Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 25, Problem 42Q

(a)

To determine

The distance between two clusters at the point in time when light was emitted from HS 1946+7658 to produce an image on Earth tonight. Suppose that the distance between two clusters of galaxies is 500 Mpc in the present-day universe. The quasar HS 1946+7658 has z=3.02. Assume ρm=2.4×1027 kg/m3, in today’s universe.

(a)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

124.34 Mpc

Explanation of Solution

Given data:

The redshift for the quasar HS 1946+7658 is 3.02.

The distance between two clusters in present day universe is 500 Mpc.

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

Formula used:

The distance between the two clusters when light was emitted from a quasar is:

d=the distance between the two clusters at present day universe1+z

Here, z is the redshift.

Explanation:

Recall the expression for the distance between the two clusters:

d=the distance between the two clusters at present day universe1+z

Substitute 3.02 for z and 500 Mpc as distance between two clusters at the present day.

d=500 Mpc1+3.02=500 Mpc4.02=124.34 Mpc

Conclusion:

Therefore, the distance between the two clusters is 124.34 Mpc.

(b)

To determine

The average density of the matter (ρm) at that point in time when light was emitted from HS 1946+7658 to produce an image on earth tonight. It is given that in today’s universe, ρm=2.4×1027 kg/m3. The quasar HS 1946+7658 has z=3.02.

(b)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

1.55×1025 kg/m3

Explanation of Solution

Given data:

The density of today’s universe is 2.4×1027 kg/m3.

The redshift for the quasar HS 1946+7658 is 3.02.

Formula used:

The relationship between redshift and average density is:

(ρm)avg=ρm(1+z)3

Here, ρm is the density of today’s universe, z is redshift, and (ρm)avg is the average density.

Explanation:

Recall the relationship between redshift and average density.

(ρm)avg=ρm(1+z)3

Substitute 3.02 for z and 2.4×1027 kg/m3 for ρm

(ρm)avg=2.4×1027 kg/m3(1+3.02)3=1.55×1025 kg/m3

Conclusion:

Therefore, the average density is 1.55×1025 kg/m3.

(c)

To determine

The temperature of the cosmic background radiation and the mass density of radiation (ρrad) at the point in time when light was emitted from HS 1946+7658 to produce an image on earth tonight t. The quasar HS 1946+7658, has z=3.02.

(c)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

1.19×1028 kg/m3

Explanation of Solution

Given data:

The redshift for the quasar HS 1946+7658 is 3.02.

Formula used:

The relationship between the mass density of radiation and the background temperature is:

ρrad=4σT4c3

Here, ρrad is the mass density of the radiation, T is the background temperature, c is the speed of light having a value of 3×108 m/s and σ is Stefan Boltzmann’s constant having a value of 5.67×108 W/m2K4.

Explanation:

The temperature of the background radiation would have been 4.02 times greater because the expansion of the universe increased by a factor 4.02 (3.02+1).

So, the background temperature is:

T=(2.72 K)4.02=10.93 K

Recall the relationship between mass density of radiation and background temperature.

ρrad=4σT4c3

Substitute 3×108 m/s for c, 5.67×108 W/m2K4 for σ and 10.93 K for T:

ρrad=4(5.67×108 W/m2K4)(10.93K)4(3×108m/s)3=3.23×1032.7×1025=1.19×1028 kg/m3

Conclusion:

Therefore, the mass density of radiation is 1.19×1028 kg/m3.

(d)

To determine

Whether the universe was matter-dominated, radiation dominated, or dark energy-dominated in the distant past, at the point in time when light was emitted from HS 1946+7658 to produce an image on earth tonight. Suppose that the distance between two clusters of galaxies is 500 Mpc in the present-day universe. The quasar HS 1946+7658, has z=3.02. Assume that ρm=2.4×1027 kg/m3, in today’s universe.

(d)

Expert Solution
Check Mark

Answer to Problem 42Q

Solution:

Matter dominated

Explanation of Solution

Given data:

The redshift for the quasar HS 1946+7658 is 3.02.

The distance between two clusters in the present-day universe is 500 Mpc.

The density of today’s universe is 2.4×1027 kg/m3.

Formula used:

The relationship between redshift and average density is:

(ρm)avg=ρm(1+z)3

Here, ρm is the density of today’s universe, z is redshift, and (ρm)avg is average density.

The relationship between mass density of radiation and background temperature is:

ρrad=4σT4c3

Here, ρrad is the mass density of the radiation, T is the background temperature, c is the speed of light with value 3×108 m/s and σ is Stefan Boltzmann’s constant having a value of 5.67×108 W/m2K4.

Explanation:

The average density (ρm)avg of the universe is 1.55×1025 kg/m3, which is calculated in part (b).

The mass density of radiation ρrad is 1.19×1028 kg/m3, which is calculated in part (c).

From these results, it is concluded that ρrad < (ρm)avg. So, it can be said that at that point in time, the universe was matter dominated.

Conclusion:

Therefore, the universe was matter dominated.

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