Canonical Momentum of Spatial Geometry

The spatial metric $\gamma_{ij}$ is the configuration variable of canonical general relativity. Its conjugate momentum is not simply proportional to the extrinsic curvature: the trace is reversed, and with the site’s convention $K_{ij}=-\tfrac12\mathcal L_n\gamma_{ij}$ the overall sign is fixed. Deriving that sign rather than importing it from another convention is essential for the constraints and Hamilton equations that follow.

Required background. Presymplectic Potentials and Currents supplies the relation between a boundary term in the variation and canonical momentum, while The Gibbons-Hawking-York Boundary Term supplies the Dirichlet action whose time-boundary variation is being decomposed.

Helpful background. The Einstein Equation supplies the covariant dynamics, Hypersurface Normals and Tangential Projectors supplies the geometry of $\Sigma_t$, and Formulations of General Relativity explains why the canonical and covariant descriptions encode the same classical theory.

Momentum from the metric velocity

The gravitational bulk Lagrangian density is

$$\mathcal L_{\mathrm{ADM}} =\frac{\alpha\sqrt\gamma}{16\pi G} \left({}^{(3)}R+K_{mn}K^{mn}-K^2-2\Lambda\right),$$

and the only dependence on $\dot\gamma_{ij}$ occurs through

$$K_{mn}=-\frac1{2\alpha} \left(\dot\gamma_{mn}-D_m\beta_n-D_n\beta_m\right).$$

At fixed $\gamma_{ij}$, lapse, and shift,

$$\frac{\partial K_{mn}}{\partial\dot\gamma_{ij}} =-\frac1{2\alpha}\delta^i{}_{(m}\delta^j{}_{n)}.$$

The derivative of the quadratic kinetic term is

$$\frac{\partial}{\partial K_{mn}} \left(K_{ab}K^{ab}-K^2\right) =2\left(K^{mn}-K\gamma^{mn}\right).$$

Combining the two signs gives

$$\boxed{ \pi^{ij} \equiv\frac{\partial\mathcal L_{\mathrm{ADM}}}{\partial\dot\gamma_{ij}} =\frac{\sqrt\gamma}{16\pi G} \left(K\gamma^{ij}-K^{ij}\right) }.$$

Thus $\pi^{ij}$ is a symmetric contravariant tensor density of weight one. This means that $\pi^{ij}d^3x$, rather than $\pi^{ij}$ alone, supplies the invariant measure factor in the canonical integral. The trace $\pi=\gamma_{ij}\pi^{ij}$ is also a density; lowering indices does not remove the factor $\sqrt\gamma$. The opposite definition of $K_{ij}$, common in some canonical references, reverses this formula. The derivation is given for the present convention in Gourgoulhon 2007, §4.5.2, pp. 67–69, PDF, while the canonical variables are developed in Arnowitt, Deser, and Misner 1962/2004, §§3–4, PDF.

Inverting the trace reversal

In $d$ spatial dimensions, trace the momentum relation:

$$\pi\equiv\gamma_{ij}\pi^{ij} =\frac{\sqrt\gamma}{16\pi G}(d-1)K.$$

For $d>1$, this determines $K$ and hence the full extrinsic curvature:

$$K_{ij} =\frac{16\pi G}{\sqrt\gamma} \left(\frac{\pi}{d-1}\gamma_{ij}-\pi_{ij}\right).$$

In three spatial dimensions,

$$K_{ij} =\frac{16\pi G}{\sqrt\gamma} \left(\frac12\pi\gamma_{ij}-\pi_{ij}\right).$$

The map fails to be invertible in $d=1$ because $K_{ij}K^{ij}-K^2$ vanishes identically there. This is the canonical trace of the fact that two-dimensional Einstein gravity has no local graviton dynamics.

For later use, the divergence of this weight-one momentum is

$$\begin{aligned} D_j\pi^{ij} &=\partial_j\pi^{ij}+\Gamma^i{}_{jk}\pi^{kj},\\ D_j\pi^j{}_i &=\partial_j\pi^j{}_i-\Gamma^k{}_{ji}\pi^j{}_k. \end{aligned}$$

The connection trace that acts on the contracted tensor index cancels the weight-one density term. Consequently the divergence of a weight-one vector density is an ordinary coordinate divergence, which is the integration by parts used in the momentum constraint.

The canonical one-form on a slice is

$$\Theta_\Sigma=\int_\Sigma d^3x\,\pi^{ij}\delta\gamma_{ij},$$

so the fundamental bracket is

$$\{\gamma_{ij}(x),\pi^{kl}(y)\} =\delta^k{}_{(i}\delta^l{}_{j)}\delta^{(3)}(x,y).$$

No momenta can be recovered from $\dot\alpha$ or $\dot\beta^i$, because the ADM Lagrangian contains neither velocity. Their conjugate momenta therefore vanish as primary constraints; this is a distinct degeneracy from the invertible trace reversal on spatial metrics.

Flat-slicing checks

On inertial slices of Minkowski spacetime, $K_{ij}=0$, so $\pi^{ij}=0$. Milne slices are also embedded in Minkowski spacetime but have $K_{ij}=-\gamma_{ij}/\tau$ and $K=-3/\tau$. They therefore give

$$\pi^{ij} =-\frac{\sqrt\gamma}{8\pi G\tau}\gamma^{ij}, \qquad \pi=-\frac{3\sqrt\gamma}{8\pi G\tau}.$$

Substitution into the inverse formula returns $K_{ij}=-\gamma_{ij}/\tau$. The nonzero momentum records how the hyperboloidal slice sits in spacetime, not four-dimensional curvature.

Common pitfalls

Importing the sign from a different convention. The sign of $\pi^{ij}$ is tied to the sign in the definition of $K_{ij}$; the pair must be changed together.

Treating $\pi^{ij}$ as an ordinary tensor. It contains $\sqrt\gamma$ and is therefore a density. This matters in its covariant divergence and in integrals over the slice.

Exercise: recover the velocity in arbitrary dimension

Starting from the $d$-dimensional momentum relation, solve for $\dot\gamma_{ij}$ in terms of $\pi^{ij}$, $\alpha$, and $\beta^i$.

Solution

The trace gives

$$K=\frac{16\pi G}{(d-1)\sqrt\gamma}\pi.$$

Substitution into the untraced relation yields

$$K_{ij}=\frac{16\pi G}{\sqrt\gamma} \left(\frac{\pi}{d-1}\gamma_{ij}-\pi_{ij}\right).$$

Finally, rearrange the defining relation for $K_{ij}$:

$$\dot\gamma_{ij} =D_i\beta_j+D_j\beta_i -\frac{32\pi G\alpha}{\sqrt\gamma} \left(\frac{\pi}{d-1}\gamma_{ij}-\pi_{ij}\right).$$

For $d=3$, this is $\dot\gamma_{ij}=\mathcal L_\beta\gamma_{ij}-2\alpha K_{ij}$, exactly the metric-evolution equation with the site’s extrinsic-curvature sign.

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.