in a few papers I came across a statement that the Kerr-NUT metric

$g_{uu}=\rho\overline{\rho}(r^{2}-2mr-l^{2}+a^{2}\cos^{2}x)$

$g_{ur}=1$

$g_{uy}=-2\rho\overline{\rho}l\cos x(r^{2}-2mr-l^{2}+a^{2})+2\rho\overline{\rho}a(mr+l^{2})\sin^{2}x$

$g_{ry}=-a\sin^{2}x-2l\cos x$

$g_{xx}=-r^{2}-(l-a\cos x)^{2}$

$g_{yy}=\rho\overline{\rho}(r^{2}-2mr-l^{2}+a^{2})(a\sin^{2}x+2l\cos x)^{2}-\rho\overline{\rho}(r^{2}+l^{2}+a^{2})^{2}\sin^{2}x$

is *not* asymptotically flat. Here, the coordinates are $(u,r,x,y)$, $a$ is the Kerr parameter and $l$ is the NUT parameter. Further, $\rho=-1/(r+il-ia\cos x)$.

In particular, it should be due to the NUT parameter $l$, i.e. it's considered as a measure of "asymptotic non-flatness" (e.g. here http://arxiv.org/pdf/gr-qc/0104027.pdf)

My question is, how does one show explicitly that its asymptotically non-flat? What kind of coordinate transformation is involved?

When I say "asymptotically flat", I mean its definition as asymptotically simple spacetime $(M,g)$ and $R_{ij}=0$ in a neighborhood of $\mathcal{I}=\partial M$.

Used definition of symptomatically simple: $(M,g)$ is called symptomatically simple if $\exists(\tilde{M},\tilde{g})$ such that $M$ is a submanifold of $\tilde{M}$ with a smooth boundary, and $\exists$ smooth scalar field $\Omega$ on $\tilde{M}$ such that $\tilde{g}=\Omega^{2}g$, with $d\Omega\neq0$ on $\partial M$.

Thank you for any help.

This post imported from StackExchange MathOverflow at 2015-10-15 08:59 (UTC), posted by SE-user GregVoit