1
$\begingroup$

I was looking at the answer to this question: The difference between comoving and proper distances in defining the observable universe and wanted to reproduce the chart in order to better understand the relationship between the Hubble Sphere and the Particle Horizon, but I'm stuck with a very basic problem:

enter image description here

How do you map Cosmic Time (left axis) to Scale Factor (right axis). Obviously it was done for this chart, but I can't find a formula anywhere.

EDIT: I found this formula on the Wiki pages. This is specifically the formula that I'm trying to solve: $$\eta(t)=\int_0^t \frac{dt\prime}{a(t\prime)}$$ Where do I get the formula for $a(t\prime)$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Cosmic Time

Let us denote the age of the universe as $t$. By using $\frac{da}{dt} = \dot{a}$ we can write

$$t = \int_{t_e}^{t_0}dt =\int_{a(t_e)}^{a(t_o)}\frac{da}{\dot{a}}~~~~(1)$$ But instead of working in $da$ we can convert it to $dz$. We know that $$1 + z = a^{-1}$$

hence

$$dz = -a^{-2}da$$ or $$da = -a^2dz = -(1+z)^{-2}dz ~~~~(2)$$

and we have

$$\frac{\dot{a}}{a} = H_0 E(z)~~~~(3)$$

Where

$$E(z) = \sqrt{\Omega_{r,0}(1+z)^4 + \Omega_{m,0}(1+z)^3 + \Omega_{\Lambda,0} + \Omega_{\kappa}(1+z)^2}$$

Let us put $(2)$ and $(3)$ into $(1)$

$$t = \int_{a(t_e)}^{a(t_o)}\frac{da}{\dot{a}} = \int_z^0 \frac{-(1+z)^{-2}dz}{aH_0E(z)}$$ Which is equal to

$$t = \frac{1}{H_0}\int_0^z \frac{dz}{(1+z)E(z)}$$

Let us do the calculation for the current age of the universe,

$$t_{uni} = \frac{1}{H_0}\int_0^{\infty} \frac{dz}{(1+z)\sqrt{\Omega_{r,0}(1+z)^4 + \Omega_{m,0}(1+z)^3 + \Omega_{\Lambda,0} + \Omega_{\kappa}(1+z)^2}}$$

For $\Omega_{r,0} = 0$, $\Omega_{m,0} = 0.31$, $\Omega_{\Lambda,0} = 0.69$ and $\Omega_{\kappa} = 1 - \Omega_0 = 0$,

$$t_{uni} = \frac{1}{H_0}\int_0^{\infty} \frac{dz}{(1+z)\sqrt{0.3(1+z)^3 + 0.7}}$$

If we use the https://www.integral-calculator.com and do the numeric calculation we obtain

$$ \int_0^{\infty}\frac{dz}{(1+z)\sqrt{0.3(1+z)^3 + 0.7}} = 0.9641$$

For $H_0 = 68km/s/Mpc$

$$t_{uni} = 0.9641 \times H_0^{-1}= 0.9641 \times 14.39~\text{Gyr} = 13.8733 ~\text{Gyr}$$

Conformal Time

In this case from the FLRW metric we can write,

$$ds^2 = -c^2dt^2 + a^2dr^2$$ for light $ds = 0$ Thus we have,

$$c^2dt^2 = a^2dr^2$$ or

$$r= c\int_{t_0}^{t_e}\frac{dt}{a(t)}$$

Conformal time represents the time it takes for the light to travel distance $r$. Mathematically

$$\eta = \frac{r}{c}$$

$$\eta = \int_{t_e}^{t_o}\frac{dt}{a(t)}$$

But you can write this equation in terms of z to make the calculations easier.

So we know that $$1 + z = a(t)^{-1}$$

hence

$$\frac{dz}{dt} = \frac{dz} {da} \frac{da} {dt}$$ $$\frac{dz}{dt} = -\frac{1}{a^2} \dot{a}$$ $$dz = -\frac{\dot{a}}{a^2}dt$$

$$\eta = \int_{t_e}^{t_o}\frac{dt}{a(t)} = \int_{z}^{0}\frac{-dza^2 / \dot{a} }{a}$$

Thus

$$\eta = \int_{0}^{z}\frac{dza}{\dot{a}} $$

And by using (3) you can write

$$\eta = \frac{1}{H_0}\int_{0}^{z}\frac{dz}{E(z)}$$

When you use the above equation to calculate the conformal time since the begining of the universe you need to write,

$$\eta = \frac{1}{H_0}\int_{0}^{\infty}\frac{dz}{E(z)}$$

When you use the integral calculator site and do the above integral you will get

$$ \eta = 3.2 \times 1/H_0 = 3.2 \times 14.38Gyr = 46.016 Gyr$$

In summary the age of the universe for a given $z$ can be calculated by

$$t = \frac{1}{H_0}\int_0^z \frac{dz}{(1+z)E(z)}$$

and the conformal time

$$\eta = \frac{1}{H_0}\int_{0}^{z}\frac{dz}{E(z)}$$

These integrals cannot be done by hand (in general). Either you need to write a computer program or more simply just you can use the site that I mentioned and select the numerical calculation box.

I did all my calculations for a given $z$ value but I only did this to show you the difference between $t$ (cosmic time) and $\eta$(conformal time). For a given $a$ you can easily find the corresponding $z$ by using $z=a^{-1}-1$ and then you can calculate $\eta$ or $t_0$.

Edit: To answer your other question. There's not a direct equation but you can find a way to do it. For instance, let us use equation $(3)$

$$\frac{\dot{a}}{a} = H_0 E(z)$$

but in this case, we will write $E(a)$ instead of $E(z)$ which is a simple thing to do. So we have

$$E(a) =\sqrt{\Omega_{r,0}a^{-4} + \Omega_{m,0}a^{-3} + \Omega_{\Lambda,0} + \Omega_{\kappa}a^{-2}}$$

$$\frac{da}{dt} = a H_0 E(a)$$

Now, $a$ goes inside of $E(a)$ as $a^2$. Thus

$$\frac{da}{dt} = H_0 \sqrt{\Omega_{r,0}a^{-2} + \Omega_{m,0}a^{-1} + \Omega_{\Lambda,0}a^2 + \Omega_{\kappa}}$$

so $$t = \frac{1}{H_0} \int_0^a \frac{da}{\sqrt{\Omega_{r,0}a^{-2} + \Omega_{m,0}a^{-1} + \Omega_{\Lambda,0}a^2 + \Omega_{\kappa}}}~~~~(4)$$

So we have a function of $t(a)$ however for some values you can take the inverse of this function (It makes problem for the closed universe models). For instance you can plot $t$ vs $a$ ($t(a)$) and then reverse the axis to get $a$ vs $t$ ($a(t)$). This is the general case.

If you are asking for our universe you can make some approximations and then find the function. For example, you can look at this site. But it's only an approximation and works for only some $\Omega$ parameters with a given time interval.

Either you have to make an approximation to equation (4) for each case and then take the inverse, or you have to plot $t(a)$ and from there by chaning the axis you can find $a(t)$. I am not sure if there's another way to do it.

$\endgroup$
16
  • $\begingroup$ Is this formula giving cosmic time or conformal time? $\endgroup$
    – Gluon Soup
    Apr 27, 2020 at 21:18
  • $\begingroup$ It gives the cosmic time $\endgroup$
    – seVenVo1d
    Apr 27, 2020 at 21:42
  • $\begingroup$ Try it for $a(t_e) = 0.8$ you ll see that t is close to $10Gyr$. Conformal time has a similar equation indeed but its written in terms of $z$ $$ \eta = -\frac{1}{H_0}\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$ There is a quite difference.. if you want to know where the cosmic time equation comes from I can show you the derivation $\endgroup$
    – seVenVo1d
    Apr 27, 2020 at 21:52
  • $\begingroup$ I don't mean to sound ungrateful - this is a really useful formula - but I really want to go in the other direction. That is, looking at the chart in the original post, I want a linear axis of time (or conformal time, I can convert) and the less-than-linear scale factor. My math skills aren't up to the task of solving this integral for $a$. $\endgroup$
    – Gluon Soup
    Apr 27, 2020 at 22:28
  • $\begingroup$ @GluonSoup I edited my post maybe this helps $\endgroup$
    – seVenVo1d
    Apr 28, 2020 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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