The ADM Action

The ADM action is the Einstein–Hilbert action reorganized as a time integral over spatial metrics. Its bulk Lagrangian contains the intrinsic scalar curvature of a slice and a quadratic kinetic term in its extrinsic curvature. Lapse and shift enter without time derivatives, foreshadowing their role as multipliers of constraints rather than propagating canonical fields.

Required background. Presymplectic Potentials and Currents supplies the passage from a Lagrangian variation to canonical data, while The Gibbons-Hawking-York Boundary Term supplies the boundary term needed when induced metrics are fixed.

Helpful background. The Einstein Equation supplies the covariant field equation, Hypersurface Normals and Tangential Projectors supplies the slice geometry, and Formulations of General Relativity places the canonical description among equivalent continuum formulations.

The spacetime action and its foliation

Consider a region bounded initially and finally by spacelike slices and, when present, by a timelike side boundary. The gravitational action starts from

$$S_{\rm grav} =\frac1{16\pi G} \int_{\mathcal M}d^4x\sqrt{-g} \left({}^{(4)}R-2\Lambda\right) +S_{\rm boundary}.$$

Use the chapter conventions

$$(\partial_t)^\mu=\alpha n^\mu+\beta^\mu, \qquad \gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu, \qquad K_{ij}=-\frac12\mathcal L_n\gamma_{ij}.$$

The metric determinant factors as

$$\sqrt{-g}=\alpha\sqrt\gamma.$$

The remaining task is to rewrite the four-dimensional scalar curvature. The Gauss relation supplies its spatial part, while the two-normal Ricci relation supplies a normal derivative of $K$ and a spatial divergence involving the lapse.

Separating bulk and divergence terms

For the selected sign of $K_{ij}$, the curvature identity can be arranged as

$$\begin{aligned} \sqrt{-g}\,{}^{(4)}R ={}&\alpha\sqrt\gamma \left( {}^{(3)}R+K_{ij}K^{ij}-K^2 \right)\\ &-2\partial_t(\sqrt\gamma K) +2\partial_i \left[ \sqrt\gamma (K\beta^i-D^i\alpha) \right]. \end{aligned}$$

Here $D_i$ is the spatial Levi–Civita derivative and $\gamma=\det(\gamma_{ij})$. The volume element is a density, so its Lie derivative is $\mathcal L_\beta\sqrt\gamma=\partial_i(\beta^i\sqrt\gamma)$, not merely $\beta^i\partial_i\sqrt\gamma$. The split can be checked from $\partial_\perp\sqrt\gamma=-\alpha K\sqrt\gamma$, where $\partial_\perp=\partial_t-\mathcal L_\beta$. The normal derivative of $K$ first produces a $+K^2$ term; integration by parts through the evolving volume element changes the surviving bulk combination to $K_{ij}K^{ij}-K^2$.

The total time derivative is precisely the second-normal-derivative content that the Gibbons–Hawking–York terms on the initial and final caps cancel for a Dirichlet variational problem. The spatial divergence combines with the action on a timelike side boundary. It may not be discarded on a finite domain or at an asymptotic end: after the Legendre transform it becomes part of the surface Hamiltonian.

With the cap term treated and the still-relevant side-boundary contribution denoted by $S_{\partial\Sigma}$, the ADM form is

$$S_{\rm grav} =\frac1{16\pi G} \int dt\int_{\Sigma_t}d^3x\, \alpha\sqrt\gamma \left( {}^{(3)}R+K_{ij}K^{ij}-K^2-2\Lambda \right) +S_{\partial\Sigma}.$$

Gourgoulhon derives the scalar-curvature split and bulk action in Gourgoulhon 2007, §3.4.2, pp. 48–50 and §4.5.1, pp. 66–67, PDF. The original canonical construction, including the role of surface terms, is developed in Arnowitt, Deser, and Misner 1962/2004, §§3–5, PDF.

Where the velocities occur

In adapted coordinates,

$$K_{ij} =-\frac1{2\alpha} \left( \dot\gamma_{ij} -D_i\beta_j-D_j\beta_i \right).$$

Therefore the ADM bulk Lagrangian depends on $\dot\gamma_{ij}$ through $K_{ij}$ but contains neither $\dot\alpha$ nor $\dot\beta^i$. The Hessian with respect to all ten metric velocities is degenerate. In canonical language,

$$p_\alpha\equiv\frac{\partial\mathcal L}{\partial\dot\alpha}=0, \qquad p_i\equiv\frac{\partial\mathcal L}{\partial\dot\beta^i}=0$$

are primary constraints. Variation with respect to lapse and shift will yield the Hamiltonian and momentum constraints rather than evolution equations for $\alpha$ and $\beta^i$.

The kinetic combination is indefinite. Decompose

$$K_{ij}=A_{ij}+\frac13K\gamma_{ij}, \qquad \gamma^{ij}A_{ij}=0.$$

Then

$$K_{ij}K^{ij}-K^2 =A_{ij}A^{ij}-\frac23K^2.$$

The negative trace direction is the conformal mode of the DeWitt metric on spatial metrics. This indefiniteness is a structural property of the gravitational kinetic term; it is not evidence that coordinate time evolution is itself an instability.

Flat and hyperboloidal checks

For inertial Minkowski slices,

$${}^{(3)}R=0, \qquad K_{ij}=0, \qquad \Lambda=0,$$

so the ADM bulk density vanishes.

Milne slices of the same flat spacetime have

$${}^{(3)}R=-\frac6{\tau^2}, \qquad K_{ij}=-\frac1\tau\gamma_{ij}, \qquad K=-\frac3\tau.$$

Hence

$$K_{ij}K^{ij}-K^2 =\frac3{\tau^2}-\frac9{\tau^2} =-\frac6{\tau^2},$$

and the bulk integrand is $-12/\tau^2$, not zero. This does not contradict ${}^{(4)}R=0$: the omitted total derivative contributes $+12/\tau^2$ locally. Milne slicing therefore gives a decisive warning that a boundary term may vanish after integration under specified conditions but is not pointwise negligible.

Common pitfalls

Discarding a divergence requires boundary conditions. A total derivative can change the variational principle and the value of the Hamiltonian even though it does not change bulk Euler–Lagrange equations.

The ADM Lagrangian does not make lapse and shift dynamical. Their absence from the velocity Hessian is the origin of primary constraints.

A nonzero bulk density need not imply spacetime curvature. On Milne slices, the divergence terms complete the identity ${}^{(4)}R=0$.

Exercise: verify the Milne curvature split

For the Milne data above with $\alpha=1$ and $\beta^i=0$, evaluate the bulk term and the time derivative $-2\partial_\tau(\sqrt\gamma K)/\sqrt\gamma$. Show that their sum vanishes.

Solution

Since $\sqrt\gamma=\tau^3\sqrt q$ and $K=-3/\tau$,

$$\sqrt\gamma K=-3\tau^2\sqrt q,$$

so

$$-\frac2{\sqrt\gamma} \partial_\tau(\sqrt\gamma K) =-\frac2{\tau^3\sqrt q}(-6\tau\sqrt q) =\frac{12}{\tau^2}.$$

The bulk curvature combination is

$${}^{(3)}R+K_{ij}K^{ij}-K^2 =-\frac6{\tau^2}+\frac3{\tau^2}-\frac9{\tau^2} =-\frac{12}{\tau^2}.$$

The spatial divergence vanishes because the lapse is spatially constant and the shift is zero. Bulk plus time derivative is therefore zero, as required by flat four-dimensional spacetime.

References

• Arnowitt, Richard, Stanley Deser, and Charles W. Misner. “The Dynamics of General Relativity.” In Gravitation: An Introduction to Current Research, edited by Louis Witten, 227–265. Wiley, 1962; reprinted as arXiv:gr-qc/0405109 (2004). arXiv.
• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, arXiv:gr-qc/0703035v1 (2007). arXiv.