ADM Metric Decomposition

The ADM metric is not a new spacetime geometry: it is the covariant metric written in variables adapted to a spacelike foliation. A positive lapse $\alpha$, spatial shift $\beta^i$, and positive-definite spatial metric $\gamma_{ij}$ reconstruct every component of $g_{\mu\nu}$. Their inverse and determinant follow algebraically, and they make clear that the coordinate time lines need not be causal even though the slices are spacelike.

Required background. Hypersurface Normals and Tangential Projectors supplies $n^\mu$ and $\gamma_{\mu\nu}$, while The Einstein Equation supplies the four-dimensional metric whose components are being reorganized.

Helpful background. The Gibbons-Hawking-York Boundary Term explains why fixing the induced metric is a natural gravitational boundary condition.

Reconstructing the line element

In coordinates $(t,x^i)$ adapted to the foliation, the coordinate basis vector splits as

$$(\partial_t)^\mu=\alpha n^\mu+\beta^\mu, \qquad \beta^\mu n_\mu=0.$$

A coordinate displacement is therefore

$$dx^\mu=(\partial_t)^\mu dt+(\partial_i)^\mu dx^i =\alpha n^\mu dt+(\partial_i)^\mu(dx^i+\beta^i dt).$$

The normal and spatial pieces are orthogonal. Since $g(n,n)=-1$ and the restriction of $g$ to a slice is $\gamma_{ij}$, taking the squared interval gives

$$ds^2=-\alpha^2dt^2 +\gamma_{ij}(dx^i+\beta^i dt)(dx^j+\beta^jdt).$$

Expanding this expression yields the covariant components

$$g_{tt}=-\alpha^2+\beta_i\beta^i, \qquad g_{ti}=\beta_i, \qquad g_{ij}=\gamma_{ij},$$

where $\beta_i=\gamma_{ij}\beta^j$. Thus $g_{tt}$ alone is not the negative lapse squared: it also contains the coordinate motion within the slice.

The same split fixes the normal covector immediately. Because a vector tangent to $\Sigma_t$ has $dt=0$ and the future normal must pair negatively with $\partial_t$, one has

$$n_\mu=(-\alpha,0,0,0), \qquad n^\mu=\frac1\alpha(1,-\beta^i).$$

The second formula follows either by raising the index or directly from $n=(\partial_t-\beta)/\alpha$. It verifies both $n^\mu n_\mu=-1$ and $n^\mu\partial_\mu t=1/\alpha$.

Inverse metric and determinant

Write the metric in block form:

$$g_{\mu\nu} = \begin{pmatrix} -\alpha^2+\beta_k\beta^k & \beta_j\\ \beta_i & \gamma_{ij} \end{pmatrix}.$$

Solving $g^{\mu\rho}g_{\rho\nu}=\delta^\mu{}_\nu$ gives

$$g^{tt}=-\frac1{\alpha^2}, \qquad g^{ti}=\frac{\beta^i}{\alpha^2}, \qquad g^{ij}=\gamma^{ij}-\frac{\beta^i\beta^j}{\alpha^2}.$$

For example, the mixed product in the $t$-$j$ block is

$$g^{tt}g_{tj}+g^{ti}g_{ij} =-\frac{\beta_j}{\alpha^2} +\frac{\beta^i\gamma_{ij}}{\alpha^2}=0,$$

while the $t$-$t$ product equals one. These two cancellations are a quick way to detect a wrong sign in $g^{ti}$.

The Schur complement of the spatial block is

$$g_{tt}-g_{ti}\gamma^{ij}g_{jt}=-\alpha^2.$$

Consequently, if $\gamma=\det(\gamma_{ij})$,

$$\det g=-\alpha^2\gamma, \qquad \sqrt{-g}=\alpha\sqrt\gamma.$$

The four-volume therefore factors into normal proper time and spatial volume: $d^4x\sqrt{-g}=dt\,\alpha\,d^3x\sqrt\gamma$. Gourgoulhon gives this component reconstruction in Gourgoulhon 2007, §4.2.3, pp. 57–59, PDF.

Causal and coordinate conditions

The ADM variables define a Lorentzian metric with spacelike $t$-slices when

$$\alpha>0 \quad\text{and}\quad \gamma_{ij}\text{ is positive definite}.$$

No inequality $\gamma_{ij}\beta^i\beta^j<\alpha^2$ is needed for the slices. That additional inequality only makes $\partial_t$ timelike:

$$g(\partial_t,\partial_t) =-\alpha^2+\gamma_{ij}\beta^i\beta^j.$$

This distinction matters near rotating-coordinate light cylinders and in coordinates with strong shift. It is also why a coordinate speed obtained from $dx^i/dt$ is not by itself a physical speed measured by the normal observers.

Minkowski and Milne checks

For inertial Minkowski coordinates,

$$\alpha=1, \qquad \beta^i=0, \qquad \gamma_{ij}=\delta_{ij}.$$

The component formulas give $g_{\mu\nu}=\operatorname{diag}(-1,1,1,1)$, $\det g=-1$, and $n=\partial_t$.

Inside the future light cone, Milne coordinates give

$$ds^2=-d\tau^2+\tau^2q_{ij}dx^idx^j,$$

where $q_{ij}$ is the unit hyperbolic metric. Here

$$\alpha=1, \qquad \beta^i=0, \qquad \gamma_{ij}=\tau^2q_{ij}, \qquad \sqrt{-g}=\tau^3\sqrt q.$$

The determinant changes because the hyperboloidal spatial volume grows, not because the four-dimensional curvature is nonzero. The coordinate transformation from $(T,R)$ to $(\tau,\chi)$ has precisely the Jacobian needed to produce this volume factor.

Common pitfalls

The lapse is not $\sqrt{-g_{tt}}$ when the shift is nonzero. The correct identity is $-g_{tt}=\alpha^2-\beta_i\beta^i$.

The inverse spatial block is not simply $\gamma^{ij}$. It contains the rank-one correction $-\beta^i\beta^j/\alpha^2$.

A spacelike coordinate thread does not spoil the foliation. The signature and slice character are controlled by positive $\alpha$ and positive-definite $\gamma_{ij}$.

Exercise: rotating Minkowski coordinates

In cylindrical coordinates set $T=t$ and $\Phi=\phi+\Omega t$. Show that the Minkowski metric has lapse $1$, shift $\beta^\phi=\Omega$, and spatial metric $dr^2+r^2d\phi^2+dz^2$. Compute its determinant and locate where $\partial_t$ becomes null.

Solution

Substitution of $d\Phi=d\phi+\Omega dt$ gives

$$ds^2 =-(1-\Omega^2r^2)dt^2 +2\Omega r^2dt\,d\phi +dr^2+r^2d\phi^2+dz^2.$$

Comparison with the ADM form gives $\alpha=1$, $\beta^\phi=\Omega$, and $\beta_\phi=r^2\Omega$. The spatial determinant is $\gamma=r^2$, so

$$\det g=-\alpha^2\gamma=-r^2,$$

independent of $\Omega$, as required for a mere coordinate rotation. Finally,

$$g(\partial_t,\partial_t)=-(1-\Omega^2r^2),$$

which vanishes at $r=1/\lvert\Omega\rvert$. The slices remain ordinary Euclidean hyperplanes on both sides of this coordinate light cylinder.

References

• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, arXiv:gr-qc/0703035v1 (2007). arXiv.