Friedman universe

Geometry of a space with constant curvature

Suppose the universe is uniformly and isotropically filled with matter. The curvature, if any, is then constant througout the universe. The surface of a sphere is apparently a space with constant curvature, where the length element in spherical coordinates is
dl2=a2(dθ2+sin2θdφ2)   ,
(1)
where a2 is the sphere's radius. Let us indtroduce polar coordinates {r,φ} on the surface. The length of an surface parallel is equal 2πasinθ, therefore if we the curcumference of a circle to equal 2πr we need to introduce r=asinθ. The length element in our polar coordinates is then
dl2= dr2
1− r2
a2
 +r2dφ2   ,
(2)
which is a general length element of a space with constant positive curvature. Another possibility1 is
dl2= dr2
1+ r2
a2
 +r2dφ2   ,
(3)
which is a general length element of a space with negative constant curvature. In spherical coordinates r=asinhθ the latter is
dl2=a2(dθ2+sinh2θdφ2)   ,
(4)
Generalising to three dimensions: the 3D space with constant curvature can have one of the three possible geometries: flat2; closed,
dl2= dr2
1− r2
a2
 +r2dΩ2 = a2(dχ2+sin2χdΩ2)   ,
(5)
where r=asinχ; or open,
dl2= dr2
1+ r2
a2
 +r2dΩ2 = a2(dχ2+sinh2χdΩ2)   ,
(6)
where r=asinhχ.

Friedman equation and solutions

Friedman metric describes a homogeneous and isotropic universe. For the closed universe the interval is
ds2 = a2(dη2 − dχ2 −sin2χ(dθ2 +sin2θdφ2)) ,
(7)
where r=asinχ, η is the scaled time coordinate, dt=adη, and a(η) is the scale parameter of the universe (the radius of the 4-sphere). The components of the Ricci tensor are
Rχχ = Rθθ = Rφφ = −( 1
a4
 )(2a2+a′2+aa")  ,
(8)
Rηη = ( 3
a4
 )(a′2−aa")  ,R = −( 6
a3
 )(a+a")  ,
(9)
where prime denotes the derivative with respect to η.
Assuming that the universe is filled with a perfect fluid the energy-momentum tensor of the matter is Tab=(ε+ p)uaub− pgab where ε is the energy density and p is the pressure. In our frame, where the matter is at rest, the 4-velosity ua=([1/(a)],0,0,0).
The Einstein's equations Rab−[1/2]Rδab = κTab will then have the ηη component
( 3
a4
 )(a2+a′2) = κε ,
(10)
called Friedman's equation, and the three identical spatial equations ([1/(a4)])(a2+2aa"−a′2)=−κp.
The Friedman's equation together with the energy conservation equation, dε = −(ε+ p)3[(da)/(a)], can be integrated for the matter dominated universe, where the pressure is zero, p=0, and the energy density ε is equal to the mass density μ. The energy conservation gives μa3=Const and the subsequent integration of the Friedmann's equation gives for a closed universe (a2 > 0) a "Big Bang → Big Crunch" scenario:
a=a0(1−cos(η)) ,  t=a0(η−sin(η))
(11)
For the open isotropic universe the Friedman's equation provides a "Big Bang → Expansion Forever" scenario:
a=a0(cosh(η)−1) ,  t=a0(sinh(η)−η)
(12)
For a flat isotropic universe ds2=dt2−b2(t)(dx2+dy2+dz2) the scenario is also "Big Bang → Expansion Forever" (see the Exercise): μb3=const, b=const t2/3.
At early stages with high densities the universe was (probably) rather radiation dominated, p=[(ε)/3]. This, however, doesn't save us from the singular point at η = 0. Indeed, we have (for η << 1) : εa4=const ,  a=const·t1/2

Footnotes:

1 the third possibility is a flat space, a=∞
2  dl2=dx2+dy2+dz2

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On 12 Oct 2007, 13:02.