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 $500\text{ Mpc}$ in the present-day universe. The quasar HS 1946+7658 has $z=3.02$. Assume ${\rho }_m=2.4\times {10}^{-27}{\text{ kg/m}}^3$, in today’s universe.

Answer to Problem 42Q

Solution:

$\text{124}\text{.34 Mpc}$

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 $500\text{ Mpc}$.

The density of matter in today’s universe is ${\rho }_m=2.4\times {10}^{-27}{\text{ kg/m}}^3$.

Formula used:

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

$d=\frac{\text{the distance between the two clusters at present day universe}}{1+z}$

Here, z is the redshift.

Explanation:

Recall the expression for the distance between the two clusters:

$d=\frac{\text{the distance between the two clusters at present day universe}}{1+z}$

Substitute 3.02 for z and $500\text{ Mpc}$ as distance between two clusters at the present day.

$\begin{array}{c}d=\frac{500\text{ Mpc}}{1+3.02}\\ =\frac{500\text{ Mpc}}{4.02}\\ =124.34\text{ Mpc}\end{array}$

Conclusion:

Therefore, the distance between the two clusters is $\text{124}\text{.34 Mpc}$.

(b)

To determine

The average density of the matter $\left({\rho }_m\right)$ 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, ${\rho }_m=2.4\times {10}^{-27}{\text{ kg/m}}^3$. The quasar HS 1946+7658 has $z=3.02$.

Answer to Problem 42Q

Solution:

$1.55\times {10}^{-25}{\text{ kg/m}}^{\text{3}}$

Explanation of Solution

Given data:

The density of today’s universe is $2.4\times {10}^{-27}{\text{ kg/m}}^3$.

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

Formula used:

The relationship between redshift and average density is:

${\left({\rho }_m\right)}_{avg}={\rho }_m{\left(1+z\right)}^3$

Here, ${\rho }_m$ is the density of today’s universe, z is redshift, and ${\left({\rho }_m\right)}_{avg}$ is the average density.

Explanation:

Recall the relationship between redshift and average density.

${\left({\rho }_m\right)}_{avg}={\rho }_m{\left(1+z\right)}^3$

Substitute 3.02 for z and $2.4\times {10}^{-27}{\text{ kg/m}}^3$ for ${\rho }_m$

$\begin{array}{c}{\left({\rho }_m\right)}_{avg}=2.4\times {10}^{-27}{\text{ kg/m}}^{\text{3}}{\left(1+3.02\right)}^3\\ =1.55\times {10}^{-25}{\text{ kg/m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the average density is $1.55\times {10}^{-25}{\text{ kg/m}}^{\text{3}}$.

(c)

To determine

The temperature of the cosmic background radiation and the mass density of radiation $\left({\rho }_{rad}\right)$ 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 $z=3.02$.

Answer to Problem 42Q

Solution:

$1.19\times {10}^{-28}{\text{ kg/m}}^{\text{3}}$

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:

${\rho }_{rad}=\frac{4\sigma T^4}{c^3}$

Here, ${\rho }_{rad}$ is the mass density of the radiation, T is the background temperature, c is the speed of light having a value of $3\times {10}^8\text{ m/s}$ and $\sigma$ is Stefan Boltzmann’s constant having a value of $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$.

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 $\left(3.02+1\right)$.

So, the background temperature is:

$\begin{array}{c}T=\left(2.72\text{ K}\right)4.02\\ =10.93\text{ K}\end{array}$

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

${\rho }_{rad}=\frac{4\sigma T^4}{c^3}$

Substitute $3\times {10}^8\text{ m/s}$ for c, $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$ and 10.93 K for T:

$\begin{array}{c}{\rho }_{rad}=\frac{4\left(5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right){\left(10.93\text{K}\right)}^4}{{\left(3\times {10}^8\text{m/s}\right)}^3}\\ =\frac{3.23\times {10}^{-3}}{2.7\times {10}^{25}}\\ =1.19\times {10}^{-28}{\text{ kg/m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the mass density of radiation is $1.19\times {10}^{-28}{\text{ kg/m}}^{\text{3}}$.

(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 $500\text{ Mpc}$ in the present-day universe. The quasar HS 1946+7658, has $z=3.02$. Assume that ${\rho }_m=2.4\times {10}^{-27}{\text{ kg/m}}^3$, in today’s universe.

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 $500\text{ Mpc}$.

The density of today’s universe is $2.4\times {10}^{-27}{\text{ kg/m}}^3$.

Formula used:

The relationship between redshift and average density is:

${\left({\rho }_m\right)}_{avg}={\rho }_m{\left(1+z\right)}^3$

Here, ${\rho }_m$ is the density of today’s universe, z is redshift, and ${\left({\rho }_m\right)}_{avg}$ is average density.

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

${\rho }_{rad}=\frac{4\sigma T^4}{c^3}$

Here, ${\rho }_{rad}$ is the mass density of the radiation, T is the background temperature, c is the speed of light with value $3\times {10}^8\text{ m/s}$ and $\sigma$ is Stefan Boltzmann’s constant having a value of $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$.

Explanation:

The average density ${\left({\rho }_m\right)}_{avg}$ of the universe is $1.55\times {10}^{-25}{\text{ kg/m}}^{\text{3}}$, which is calculated in part (b).

The mass density of radiation ${\rho }_{rad}$ is $1.19\times {10}^{-28}{\text{ kg/m}}^{\text{3}}$, which is calculated in part (c).

From these results, it is concluded that ${\rho }_{rad}$ < ${\left({\rho }_m\right)}_{avg}$. So, it can be said that at that point in time, the universe was matter dominated.

Conclusion:

Therefore, the universe was matter dominated.