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 Incoming links: Open dust distance-redshift relation

For radial null propagation,

d r / 1 - k r^{2} = ∣ d t ∣/ a

. Since

d z / d t = - (1 + z) H (z)

and

a = a_{0} / (1 + z)

d_{S} = a_{0} \int_{t_{e}}^{t_{0}} \frac{d t}{a ( t )} = \int_{0}^{z} \frac{d z ^{'}}{H ( z ^{'} )} .

Thus the printed extra

a_{0}

before the redshift integral is inappropriate when

H

is the physical Hubble parameter; it disappears only if

a_{0}

has been set to one.

Dust conservation and the curvature relation give

H^{2} (z) = H_{0}^{2} [Ω_{0} (1 + z)^{3} + (1 - Ω_{0}) (1 + z)^{2}], d_{S} = \frac{1}{H _{0}} \int_{0}^{z} \frac{d z ^{'}}{( 1 + z ^{'} ) 1 + Ω _{0} z ^{'}} .

To evaluate it, put

s = 1 - Ω_{0}

and

v = 1 + Ω_{0} z

. The dimensionless integral

I = H_{0} d_{S}

I = 2 \int_{1}^{v} \frac{d v ^{'}}{v ^{'2} - s ^{2}} = \frac{1}{s} lo g [\frac{v - s}{v + s} \frac{1 + s}{1 - s}] .

Inverting the proper-distance expression in part b gives

r = sinh (s I) / (a_{0} H_{0} s)

. Substituting the logarithm, writing the hyperbolic sine as half the difference of its exponential and inverse, and using

v^{2} - s^{2} = Ω_{0} (1 + z)

yields

r (z) = \frac{2}{a _{0} H _{0}} \frac{Ω _{0} z + ( 2 - Ω _{0} ) [ 1 - 1 + Ω _{0} z ]}{Ω _{0}^{2} ( 1 + z )} .

z

; its literal form would give a negative, nonzero coordinate at

z = 0

. The corrected expression has

r \sim z / (a_{0} H_{0})

near zero. At fixed positive

Ω_{0}

and

z ≫ Ω_{0}^{- 1}

, the term linear in

z

dominates, giving

r ⟶ \frac{2}{a _{0} H _{0} Ω _{0}} .

The zero-density limit is nonuniform: a pure Milne model has no such finite limiting radial coordinate.