Добавил:
Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
FREGRAVOL2.pdf
Скачиваний:
75
Добавлен:
05.06.2015
Размер:
8.65 Mб
Скачать

5.6 General Consequences of Friedman Equations

171

5.6.3 Red-Shift Distances

In our observation of the sky at extremely large distances, which means looking at very ancient objects, the only available method to measure both the space and the time separation of these sources from the vantage point of our observatory consists of determining their red-shift factor:

z

 

λ0 λe

a(t0)

(5.6.39)

 

 

 

 

λe

= a(te)

 

 

where t0 is the current cosmic time when the observed photons are absorbed and te is the remote time when they were emitted from their source.

It is therefore quite useful to use the red-shift factor as a label both for time and space distances. Indeed in the conformal coordinate system centered in our laboratory, the radial coordinate χ (z) of a distant source at red-shift z is unambiguously defined as:

 

t0

dt

1

 

z dz

 

 

χ (z) = η0 ηe =

 

 

=

 

 

 

 

 

 

(5.6.40)

te

a(t)

a

0

H (z

)

 

 

 

 

0

 

 

 

and the Hubble function is expressed in terms of z through an immediate manipulation of the Friedman equation (5.6.7):

H (z) = H0 Ω0(z + 1)3(w+1) + (1 Ω0)(z + 1)2 (5.6.41)

Let us also observe that for non-spatially flat universes the current value of the scale factor can be determined from Friedman equation as:

a0 = H0

 

 

1

 

|1 Ω0|

(5.6.42)

Hence for the case of a closed or open universe the physical distance of a source at red-shift z can be written as follows:

D±(z) = H0

 

 

1R± χ±(z)

 

|1 Ω0|

(5.6.43)

whose explicit form we shall presently evaluate for matter dominated universes. For the case of a matter dominated flat universe we can make a separate very simple calculation. Setting Ω0 = 1 and w = 0 in (5.6.41) we get:

 

1

 

 

 

 

z

dz

 

 

 

1

 

z

dz

 

 

 

χ0(z) =

 

 

 

 

 

 

 

 

 

 

= a

 

 

 

 

 

 

a

0

0

 

H (z

 

)

 

 

 

 

3

H0

 

 

 

 

 

 

 

 

 

1

 

0 0

 

(z + 1)

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

a

0

H

0

 

1

z + 1

 

 

 

 

 

(5.6.44)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 − √

 

1

 

 

 

 

 

 

 

D0

(z) = H

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

z + 1

 

 

 

 

 

 

 

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]