0
$\begingroup$

I am trying to obtain the expression for the scale factor $a(t)$ in the Einstein-de Sitter universe, that is, in a universe with a single fluid component, which corresponds to matter. To do that, I start with the Friedmann equation, simplified for the case of a single fluid component, which is the following:

$$\dfrac{\dot{a}}{a}=H_0\sqrt{\Omega_i}\ a^{-\frac{3}{2}(1+w_i)}$$

where $\Omega_i\equiv\Omega_{i,0}=\rho_{i,0}/\rho_{crit,0}$ is the density of the fluid $i$ today divided by the critical density today, considering $i=r,m,\Lambda$ (corresponding respectively to radiation, matter and dark energy). In this case we take $i=m$ (a matter dominated universe), for which $w_i\equiv w_m=0$. I am trying to reproduce the result given in the book Cosmology by Daniel Baumann as the solution to this equation, which is the following:

$$a(t)\propto t^{2/3}$$

Since in this context we assume the condition $a(t_0)=1$, the book mentions later on that, in fact:

$$a(t)=\bigg(\dfrac{t}{t_0}\bigg)^{2/3}$$

However, I see no way to arrive to this conclusion. My calculations are the following:

$$\dfrac{\dot{a}}{a}=H_0 a^{-3/2}\ \ \ \Rightarrow\ \ \ \dot{a}a^{1/2}=H_0\ \ \ \ \Rightarrow\ \ \ \dfrac{da}{dt}a^{1/2}=H_0\ \ \ \Rightarrow\ \ \ \int a^{1/2}da=\int H_0dt\ \ \ $$

$$\Rightarrow\ \ \ \dfrac{a^{3/2}}{3/2}=H_0t+C\ \ \ \Rightarrow a(t)=\bigg(\dfrac{3}{2}H_0t+\dfrac{3}{2}C\bigg)^{2/3}=\bigg(\dfrac{3}{2}H_0t+C_0\bigg)^{2/3}$$

If I impose the condition $a(t_0)=1$, what I obtain is:

$$a(t_0)=1\ \ \ \Rightarrow\ \ \ 1=\bigg(\dfrac{3}{2}H_0t_0+C_0\bigg)^{2/3}\ \ \ \Rightarrow\ \ \ \dfrac{3}{2}H_0t_0+C_0=1\ \ \ \Rightarrow\ \ \ C_0=1-\dfrac{3}{2}H_0t_0$$

Therefore, the final result would be:

$$a(t)=\bigg(\dfrac{3}{2}H_0t-\dfrac{3}{2}H_0t_0+1\bigg)^{2/3}$$

But this is not what the book proposes, and honestly, it looks ugly. I have no idea what is happening Is there any way to obtain what the book states, or maybe the formulas in the book are just approximations used in cosmology? They seem to be neglecting the additive constant, and I am puzzled as to how I can get a multiplicative integration constant instead of an additive one, so that I have something of the form $a(t)=(Ct)^{2/3}$ to later get $C=1/t_0$ by imposing the condition $a(t_0)=1$.

Any help would be greatly appreciated to solve this mystery!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

The idea is that you start with an Ansatz that $a$ is a power of $(t/t_0)$. Then you have to solve for $t_0$ which gives $t_0 = \frac{2}{3H_0}$. Plugging that into your solution precisely cancels the constant!

The fact that you can make this Ansatz is because in this case the Friedman equations are of the form $ \dot a = a^q $ for some power q.

Improve this answer
$\endgroup$
6
  • $\begingroup$ I don't quite understand. The fact that $\dot{a}=a^q$ does no imply that $a$ can be written as a power of $t/t_0$, because there could be an additive constant. Also, $t_0$ is the age of the universe, so it shouldn't be a free parameter for which we can solve, but rather a fixed number. $\endgroup$
    – Wild Feather
    Commented Apr 10, 2023 at 8:19
  • $\begingroup$ @WildFeather No it is not a free parameter. It is closely related to $H_0$. How those two are related depends on the type on universe, but in this case you find the relation that I gave. Given that the universe contains a certain amount of matter and given that $H_0$ is of a given value, there is only one possible age which does not violate the Friedmann equation. $\endgroup$
    – Tiagotje
    Commented Apr 10, 2023 at 9:58
  • $\begingroup$ In the book that I am using (Cosmology, by Baumann), the author first obtains the expression $a(t)=(t/t_0)^{2/3}$ without explaining how, and then says that this implies $t_0=\frac{2}{3H_0}$. So I suppose he is not using this when obtaining the solution. And as I said, I don't see why you can eliminate the aditive constant. Maybe I'm being dumb but I'm completely confused. $\endgroup$
    – Wild Feather
    Commented Apr 10, 2023 at 10:11
  • $\begingroup$ @WildFeather If you plug in $(t/t_0)^{2/3}$ into the Friedmann equation you automatically obtain $t_0 = \frac{2}{3 H_0}$. Have you tried it yet? Another way to obtain that $t = \frac{2}{3H_0}$ is to require that $a(0) = 0$. If you plug that into your solution, then you will see that $ H_0 $ and $ t_0 $ are related in the way that I said. $\endgroup$
    – Tiagotje
    Commented Apr 10, 2023 at 12:16
  • $\begingroup$ Yes, and you are right, we obtain that relation between $t_0$ and $H_0$, but only because we assumed that the solution takes that form. If we put $(t/t_0)^{2/3}$ instead of $\big(\frac{3}{2}H_0t\big)^{2/3}$, it totally makes sense that we would obtain $t_0=\frac{2}{3H_0}$, we are imposing it. What I don't see is why we do that, or why we don't consider an additive constant, that is, a solution of the form $(t/t_0)^{2/3}+C$ instead, which would also satisfy Friedmann's equation... Sorry if I'm taking up too much of your time, but honestly I still don't understand the reasoning behind this. $\endgroup$
    – Wild Feather
    Commented Apr 10, 2023 at 12:39

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.