Question

Verify that if the universe were critical $\mathbf{(}{\mathbf{\Omega }}_{\mathbf{M}}\mathbf{=}\mathbf{1}\mathbf{)}$. "flat”$\mathbf{(}{\mathbf{K}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathbf{0}\mathbf{)}$, and free of a cosmological constant$\mathbf{(}{\mathbf{\Omega }}_{\mathbf{\Lambda }}\mathbf{=}\mathbf{0}\mathbf{)}$, then equation ${\left(\frac{\mathbf{dR}\mathbf{/}\mathbf{dt}}{\mathbf{R}}\right)}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}^{\mathbf{3}}}$would be satisfied by a scale factor$\mathbf{R}\left(t\right)$ of${\left[\frac{3}{2}\left(t-\frac{1}{3}\right)\right]}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{.}$

Short answer

The equation ${\left(\frac{\mathrm{dR}/\mathrm{dt}}{\mathrm{R}}\right)}^2=\frac{1}{{\mathrm{R}}^3}$ is proved.

Step-by-step solution

Given data

The Friedmann equation can be expressed as ${\left(\frac{\mathrm{dR}/\mathrm{dt}}{\mathrm{R}}\right)}^2=\frac{{\mathrm{\Omega }}_{\mathrm{M}}}{{\mathrm{R}}^3}+{\mathrm{\Omega }}_{\text{A}}-\frac{{\mathrm{K}}^{\text{'}}}{{\mathrm{R}}^2}$.

Concept of Friedmann equation

If the universe were critical, flat and free of cosmological constant, the Friedmann equation is,

${\left(\frac{\mathrm{dR}/\mathrm{dt}}{R}\right)}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}^{\mathbf{3}}}$.

Differentiate the expression 

The proposed solution is$\mathrm{R}={\left(\frac{3}{2}\left(\mathrm{t}-\frac{1}{3}\right)\right)}^{2/3}$.

Differentiate the equation on both sides with respect to $t$.

$\begin{array}{l}\frac{\mathrm{dR}}{\mathrm{dt}}={\left(\frac{3}{2}\right)}^{2/3}\frac{2}{3}{\left(\mathrm{t}-\frac{1}{3}\right)}^{-1/3}\\ \frac{\mathrm{dR}}{\mathrm{dt}}={\left(\frac{3}{2}\left(\mathrm{t}-\frac{1}{3}\right)\right)}^{-1/3}\end{array}$

Substitute ${\left(\frac{3}{2}\left(\mathrm{t}-\frac{1}{3}\right)\right)}^{-1/3}$ for $\frac{\mathrm{dR}}{\mathrm{dt}}$ and ${\left(\frac{3}{2}\left(\mathrm{t}-\frac{1}{3}\right)\right)}^{2/3}$ for $\mathrm{R}$.

$\begin{array}{c}{\left(\frac{dR/dt}{R}\right)}^2={\left[{\left(\frac{3}{2}\left(t-\frac{1}{3}\right)\right)}^{-1/3}/{\left(\frac{3}{2}\left(t-\frac{1}{3}\right)\right)}^{2/3}]\right]}^2\\ {\left(\frac{dR/dt}{R}\right)}^2={\left[{\left(\frac{3}{2}\left(t-\frac{1}{3}\right)\right)}^{-1}\right]}^2\\ {\left(\frac{dR/dt}{R}\right)}^2={\left(\frac{3}{2}\left(t-\frac{1}{3}\right)\right)}^{-2}\\ {\left(\frac{dR/dt}{R}\right)}^2={\left(\frac{3}{2}\left(t-\frac{1}{3}\right)\right)}^{\left(\frac{2}{3}\right)(-3)}\end{array}$

Similarly, calculate further as shown below:

$\begin{array}{l}{\left(\frac{dR/dt}{R}\right)}^2=R^{-3}\\ {\left(\frac{dR/dt}{R}\right)}^2=\frac{1}{R^3}\end{array}$

Therefore, the equation${\left(\frac{\mathrm{dR}/\mathrm{dt}}{\mathrm{R}}\right)}^2=\frac{1}{{\mathrm{R}}^3}$ is proved.