Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
bartleby

Concept explainers

Question
Book Icon
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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
a cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?
Calculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were:  222.22 800.00 61.11 641.67 0.00 588.89 11.11 588.89 8.33 588.89 11.11 588.89 5.56 586.11 2.78 583.33   Give in the answer window the calculated repeated experiment variance in m/s2.
No chatgpt pls will upvote
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning