Universe: Stars And Galaxies
Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 25, Problem 42Q
To determine

(a)

The separation between the two clusters at the time when the light was emitted from HS1946+7658 to produce an image on Earth.

Expert Solution
Check Mark

Answer to Problem 42Q

The separation between the two clusters at the time when the light was emitted from HS1946+7658 is 124.38Mpc.

Explanation of Solution

Given:

The redshift is, z=3.02.

The separation between the two clusters at present is d0=500Mpc.

Formula Used:

The expression for the separation between the two clusters at the time when the light was emitted from the quasar is given by,

d=d01+z

Calculation:

The expression for the separation between the two clusters at the time when the light was emitted from the quasar is calculated as,

d=d01+z=500Mpc1+3.02=124.38Mpc

Conclusion:

The separation between the two clusters at the time when the light was emitted from HS1946+7658 is 124.38Mpc.

To determine

(b)

The average density of matter at the time when the light was emitted from HS1946+7658 to produce an image on Earth.

Expert Solution
Check Mark

Answer to Problem 42Q

The average density of matter at the time when the light was emitted from HS1946+7658 is 1.56×1025kg/m3.

Explanation of Solution

Given:

The redshift is, z=3.02.

The average density of matter in today’s universe is, ρm=2.4×1027kg/m3.

Formula Used:

The expression for average density of matter is given by,

(ρm)avg=ρm(1+z)3

Calculation:

The average density of matter is calculated as,

( ρ m)avg=ρm(1+z)3=(2.4× 10 27kg/ m 3)(1+3.02)3=1.56×1025kg/m3

Conclusion:

The average density of matter at the time when the light was emitted from HS1946+7658 is 1.56×1025kg/m3.

To determine

(c)

The temperature of the cosmic background radiation and the mass density of radiation at the time when the light was emitted from HS1946+7658.

Expert Solution
Check Mark

Answer to Problem 42Q

The temperature of the cosmic background radiation at the time when the light was emitted from HS1946+7658 was 10.95K and the mass density of radiation was 1.21×1028kg/m3.

Explanation of Solution

Given:

The redshift is, z=3.02.

Formula Used:

The expression for the radiation temperature is given by,

T0=T(1+z)

Here, T is the temperature of the cosmic microwave background.

The expression for the mass density of radiation is given by,

ρrad=4σT4c3

Here, σ is the Stefan Boltzmann’s constant.

Calculation:

The cosmic microwave background has a temperature of T=2.725K.

The radiation temperature is calculated as

T0=T(1+z)=(2.725K)(1+3.02)=10.95K

The mass density of radiation is calculated as

ρrad=4σ ( T 0 )4c3=4( 5.67× 10 8 W/ m 2 K 4 ) ( 10.95K )4 ( 3× 10 8 m/s )3=1.21×1028kg/m3

Conclusion:

The temperature of the cosmic background radiation at the time when the light was emitted from HS1946+7658 was 10.95K and the mass density of radiation was 1.21×1028kg/m3.

To determine

(d)

Whether the universe was matter-dominated, radiation-dominated or dark-energy-dominated at the time when the light was emitted from HS1946+7658.

Expert Solution
Check Mark

Explanation of Solution

Introduction:

Consider part (b). The average density of matter at the time when the light was emitted from HS1946+7658 is 1.56×1025kg/m3.

Consider part (c). The mass density of radiation at the time when the light was emitted from HS1946+7658 is 1.21×1028kg/m3.

The mass density of radiation is less than the average density of matter at the time when the light was emitted from HS1946+7658. Therefore, at that time, the universe was matter-dominated.

Conclusion:

The universe was matter-dominated at the time when the light was emitted from HS1946+7658.

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