Greg Egan wrote:(adsbygoogle = window.adsbygoogle || []).push({});

> Ralph Hartley wrote:

>

>> Consider a polyhedron inscribed in a sphere of radius 1, centered at the

>> origin. Let the surface of the polyhedron inherit the metric from R^3

>> (which will be flat except at the vertexes).

>>

>> For any point p other than the origin, let p_1 be the intersection of

>> the polyhedron with the ray from the origin through p. Let t(p) = |p|/|p_1|.

>>

>> The metric (on R^3-O) ds^2 = -dt(p)^2 + dp_1^2 is flat except at the

>> rays from the origin through the vertexes, and any [space]like surface has

>> total deficit 4Pi.

>

> That's a really elegant construction, but (at least in the static case) I

> think you can get rid of your Big Bang at t=0.

Yes, I realized, just after sending, that the area is constant, and the

space is just R x Polyhedron. The sphere was not needed either, any

convex polyhedron would do.

I'm pretty sure that the edges of the polyhedron are purely artifacts of

the construction as well. The space is flat except at the vertexes.

Given a reference frame on one face, you can extend it to the whole

space in many different ways, one for each way of cutting the polyhedron

up and spreading it out flat. How many there are depends on what you

count as a "different" cutting.

> This would generalise to "polyhedra" formed by triangulating any compact

> 2-manifold and putting flat metrics on the triangles. The total deficit

> in the general case will be 2pi*chi, where chi is the Euler

> characteristic of the manifold.

If you require the manifold to be orientable, and don't allow negative

masses, I think the only new case is a flat torus (which does have some

parameters).

If you don't require that it be compact (and allow some infinite

triangles), you get the cases with deficit between 0 and 2Pi.

Lets see, how many more static solutions are there? If you allow

timelike loops, quite a few more I think.

Ralph Hartley

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# Re: This Week's Finds in Mathematical Physics (Week 232)

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