Kasner Spacetime

Kasner spacetime is the simplest anisotropic vacuum cosmology. Its spatial slices are flat, yet unequal directional scale factors generally generate nonzero Weyl curvature. The two Kasner constraints leave one essential parameter. At a generic nonflat Kasner point, $t\to0^+$ is a curvature singularity: the two directions with positive exponents contract while the direction with the negative exponent expands, and the behavior reverses when the solution is followed away from the singularity. The exceptional permutations of $(1,0,0)$ are flat; there $t=0$ is a Milne-type coordinate edge rather than a curvature singularity.

Required background. Einstein’s Equation supplies the vacuum equations; Continuum Kinematics and Balance Laws supplies expansion and shear.

Helpful background. The Riemann Curvature Tensor supplies curvature diagnostics; Exact Solutions of Einstein’s Equations supplies the homogeneous-family context; Killing Fields and Isometry Groups supplies Bianchi I translations.

Metric and vacuum constraints

On $t>0$, take

$$ds^2=-dt^2+t^{2p_1}dx^2+t^{2p_2}dy^2+t^{2p_3}dz^2,$$

where the exponents are constant. The directional Hubble rates are $H_i=p_i/t$. Direct calculation gives

$$R_{tt}=\frac{\sum_i p_i-\sum_i p_i^2}{t^2},$$

and, with no sum on $i$,

$$\frac{R_{ii}}{g_{ii}} =\frac{p_i(\sum_jp_j-1)}{t^2}.$$

For the nonstatic Kasner branch, $R_{ab}=0$ therefore requires

$$p_1+p_2+p_3=1, \qquad p_1^2+p_2^2+p_3^2=1.$$

These equations are the intersection of a plane and a unit sphere: the Kasner circle. Vacuum has not forced isotropy because the trace-free extrinsic curvature, or shear, is free gravitational data. The vanishing spatial Ricci tensor and nonzero shear jointly satisfy the Hamiltonian constraint.

Exponents and parameterization

From the constraints,

$$p_1p_2+p_2p_3+p_3p_1=0.$$

Except at permutations of $(1,0,0)$, one exponent is negative and two are positive. If all three were nonnegative, the vanishing sum of pairwise products would force at most one nonzero exponent, giving precisely the exceptional point. After ordering $p_1\leq p_2\leq p_3$, a useful parameter $u\geq1$ is

$$p_1=-\frac{u}{1+u+u^2}, \qquad p_2=\frac{1+u}{1+u+u^2}, \qquad p_3=\frac{u(1+u)}{1+u+u^2}.$$

Permuting spatial axes covers the entire circle. The locally rotationally symmetric point $(-1/3,2/3,2/3)$ occurs at $u=1$ up to permutation. Ellis and van Elst place Kasner within the Bianchi I dynamics in Ellis and van Elst 1999, §§4, Open PDF.

The spatial volume density scales especially simply:

$$\sqrt{\det h}=t^{p_1+p_2+p_3}=t.$$

Kasner is also self-similar. The vector

$$X=t\partial_t+\sum_i(1-p_i)x^i\partial_{x^i}$$

obeys $\mathcal L_Xg=2g$, so it is homothetic rather than Killing. This scaling explains why dimension-four curvature invariants carry a fixed $t^{-4}$ power.

Curvature and the flat exceptions

In an orthonormal frame, the independent tidal components scale as

$$R_{0i0i}=-\frac{p_i(p_i-1)}{t^2}, \qquad R_{ijij}=\frac{p_ip_j}{t^2}.$$

Hence

$$R_{abcd}R^{abcd} =\frac{4}{t^4}\left[ \sum_i p_i^2(p_i-1)^2 +\sum_{i<j}p_i^2p_j^2 \right].$$

For a generic point on the Kasner circle, this invariant diverges as $t\to0^+$, confirming a curvature singularity. At $(1,0,0)$ and its permutations, every term vanishes. The metric $-dt^2+t^2dx^2+dy^2+dz^2$ is a Milne wedge times a flat plane and is locally Minkowski; $t=0$ is then only the edge of the chosen coordinates.

The comoving congruence has

$$\theta=\sum_iH_i=\frac1t, \qquad \sigma^2=\frac12\sum_i(H_i-\theta/3)^2 =\frac{1}{3t^2}.$$

The fixed ratio $\sigma^2/\theta^2=1/3$ shows that generic Kasner expansion does not isotropize. Exact solvability also says nothing about the stability of a Kasner epoch once Bianchi spatial curvature or matter becomes important; type IX evolution, for example, can pass through sequences of approximate Kasner regimes.

Because $R_{ab}=0$, all nonzero curvature here is Weyl curvature. Comoving worldlines with fixed $(x,y,z)$ are timelike geodesics and reach $t=0$ after finite proper time. At a generic Kasner point, the divergent Kretschmann scalar forbids a $C^2$ extension through that endpoint. At the flat exceptional points, the invariant obstruction vanishes and ordinary Minkowski coordinates extend the Milne factor. Geodesic incompleteness and curvature singularity therefore agree generically but separate at the exceptional coordinate wedge.

Anisotropic coordinate rescalings change constant prefactors multiplying $t^{2p_i}$ but not the exponents. Permuting axes changes their order, not the local isometry class. These simple equivalences should be removed before treating two exponent triples as distinct solutions.

Exercise

Verify that the $u$-parameterization satisfies both Kasner constraints.

Solution

With $D=1+u+u^2$, the numerator of the sum is $-u+(1+u)+u(1+u)=D$. For the sum of squares, the numerator is $u^2+(1+u)^2+u^2(1+u)^2=D^2$. Division by $D$ and $D^2$ gives one in each case.

References

• Ellis, George F. R., and Henk van Elst. “Cosmological Models.” In Theoretical and Observational Cosmology, edited by Marc Lachièze-Rey, 1–116. Dordrecht: Springer, 1999. DOI; Open PDF.
• Stephani, Hans, Dietrich Kramer, Malcolm MacCallum, Cornelius Hoenselaers, and Eduard Herlt. Exact Solutions of Einstein’s Field Equations. 2nd ed. Cambridge: Cambridge University Press, 2003, §§13.1. DOI.