both are defined in the fiducial flat space that one obtains setting the conformal factor to unity.

Without loss of generality we may assume that $P_a$ points along the positive z-axis, and has magnitude $P$. If we write $\widehat{K}_{ab}$ in spherical coordinates, the only non vanishing components are $$\begin{aligned}
\label{Ksphe}
\widehat{K}_{RR} &=& {3 P  \over R^2} \cos(\theta) \nonumber \\
\widehat{K}_{R\theta} &=& -{3 P  \over 2 R} \sin(\theta) = 
\widehat{K}_{\theta R}\nonumber \\
\widehat{K}_{\theta\theta} &=& -{3 P  \over 2} \cos(\theta) \nonumber \\
\widehat{K}_{\phi \phi} &=&  \sin(\theta)^2 \widehat{K}_{\theta\theta} 
\label{extriseca}\end{aligned}$$

If we write the extrinsic curvature in terms of tensor spherical harmonics [@ReWh], we see that it consists of a pure $\ell=1$ even term. This is reasonable, since the presence of momentum in an initial slice configuration is determined by the presence of a “dipole” term asymptotically, that makes the ADM integral $$P_i = {1 \over 8 \pi } \int_\infty K_{ij} d^2 S_j$$ nonvanishing.

From (\[extriseca\]), we have, $$\label{Kcuadrado}
\widehat{K}_{ab}\widehat{K}^{ab} = {9 P^2 \over 2 R^4} \left( 1 + 2
\cos^2 
\theta \right)$$

We now need to solve (\[Hamil\]), and this involves imposing boundary conditions on $\phi$. One possibility is that given by Bowen and York [@BoYo][^1]. In this case one chooses a certain 