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):

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


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

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

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

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


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

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,



1 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, $$ H(t) = H(a(t)). $$

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.

  • $\begingroup$ Many thanks for the thorough answer. The resolution really makes sense. But I'm still a bit confused about w<=-1 because I'm interested on the fact that the latest measurements from Type 1a SN give a w=-1.18. How can we get 'w' back into the mix and still have a big bang? Where do we get the equation that satisfies such a value? Is there no such thing? $\endgroup$
    – Luis
    Commented Jan 6, 2014 at 6:27
  • $\begingroup$ @Luis We have a big bang because $\Omega_{R,0}>0$ and $\Omega_{M,0}>0$, and these terms become dominant when $a$ is very small, because then $a^{1-3w}\ll a$. So the integrand in $t(a)$ approaches $a/(\sqrt{\Omega_{R,0} + \Omega_{M,0}a})$ as $a$ approaches 0, which is well-behaved. $\endgroup$
    – Pulsar
    Commented Jan 6, 2014 at 8:58
  • $\begingroup$ OK, I see that everything works going backwards to the BB, but still not clear about going forward in a model where we would have an accelerated expansion. You correctly reminded me that in the Standard Model DE remains constant. That would not hold for an accelerated expansion, would it? Wouldn't be DE increasing in relation to 'a'? Wouldn't we have to deviate from the Standard Model? And again, how would I account for when w<-1? Are we not then forced to use a quintessence model? Look forward your reply. $\endgroup$
    – Luis
    Commented Jan 6, 2014 at 17:32
  • $\begingroup$ @Luis A model with constant DE has accelerated expansion. You can see this in the expression for $\ddot{a}$: its current value ($a=1$) is $\ddot{a}\approx (1/2)H_0^2(2\Omega_{\Lambda,0}-\Omega_{M,0}) > 0$. In fact, when $a$ becomes very large, we get $\dot{a}\approx aH_0\sqrt{\Omega_{\Lambda,0}}$, which approximates an exponential expansion. $\endgroup$
    – Pulsar
    Commented Jan 7, 2014 at 9:26
  • $\begingroup$ Let me make sure I understand. DE is constant in relation to the volume whilst H is increasing with 'a'. However, DE is changing in relation to matter and radiation. That is, DE will become dominant in a distant future whilst matter and radiation will become minuscule. Correct? $\endgroup$
    – Luis
    Commented Jan 7, 2014 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.