Question

Equation (12-7) assumes a matter-dominated universe in which the energy density of radiation is insignificant. This situation prevails today and has to do with the different rates at which the densities of matter and radiation vary with the size of the universe. Matter density is simply inversely proportional to the volume, obeying ${\mathbf{\rho }}_{\mathbf{m}}{\mathbf{IR}}^{\mathbf{3}}$ , where${\mathbf{\rho }}_{\mathbf{m}}$ is the matter density now. Radiation density, however, would be proportional to $\mathit{I}\mathbf{/}{\mathit{R}}^{\mathbf{4}}$(Not only does the volume increase, but also all wavelengths are stretched in proportion to R. lowering the energy density by the extra factor.) This density drops faster As the universe grows, but it also grows more quickly in the backward time direction. In other words, long ago, the universe would have been radiation dominated. Show that if the function used for matter density in equation (12-7) is replaced by one appropriate to radiation, but retaining the assumption that K' and${\mathbf{\Omega }}_{\mathbf{A}}$ are both 0, then the scale factor $\mathit{R}$would grow as${\mathit{t}}^{\mathbf{1}\mathbf{/}\mathbf{2}}$

Short answer

The time dependence of the left-hand side and the right-hand side is the same when $R$ grows as $t^{-1/2}$.

For pure radiation, the constant should be $t^{1/2}$.

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

Calculate the derivative of the expression

Matter density is inversely proportional to the volume which means $\frac{{\rho }_M}{R^3}$.

On the other- hand the radiation density is proportional directly to $\frac{1}{{\mathrm{R}}^4}$.

${\left(\frac{\text{dR}/\text{dt}}{R}\right)}^2=\frac{1}{R^4}$

If the universe were critical, flat and free of cosmological constant.

The proposed solution is $R=\text{const}\times t^{1/2}$.

Perform the derivative, $\frac{dR}{dt}=\left(\frac{t^{-1/2}}{2}\right)\text{const}$.

So, obtain:

$\begin{array}{l}\frac{dR/dt}{R}=\frac{\left(t^{-12}\right)\text{const}}{2\left(t^{12}\right)\text{const}}\\ \frac{dR/dt}{R}=\frac{1}{2}{t}^{-1}\end{array}$

Then calculate as shown below

$\begin{array}{c}{\left(\frac{dR/dt}{R}\right)}^2={\left(\frac{1}{2}{t}^{-1}\right)}^2\\ {\left(\frac{dR/dt}{R}\right)}^2=\frac{1}{4}{t}^{-2}\\ {\left(\frac{dR/dt}{R}\right)}^2\left(Const.\right)=\left(t^{-2}\right)\\ {\left(\frac{dR/dt}{R}\right)}^2=\frac{1}{R^4}\end{array}$

Thus, the scale factor R is $t^{1/2}$