research32 min studyLesson 842 in Physics

ADM Hamiltonian and momentum constraints

Test admissible spatial data and distinguish local constraint residuals from numerical convergence.

Before you begin

On this page

The question for today

What you’ll understand

Evaluate Hamiltonian and momentum residuals in the fixed curvature convention.
Explain why freely assigning metric and curvature generally fails the constraints.
Use a small derivative stencil without claiming a full Einstein solution.

A spatial metric and extrinsic curvature cannot be chosen independently at every grid point. Four projections of Einstein's equation constrain them on each slice. These Hamiltonian and momentum constraints are the relativistic counterpart of compatibility conditions in other field theories, but their nonlinear geometric terms make the initial-data problem especially coupled. Evolving plausible-looking fields is insufficient: their consistency must be checked against both the constraints and the matter assumptions.

Project the stress tensor and Einstein equation

Let E be the energy density measured by the future normal and S the spatial momentum density obtained by projecting minus the stress tensor contracted with that normal. With the module's negative Lie-derivative convention for K, the Hamiltonian constraint combines spatial scalar curvature, the square of the trace K, and the tensor squared norm. The momentum constraint differentiates the trace-adjusted extrinsic curvature using the spatial covariant derivative.

In geometric units E and S have inverse-length-squared units, matching both intrinsic curvature and a derivative of K. The two different squares in the Hamiltonian expression must not be interchanged: the square of a sum includes cross terms, whereas the squared tensor norm sums products with raised indices. Even constant extrinsic curvature can contribute nontrivially to the Hamiltonian constraint. Flat spatial coordinates therefore do not make the constraint automatically vanish.

\mathcal H=R+K^2-K_{ij}K^{ij}-16\pi E=0,\qquad \mathcal M^i=D_j(K^{ij}-\gamma^{ij}K)-8\pi S^i=0

R is the intrinsic three-dimensional scalar curvature. E=T(n,n) and S is minus the spatial projection of T contracted with n; these signs use signature minus-plus-plus-plus.↔ Scroll wide equations horizontally.

Separate continuum propagation from discrete residuals

In the continuum, the contracted Bianchi identity and compatible matter equations give a homogeneous propagation system for the constraints when the evolution equations hold. Satisfying the constraints initially and imposing suitable incoming boundary conditions can then preserve them in the appropriate well-posed setting. This argument does not say every formulation or arbitrary boundary prescription has satisfactory numerical behavior. The propagation system and its characteristic fields must actually be analyzed.

A finite grid introduces truncation, interpolation, roundoff, and iterative-solver errors. A computed constraint residual includes those effects and the derivative operator used to measure it. Refining the grid while keeping the same inconsistent continuum data can leave a nonzero limiting residual. Conversely, a small residual in one norm need not certify accurate wave phase, correct boundary conditions, or the absence of unresolved localized errors. Constraint monitoring is necessary evidence with a stated normalization, rather than a complete accuracy certificate.

K_{ij}=k\gamma_{ij}\ \Longrightarrow\ K=3k,\qquad K_{ij}K^{ij}=3k^2,\qquad K^2-K_{ij}K^{ij}=6k^2

The example is locally isotropic extrinsic curvature in three spatial dimensions. It is an algebraic identity, not a complete matter model or cosmological solution.↔ Scroll wide equations horizontally.

Use local stencils to diagnose which condition fails

In a flat Cartesian metric, the covariant derivative reduces to the ordinary derivative, making a small diagnostic stencil easy to audit. At the center of a trace-free diagonal curvature profile the Hamiltonian residual can vanish because every curvature component is zero there. A nonzero derivative through neighboring points can nevertheless violate the momentum constraint. Checking only the Hamiltonian projection would miss this inconsistency completely.

A discrete derivative also needs its grid spacing and coordinate units. The centered difference uses two neighboring values divided by twice the spacing; forgetting that factor produces a momentum residual wrong by two. At boundaries the stencil changes and can have a different error order. In an actual curved metric, connection terms and index raising must be included consistently. The following finite examples expose algebra and stencil logic while deliberately stopping short of constructing globally admissible data.

\partial_x f(0)\approx\frac{f(h)-f(-h)}{2h},\qquad \|\mathcal H\|_\infty=\max_{\rm grid}|\mathcal H|

The first formula is a centered derivative with second-order truncation error for a smooth profile. The second illustrates one specified residual norm; coordinate units and computational domain must also be stated.↔ Scroll wide equations horizontally.

Work it through

Flat geometry with isotropic extrinsic curvature

At a flat Cartesian point let $K_{ij}=k\delta_{ij}$ with constant $k=0.5/L$ and spatially constant matter momentum zero. Find the energy density required by the Hamiltonian constraint and the momentum residual. What happens if the same data are called vacuum?

The trace is $K=1.5/L$ and the squared norm is $3(0.5)^2/L^2=0.75/L^2$ . Thus the difference of the two squared terms is $2.25-0.75=1.5$ in inverse-L-squared units.
With R zero, the Hamiltonian equation requires $16\pi E=1.5/L^2$ , so $E=3/(32\pi L^2)$ . Constant curvature and zero matter momentum give zero momentum residual in this flat Cartesian diagnostic.
Calling the same geometry vacuum sets E to zero and leaves $\mathcal H=1.5/L^2$ . Supplying the required local energy is a constraint calculation; it does not establish a global matter solution or its subsequent evolution.

The result

The required local density is three over 32 pi L squared, the momentum residual vanishes, and the vacuum interpretation fails the Hamiltonian condition.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A derivative can carry independent information

Reset and set coefficient one, position zero; predict quadratic output zero and slope zero. Compare a fresh Reset run at coefficient one, position one; predict output one and slope two. The displayed derivative is ordinary scalar calculus, not the covariant momentum constraint or a test of initial spacetime data.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forValues and derivatives supply different diagnostics; the lab does not evaluate Einstein constraints.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The momentum constraints contain spatial derivatives of the trace-adjusted curvature and matter momentum. A pointwise Hamiltonian cancellation supplies none of that derivative information.

Your turn

Try it yourself

On a flat Cartesian stencil with $h=0.5L$ , set $K_{xx}=f$ , $K_{yy}=-f$ , and other components zero. Let f at minus h, zero, plus h be respectively $-0.1/L,0,0.1/L$ . In vacuum, find the centered momentum residual in x and the Hamiltonian residual at the center.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The trace vanishes throughout the stencil. At its center all curvature components vanish, but their derivative need not.

Compare with the solution

The trace is zero. The centered derivative of K-xx is $(0.1-(-0.1))/(2\times0.5)L^{-2}=0.2/L^2$ , so $\mathcal M^x=0.2/L^2$ . At the central point R and every K component are zero, giving Hamiltonian residual zero. These are prescribed toy samples, not vacuum initial data: the independent momentum diagnostic exposes their inconsistency even though the central Hamiltonian test passes.

Keep the idea

Take it with you

The four constraints restrict allowable initial geometry and matter.
Trace squared and tensor norm squared are different contractions.
Constraint preservation and numerical accuracy require more than one small residual.

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