38 Special Relativity
Similarly, the area of a three-sphere centered at the origin and going through the
galaxy at휎is
퐴= 4 휋푎^2 휎^2 = 4 휋푎^2 sin^2 (푑P∕푎). (2.40)Clearly,퐴goes through a maximum when푑P=^12 휋푎, and decreases back to 0 when푑P
reaches휋푎.Notethat퐴∕4 equals the area enclosed by the circle formed by intersecting
a two-sphere of radius푅with a horizontal plane, as shown in Figure 2.1. The intersec-
tion with an equatorial plane results in the circle enclosing maximal area,퐴∕ 4 =휋푅^2 ,
all other intersections making smaller circles. A plane tangential at either pole has no
intersection, thus the corresponding ‘circle’ has zero area.
The volume of the three-sphere in Equation (2.28) can then be written in analogy
with Equation (2.23),
푉= 2
∫2 휋0d휙
∫휋0d휃
∫10d휎√
detgRW, (2.41)where the determinant of the spatial part of the Robertson–Walker metric matrixgRW
is now
detgRW=푎^6휎^4
1 −휎^2
sin^2 휃. (2.42)The factor 2 in Equation (2.41) comes from the sign ambiguity of푤in Equa-
tion (2.28). Both signs represent a complete solution. Inserting Equation (2.42) into
Equation (2.41) one finds the volume of the three-sphere:
푉= 2 휋^2 푎^3. (2.43)
The hyperbolic case is different. Setting푘=−1, the function in Equation (2.39) is
i−^1 sin−^1 i휎≡sinh−^1 휎, thus
푑P=푎휒=푎sinh−^1 휎 or 휎=sinh(푑P∕푎). (2.44)Clearly this space is open because휎grows indefinitely with푑P.Theareaofthe
three-hyperboloid through the galaxy at휎is
퐴= 4 휋푎^2 휎^2 = 4 휋푎^2 sinh^2 (푑P∕푎). (2.45)
Let us differentiate푑Pin Equation (2.39) with respect to time, noting that휎is a con-
stant since it is a comoving coordinate. We then obtain the Hubble flow푣experienced
by a galaxy at distance푑P:
푣=푑̇P=푎̇(푡)
∫휎0d휎
√
1 −푘휎^2=
푎̇(푡)
푎(푡)
푑P. (2.46)
Thus the Hubble flow is proportional to distance, and Hubble’s law emerges in a form
more general than Equation (1.20):
퐻(푡)=푎̇(푡)
푎(푡)
. (2.47)
Recall that푣is the velocity of expansion of the space-time geometry. A galaxy with
zero comoving velocity would appear to have a radial recession velocity푣because of
the expansion.