Particle horizon

In terms of comoving distance, the particle horizon is equal to the conformal time ${\displaystyle \eta }$ that has passed since the Big Bang, times the speed of light ${\displaystyle c}$. In general, the conformal time at a certain time ${\displaystyle t}$ is given by

${\displaystyle \eta =\int _{0}^{t}{\frac {dt'}{a(t')}},}$

where ${\displaystyle a(t)}$ is the scale factor of the Friedmann–Lemaître–Robertson–Walker metric, and we have taken the Big Bang to be at ${\displaystyle t=0}$. By convention, a subscript 0 indicates "today" so that the conformal time today ${\displaystyle \eta (t_{0})=\eta _{0}=1.48\times 10^{18}{\text{ s}}}$. Note that the conformal time is not the age of the universe, which is estimated around ${\displaystyle 4.35\times 10^{17}{\text{ s}}}$. Rather, the conformal time is the amount of time it would take a photon to travel from where we are located to the furthest observable distance, provided the universe ceased expanding. As such, ${\displaystyle \eta _{0}}$ is not a physically meaningful time (this much time has not yet actually passed); though, as we will see, the particle horizon with which it is associated is a conceptually meaningful distance.

The particle horizon recedes constantly as time passes and the conformal time grows. As such, the observed size of the universe always increases.[1][3] Since proper distance at a given time is just comoving distance times the scale factor[4] (with comoving distance normally defined to be equal to proper distance at the present time, so ${\displaystyle a(t_{0})=1}$ at present), the proper distance to the particle horizon at time ${\displaystyle t}$ is given by[5]

${\displaystyle a(t)H_{p}(t)=a(t)\int _{0}^{t}{\frac {c\,dt'}{a(t')}}}$

and for today ${\displaystyle t=t_{0}}$

${\displaystyle H_{p}(t_{0})=c\eta _{0}=14.4{\text{ Gpc}}=46.9{\text{ billion light years}}.}$

In this section we consider the FLRW cosmological model. In that context, the universe can be approximated as composed by non-interacting constituents, each one being a perfect fluid with density ${\displaystyle \rho _{i}}$, partial pressure ${\displaystyle p_{i}}$ and state equation ${\displaystyle p_{i}=\omega _{i}\rho _{i}}$, such that they add up to the total density ${\displaystyle \rho }$ and total pressure ${\displaystyle p}$.[6] Let us now define the following functions:

• Hubble function ${\displaystyle H={\frac {\dot {a}}{a}}}$
• The critical density ${\displaystyle \rho _{c}={\frac {3}{8\pi G}}H^{2}}$
• The i-th dimensionless energy density ${\displaystyle \Omega _{i}={\frac {\rho _{i}}{\rho _{c}}}}$
• The dimensionless energy density ${\displaystyle \Omega ={\frac {\rho }{\rho _{c}}}=\sum \Omega _{i}}$
• The redshift ${\displaystyle z}$ given by the formula ${\displaystyle 1+z={\frac {a_{0}}{a(t)}}}$

Any function with a zero subscript denote the function evaluated at the present time ${\displaystyle t_{0}}$ (or equivalently ${\displaystyle z=0}$). The last term can be taken to be ${\displaystyle 1}$ including the curvature state equation.[7] It can be proved that the Hubble function is given by

${\displaystyle H(z)=H_{0}{\sqrt {\sum \Omega _{i0}(1+z)^{n_{i}}}}}$

where the dilution exponent ${\displaystyle n_{i}=3(1+\omega _{i})}$. Notice that the addition ranges over all possible partial constituents and in particular there can be countably infinitely many. With this notation we have:[7]

${\displaystyle {\text{The particle horizon }}H_{p}{\text{ exists if and only if }}N>2}$

where ${\displaystyle N}$ is the largest ${\displaystyle n_{i}}$ (possibly infinite). The evolution of the particle horizon for an expanding universe (${\displaystyle {\dot {a}}>0}$) is:[7]

${\displaystyle {\frac {dH_{p}}{dt}}=H_{p}(z)H(z)+c}$

where ${\displaystyle c}$ is the speed of light and can be taken to be ${\displaystyle 1}$ (natural units). Notice that the derivative is made with respect to the FLRW-time ${\displaystyle t}$, while the functions are evaluated at the redshift ${\displaystyle z}$ which are related as stated before. We have an analogous but slightly different result for event horizon.

The concept of a particle horizon can be used to illustrate the famous horizon problem, which is an unresolved issue associated with the Big Bang model. Extrapolating back to the time of recombination when the cosmic microwave background (CMB) was emitted, we obtain a particle horizon of about

${\displaystyle H_{p}(t_{\text{CMB}})=c\eta _{\text{CMB}}=284{\text{ Mpc}}=8.9\times 10^{-3}H_{p}(t_{0})}$

which corresponds to a proper size at that time of:

${\displaystyle a_{\text{CMB}}H_{p}(t_{\text{CMB}})=261{\text{ kpc}}}$

Since we observe the CMB to be emitted essentially from our particle horizon (${\displaystyle 284{\text{ Mpc}}\ll 14.4{\text{ Gpc}}}$), our expectation is that parts of the cosmic microwave background (CMB) that are separated by about a fraction of a great circle across the sky of

${\displaystyle f={\frac {H_{p}(t_{\text{CMB}})}{H_{p}(t_{0})}}}$

(an angular size of ${\displaystyle \theta \sim 1.7^{\circ }}$)[8] should be out of causal contact with each other. That the entire CMB is in thermal equilibrium and approximates a blackbody so well is therefore not explained by the standard explanations about the way the expansion of the universe proceeds. The most popular resolution to this problem is cosmic inflation.

• Cosmological horizon
• Observable universe