# Density parameter as a function of scale factor (cosmology)

I'm trying to recreate the following plot of the density parameters $$\Omega_i$$ as a function of the scale factor:

(Taken from here)

which uses current values for $$\Omega_{R0}=5\times10^{-5}$$, $$\Omega_{M0} = 0.3$$, and $$\Omega_{\Lambda0}=0.7$$.

I know the scale factor is defined as $$\Omega_i = \frac{8\pi G}{3H^2}\rho_i.$$ From the Friedman equation, we also have $$\Omega_R + \Omega_M + \Omega_\Lambda = 1$$ which is clear from the plot.

However, I'm still confused as to where the $$a$$ dependence is coming from. I know part of it comes from writing the densities as power laws: $$\rho_i \sim a^{-n}$$ where $$n=0,3,$$ or $$4$$ for $$\Lambda, M$$ and $$R$$ respectively. But clearly this is not enough to reproduce the plot. I thought then maybe it is from the $$H=\dot{a}/a$$ but if I just want to vary $$a$$ what would happen with the $$\dot{a}$$.

It seems I'm missing something. Should I not be trying to derive $$\Omega_i(a)$$ from the definition of $$\Omega$$? It seems like there should be some sort of differential equation I should be able to solve to get $$\Omega_i(a)$$ using $$\Omega_i(1)=\Omega_{i0}$$ as an initial condition.

Sorry if this is asking too much but I am just extremely confused.

Using the Friedmann equation (in terms of $$H$$) and the respective scale-factor dependence of energy densities ($$\rho_i$$, where $$i = M, R, \Lambda$$), we can obtain the scale-factor dependence of density parameters as follows: \begin{align*} \Omega_M(a) &= \frac{8\pi G}{3 (H(a))^2}\left(\rho_M(a)\right)^{-3} \\ &= \frac{8\pi G}{3 H_0^2}\frac{\rho_{M0}a^{-3}}{(\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4})} \\ \implies\Omega_M(a) &= \frac{\Omega_{M0}a^{-3}}{\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4}} \end{align*}

\begin{align*} \Omega_R(a) &= \frac{8\pi G}{3 (H(a))^2}\left(\rho_R(a)\right)^{-4} \\ &= \frac{8\pi G}{3 H_0^2}\frac{\rho_{R0}a^{-4}}{(\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4})} \\ \implies\Omega_R(a) &= \frac{\Omega_{R0}a^{-4}}{\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4}} \end{align*}

\begin{align*} \Omega_\Lambda(a) &= \frac{\Lambda}{3 (H(a))^2} \\ &= \frac{\Lambda}{3 H_0^2}\frac{1}{(\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4})} \\ \implies\Omega_\Lambda(a) &= \frac{\Omega_{\Lambda0}}{\Omega_{\Lambda0} + (1 - \Omega_{\Lambda0} - \Omega_{M0} - \Omega_{R0})a^{-2} + \Omega_{M0}a^{-3} + \Omega_{R0}a^{-4}} \end{align*}

Using these expressions and the given current values of density parameters: $$(\Omega_{R0},\,\Omega_{M0},\,\Omega_{\Lambda0}) \equiv (5\times10^{-5},\,0.3,\,0.7)$$, we can plot these functions in Mathematica to arrive at the following semi-log graph:

The code for the plot is given below:

oR0 = 5*10^-5; oM0 = 0.3; oL0 = 0.7;

oR[a_] := (oR0 a^-4)/(
oL0 + (1 - oM0 - oL0 - oR0) a^-2 + oM0 a^-3 + oR0 a^-4);

oM[a_] := (oM0 a^-3)/(
oL0 + (1 - oM0 - oL0 - oR0) a^-2 + oM0 a^-3 + oR0 a^-4);

oL[a_] := oL0/(oL0 + (1 - oM0 - oL0 - oR0) a^-2 + oM0 a^-3 + oR0 a^-4);

oRPlot = LogLinearPlot[oR[a], {a, 10^-30, 10^10}, PlotRange -> All,
PlotStyle -> Magenta];
oMPlot = LogLinearPlot[oM[a], {a, 10^-30, 10^10}, PlotRange -> All,
PlotStyle -> Blue];
oLPlot = LogLinearPlot[oL[a], {a, 10^-30, 10^10}, PlotRange -> All,
PlotStyle -> Orange];

plot = Show[oMPlot, oLPlot, oRPlot, Frame -> True, GridLines -> {{0}, {1}},
AxesStyle -> Opacity[1],
FrameLabel -> {{"\!$$\*SubscriptBox[\(\[CapitalOmega]$$, \
$$i$$]\)(a)", None}, {"a", None}}, ImageSize -> Large];
Legended[plot,
LineLegend[{Magenta, Blue,
Orange}, {"\!$$\*SubscriptBox[\(\[CapitalOmega]$$, $$R$$]\)",
"\!$$\*SubscriptBox[\(\[CapitalOmega]$$, $$M$$]\)",
"\!$$\*SubscriptBox[\(\[CapitalOmega]$$, \
$$\[CapitalLambda]$$]\)"}]]

• Here, $c$ is taken to be 1.
– P_0
Sep 29, 2021 at 10:46