Chapter 25, Problem 52Q

(a)

To determine

To analyze: The slowing down of the expansion of the universe in the absence of a cosmological constant. The expansion of the universe can be expressed in terms of the deceleration parameter, given by $q_0$. If $q_0$ is positive, it is slowing down; if $q_0$ is negative, it is speeding up; if $q_0=0$, the rate of expansion is constant. It is given that the dark energy is the result of a cosmological constant and the parameter corresponding to deceleration is evaluated using $q_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }$. (The density parameter ${\Omega }_0$ is equal to ${\Omega }_m+{\Omega }_{\Lambda }$.)

Explanation of Solution

Given data:

The expansion of the universe can be expressed in terms of the deceleration parameter. It is denoted by $q_0$, if $q_0$ is positive, it is slowing down; if $q_0$ is negative, it is speeding up; if $q_0=0$, the rate of expansion is constant.

Formula used:

The given formula is written as,

$q_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }$

Here, ${\Omega }_m$ is the matter density parameter, ${\Omega }_{\Lambda }$ is the dark energy density parameter, and $q_0$ is the deceleration parameter.

Explanation:

If the dark density is zero, then there is no cosmological constant, ${\Omega }_0=1$ from table 25-2 and ${\Omega }_{\Lambda }=0$.

Recall the given formula.

$q_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }$

Substitute 0 for ${\Omega }_{\Lambda }$ and 1 for ${\Omega }_m$.

$\begin{array}{c}{q}_0=\frac{1}{2}\left(1\right)-\frac{3}{2}\left(0\right)\\ =0.50\end{array}$

Conclusion:

Therefore, the expansion of the universe has slowed down because $q_0$ is positive.

(b)

To determine

The present day universe has a deceleration parameter which can be evaluated using the value of ${\Omega }_m$ and ${\Omega }_{\Lambda }$. The value of ${\Omega }_m$ and ${\Omega }_{\Lambda }$ is used from the combined observations given in table 25-2. Also, based on this, it is to be determined whether the universe is expanding slowly or speedily. It is given that the expansion of universe can be expressed in terms of the deceleration parameter, which is represented by $q_0$. If $q_0$ is positive, it is slowing down; if $q_0$ is negative, it is speeding up; if $q_0=0$, the rate of expansion proceeds is constant. It is given that the dark energy is the result of a cosmological constant and the parameter corresponding to deceleration is evaluated using $q_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }$. (The density parameter ${\Omega }_0$ is equal to ${\Omega }_m+{\Omega }_{\Lambda }$.)

Answer to Problem 52Q

Solution:

$-0.64$, speeding up.

Explanation of Solution

Given data:

Use table 25-2 to get,

$\begin{array}{l}{\Omega }_m=.24\\ {\Omega }_{\Lambda }=.76\end{array}$

Formula used:

The provided relation can be written as:

$\begin{array}{c}{q}_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }\\ =\frac{1}{2}\left({\Omega }_m+{\Omega }_{\Lambda }\right)-\frac{3}{2}{\Omega }_{\Lambda }\end{array}$

Here, ${\Omega }_m$ is the matter density parameter, ${\Omega }_{\Lambda }$ is density parameter of the dark energy and so the density parameter ${\Omega }_0$ is equal to ${\Omega }_m+{\Omega }_{\Lambda }$. And, $q_0$ is the deceleration parameter.

Explanation:

Recall the provided relation.

$q_0=\frac{1}{2}\left({\Omega }_m+{\Omega }_{\Lambda }\right)-\frac{3}{2}{\Omega }_{\Lambda }$

Substitute 0.76 for ${\Omega }_{\Lambda }$ and 0.24 for ${\Omega }_m$.

$\begin{array}{c}{q}_0=\frac{1}{2}\left(.24+.76\right)-\frac{3}{2}\left(.76\right)\\ =0.5-1.14\\ =-0.64\end{array}$

Conclusion:

Therefore, the expansion of the universe is speeding up because $q_0$ is negative.

(c)

To determine

The value of ${\Omega }_{\Lambda }$ for a universe having the value of ${\Omega }_m$ same as that of our universe. Such a universe expands at a constant rate. Also, determine whether the universe is matter dominant or dark energy dominant. It is given that the expansion of the universe can be written in terms of the deceleration parameter, which is given by $q_0$. If $q_0$ is positive, it is slowing down; if $q_0$ is negative, it is speeding up; if $q_0=0$, the rate of expansion is constant. It is also given that dark energy is the result of a cosmological constant and the parameter corresponding to deceleration is evaluated using $q_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }$. (The density parameter ${\Omega }_0$ is equal to ${\Omega }_m+{\Omega }_{\Lambda }$).

Answer to Problem 52Q

Solution:

0.12, matter dominant.

Explanation of Solution

Given data:

The universe is expanding at a constant rate.

Formula used:

The provided relation can be written as:

$\begin{array}{c}{q}_0=\frac{1}{2}{\Omega }_0-\frac{3}{2}{\Omega }_{\Lambda }\\ =\frac{1}{2}\left({\Omega }_m+{\Omega }_{\Lambda }\right)-\frac{3}{2}{\Omega }_{\Lambda }\end{array}$

Here, ${\Omega }_m$ is the matter density parameter, ${\Omega }_{\Lambda }$ is density parameter of the dark energy and so the density parameter ${\Omega }_0$ is equal to ${\Omega }_m+{\Omega }_{\Lambda }$. And, $q_0$ is the deceleration parameter.

Explanation:

The expansion of the universe is happening at a constant rate. So, the value of $q_0=0$.

Recall the provided relation.

$q_0=\frac{1}{2}\left({\Omega }_m+{\Omega }_{\Lambda }\right)-\frac{3}{2}{\Omega }_{\Lambda }$

Rearrange as,

${\Omega }_{\Lambda }=\frac{{\Omega }_m-2q_0}{2}$

Substitute 0 for $q_0$ and 0.24 for ${\Omega }_m$.

$\begin{array}{c}{\Omega }_{\Lambda }=\frac{.24-2\left(0\right)}{2}\\ =0.12\end{array}$

Conclusion:

Therefore, the dark density parameter is small. So, the universe is matter dominant.