Chapter 26, Problem 37Q

To determine

To show: The Jeans length of the early universe was 100 ly and the total mass of the universe was $4\times {10}^5{\text{ M}}_{\odot }$, if the universe was contained in a sphere. The temperature of the universe was 3000 K and the density of the universe was ${10}^{-18}{\text{ kg/m}}^3$ in the early universe.

Explanation of Solution

Given data:

Temperature of the early universe is $3000\text{ K}$ and ${\text{ρ}}_{\text{m}}$ is the density of the universe is ${10}^{-18}{\text{ kg/m}}^3$.

Formula used:

According to Jeans, an object will only grow if the fluctuation in density crosses the density that is called Jean Length. Jeans length is calculated by the expression.

$L_{\text{J}}=\sqrt{\frac{\pi kT}{mG{\rho }_{\text{m}}}}$

Here, k is Boltzmann constant, T is the temperature, m is the mass of a single particle in the gas, G is the universal gravitational constant, and ${\rho }_{\text{m}}$ is the average density of matter in gas.

The expression for mass is:

$m=V\rho$

Here, m is the mass, V is the volume, and $\rho$ is the density.

The volume of a sphere is:

$V=\frac{4}{3}\pi r^3$

Here, r is the radius of the sphere.

The relation between light year and meter is:

$1\text{ ly}=9.46\times {10}^{15}\text{ m}$

The mass of the sun is denoted as:

$1{\text{ M}}_{\odot }=1.99\times {10}^{30}\text{ kg}$

Explanation:

Consider the value of Boltzmann constant, universal gravitational constant, and mass of hydrogen as $1.38\times {10}^{-23}\text{ J/K}$, $6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$, and $1.67\times {10}^{-27}\text{ kg}$, respectively.

Recall the expression for calculating the Jean length.

$L_{\text{J}}=\sqrt{\frac{\pi kT}{mG{\rho }_{\text{m}}}}$

Substitute $1.38\times {10}^{-23}\text{ J/K}$ for k, $6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ for G, $1.67\times {10}^{-27}\text{ kg}$ for m, ${10}^{-18}{\text{ kg/m}}^3$ for ${\rho }_{\text{m}}$, and 3000 K for T.

$\begin{array}{c}{L}_{\text{J}}=\sqrt{\frac{\pi \times \left(1.38\times {10}^{-23}\text{ J/K}\right)\times 3000\text{ K}}{\left(1.67\times {10}^{-27}\text{ kg}\right)\times \left(6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2\right)\times \left({10}^{-18}{\text{ kg/m}}^3\right)}}\\ =1.0805\times {10}^{18}\text{ m}\end{array}$

Since the length of the universe is given as 100 ly, convert the above Jeans length into ly by using the conversion formula.

$\begin{array}{c}{L}_{\text{J}}=1.0805\times {10}^{18}\text{ m}\times \left(\frac{1\text{ ly}}{9.46\times {10}^{15}\text{ m}}\right)\\ =114.21\text{ ly}\end{array}$

Consider the sphere of diameter equal to Jean length. Recall the expression for volume.

$V=\frac{4}{3}\pi r^3$

Substitute $\left(\frac{1.0805\times {10}^{18}\text{ m}}{2}\right)$ for r.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(\frac{1.0805\times {10}^{18}\text{ m}}{2}\right)}^3\\ =6.61\times {10}^{53}{\text{ m}}^3\end{array}$

Recall the expression of mass.

$m=V\rho$

Substitute $6.61\times {10}^{53}{\text{ m}}^3$ for V and ${10}^{-18}{\text{ kg/m}}^3$ for ${\rho }_{\text{m}}$, and use conversion expression to express it in terms of the mass of the sun.

$\begin{array}{c}m=6.61\times {10}^{53}{\text{ m}}^3\times {10}^{-18}{\text{ kg/m}}^3\\ =\left(0.66\times {10}^{36}\text{ kg}\right)\left(\frac{1{\text{ M}}_{\odot }}{1.99\times {10}^{30}\text{ kg}}\right)\\ =3.3\times {10}^5{\text{ M}}_{\odot }\end{array}$

Conclusion:

Therefore, by substituting the condition prevalent in the early universe, it can be concluded that the Jeans length for the early universe is nearly 100 ly and the total mass nearly is $4\times {10}^5{\text{ M}}_{\odot }$.