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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?

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  • $\begingroup$ Depends on what you're putting into the universe. I doubt there's a closed form solution for the general case (matter + radiation + cosmological constant + curvature), though there are many formulas given for the case of one or two components, which are listed in most cosmology textbooks. $\endgroup$ – knzhou Jan 21 at 16:49
  • $\begingroup$ $a(t)$ comes from solving the Friedmann equations. For example, for a matter-dominated universe, it is proportional to $t^{2/3}$. $\endgroup$ – G. Smith Jan 21 at 16:53
  • $\begingroup$ I want to be able to compute the conformal time to some arbitrary time in the universe. For example, conformal time to recombination. Conformal time to reionization, conformal time to the present. $\endgroup$ – Quarkly Jan 21 at 18:04
  • $\begingroup$ @G.Smith - Yes, I've seen in several textbooks the $t^{2/3}$ relation, but I'm not interested in the proportion, I'm interested in the exact solution. I can't put a proportion into the limits of an definite integral. $\endgroup$ – Quarkly Jan 21 at 18:14
  • $\begingroup$ Are you looking for something like $$\eta(z)=H_0^{-1}\int \frac {da} {a^2\sqrt{E(a)}}$$ for $$E(a)=\sqrt {\Omega_{\Lambda}+\Omega_m a^{-3}+\Omega_ra^{-4}+\Omega_{\kappa} a^{-2}}$$ $\endgroup$ – Reign Jan 21 at 20:03
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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 $z$.

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  • $\begingroup$ That's very helpful, and I think most of the way there, but how do you integrate using Cosmic time (t)? For instance, my first test will be to use this formula to calculate the Conformal Time at the present moment, $t_0$. $\endgroup$ – Quarkly Jan 21 at 22:18
  • $\begingroup$ @DonaldAirey I calculated it, And it seems correct $\endgroup$ – Reign Jan 22 at 16:36
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$a(t)$ 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.

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  • $\begingroup$ I'm interested in a formula (or formulas) that work for all time. One of the calculations I'm trying to understand is the Sound Horizon size at recombination which involves integrating $\eta$ from 0 to $\eta_{CMB}$. How do I do this if $\eta$=<it dependes>? $\endgroup$ – Quarkly Jan 21 at 18:12
  • $\begingroup$ You seem to have an unjustified expectation that every problem should be solvable analytically. In this case, you have to numerically solve the Friedmann equations and numerically integrate to get the conformal time. $\endgroup$ – G. Smith Jan 21 at 18:14
  • $\begingroup$ As I mentioned in my original post, numerical or analytics, I don't care. I do have a possibly unreasonable expectation that there's an unambiguous answer. $\endgroup$ – Quarkly Jan 21 at 18:16
  • $\begingroup$ I don’t see any reason why the answer would be ambiguous. $\endgroup$ – G. Smith Jan 21 at 18:18
  • $\begingroup$ Are you concerned about how $a(t)$ gets normalized? It is normalized so that at the present time (i.e. when $\dot{a}/a$ is $H_0$), $a$ is 1. This removes any ambiguity. For example, in the only-matter case, the $t^{2/3}$ solution is really $(3H_0 t/2)^{2/3}$. The matter-plus-dark-energy solution in my answer is normalized like this. (To prove it, you have to use $\Omega_m+\Omega_\Lambda=1$.) $\endgroup$ – G. Smith Jan 21 at 18:58
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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.

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