Kasner metric

Example

On $(0, \infty) \times \R^3$ with coordinates $(t, x, y, z)$, the Kasner metric with exponents $p_1, p_2, p_3 \in \R$ is $$g = -dt^2 + t^{2p_1} dx^2 + t^{2p_2} dy^2 + t^{2p_3} dz^2.$$ Apart from the trivial case $p_1 = p_2 = p_3 = 0$, which is flat Minkowski space, it is a vacuum solution exactly when the Kasner conditions hold: $$p_1 + p_2 + p_3 = 1, \qquad p_1^2 + p_2^2 + p_3^2 = 1.$$

Check: a computation of the Ricci curvature gives $$R_{tt} = \frac{(p_1 + p_2 + p_3) - (p_1^2 + p_2^2 + p_3^2)}{t^2}, \qquad R_{xx} = p_1 (p_1 + p_2 + p_3 - 1)\, t^{2p_1 - 2},$$ and similar formulas for $R_{yy}$ and $R_{zz}$, while the off-diagonal components vanish. All of them vanish under the Kasner conditions. Conversely, if all $R_{ii}$ vanish, then either every $p_i = 0$ or $p_1 + p_2 + p_3 = 1$, and then $R_{tt} = 0$ gives $p_1^2 + p_2^2 + p_3^2 = 1$.

Example: $(p_1, p_2, p_3) = (\frac23, \frac23, -\frac13)$. The sum is $1$ and the sum of squares is $\frac49 + \frac49 + \frac19 = 1$. Two directions expand while the third contracts. The exponents $(1, 0, 0)$ also satisfy the conditions; this case is flat, a piece of Minkowski space in Milne-type coordinates. If no exponent equals $1$, one exponent is negative and the other two are positive, and the curvature blows up as $t \to 0$, a gravitational singularity.