Chapter 33, Problem 32P

(a)

To determine

The size of the universe, compared to today, at given time.

Answer to Problem 32P

The size of the universe, compared to today, at $t={\text{10}}^6\text{ yr}$ is $3\times {10}^{-3}$ .

Explanation of Solution

Given:

Time of universe is $t{\text{ =10}}^6\text{ yr}$ .

Assuming universe has a current size of ${10}^{10}\text{ ly}$

Formula used:

Since scale of the Universe is inversely proportional to the absolute temperature. Considering the current scale to be 1 and current temperature to be $3\text{ K}$ . Product of both scale and the temperature is 3. Taking help from the figure below:

  

Calculation:

As shown in the figure, at $t{\text{ =10}}^6\text{ yr}$ , temperature is about $1000\text{ K}$ . Now, calculating the scale

  $\begin{array}{l}(\text{Scale})(\text{Temperature})=3\\ \text{Scale =}\frac{3}{\text{Temperature}}=\frac{3}{1000}=3\times {10}^{-3}\approx 3\times {10}^7\text{ ly}\end{array}$

Conclusion:

The size of the universe, compared to today, at given time is $3\times {10}^{-3}$ .

(b)

To determine

The size of the universe, compared to today, at given time.

Answer to Problem 32P

The size of the universe, compared to today, at $t\text{ =1 s}$ is $3\times {10}^{-10}$ .

Explanation of Solution

Given:

Time of the universe is $t\text{ =1 s}$ . Assuming universe has a current size of ${10}^{10}\text{ ly}$

Formula used:

Since scale of the universe is inversely proportional to the absolute temperature. Considering the current scale to be 1 and current temperature to be $3\text{ K}$ . Product of both scale and the temperature is 3. Taking help from the figure below:

  

Calculation:

As shown in the figure, at $t\text{ =1 s}$ , temperature is about ${10}^{10}\text{ K}$ . Now, calculating the scale

  $\begin{array}{l}(\text{Scale})(\text{Temperature})=3\\ \text{Scale =}\frac{3}{\text{Temperature}}=\frac{3}{{10}^{10}}=3\times {10}^{-10}\approx 3\text{ ly}\end{array}$

Conclusion:

The size of the universe, compared to today, at given time is $3\times {10}^{-10}$ .

(c)

To determine

The size of the universe, compared to today, at given time.

Answer to Problem 32P

The size of the universe, compared to today, at $t{\text{ =10}}^{-6}\text{ s}$ is $3\times {10}^{-13}$ .

Explanation of Solution

Given:

Time of the universe is $t{\text{ =10}}^{-6}\text{ s}$ .

Assuming universe has a current size of ${10}^{10}\text{ ly}$

Formula used:

Since scale of the universe is inversely proportional to the absolute temperature. Considering the current scale to be 1 and current temperature to be $3\text{ K}$ . Product of both scale and the temperature is 3. Taking help from the figure below:

  

Calculation:

As shown in the figure, at $t{\text{ =10}}^{-6}\text{ s}$ , temperature is about ${10}^{13}\text{ K}$ . Now, calculating the scale

  $\begin{array}{l}(\text{Scale})(\text{Temperature})=3\\ \text{Scale =}\frac{3}{\text{Temperature}}=\frac{3}{{10}^{13}}=3\times {10}^{-13}\approx 3\times {10}^{-3}\text{ ly}\approx 3\times {10}^{13}\text{ m}\end{array}$

Conclusion:

The size of the universe, compared to today, at given time is $3\times {10}^{-13}$ .

(d)

To determine

The size of the universe, compared to today, at given time.

Answer to Problem 32P

The size of the universe, compared to today, at $t={\text{10}}^{-35}\text{ s}$ is $3\times {10}^{-27}$ .

Explanation of Solution

Given:

Time of universe is $t{\text{ =10}}^{-35}\text{ s}$ .

Assuming universe has a current size of ${10}^{10}\text{ ly}$

Formula used:

Since scale of the universe is inversely proportional to the absolute temperature. Considering the current scale to be 1 and current temperature to be $3\text{ K}$ . Product of both scale and the temperature is 3. Taking help from the figure below:

  

Calculation:

As shown in the figure, at $t={\text{10}}^{-35}\text{ s}$ , temperature is about ${10}^{27}\text{ K}$ . Now, calculating the scale

  $\begin{array}{l}(\text{Scale})(\text{Temperature})=3\\ \text{Scale =}\frac{3}{\text{Temperature}}=\frac{3}{{10}^{27}}=3\times {10}^{-27}\approx 3\times {10}^{-17}\text{ ly}\approx 0.3\text{ m}\end{array}$

Conclusion:

The size of the universe, compared to today, at given time is $3\times {10}^{-27}$ .