# Bowen-York Solution

The freely specifiable data in the conformal transverse-traceless decomposition of the constraints is

$\begin{array}{rcl} {\tilde g}_{ab} & & {\rm conformal\ metric} \\ {\tilde M}_{ab} & & {\rm symmetric\ trace\ free\ tensor} \\ K & & {\rm trace\ of\ the\ extrinsic\ curvature} \end{array}$

The physical metric is $g_{ab} = \psi^4{\tilde g}_{ab}$. Let ${\tilde D}_a$ denote the covariant derivative built from the conformal metric. (Indices are raised with the inverse conformal metric.) Let $D_a$ denote the covariant derivative build from the physical metric.

Choose maximal slicing ($K=0$), conformal flatness ($\tilde g_{ab}$ is a flat metric) and $\tilde M_{ab} = 0$. With

$K^{ab} = \psi^{-10} {\tilde A}^{ab}$

the momentum constraint $D_a K^{ab} = 0$ can be written as

${\tilde D}_a {\tilde A}^{ab} = 0$

This is solved by

${\tilde A}^{ab} = {\tilde D}^a V^b + {\tilde D}^b V^a - \frac{2}{3} {\tilde g}^{ab} {\tilde D}_c V^c$

if the vector $V^a$ satisfies

${\tilde D}_b{\tilde D}^b V^a + \frac{1}{3} {\tilde D}^a {\tilde D}_b V^b = 0$

The Bowen-York solution is

$V^a = -\frac{1}{4r} (7P^a + n^a n_b P^b) + \frac{1}{r^2} \epsilon^{abc}n_b S_c$

which yields

${\tilde A}^{ab} = \frac{3}{2r^2} [P^a n_b + P^b n^a - (g^{ab} - n^a n^b) n_c P^c ] + \frac{3}{r^3} [\epsilon^{acd}S_c n_d n^b + \epsilon^{bcd} S_c n_d n^a ]$

In these expressions the components of $P^a$ and $S^a$ are constants in Cartesian coordinates. Also, $r = \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2}$ is the coordinate distance from the origin, $n^a = x^a/r$ is the radial normal vector, and $\epsilon^{abc}$ is the alternating symbol. Indices on these quantities are lowered with the conformal metric.