Extrinsic Curvature and Metric Evolution

Extrinsic curvature is the normal velocity of the spatial metric, with a sign fixed by convention. For the convention used here, expansion toward the future gives negative $K_{ij}$. Once lapse and shift are restored, the coordinate-time change of $\gamma_{ij}$ separates cleanly into normal geometric evolution and a spatial Lie derivative caused only by relabeling.

Required background. Hypersurface Normals and Tangential Projectors supplies the induced connection and projections, while The Einstein Equation supplies the dynamics for which $K_{ij}$ becomes initial and evolved data.

Helpful background. The Gibbons-Hawking-York Boundary Term shows why normal metric derivatives and extrinsic curvature enter the gravitational variational problem.

Extrinsic curvature as an embedding tensor

Let $n^\mu$ be the future unit normal and $\gamma^\mu{}_\nu=\delta^\mu{}_\nu+n^\mu n_\nu$ the spatial projector. This volume fixes

$$K_{\mu\nu} =-\gamma_\mu{}^\rho\gamma_\nu{}^\sigma\nabla_\rho n_\sigma.$$

Both indices are spatial. Hypersurface orthogonality makes $K_{\mu\nu}$ symmetric. For tangent vectors $X^\mu$ and $Y^\mu$, metric compatibility gives

$$(\mathcal L_n g)(X,Y) =g(\nabla_Xn,Y)+g(X,\nabla_Yn).$$

The normal terms in $\gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu$ vanish when evaluated on $X$ and $Y$. Symmetry of $K$ then yields

$$(\mathcal L_n\gamma)(X,Y)=-2K(X,Y),$$

or equivalently

$$K_{ij}=-\frac12\mathcal L_n\gamma_{ij}.$$

This equation fixes the sign before any coordinate calculation. Sources that define $K_{ij}=+\tfrac12\mathcal L_n\gamma_{ij}$ describe the same embedding with the opposite tensor; every term odd in $K_{ij}$ must then be translated, not just the definition.

The trace and trace-free parts are

$$K=\gamma^{ij}K_{ij}, \qquad A_{ij}=K_{ij}-\frac13K\gamma_{ij}.$$

The trace measures local volume expansion of the normal congruence, while $A_{ij}$ measures shear of the slice geometry.

Coordinate-time metric evolution

The coordinate evolution vector is fixed as

$$(\partial_t)^\mu=\alpha n^\mu+\beta^\mu.$$

After projecting all free indices onto a slice, normal evolution is represented by $\partial_t-\mathcal L_\beta=\alpha\operatorname{proj}_{\Sigma}\mathcal L_n$. For the spatial metric, with its indices pulled back to the slice, this gives the kinematic evolution equation

$$(\partial_t-\mathcal L_\beta)\gamma_{ij} =-2\alpha K_{ij}.$$

Because $D_k\gamma_{ij}=0$,

$$(\mathcal L_\beta\gamma)_{ij} =D_i\beta_j+D_j\beta_i.$$

Therefore the coordinate formula for extrinsic curvature is

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

The shift-dependent terms remove coordinate deformation within a slice. It is incorrect to infer nonzero extrinsic curvature from $\partial_t\gamma_{ij}$ alone when the shift is nonzero.

Taking the trace gives the spatial-volume law. For the volume form $\boldsymbol\epsilon_\Sigma=\sqrt\gamma\,d^3x$,

$$(\partial_t-\mathcal L_\beta)\boldsymbol\epsilon_\Sigma =-\alpha K\boldsymbol\epsilon_\Sigma.$$

In a fixed coordinate chart this is

$$\partial_t\ln\sqrt\gamma =-\alpha K+D_i\beta^i.$$

Thus $K<0$ means that normal observers see the local spatial volume expand toward the future under the convention selected here. Gourgoulhon derives the Lie-derivative identity and its adapted-coordinate form in Gourgoulhon 2007, §3.3.5, pp. 45–46 and §4.3.1, pp. 59–60, PDF.

Flat slices and curved slices

For inertial Minkowski slices,

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

so $\partial_t\gamma_{ij}=0$ and hence $K_{ij}=0$. The same conclusion holds for a rigidly rotating spatial grid: $\partial_t\gamma_{ij}=0$ and $\mathcal L_\beta\gamma_{ij}=0$ because the rotation is a spatial Killing field.

By contrast, the Milne foliation of the same flat spacetime has

$$\gamma_{ij}=\tau^2q_{ij}, \qquad \alpha=1, \qquad \beta^i=0,$$

where $q_{ij}$ is the unit hyperbolic metric. Direct differentiation gives

$$\partial_\tau\gamma_{ij}=2\tau q_{ij} =\frac2\tau\gamma_{ij},$$

and therefore

$$K_{ij}=-\frac1\tau\gamma_{ij}, \qquad K=-\frac3\tau.$$

The volume element satisfies $\sqrt\gamma=\tau^3\sqrt q$ and $\partial_\tau\ln\sqrt\gamma=3/\tau=-K$, exactly as the trace equation requires. Nonzero $K_{ij}$ here records how hyperboloids sit inside Minkowski spacetime; it is not four-dimensional curvature.

A complementary warning comes from static Schwarzschild-time slices. Their spatial metric is intrinsically curved, but with zero shift and no time dependence they have $K_{ij}=0$. Intrinsic curvature and extrinsic curvature answer different questions.

Common pitfalls

Do not import the opposite sign selectively. Changing the sign of $K_{ij}$ also changes the visible signs of momentum constraints, canonical momenta, junction formulas, and every term linear in $K_{ij}$.

Coordinate-time change is not normal change. The spatial Lie derivative $\mathcal L_\beta\gamma_{ij}$ must be removed before reading off the embedding velocity.

Nonzero extrinsic curvature does not imply curved spacetime. Milne hyperboloids have $K_{ij}\ne0$ in a flat ambient geometry.

Exercise: an expanding flat spatial metric

Consider

$$ds^2=-\alpha(t)^2dt^2+a(t)^2\delta_{ij}dx^idx^j$$

with zero shift. Derive $K_{ij}$ and $K$, and verify the volume-evolution equation. Do not set $\alpha=1$ during the derivation.

Solution

The spatial metric is $\gamma_{ij}=a^2\delta_{ij}$, so

$$\partial_t\gamma_{ij} =2a\dot a\,\delta_{ij} =2\frac{\dot a}{a}\gamma_{ij}.$$

With $\beta^i=0$,

$$K_{ij} =-\frac1{2\alpha}\partial_t\gamma_{ij} =-\frac{\dot a}{\alpha a}\gamma_{ij}, \qquad K=-3\frac{\dot a}{\alpha a}.$$

Since $\sqrt\gamma=a^3$,

$$\partial_t\ln\sqrt\gamma=3\frac{\dot a}{a} =-\alpha K.$$

The lapse converts coordinate-time expansion into expansion per unit normal proper time. For $a(t)=t$ and $\alpha=1$ with hyperbolic rather than Euclidean unit spatial metric, this sign agrees with the Milne result.

References

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