Chapter 25, Problem 33Q

(a)

To determine

${\rho }_{rad}$, if it is supposed that the present-day temperature of the cosmic background radiation is increased by a factor of 100. The temperature will get increased from 2.725 K to 272.5 K.

Answer to Problem 33Q

Solution:

$4.6\times {10}^{-3}{\text{ kg/m}}^3$

Explanation of Solution

Given data:

The temperature is increased by a factor of 100.

The temperature is increased from 2.725 K to 272.5 K.

Formula used:

The expression for mass density of radiation is:

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

Here, ${\rho }_{rad}$ is the mass density of radiation, $T$ is the temperature of the radiation, c is the speed of light having value $3\times {10}^8\text{ m/s}$, and $\sigma$ is the Stefan-Boltzmann constant having value $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$.

Explanation:

Recall the formula for mass density of radiation:

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

Substitute 272.5 K for T, $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$ and $3\times {10}^8\text{ m/s}$ for c.

$\begin{array}{c}{\rho }_{rad}=\frac{4\left(5.67\times {10}^{-8}{\text{ W/m}}^2\cdot {\text{K}}^4\right){\left(272.5\text{ K}\right)}^4}{{\left(3\times {10}^8\text{ m/s}\right)}^3}\\ =\frac{1250.57}{2.7\times {10}^{25}}\\ =4.6\times {10}^{-23}{\text{ kg/m}}^3\end{array}$

Conclusion:

Therefore, the density of the radiation, ${\rho }_{rad}$, is $4.6\times {10}^{-23}{\text{ kg/m}}^3$.

(b)

To determine

Whether it would be more accurate to describe our universe as matter-dominated or radiation-dominated. If the average density of matter, ${\rho }_m$, remains unchanged, it is supposed that the present-day temperature of the cosmic background radiation is increased by a factor of 100. The temperature is increased from 2.725 K to 272.5 K.

Answer to Problem 33Q

Solution:

Radiation-dominated.

Explanation of Solution

Given data:

The temperature is increased by a factor of 100 from 2.725 K to 272.5 K.

Formula used:

The expression for mass density of radiation is:

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

Here, ${\rho }_{rad}$ is the mass density of a radiation, $T$ is the temperature of the radiation, c is the speed of light having value $3\times {10}^8\text{ m/s}$ and $\sigma$ is the Stefan-Boltzmann constant having value $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$.

Explanation:

Recall the expression for mass density of radiation.

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

Substitute 272.5 K for T, $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$ and $3\times {10}^8\text{ m/s}$ for c.

$\begin{array}{c}{\rho }_{rad}=\frac{4\left(5.67\times {10}^{-8}{\text{ W/m}}^2\cdot {\text{K}}^4\right){\left(272.5\text{ K}\right)}^4}{{\left(3\times {10}^8\text{ m/s}\right)}^3}\\ =\frac{1250.57}{2.7\times {10}^{25}}\\ =4.6\times {10}^{-23}{\text{ kg/m}}^3\end{array}$

From the calculation, ${\rho }_{rad}$ is greater than ${\rho }_m$ (since ${\rho }_m$ is unchanged even after the temperature is increased by a factor of 100 while ${\rho }_{rad}$ increases as the temperature increases). So, the universe would be radiation-dominated.

Conclusion:

Therefore, the universe would be radiation-dominated.