trace-free and trace parts, $K_{ij}=A_{ij}+\tfrac13\gamma_{ij}K$ . Then a short computation gives $K^2-K_{ij}K^{ij}=\tfrac23 K^2 - A_{ij}A^{ij}$ , so the constraint reads ${}^{(3)}R + \tfrac23 K^2 - A_{ij}A^{ij}=16\pi\rho$ . Why: separating the trace is what produces the characteristic $\tfrac23$ coefficient and isolates the shear $A_{ij}$ , the quantity the momentum constraint controls.

Step 2. Insert the conformal identity for the three-dimensional Ricci scalar, ${}^{(3)}R = \psi^{-4}{}^{(3)}\tilde R - 8\psi^{-5}\tilde D^2\psi$ , and the York rescaling $A_{ij}A^{ij}=\psi^{-12}\tilde A_{ij}\tilde A^{ij}$ (from $A^{ij}=\psi^{-10}\tilde A^{ij}$ ). Why: these two substitutions move all the $\psi$ -dependence into explicit powers, so the unknown scalar appears through a Laplacian and a set of source powers rather than buried inside curvature.

Step 3. Multiply through by $-\psi^5$ and collect. The result is $8\tilde D^2\psi - \psi\,{}^{(3)}\tilde R - \tfrac23\psi^5 K^2 + \psi^{-7}\tilde A_{ij}\tilde A^{ij} = -16\pi\psi^5\rho$ , equation (5). On a maximal slice $K=0$ and, choosing a flat background $\tilde\gamma_{ij}=\delta_{ij}$ in vacuum, this reduces to Laplace's equation $\nabla^2\psi=0$ . Why: this final form has a positive-definite structure guaranteeing (under standard conditions) a unique positive solution $\psi$ , and the vacuum reduction to $\nabla^2\psi=0$ is exactly the equation whose solution $\psi=1+M/(2r)$ is the Schwarzschild slice of the worked example below. *The $\tfrac23$ coefficient and the $\psi^{-7}$ , $\psi^5$ powers in equation (5) follow directly from the conformal splitting carried out in Steps 1–3 above.*

Limiting cases

The four constraints collapse to something far simpler in the slicings and symmetries that dominate practice.

Case	Result	Meaning
Moment of time symmetry	$K_{ij}=0\ \Rightarrow\ {}^{(3)}R = 16\pi\rho$	Momentum constraint holds identically; the Hamiltonian constraint becomes a single equation for $\gamma_{ij}$ . Time-symmetric Schwarzschild and Brill data live here.
Vacuum	$\rho=0,\ S^i=0$	Sources vanish: ${}^{(3)}R+K^2-K_{ij}K^{ij}=0$ and $D_j(K^{ij}-\gamma^{ij}K)=0$ . Black-hole and gravitational-wave initial data.
Maximal slicing	$K\equiv\gamma^{ij}K_{ij}=0$	The conformal momentum constraint becomes linear and decouples from $\psi$ ; foundation of the conformal transverse-traceless method.
Conformally flat, momentarily static	$\gamma_{ij}=\psi^4\delta_{ij},\ K_{ij}=0$	Hamiltonian constraint reduces to $\nabla^2\psi=0$ ; Brill–Lindquist punctures $\psi=1+\sum_a M_a/(2\lvert\mathbf x-\mathbf x_a\rvert)$ are exact solutions.

Common trapThe constraints are just a gauge choice you can ignore

They are not, on either count. First, the constraints are physical requirements, not coordinate conventions. The genuine gauge freedom of the 3+1 split lives entirely in the lapse $N$ and shift $N^i$ — four functions you may pick at will. The constraints are the other four equations, and they bind the physical data $(\gamma_{ij},K_{ij})$ ; hand a code an arbitrary pair and you are not evolving Einstein's theory, you are evolving a different, unphysical spacetime. Second, "evolution preserves them so I can forget them" is only half true. Analytically the Bianchi identities guarantee that if $\mathcal C={}^{(3)}R+K^2-K_{ij}K^{ij}-16\pi\rho$ and $\mathcal C^i$ vanish initially they vanish forever — the constraint residuals obey a closed homogeneous system. But a finite-difference code seeds tiny nonzero residuals every step, and in many naive formulations these grow exponentially and wreck the run. This is exactly why constraint-damping formulations (BSSN, Z4c, generalized harmonic) exist: they add terms that drive $\mathcal C$ and $\mathcal C^i$ back toward zero rather than letting them blow up.

Worked examples

1 · The Schwarzschild slice as a puncture (fully worked)

Take a moment of time symmetry, $K_{ij}=0$ , in vacuum, and look for conformally flat data $\gamma_{ij}=\psi^4\delta_{ij}$ . Solve the constraints, find the areal radius of the throat, read off the mass, and convert to a solar-mass hole ( $G=c=1$ ).

Step 1. With $K_{ij}=0$ the momentum constraint $D_j(K^{ij}-\gamma^{ij}K)=8\pi S^i$ reads $0=0$ — satisfied identically. Only the Hamiltonian constraint remains, and with $\rho=0$ , $\tilde\gamma_{ij}=\delta_{ij}$ (so ${}^{(3)}\tilde R=0$ ) and $K=0$ , equation (5) collapses to $8\nabla^2\psi=0$ , i.e. the flat-space Laplace equation $\nabla^2\psi=0$ .

Step 2. The spherically symmetric harmonic solution that tends to $1$ at infinity and carries a single point source is $\psi=1+\dfrac{M}{2r}$ , with $r$ the isotropic radius. It is harmonic everywhere except the puncture $r=0$ : $\nabla^2(1)=0$ and $\nabla^2(1/r)=0$ for $r\neq0$ . The full initial data are $\gamma_{ij}=(1+M/2r)^4\,\delta_{ij}$ , $K_{ij}=0$ — a slice of the Schwarzschild geometry.

Step 3. Find the throat. The areal (Schwarzschild) radius is $R(r)=\psi^2 r=(1+M/2r)^2 r=r+M+\dfrac{M^2}{4r}$ . Minimizing, $\dfrac{dR}{dr}=1-\dfrac{M^2}{4r^2}=0\Rightarrow r=\dfrac{M}{2}$ . At that radius $\psi=1+\dfrac{M}{2\cdot M/2}=\mathbf{2}$ and $R_{\min}=\dfrac{M}{2}+M+\dfrac{M}{2}=\mathbf{2M}$ . The minimal surface sits at areal radius $2M$ — exactly the Schwarzschild horizon, with area $A=4\pi R_{\min}^2=\mathbf{16\pi M^2}$ .

Step 4. Read the mass and scale it. As $r\to\infty$ , $\psi=1+\dfrac{M}{2r}+\cdots$ , and the ADM mass is twice the coefficient of $1/r$ : $M_{\rm ADM}=2\cdot\dfrac{M}{2}=\mathbf{M}$ . For one solar mass the geometric length is $GM_\odot/c^2\approx1.477\ \mathrm{km}$ , so the throat areal radius is $2M\approx\mathbf{2.95\ km}$ — the familiar Schwarzschild radius, recovered purely from solving the constraints. Time-symmetric conformally-flat vacuum data reduce the four constraints to $\nabla^2\psi=0$ , with $\psi=1+M/2r$ , a throat at areal radius $2M$ , area $16\pi M^2$ , and ADM mass $M$ .

2 · Counting the constraints and the degrees of freedom

Show that the ten Einstein equations leave general relativity with two propagating degrees of freedom, and check the count against electromagnetism.

Step 1. Start from the ten components of $G_{\mu\nu}=8\pi T_{\mu\nu}$ . By Derivation 2 the four projections with a normal index carry no second time derivative — one Hamiltonian and three momentum constraints — so only six are evolution equations for $(\gamma_{ij},K_{ij})$ .

Step 2. Subtract the gauge freedom. The lapse $N$ and shift $N^i$ are four functions freely chosen at every point; they are pure coordinate freedom and carry no physics. Removing them: $10 - 4\,(\text{gauge}) - 4\,(\text{constraints}) = \mathbf{2}$ genuine dynamical degrees of freedom — the two polarizations $h_+$ and $h_\times$ of a gravitational wave.

Step 3. Check against electromagnetism, which runs on the identical logic. The potential $A_\mu$ has four components; Gauss's law $\nabla\!\cdot\!\mathbf E=\rho$ is one constraint, the gauge freedom $A_\mu\to A_\mu+\partial_\mu\lambda$ removes one more, leaving $4-1-1=\mathbf{2}$ photon polarizations. Gravity has four constraints and four gauge functions where electromagnetism has one and one — but both land on two physical polarizations, and in both the "extra" equations are constraints, not dynamics.

Physics of the portrait

The constraints have two histories running in parallel: the canonical one, in which Dirac and then Arnowitt, Deser and Misner recast gravity as a Hamiltonian system whose Hamiltonian is literally a sum of the constraints; and the geometric one, in which the same four equations are the embedding conditions for initial data — and it is the geometric line that made them solvable.

Origins · 1 minAndré Lichnerowicz and how this was discovered

André Lichnerowicz

1915–1998 · geometry of the Einstein initial-value constraints

André Lichnerowicz was a French differential geometer, trained in the tradition of Élie Cartan, who did more than anyone to turn the initial-value constraints of general relativity from an obstacle into a tractable problem. In 1944 he showed that the four constraints, opaque as a coupled nonlinear system in the raw metric, become manageable once the spatial metric is factored into a background times a positive conformal factor, $\gamma_{ij}=\psi^4\tilde\gamma_{ij}$ : the Hamiltonian constraint then reduces to a single quasi-linear elliptic equation for $\psi$ , the equation that now bears his name and sits at the heart of equation (5). His deeper contribution was a viewpoint — the constraints are not awkward leftovers but the precise conditions under which a Riemannian three-geometry can be embedded, with prescribed extrinsic curvature, into a spacetime solving Einstein's equations. That framing was completed by his student Yvonne Choquet-Bruhat, whose 1952 theorem proved constrained data evolve to a unique spacetime, and by James York in the 1970s, whose conformal transverse-traceless decomposition turned the insight into the algorithm every initial-data code still runs. Beyond relativity Lichnerowicz left his name on the Weitzenböck–Lichnerowicz formula and the Lichnerowicz Laplacian, and as a Bourbaki member chaired the commission that reshaped French mathematics teaching. The modern recipe — solve an elliptic equation for $\psi$ , then evolve — is his, eight decades on.

History

1944

— Lichnerowicz recasts the initial-value constraints as an elliptic equation for a conformal factor

1958

— Dirac casts general relativity in Hamiltonian form and isolates its four constraints

1962

— Arnowitt, Deser and Misner complete the 3+1 (ADM) formulation of the dynamics

1979

— York's conformal transverse-traceless method makes the constraints routinely solvable

2005

— the first stable binary-black-hole mergers evolve constraint-satisfying puncture data

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §21.4–21.6 · Wald, General Relativity (1984), §10.2 · Carroll, Spacetime and Geometry (2004), Appendix E (3+1) · Baumgarte & Shapiro, Numerical Relativity (2010), Ch. 2–3 · Gourgoulhon, 3+1 Formalism in General Relativity (2012), Ch. 4 · Arnowitt, Deser & Misner, "The Dynamics of General Relativity" (1962). Key consistency checks: equations (1)–(2) follow from double-contracting the Gauss equation and contracting the Codazzi equation; the degree-of-freedom count gives $10-4-4=2$ ; setting $K_{ij}=0$ reduces the Hamiltonian constraint to ${}^{(3)}R=16\pi\rho$ and, conformally flat in vacuum, to $\nabla^2\psi=0$ with $\psi=1+M/2r$ , throat at areal radius $2M$ , area $16\pi M^2$ , ADM mass $M$ ; the $\tfrac23$ coefficient in equation (5) comes from splitting the trace of $K_{ij}$ . Sign conventions throughout: $n_\mu n^\mu=-1$ , $K_{ij}=-\gamma^{\alpha}{}_i\gamma^{\beta}{}_j\nabla_\alpha n_\beta$ , $G=c=1$ ; the only numerical constant beyond the textbook $8\pi$ is that $\tfrac23$ .

Prerequisites. ADM 3+1 Decomposition of Spacetime · Gauss–Codazzi–Mainardi Equations · Einstein Field Equations. Leads to. Initial-Value (Cauchy) Problem in General Relativity · Conformal Method for Constraint Solving · ADM Hamiltonian Formulation · BSSN Formulation · Numerical Relativity and Binary Black Hole Mergers.

Ask the tutor: Ask me to walk through the Bowen–York momentum constraint, the Wheeler–DeWitt equation, or how BSSN actually damps constraint violations in a running simulation.

Are the Hamiltonian and momentum constraints the same thing as the ADM Hamiltonian?

Related but not identical. The ADM Hamiltonian is $H=\int (N\,\mathcal H + N^i\,\mathcal H_i)\,d^3x$ , with $\mathcal H$ the Hamiltonian-constraint density and $\mathcal H_i$ the momentum-constraint density. Because the lapse $N$ and shift $N^i$ are arbitrary, stationarity forces $\mathcal H=0$ and $\mathcal H_i=0$ — the constraints. So the constraints are the pieces the Hamiltonian is built from, and the striking consequence is that the bulk Hamiltonian of general relativity vanishes on physical solutions; all the energy lives in a boundary term at infinity, the ADM mass. This is the canonical route to the same four equations we derived from Gauss–Codazzi.

If evolution preserves the constraints, why do numerical codes worry about them at all?

Analytically the Bianchi identities guarantee preservation: the constraint residuals satisfy a closed homogeneous system, so zero stays zero. Numerically, though, every finite-difference or spectral step introduces truncation error that acts as a small nonzero source for those residuals. In many otherwise reasonable formulations — bare ADM is the classic example — these seeded violations grow exponentially and destroy the simulation within a few light-crossing times. The fix is not more resolution alone but a better formulation: BSSN, Z4c, and generalized harmonic gauge add terms that make the constraint-violating modes decay rather than grow. Codes monitor $\mathcal C$ and $\mathcal C^i$ at every step and trust a run only if they converge to zero at the expected order as the grid is refined.

Why is there a Hamiltonian constraint and a momentum constraint, rather than one kind?

Because the normal index of the Einstein tensor can be contracted in two geometrically distinct ways. Project $G_{\mu\nu}$ with two normals, $G_{\mu\nu}n^\mu n^\nu$ , and you get a single scalar equation — the Hamiltonian constraint, tied to energy density. Project with one normal and one tangential index, $\gamma^{i}{}_\mu G^{\mu}{}_\nu n^\nu$ , and you get a spatial vector — the three momentum constraints, tied to momentum density. The scalar controls the "amount" of curvature the slice must carry; the vector controls how that curvature shears and flows. They come as a set of four because the normal direction plus the three tangential directions exhaust the ways a single index can point relative to the slice.

What exactly is the conformal method, and why rescale the metric?

The raw constraints entangle all six components of $\gamma_{ij}$ and six of $K_{ij}$ nonlinearly, and it is unclear which pieces are free and which are determined. The conformal method, due to Lichnerowicz and completed by York, disentangles them. You freely specify a background conformal geometry $\tilde\gamma_{ij}$ and a trace-free tensor $\tilde A_{ij}$ — the "shape" and the gravitational-wave content — and let the constraints determine the rest: a scalar $\psi$ from the Hamiltonian constraint and a vector potential from the momentum constraint. Writing $\gamma_{ij}=\psi^4\tilde\gamma_{ij}$ is what makes the Hamiltonian constraint the well-posed elliptic equation (5). On a maximal slice ( $K=0$ ) the momentum part goes linear and decouples, so you solve it first, then feed it into the nonlinear equation for $\psi$ .

How do the constraints give the ADM mass and momentum?

Integrate a constraint over the slice and use the divergence theorem. The integrated Hamiltonian constraint becomes a surface integral at spatial infinity of metric derivatives — the ADM mass, the total energy measured by the far-field gravity. The momentum constraint likewise integrates to the ADM linear (and, with a rotational weighting, angular) momentum. So the constraints do double duty: locally they restrict the data, and globally, through their boundary terms, they define the conserved energy-momentum. This is why a merger's radiated energy is quoted as an initial-minus-final ADM mass — both endpoints are constraint integrals.

Does a "naked" pair of γ_ij and K_ij ever fail to be valid initial data?

Yes, and that is the whole point. Pick a random three-metric and a random symmetric $K_{ij}$ , and almost certainly ${}^{(3)}R+K^2-K_{ij}K^{ij}\neq16\pi\rho$ somewhere — the pair cannot be embedded in any spacetime solving Einstein's equations. The constraints are the exact obstruction. Unlike Newtonian gravity, where positions and velocities are free, general relativity's initial data live on a constraint submanifold. So you never guess $(\gamma_{ij},K_{ij})$ directly — you specify the freely choosable conformal pieces and solve the constraints for the rest, guaranteeing a valid slice.