10 From Newton to Hubble
volume element,Anhas the dimensions of number of stars per linear distance. The
inverse of this,
퓁= 1 ∕An, (1.4)
is the mean radial distance between stars, or themean free pathof photons emitted
from one star and being absorbed in collisions with another. We can also define a
mean collision time:
휏=퓁∕푐. (1.5)
The value of휏can be roughly estimated from the properties of the Sun, with radius
푅⊙and density휌⊙. Let the present mean density of luminous matter in the Universe
be휌 0 and the distance to the farthest visible star푟∗. Then the collision time inside this
volume of size^43 휋푟^3 ∗is
휏≃휏⊙=
1
퐴⊙nc=
1
휋푅^2 ⊙
4 휋푟^3 ∗
3 Nc=
4 휌⊙푅⊙
3 휌 0 푐
. (1.6)
Taking the solar parameters from Table A.2 in the appendix we obtain approximately
1023 yr.
The probability that a photon does not collide but arrives safely to be observed by us
after a flight distance푟can be derived from the assumption that the photon encounters
obstacles randomly, that the collisions occur independently and at a constant rate퓁−^1
per unit distance. The probability푃(푟)that the distance to the first collision is푟is then
given by the exponential distribution
푃(푟)=퓁−^1 e−푟∕퓁. (1.7)Thus flight distances much longer than퓁are improbable.
Applying this to photons emitted in a spherical shell of thickness d푟, and integrating
the spherical shell from zero radius to푟∗, the fraction of all photons emitted in the
direction of the center of the sphere and arriving there to be detected is
푓(푟∗)=
∫푟∗0퓁−^1 e−푟∕퓁d푟= 1 −e−푟∗∕퓁. (1.8)Obviously, this fraction approaches 1 only in the limit of an infinite universe. In
that case every point on the sky would be seen to be emitting photons, and the sky
would indeed be as bright as the Sun at night. But since this is not the case, we must
conclude that푟∗∕퓁is small. Thus the reason why the whole field of vision is not filled
with stars is that the volume of the presently observable Universe is not infinite, it is
in fact too small to contain sufficiently many visible stars.
Lord Kelvin’s original result follows in the limit of small푟∗∕퓁, in which case
푓(푟∗)≈푟∕퓁.The exponential effect in Equation (1.8) was neglected by Lord Kelvin.
We can also replace the mean free path in Equation (1.8) with the collision time
[Equation (1.5)], and the distance푟∗with the age of the Universe푡 0 , to obtain the
fraction
푓(푟∗)=푔(푡 0 )= 1 −e−푡^0 ∕휏. (1.9)