EBK LOOSE-LEAF VERSION OF UNIVERSE
EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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Chapter 25, Problem 53Q

(a)

To determine

The value of qz at z=0.5, if the value of the deceleration parameter varies with time. It is given that the formula for the deceleration parameter at a redshift, z, for a flat universe, is qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]. Here, ΩΛ is the dark energy density parameter. Use the value of ΩΛ from table 25-2.

(a)

Expert Solution
Check Mark

Answer to Problem 53Q

Solution:

0.22

Explanation of Solution

Given data:

The value of the redshift is 0.5.

Using table 25-2, the value of ΩΛ is found to be 0.76.

The formula for the deceleration parameter at redshift z is qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3].

Formula used:

The given formula is written as,

qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]

Here, ΩΛ is the dark energy density parameter and qz is the deceleration parameter at the redshift, z.

Explanation:

Recall the provided formula.

qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]

Substitute 0.5 for z and 0.76 for ΩΛ.

qz=1232[.76.76+(1.76)(1+.5)3]=1232[.761.57]=0.50.7261=0.22

Conclusion:

Therefore, the value of qz is 0.22.

(b)

To determine

The value of qz at z=1, if the value of the deceleration parameter varies with time. It is given that the formula for the deceleration parameter at the redshift, z, for a flat universe, is qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]. Here, ΩΛ is the dark energy density parameter. Use the value of ΩΛ from table 25-2.

(b)

Expert Solution
Check Mark

Answer to Problem 53Q

Solution:

0.074

Explanation of Solution

Given data:

The value of the redshift is 1.

Using table 25-2, the value of ΩΛ is found to be 0.76.

Formula used:

The provided formula is written as:

qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]

Here, ΩΛ is the dark energy density parameter and qz is the deceleration parameter at the redshift, z.

Explanation:

Recall the provided formula.

qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]

Substitute 1 for z and 0.76 for ΩΛ.

qz=1232[.76.76+(1.76)(1+1)3]=1232[.762.68]=0.50.4253=0.074

Conclusion:

Therefore, the value of qz is 0.074.

(c)

To determine

An explanation for the result that shows that the expansion of the universe was decreasing at z=1. The expansion of the universe has changed from deceleration to acceleration between z=1 and z=0.5. It is given that the formula for the deceleration parameter at the redshift, z for a flat universe, is qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]. Here, ΩΛ is the dark energy density parameter.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data:

The values of the redshift are 1 and 0.5.

Using table 25-2, the value of ΩΛ is found to be 0.76.

Formula used:

The provided formula is written as:

qz=1232[ΩΛΩΛ+(1ΩΛ)(1+z)3]

Here, ΩΛ is the dark energy density parameter and qz is the deceleration parameter at the redshift, z.

Explanation:

The value of qz at z=.5 is 0.22, which is calculated in part (a) and value of qz at z=1 is 0.074, which is calculated in part (b). From this, it can be seen that the value at redshift .5 is negative and at redshift 1 is positive.

A positive value of qz indicates that the expansion of the universe is decelerated, which is at redshift 1. On the other hand, the value qz is negative at redshift 0.5, so the expansion of the universe is accelerated. So, it can be said that the universe must have accelerated between the redshifts, 1 and 0.5.

Conclusion:

Therefore, the universe accelerated between redshift 1 and 0.5.

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