Theparticle horizon (also called thecosmological horizon, thecomoving horizon (inScott Dodelson's text), or thecosmic light horizon) is the maximum distance from which light fromparticles could have traveled to theobserver in theage of the universe. Much like the concept of aterrestrial horizon, it represents the boundary between the observable and the unobservable regions of the universe,^[1] so its distance at the present epoch defines the size of theobservable universe.^[2] Due to the expansion of the universe, it is not simply theage of the universe times thespeed of light (approximately 13.8 billion light-years), but rather the speed of light times the conformal time. The existence, properties, and significance of a cosmological horizon depend on the particularcosmological model.

The particle horizon is a distance in a comoving coordinate system, a system that has the expansion of the universe built-in. The expansion is defined by a (dimensionless)scale factor${\displaystyle a(t)}$ set to have a value of one today. The time that light takes to travel a distancedx in the comoving coordinate system will be${\displaystyle dx=dt/a(t)}$ in units of light years (${\displaystyle c=1}$). The total distance light can travel in the timet since theBig Bang at${\displaystyle t=0}$ sums all the incremental distances:^[3]: 34

$${\displaystyle \eta ={\int }_0^t\frac{d{t}^{\prime }}{a(t^{\prime })}}$$

Thecomoving horizon${\displaystyle \eta }$ increases monotonically and thus can be used a time parameter: the particle horizon is equal to theconformal time${\displaystyle \eta }$ that has passed since theBig Bang, times thespeed of light${\displaystyle c}$.^[3]: 34

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 isnot theage of the universe as generally understood.That age refers instead to a time as defined by the Robertson-Walker form of the cosmological metric, which time is presumed to be measured by a traditional clock and estimated to be around${\displaystyle 4.35\times {10}^{17}\text{ s}}$. By contrast${\displaystyle {\eta }_0}$ is the age of the universe as measured by a Marzke-Wheeler "light clock".^[4]

The particle horizon recedes constantly as time passes and the conformal time grows. As such, the observed size of the universe always increases.^[1][5] Since proper distance at a given time is just comoving distance times the scale factor^[6] (withcomoving distance normally defined to be equal to proper distance at the present time, so${\displaystyle a(t_0)=1}$ at present), the proper distance,${\displaystyle d_p(t),}$ to the particle horizon at time${\displaystyle t}$ is given by^[7]: 417

$${\displaystyle d_p(t)=a(t){\int }_0^t\frac{cd{t}^{\prime }}{a(t^{\prime })}}$$

The value of the distance to the horizon depends on details in${\displaystyle a(t)}$.

In this section we consider theFLRW cosmological model. In that context, the universe can be approximated as composed by non-interacting constituents, each one being aperfect fluid with density${\displaystyle {\rho }_i}$, partial pressure${\displaystyle p_i}$ andstate equation${\displaystyle p_i={\omega }_i{\rho }_i}$, such that they add up to the total density${\displaystyle \rho }$ and total pressure${\displaystyle p}$.^[8] Let us now define the following functions:

• Hubble function${\displaystyle H=\frac{\dot{a}}{a}}$
• The critical density${\displaystyle {\rho }_c=\frac{3}{8\pi G}{H}^2}$
• Thei-th dimensionless energy density${\displaystyle {\mathrm{\Omega }}_i=\frac{{\rho }_i}{{\rho }_c}}$
• The dimensionless energy density${\displaystyle \mathrm{\Omega }=\frac{\rho }{{\rho }_c}=\sum {\mathrm{\Omega }}_i}$
• 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.^[9] It can be proved that the Hubble function is given by

$${\displaystyle H(z)=H_0\sqrt{\sum {\mathrm{\Omega }}_{i0}(1+z{)}^{n_i}}}$$

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:^[9]

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

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:^[9]

$${\displaystyle \frac{d}{dt}{d}_p=d_p(z)H(z)+c}$$

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 forevent horizon.

The concept of a particle horizon can be used to illustrate thehorizon problem, which is an unresolved issue associated with theBig Bang model. Extrapolating back to the time ofrecombination when thecosmic 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 thecosmic microwave background (CMB) that are separated by about a fraction of agreat circle across the sky of

$${\displaystyle f=\frac{H_p(t_{\text{CMB}})}{H_p(t_0)}}$$

(anangular size of${\displaystyle \theta \sim {1.7}^{\circ }}$)^[10] should be out ofcausal contact with each other. That the entire CMB is inthermal equilibrium and approximates ablackbody so well is therefore not explained by the standard explanations about the way theexpansion of the universe proceeds. The most popular resolution to this problem iscosmic inflation.

• Cosmological horizon
• Observable universe

• Physical cosmology
• Physical horizons

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