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What does it mean when a light cone intersects a particle horizon? Like here, about 2 billion years after the big bang? Does it have any significance? I went through a few simple equations and the outcome was about this point (in Planck times) and I'm wondering about that. enter image description here

The image has been taken from here: https://i.stack.imgur.com/Uzjtg.png

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  • $\begingroup$ Interesting question. But can you please edit the question in order to insert a reference and possibly link you took the figure from? $\endgroup$ Sep 6, 2022 at 12:32
  • $\begingroup$ That looks like Pulsar's diagram from physics.stackexchange.com/a/63780/123208 $\endgroup$
    – PM 2Ring
    Sep 6, 2022 at 13:22

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In comoving coordinates, the distance $d_\mathrm{PH}$ to the particle horizon (PH) is equal to the conformal time $\eta$ that has passed since the Big Bang, times the speed of light $c$. That means that if you draw a spacetime diagram with $\eta$ as a function of comoving distance $d$, light cones become straight lines with a slope $c$. For a suitable aspect ratio of the plot, you can make those line have a 45º slope.

This is exactly what your example has done, although the $y$ axis labels have been translated from $\eta$ to time $t$ and scale factor $a$ for the left and right axis, respectively.

Now since the PH is actually the future light cone of an event at Big Bang, then the past light cone of any event at a given conformal time $\eta\,'$ will always intersect the PH at exactly $\eta\,'/2$.

For us, today at time $t_0=13.8\,\mathrm{Gyr}$, the conformal time is $$ \eta_0 = \int_0^{t_0}\frac{dt'}{a(t')} = 46.2\,\mathrm{Gyr}, $$ where I've used a Planck (2020) cosmology to evaluate the integral. The conformal time $\eta_x$ at the time $t_x$ of the crossing is then half of this, i.e. $\eta_x = 23.1\,\mathrm{Gyr}$. The age of the Universe at this time is thus given by $$ 23.1\,\mathrm{Gyr} = \int_0^{t_x} \frac{dt'}{a(t')}. $$ The solution to this equation $t_x = 1.7\,\mathrm{Gyr}$, which is the "about 2 billion years" you mention.

Here's a Python snippet to do the calculations yourself. If you want to play around with various cosmologies, uncomment the line starting with # cosmo = Flat... (i.e. remove the #).

from astropy import units as u
from astropy.cosmology import Planck18, FlatLambdaCDM, z_at_value
import astropy.constants as cc

cosmo = Planck18
# cosmo = FlatLambdaCDM(H0=67.3, Om0=.315) # Uncomment for custom cosmology

c    = cc.c                         # Speed of light
dPH0 = cosmo.comoving_distance(1e7) # Particle horizon today (1e7 is almost inf)
eta0 = dPH0 / c                     # Conformal time today
etax = eta0 / 2                     # Conformal time at crossing
dPHx = c * etax                     # Particle horizon at crossing (in comoving coords)
zx   = z_at_value(cosmo.comoving_distance,dPHx) # Redshift at crossing
tx   = cosmo.age(zx)                # Age of Universe at crossing
dH0  = c / cosmo.H0                 # Hubble radius today
dHx  = c / cosmo.H(zx) * (1+zx)     # Hubble radius at crossing in comoving coords

print('For H0 = {:.1f} and Omega_m0 = {:.3f}:'.format(cosmo.H0,cosmo.Om0))
print('  Distance to PH today:        {:.2f}'.format(dPH0.to(u.Glyr)))
print('  Hubble radius today:         {:.2f}'.format(dH0.to(u.Glyr)))
print('  Distance to PH at crossing:  {:.2f}'.format(dPHx.to(u.Glyr)))
print('  Hubble radius at crossing:   {:.2f}'.format(dHx.to(u.Glyr)))
print('  Corresponding redshift:      {:.2f}'.format(zx))
print('  Age of Universe at crossing: {:.2f}'.format(tx.to(u.Gyr)))
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  • $\begingroup$ Using the slightly different cosmology that Pulsar used for that spacetime diagram, I get the slightly smaller value of 1.6 Gyr. $\endgroup$
    – pela
    Sep 6, 2022 at 15:01
  • $\begingroup$ I got 1.8 billion years and Hubble radius of 11.5 Gly, is this correct for that time? And thanks for that answer. $\endgroup$ Sep 6, 2022 at 15:27
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    $\begingroup$ @KontrolaFaktů You're welcome :) Your values sound about right, depending on which cosmology you used. I added my Python code if you want to check. $\endgroup$
    – pela
    Sep 6, 2022 at 20:17

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