How exactly is conformal time, $\eta$, calculated?

Question

As I understand it, Conformal Time is basically the comoving distance divided by the speed of light. $$\eta=\int_0^t\frac {dt'}{a(t')}$$ I can make the connection between the scale factor and redshift:$$\frac {1}{a}=(z+1)$$but this is where I'm stuck. I want to perform the actual integration and get a concrete value for the conformal time as a function of cosmological time (e.g. $f(t) = \eta$), but I can't find a formula relating t to the scale factor or redshift.

Could someone show me step by step the derivation of the formula?

Answer 1

I am not sure this is what you want but I want to give it a try,

$$\eta=\int \frac {dt} {a}=\int\frac {da} {a\dot{a}}=\int\frac {da} {a^2H}$$ and we can write $$H(z)=H_0E(z)$$ $$E(z)=\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}$$

so we have,

$$\eta=\int\frac {da} {a^2H_0E(z)}$$

and $dz=-da/a^2$ so we can write,

$$\eta=-H_0^{-1}\int\frac {dz} {E(z)}$$

And by taking initial coniditon as $z=\infty$, and due to the minus sign the integral becomes,

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

To find the current($t_0$) conformal time, we can use the above equation, for $z=0$

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

$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}}$$

For the current values of $\Omega_{\Lambda}=0.69$, $\Omega_m=0.31$,$\Omega_{\kappa}= \Omega_r=0$

we have,

$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3}}$$

$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}$$

If we take $H_0=70km/s/Mpc$ then $1/H_0=1/(70\times 3.2408\,10^{-20})=4.4133353\,10^{17}s$

And the integral gives, $$\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}=3.266054427285631$$

so $$\eta(t_0)=3.266054427285631\times 4.4133353\,10^{17}s=1.4414193\,10^{18}=45.70 \,\text {Gigayear}$$

To calculate the integral you can use, this site

I write the integral in terms of $z$ but, its also possible to write the equation in terms of $a(t)$ (the begining part of the derivation). But $z$ is the observable value so I prefer to write in that form.

For a given $t$ you can turn easily $a(t)$ to $z$.

Answer 2

$a(t)$ comes from solving the Friedmann equations. If you want to take matter, radiation, and dark energy into account you have to do this numerically.

However, if you don’t care about the first 10 million years after the Big Bang, you can ignore radiation and get a nice analytic solution taking matter (including dark matter) and dark energy into account:

$$a(t)=\left(\frac{\Omega_m}{\Omega_\Lambda}\right)^{1/3}\left(\sinh{\frac{t}{t_\Lambda}}\right)^{2/3}$$

where

$$t_\Lambda=\frac{2}{3H_0 \Omega_\Lambda^{1/2}}.$$

Here $\Omega_m$ is the current fraction of the critical density which is matter, $\Omega_\Lambda$ is the current fraction of the critical density which is dark energy, and $H_0$ is the current Hubble constant. This is for a flat universe, which is what we observe.

This equation, and values for these three numbers, can be found in Wikipedia’s Lambda-CDM model article:

$\Omega_m=0.3089$

$\Omega_\Lambda=0.6911$

$H_0=67.74$ km/s/Mpc

This solution shows how the scale factor first grew as $t^{2/3}$ when matter dominated and later grows exponentially as dark energy dominates.

Answer 3

There's not going to be a way to relate $\eta$ to $t$ unless you already know the function $a(t)$, which is going to depend on your cosmological model in question. The integral representation given is going to be the closest you can get to a closed-form expression without explicitly being given an explicit value of $a$.

For example, for an inflationary era, we have $a=a_0e^{Ht}$, where $H$ is the constant Hubble parameter. In this case, we have

$$\eta(t)=\int\frac{\mathrm{d}t}{a(t)}=\frac{1}{a_0}\int\mathrm{d}t\,e^{-Ht}=C-\frac{1}{a_0H}e^{-Ht}.$$

This can be inverted to give

$$a(t)=a_0e^{Ht}=\frac{1}{H(C-\eta)}.$$

Picking $C=0$ gives

$$a(\eta)=-\frac{1}{H\eta}.$$

Both $\eta(t)$ and $a(\eta)$ are nice, simple closed-form expressions, but only hold during an inflationary era. Other eras (matter-dominated, radiation, cosmological constant, combinations of the three, etc.) will have different values of $a(t)$, and thus the definition of $\eta$ will change, and no closed-form (i.e. without an integral sign) expression can encompass every case.