Proof of asymptotic non-flatness

Question

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

Comments

I think the idea is that a spacetime with a NUT parameter always has one or more semi-infinite wire-like singularities (which I think can be either a curvature or a non-curvature singularity, depending on the specific example). Since this singularity extends to infinite distances, clearly the spacetime doesn't look like Minkowski space far away. As far as formal proof, isn't the topology just wrong for an asymptotically flat space? Since the singularity is like a line removed from the manifold, you should have non-contractible curves going around it.

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

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One thing that I don't quite understand about your definition of asymptotic flatness, if it's intended to be a rigorous one rather than a schematic outline, is that you refer to $\partial M$, but normally M is going to be a manifold (not a manifold-with-boundary). As far as I know, the standard definition these days is the one in Wald, ch. 11, and it's much more complicated than the one you've stated. As an example, $\Omega$'s derivatives have to behave in a certain way at $i^0$, whereas different conditions hold at scri+ and scri-.

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

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@BenCrowell thank you.Since I"m from a purely physics background, could you please elaborate more on your first comment, i.e. when you say "…topology is just wrong for an asymptotically flat space". How do I see that?

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

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Re your definition of asymptotic flatness, Wald describes on p. 282 an earlier, simpler version by Penrose, which sounds like it may be what you're using...? Re the definition of $\partial M$, see p. 276; this is the result of a nontrivial construction, not just the ordinary notion of the boundary of a point-set.

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

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Re the topology, I'm not 100% sure that I'm even right, but I also can't really tell from your comment what in particular it is that you are asking for explanation of. There definitely are topological properties implied by asymptotic flatness, though. See, e.g., Wald, p. 279, where he discusses the implied topology of scri+. Another way of proving that NUTty spacetimes aren't asymptotically flat may be that they don't have the asymptotic symmetries that are implied by asymptotic flatness. See Wald p. 284ff. The singularity stretches to i0 and would break rotational symmetry.

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

Answer 1

A manifold is transformable to Minkowski in a certain region iff $R^{\mu}_{\; \nu \kappa \lambda}=0$ iff all parallel transport is "integrable" (you parallel transport on any loop in the region and get the same vector). Either of these are used as a definition of flatness. Most often, one is working with a metric in a form which reduces to the Minkowski metric in some known coordinate system but here you have to be slightly more careful.

The Kerr-NUT metric is not asymptotically flat because of the half-infinite string singularity on the axis. You can verify this by making a small loop around the top ($x=0$) part of the axis at infinity (i.e. only with asymptotically surviving metric coefficients) and finding out that parallel transport along the loop changes vectors. Apart from the upper half-axis, the metric is everywhere asymptotically flat.

Comments

@Void thank you for the comment. Could you please elaborate more on "You can verify this by making a small loop around the top (x=0) part of the axis at infinity ...and finding out that parallel transport along the loop changes vectors"? I have limited experience in the field, and I don't know how to attempt what you wrote