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

1. Jun 16, 2006

Greg Egan

[SOLVED] Re: This Week's Finds in Mathematical Physics (Week 232)

I wrote:

>magine 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.

Amazingly enough, this does work! The geometry of the hyperboloids in
Minkowski spacetime is changed, by the excision of wedges, to exactly
that of the corresponding FRW-solution spacelike slices.

Calculations follow, for anyone interested in the grungy details ...

If a deficit angle A originates around a particle of velocity v on the
unit hyperboloid, then the amount of the azimuthal coordinate that this
steals further out at velocity w is, to first order in A:

A (w-v) / ( w sqrt(1-v^2) )

Rephrasing this in terms of geodesic distances along the unit hyperboloid
from the "pole" at (1,0,0), if the distances are r for v and q for w, we
get:

A (cosh(q) - sinh(q)/tanh(r))

To find T(r), the total angular deficit at distance r, we need to
integrate the deficit angles for all the matter out to r, weighted by
this adjustment factor. We can't actually know the amount of matter at
distance q without knowing the circumference of a circle at distance q
... which depends on the angular deficit. This leads to an integral
equation, but we don't need to solve it from first principles, because
the guess that the geometry remains hyperbolic but just changes radius
turns out to be correct.

On a hyperboloid of radius a, the circumference of a circle of radius r
is:

c_a(r) = 2pi a sinh(r/a)

On the unit hyperboloid with a deficit angle T(r), the circumference is:

c_{1,d}(r) = (2pi - T(r)) sinh(r)

If the deficit angle T(r) is to mimic a change of hyperboloid radius from
1 to a, making these two circumference formulas equal, we must have:

T(r) = 2pi (1 - a sinh(r/a) / sinh(r) )

If we integrate a matter distribution based on a constant density rho, we
get:

T(r) = integral_{0,r} 8pi rho c_a(q) (cosh(q) - sinh(q)/tanh(r)) dq

If we evaluate this, then put a=sqrt(1+2C), rho=C/(4pi(1+2C)) -- where C
is a constant due to conservation of mass, discussed in my previous post
on the FRW solutions -- then these two ways of computing T(r) are in
perfect agreement.

Of course, although our particle cloud is homogeneous on the FRW
spacelike slices, it won't be homogeneous on the *uncut* hyperboloids in
Minkowski spacetime.