Chapter 25, Problem 42Q

To determine

(a)

The separation between the two clusters at the time when the light was emitted from $\text{HS}1946+7658$ to produce an image on Earth.

Answer to Problem 42Q

The separation between the two clusters at the time when the light was emitted from $\text{HS}1946+7658$ is $124.38\text{Mpc}$.

Explanation of Solution

Given:

The redshift is, $z=3.02$.

The separation between the two clusters at present is $d_0=500\text{Mpc}$.

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,

$d=\frac{d_0}{1+z}$

Calculation:

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

$\begin{array}{c}d=\frac{d_0}{1+z}\\ =\frac{500\text{Mpc}}{1+3.02}\\ =124.38\text{Mpc}\end{array}$

Conclusion:

The separation between the two clusters at the time when the light was emitted from $\text{HS}1946+7658$ is $124.38\text{Mpc}$.

To determine

(b)

The average density of matter at the time when the light was emitted from $\text{HS}1946+7658$ to produce an image on Earth.

Answer to Problem 42Q

The average density of matter at the time when the light was emitted from $\text{HS}1946+7658$ is $1.56\times {10}^{-25}\text{kg}/{\text{m}}^3$.

Explanation of Solution

Given:

The redshift is, $z=3.02$.

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

Formula Used:

The expression for average density of matter is given by,

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

Calculation:

The average density of matter is calculated as,

$\begin{array}{c}{\left({\rho }_m\right)}_{\text{avg}}={\rho }_m{\left(1+z\right)}^3\\ =\left(2.4\times {10}^{-27}\text{kg}/{\text{m}}^3\right){\left(1+3.02\right)}^3\\ =1.56\times {10}^{-25}\text{kg}/{\text{m}}^3\end{array}$

Conclusion:

The average density of matter at the time when the light was emitted from $\text{HS}1946+7658$ is $1.56\times {10}^{-25}\text{kg}/{\text{m}}^3$.

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 $\text{HS}1946+7658$.

Answer to Problem 42Q

The temperature of the cosmic background radiation at the time when the light was emitted from $\text{HS}1946+7658$ was $10.95\text{K}$ and the mass density of radiation was $1.21\times {10}^{-28}\text{kg}/{\text{m}}^3$.

Explanation of Solution

Given:

The redshift is, $z=3.02$.

Formula Used:

The expression for the radiation temperature is given by,

$T_0=T\left(1+z\right)$

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

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

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

Here, $\sigma$ is the Stefan Boltzmann’s constant.

Calculation:

The cosmic microwave background has a temperature of $T=2.725\text{K}$.

The radiation temperature is calculated as

$\begin{array}{c}{T}_0=T\left(1+z\right)\\ =\left(2.725\text{K}\right)\left(1+3.02\right)\\ =10.95\text{K}\end{array}$

The mass density of radiation is calculated as

$\begin{array}{c}{\rho }_{\text{rad}}=\frac{4\sigma {\left(T_0\right)}^4}{c^3}\\ =\frac{4\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right){\left(10.95\text{K}\right)}^4}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^3}\\ =1.21\times {10}^{-28}\text{kg}/{\text{m}}^3\end{array}$

Conclusion:

The temperature of the cosmic background radiation at the time when the light was emitted from $\text{HS}1946+7658$ was $10.95\text{K}$ and the mass density of radiation was $1.21\times {10}^{-28}\text{kg}/{\text{m}}^3$.

To determine

(d)

Whether the universe was matter-dominated, radiation-dominated or dark-energy-dominated at the time when the light was emitted from $\text{HS}1946+7658$.

Explanation of Solution

Introduction:

Consider part (b). The average density of matter at the time when the light was emitted from $\text{HS}1946+7658$ is $1.56\times {10}^{-25}\text{kg}/{\text{m}}^3$.

Consider part (c). The mass density of radiation at the time when the light was emitted from $\text{HS}1946+7658$ is $1.21\times {10}^{-28}\text{kg}/{\text{m}}^3$.

The mass density of radiation is less than the average density of matter at the time when the light was emitted from $\text{HS}1946+7658$. Therefore, at that time, the universe was matter-dominated.

Conclusion:

The universe was matter-dominated at the time when the light was emitted from $\text{HS}1946+7658$.