Chapter 26, Problem 37Q

To determine

The proof that the Jeans length for the universe at the era of recombination was about $100\text{ly}$ and the total mass contained in a sphere with this diameter was about $4\times {10}^5{\text{M}}_{\odot }$.

Answer to Problem 37Q

The proof that the that the Jeans length for the universe at the era of recombination was about $100\text{ly}$ and the total mass contained in a sphere with this diameter was about $4\times {10}^5{\text{M}}_{\odot }$ is as stated below.

Explanation of Solution

Given:

The temperature of the early universe is, $T=3000\text{K}$.

The density of the universe is, ${\rho }_{\text{m}}={10}^{-18}\text{kg}/{\text{m}}^3$.

Formula Used:

The expression for Jeans length is given by,

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

Here, $k$ is the Boltzmann constant and $m$ is the mass of a single particle in the gas.

The expression for volume is given by,

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

The expression for mass density is given by,

$\rho =\frac{m}{V}$

Rewrite the above expression for mass.

$\begin{array}{c}m=\rho V\\ =\rho \left(\frac{4}{3}\pi r^3\right)\end{array}$

Calculation:

The Jeans length is calculated as,

$\begin{array}{c}{L}_{\text{J}}=\sqrt{\frac{\pi kT}{mG{\rho }_{\text{m}}}}\\ =\sqrt{\frac{\pi \left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(3000\text{K}\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^3\right)\left({10}^{-18}\text{kg}/{\text{m}}^3\right)}}\\ =\sqrt{\frac{1.30\times {10}^{-19}}{1.11\times {10}^{-55}}}\text{m}\\ =1.08\times {10}^{18}\text{m}\end{array}$

Solve further,

$\begin{array}{c}{L}_{\text{J}}=\left(1.08\times {10}^{18}\text{m}\times \frac{1\text{ly}}{9.46\times {10}^{15}\text{m}}\right)\\ =114.16\text{ly}\\ \approx 100\text{ly}\end{array}$

The mass contained in the sphere is calculated as

$\begin{array}{c}m=\rho \left(\frac{4}{3}\pi r^3\right)\\ =\left({10}^{-18}\text{kg}/{\text{m}}^3\right)\left(\frac{4}{3}\pi {\left(\frac{1.08\times {10}^{18}\text{m}}{2}\right)}^3\right)\\ =\left({10}^{-18}\text{kg}/{\text{m}}^3\right)\left(6.60\times {10}^{53}{\text{m}}^3\right)\\ =6.60\times {10}^{35}\text{kg}\end{array}$

Solve further,

$\begin{array}{c}m=\left(6.60\times {10}^{35}\text{kg}\times \frac{1{\text{M}}_{\odot }}{1.99\times {10}^{30}\text{kg}}\right)\\ =3.32\times {10}^5{\text{M}}_{\odot }\end{array}$

This value is nearly equal to $4\times {10}^5{\text{M}}_{\odot }$.

Conclusion:

Therefore, it is proved that the Jeans length for the universe at the era of recombination was about $100\text{ly}$ and the total mass contained in a sphere with this diameter was about $4\times {10}^5{\text{M}}_{\odot }$.