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

1. Jun 14, 2006

### Greg Egan

I've worked out the FRW dust solutions in 2+1 gravity, that is the
homogeneous, isotropic solutions where the universe contains a continuum
of matter that approximates the presence of pressure-free particles at
the same density everywhere.

As in 3+1 gravity there can be negative, zero, or positive curvature for
the spacelike slices (i.e. hyperboloids, planes or spheres), but the time
evolution is extremely simple: either the solution is static (which is
only possible for spherical spatial geometry or zero density), or it is
expanding at a constant rate.

The metric is:

ds^2 = -dt^2 + R(t)^2 dw^2

where dw^2 is the metric on the unit hyperboloid, plane, or unit sphere.
If we put k=-1,0,1 for these alternatives, the relationship between R(t)
and the density of matter is:

R'(t)^2 + k = 8 pi rho R(t)^2

while the lack of pressure gives:

R''(t) = 0

For spherical geometry, the radius of the sphere is R(t), so the total
amount of matter present is M = 4 pi rho R(t)^2, and:

R'(t)^2 + 1 = 2 M

This implies that a static solution is possible iff:

M = 1/2 (in units where G=c=1).

In these units, the angular deficit associated with a mass of M is 8piM,
so this static solution has a total angular deficit of 4pi, in agreement
with the finite particle polyhedron solutions Ralph Hartley has described.

The only other *homogeneous, isotropic* static solution is the vacuum
solution. If we put rho=0 in the planar FRW solution, we get
R(t)=constant, and that's obviously Minkowski space, but if we put rho=0
in the hyperboloid solution we get R(t)=t, and the solution is the
interior of the light cone in Minkowski space, i.e. a portion of the same
vacuum solution.

For the non-static solutions, we note that

4 pi rho R(t)^2 = some constant C

by conservation of mass; for the spherical geometry we have C=M, the
total mass in the universe, but for the other geometries where there's an
infinite amount of matter in the universe C will still be constant. So
we have:

R(t) = sqrt(2C - k) t
rho(t) = C / [4pi (2C-k) t^2 ]

It'd be nice to be able to relate this to various discrete particle
solutions. I previously posted a solution with a ring of n particles
moving away from the origin with equal speed v in a symmetrical manner,
and noted that if we look at what's happening on the unit hyperboloid in
Minkowski space, the angular deficit associated with each particle
becomes much less at hyperbolic infinity, i.e. if we cut out a wedge by
arranging two planes to intersect along each particle's world line, the
sum of the angles between these planes can be much more than 2pi because
they take much smaller "bites" out of the light cone than the angles they
subtend around the world lines.

Specifically, if the angle of the "bite" is B, it turns out that for
small deficit angles A around the world line we have:

B = A sqrt( (1-v) / (1+v) )

Now, imagine a cloud of particles with world lines starting from the
origin of Minkowski space with all possible velocities, spreading out at
all angles. By cutting out wedges around their world lines, we ought to
be able to get one of the expanding hyperboloid FRW solutions, where each
spacelike slice at time t is isometric to the unit hyperboloid multiplied
by sqrt(1+2C) t, and the world lines of all the particles are normal to
every spacelike slice.

In hyperbolic geometry the area and circumference of a circle increase
with distance faster than in the Euclidean plane, but as we move out
through our cloud of particles in Minkowski space with wedges cut out,
we're *losing* area and circumference. So all the hyperboloids in the
original Minkowski space get flattened out. I guess the trick is to find
a way to make sure that their geometry is still that of a hyperboloid,
but a slightly larger, hence flatter, one ... by a factor of sqrt(1+2C).
But I haven't been able to figure out yet how to get the particles
distributed in the original Minkowski spacetime in such a way that
everything works out nicely.