Equation for Hubble Value as a function of time

Question

I am trying to write the equation for the situation where the Hubble parameter $H$ would be changing over time. In other words, it would represent an accelerated expansion of the Universe. That is, $H$ can no longer be the simple $H=1/t$. In the new equation, I should be able to plug a future time and see what the Hubble Value will be in that future.

I think I got most of the concepts right. First of all I understand that the key to the problem is in $H=a˙(t)/a(t)$. Where $a(t)$ is the scale factor from the Friedman equations. I also understand that if $H$ is changing then $a¨(t)> 0$ and also $H'(a)>0$. But I'm still very uncomfortable when my pencil meets the paper. The Friedman equations are not stated as a function of $t$, but as a function of $a$ where $a$ is the time scale, and frankly, I don't know how to use the time scale factor.

In any event, here is my poor attempt to do it. According to Wikipedia, one of the solutions of Friedman equation is (assume flat space k=0):

$a(t) = a_0 t ^{2/3(w+1)}$

Therefore:

$a'(t) = d(a_0 t ^{2/3(w+1)}) / dt$

$a'(t) = (2a_0/3(w+1)) t^{ -(1+3w)/3(w+1)}$

And I suppose that we could now substitute: $H=a˙(t)/a(t)$ with the above:

$H = (2a_0/3(w+1)) t^{ -(1+3w)/3(w+1)} / a_0 t ^{2/3(w+1)}$

Simplified:

$H = (2/3(w+1)) t^{-(3w+2)/3(w+1)}$

And $w$ is typically known from observation.

I will appreciate if someone can let me know if I am in the right path or totally derailed. I have a feeling that $a(t) = a_0 t ^{2/3(w+1)}$ was not the right place to start because if $w=-1$, then everything goes down the drain. But then again, in an accelerated expansion, $w$ would not equal -1. It would always be less than -1. Also, in the final equation, if $w<-1$ then H<0, which could not be right. So I'm not sure what to think.

Many thanks in advance,

Luis

Answer 1

The general solution works as follows:

We start with the Friedmann equation $$ \dot{a}^2 - \frac{8\pi G}{3}\rho a^2 = -kc^2, $$ with $k=0,\ 1,\ $or $-1$, and $\rho$ the total density. Since the right-hand side is constant, we can write $$ \dot{a}^2 - \frac{8\pi G}{3}\rho a^2 = \dot{a}_0^2 - \frac{8\pi G}{3}\rho_0 a_0^2, $$ where the subscript 0 denotes the present-day values. If we introduce the Hubble constant $$ H_0 = \frac{\dot{a}_0}{a_0} $$ and the present-day critical density $$ \rho_{c,0} = \frac{3H_0^2}{8\pi G}, $$ we get $$ \frac{\dot{a}^2}{a_0^2} - H_0^2\frac{\rho}{\rho_{c,0}} \frac{a^2}{a_0^2} = H_0^2 - H_0^2\frac{\rho_0}{\rho_{c,0}} $$ or $$ H^2 = \frac{\dot{a}^2}{a^2} = H_0^2\left[\frac{\rho}{\rho_{c,0}} + \frac{a_0^2}{a^2}\left(1 - \frac{\rho_0}{\rho_{c,0}}\right)\right]. $$ Now, there are three contributions to the total density: radiation, matter (normal and dark) and dark energy: $$ \rho = \rho_R + \rho_M + \rho_{\Lambda}. $$ These densities change over time as follows: the matter density decreases as the volume of the universe increases, so $\rho_M\sim a^{-3}$, as you'd expect. The radiation falls off as $\rho_R\sim a^{-4}$ (the extra factor is due to redshift). And in the Standard Model, the dark energy remains constant: $\rho_{\Lambda} = \text{const}$. In other words, $$ \begin{align} \rho_R a^4 &= \rho_{R,0}\, a_0^4,\\ \rho_M a^3 &= \rho_{M,0}\, a_0^3,\\ \rho_\Lambda &= \rho_{\Lambda,0}, \end{align} $$ and finally, with the notations $$ \Omega_{R,0} = \frac{\rho_{R,0}}{\rho_{c,0}},\quad \Omega_{M,0} = \frac{\rho_{M,0}}{\rho_{c,0}},\quad \Omega_{\Lambda,0} = \frac{\rho_{\Lambda,0}}{\rho_{c,0}},\\ \Omega_{K,0} = 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0}, $$ we find $$ H(a) = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}, $$ where we used the convention $a_0=1$. Also note that $$ \dot{a} = H_0\sqrt{\Omega_{R,0}\,a^{-2} + \Omega_{M,0}\,a^{-1} + \Omega_{K,0} + \Omega_{\Lambda,0}\,a^2},\\ \ddot{a} = -\frac{1}{2}H_0^2\left(2\,\Omega_{R,0}\,a^{-3}+\Omega_{M,0}\,a^{-2} -2\,\Omega_{\Lambda,0}\,a\right). $$ The latest values of the parameters, obtained from the Planck mission, are $$ H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} = 9.24\times 10^{-5},\qquad\Omega_{M,0} = 0.315,\\ \Omega_{\Lambda,0} = 0.685,\qquad\Omega_{K,0} = 0. $$ So now we have the Hubble parameter as a function of the scale radius $a$. How can we convert this into a function of time? From $$ \dot{a} = \frac{\text{d}a}{\text{d}t} $$ we get $$ \text{d}t = \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}, $$ so that $$ \begin{align} t(a) &= \int_0^a \frac{a'\,\text{d}a'}{a'^2H(a')}\\ &= \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}}. \end{align} $$ Inverting this relation, we get $a(t)$. Unfortunately, this inversion has to be done numerically. And finally, $$ H(t) = H(a(t)). $$

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P.S. The solution that you mentioned is the case where radiation and matter are negligible, and dark energy has a more general form (called quintessence): $$ \rho_R=\rho_M=0,\quad \rho_\Lambda = \rho_{\Lambda,0}\,a^{-3(1+w)}, $$ where $w=-1$ corresponds with the normal case of a cosmological constant. In this case, for a universe with no curvature, $$ H^2 = H_0^2\,a^{-3(1+w)},\qquad t(a) = \frac{1}{H_0}\int_0^a a'^{(1+3w)/2}\,\text{d}a', $$ with solution $a\sim t^{2/(3+3w)}$, for $w>-1$. Solutions with $w\leqslant-1$ have no big bang, i.e. the lower bound in the integral $t(a)$ cannot be zero.

In any case, these are not accurate descriptions of our universe, since they ignore the contributions of matter and radiation.