# If the expansion of the Universe accelerates, why its horizon does not shrink?

In de Sitter (expanding) universe to which our universe asymptotically approaches, the higher the rate of space expansion, the smaller the radius of the cosmic horizon. Why then our universe is thought of being accelerating expansion yet its radius is not thought to reduce?

• My understanding is that it is thought to reduce. Do you have a reference that says otherwise? – Asher Oct 12 '15 at 6:06
• @Asher Holographic principle usually assumes that the total entropy of the universe is equal to the area of its boundary. Since entropy of universe grows, the horizon area also should grow. For instance, this answer by Ron Maimon claims the entropy of the universe is equal to the area of the cosmological horizon in Planck units: physics.stackexchange.com/a/36121/1186 – Anixx Oct 12 '15 at 6:31
• @Anixx some people think that the area of the cosmic horizon puts a bound on the total entropy of the universe (though people argue about that point), but the actual entropy of the universe is much lower than that bound. – Nathaniel Oct 17 '15 at 1:00
• it expands in meters since the rate of expansion is positive, but the matter moving inside the horizon is becoming increasingly redshifted – diffeomorphism Oct 20 '15 at 21:43

A de Sitter universe is a cosmological solution to Einstein's field equations of General Relativity which is named after Willem de Sitter. It models the universe as spatially flat and neglects ordinary matter, so the dynamics of the universe are dominated by the cosmological constant, thought to correspond to dark energy in our universe or the inflaton field in the early universe.

Mathematically A de Sitter universe has no ordinary matter content but with a positive cosmological constant ($\Lambda$) which sets the expansion rate, H. A larger cosmological constant leads to a larger expansion rate:

$H \propto \sqrt{\Lambda}$,

where the constants of proportionality depend on conventions.

It is common to describe a patch of this solution as an expanding universe of the FLRW form where the scale factor is given by

$a(t) = e^{Ht}$,

where the constant $H$ is the Hubble expansion rate and $t$ is time. As in all FLRW spaces, $a(t)$, the scale factor, describes the expansion of physical spatial distances.

To examine the de Sitter Universe to Cosmic Horizon further we need to define Cosmic or Cosmological horizon.

https://en.wikipedia.org/wiki/Cosmological_horizon

A cosmological horizon is a measure of the distance from which one could possibly retrieve information. This observable constraint is due to various properties of general relativity, the expanding universe, and the physics of Big Bang cosmology. Cosmological horizons set the size and scale of the observable universe.

There are a number of different horizons referenced in the Wiki link above. {Interesting reading} Let's look at a couple the Hubble Horizon and Event Horizon here :

1 Hubble Horizon

One can define a so-called "Hubble Horizon" which shows roughly how far light would travel if space were not expanding. This size is

$\chi = c t$

where t is the lookback time since the Big Bang (otherwise known as the age of the universe) which, according to the Friedmann Equations, is:

$t = \int^{a}_{0}{\frac{da}{H_0 \sqrt{\Omega_R a^{-2} + \Omega_m a^{-1} + \Omega_k +\Omega_\Lambda a^2}}}$

where $H_0$ is the Hubble Constant and the $\Omega$ density parameters are, in order, the density of radiation, matter, curvature, and dark energy scaled to the critical density of the universe.

Today, roughly:

$\chi_0 = \frac{c}{H_0}$

yielding a Hubble horizon of some 4.2 Gpc. This horizon is not really a physical size, but it is often used as useful length scale as most physical sizes in cosmology can be written in terms of those factors.

2 Event Horizon

The particle horizon differs from the cosmic event horizon, in that the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time, while the event horizon is the largest comoving distance from which light emitted now can ever reach the observer in the future. The current distance to our cosmic event horizon is about 5 Gpc, well within our observable range given by the particle horizon.

In general, the proper distance to the event horizon at time t is given by

$d_e(t) = a(t) \int_{t}^{t_{max}} \frac{cdt'}{a(t')}$

where $t_{max}$ is the time-coordinate of the end of the universe, which would be infinite in the case of a universe that expands forever.

For our case, assuming that dark energy is due to a cosmological constant, $d_e(t_0)$ < $\infty$.

For perspective : {Because our Universe entered the Dark Energy Dominated Era a few billion years ago, our universe is probably approaching a de Sitter universe in the infinite future. If the current acceleration of our universe is due to a cosmological constant then as the universe continues to expand all of the matter and radiation will be diluted. Eventually there will be almost nothing left but the vacuum energy, tiny thermal fluctuations, quantum fluctuations and our universe will have become a de Sitter universe.}

The exponential expansion of the scale factor means that the physical distance between any two non-accelerating observers will eventually be growing faster than the speed of light. At this point those two observers will no longer be able to make contact. Therefore any observer in a de Sitter universe would see event horizons beyond which that observer can never see nor learn any information. If our universe is approaching a de Sitter universe then eventually we will not be able to observe any galaxies other than our own Milky Way (and any others in the gravitationally bound Local Group, assuming they were to somehow survive to that time without merging).

So what do we glean from the above "de Sitter" Model of our Universe. 1. It is expanding exponentially, the space within the mater is growing exponentially so the mater is finding itself exponentially far away from other non-gravitationally attracted matter. Moving much faster then the speed of light. $a(t) = e^{Ht}$ 2. The mathematics of this model provides for a spatially flat universe.

Now let's examine your statement and question.

1. In de Sitter (expanding) universe to which our universe asymptotically approaches, the higher the rate of space expansion, the smaller the radius of the cosmic horizon.

The de Sitter universe is expanding at $e^{Ht}$ Where $H$ is the Hubble Constant and $t$ is time.

The distance of Hubble Horizon {Cosmic Horizon} is given by

$\chi_0 = \frac{c}{H_0}$ {keeping in mind $H$ is approaching a constant}

Now assuming that the distance $\chi_0$ is equal to the distance in all directions to this horizon we can conclude that $\chi_0$ is equivalent to the radius of a spherical event horizon, from the point of view of the observer, the radius of the "Hubble Sphere". {See Hubble Sphere below}

Since $\chi_0$ = $c/H_0$ ,is a quotient of two constants $c/H_0$, it too is a constant.

Therefore the radius of the cosmic horizon is {approaching} a constant {for the current value of $H_0$ and in the general sense, as $H$ approaches a constant.}

{See reference below $H$ is approaching a constant}

If the expansion of the Universe accelerates, why its horizon does not shrink?

Since $r_{HS}$ is a approaching a constant, the horizon approaches a constant as the Universe expands at an accelerating rate, $e^{Ht}$.

1. Why then our universe is thought of being accelerating expansion yet its radius is not thought to reduce?

Now to answer your next question. The radius of the spatially flat and expanding Universe is not the same as the radius of the Cosmic Horizon we just examined. The Cosmic Horizon is about a 4.2 GPC - 5 GPC radius and is governed by the speed of light and the hubble constant. This is a lot smaller than the entire universe which is expanding exponentially away by space filling in the matter of the universe that is non-gravitationally bound. We conclude that the Cosmic horizon radius is not equal to the de Sitter Universe's radius, it is that simple. They are different radii.

The cosmic radius is {approaching a} constant {as to $\frac{c}{H_{later}}$} while the other the radius of the de Sitter Universe is increasing as the spatially flat universe is expanding at $e^{Ht}$.

Value of Hubbles Constant -

As of 2013 - https://en.wikipedia.org/wiki/Hubble%27s_law

67.80±0.77 km/s/Mpc.

Best value for the local $H_0$ determination is around 71 (+/- 7) km/s/Mpc.

$H_0$ is Hubble's constant and corresponds to the value of H (often termed the Hubble parameter which is a value that is time dependent and which can be expressed in terms of the scale factor) in the Friedmann equations taken at the time of observation denoted by the subscript 0.

Hubble Sphere - In cosmology, a Hubble volume, or Hubble sphere, is a spherical region of the Universe surrounding an observer beyond which objects recede from that observer at a rate greater than the speed of light due to the expansion of the Universe. The Hubble volume is approximately 1031 cubic light years.

The proper radius of a Hubble sphere (known as the Hubble radius or the Hubble length) is $c/H_0$, where $c$ is the speed of light and $H_0$ is the Hubble constant. The surface of a Hubble sphere is called the microphysical horizon, the Hubble surface, or the Hubble limit.

Hubble's law is considered a fundamental relation between recessional velocity and distance.

For distances D larger than the radius of the Hubble sphere $r_{HS}$ , objects recede at a rate faster than the speed of light (See Uses of the proper distance for a discussion of the significance of this): $r_{HS} = \frac{c}{H_0}$

Hubble Parameter approaching a Constant Value over time:

Let's first derive Hubble's Law. Consider a galaxy at present distance $x$ from us. Then, if the local velocity of the galaxy within its cluster is ignored, its cosmic distance will change over time as $$D(t) = a(t)\,x,$$ where $a(t)$ is the so-called scale factor. If we take the time derivative of this equation, we get $$\dot{D} = \dot{a}\,x = \left(\frac{\dot{a}}{a}\right)\,ax,$$ which we can write as $$v = HD,$$ where $v = \dot{D}$ is the recession velocity and $H = \dot{a}/a$ is the Hubble parameter. This is the famous Hubble Law. In other words, the Hubble parameter is a ratio of two quantities. The expansion of the universe implies that $a(t)$ increases over time, and $\dot{a}>0$. In fact, the expansion of the universe is accelerating, which means that $\dot{a}(t)$ increases over time as well. However, the ratio $H = \dot{a}/a$ decreases, because $a(t)$ increases more rapidly than $\dot{a}(t)$. The expansion rate of the universe is not high enough for $H(t)$ to increase.

Note that if the expansion were exponential, $a(t)\sim\exp(Ht)$, then $H$ is constant. In the standard cosmological model, the universe is evolving towards a state of exponential expansion, dominated by a constant dark energy density. In other words, if the standard model is correct, the Hubble parameter is slowly converging towards a constant value, and the expansion rate of the universe is increasing towards an exponential rate.

The confusion between shrinking, constant, and expanding horizons is mainly caused by the fact that the "spatial radius of horizon" or "spatial slice of space-time" are simply not invariantly defined concepts. There is one coordinate system in which the horizon has a constant diameter, but in the coordinate systems to which the $\Lambda$-FLRW coordinates converge it is not so.

First, let us see the "static" coordinate system: $$d s^2 = -(1-\Lambda R^2/3) d T^2 + (1-\Lambda R^2/3)^{-1} dR^2 + R^2 d \Omega^2$$ I.e. the horizon is at $R=\alpha=\sqrt{3/\Lambda}$. The position of the horizon is static as is the whole metric. These coordinates are in fact "Schwarzschild-like" because surfaces of constant $R,T$ are spheres with constant curvature $1/R^2$ quite like in the Schwarzschild-coordinate case.

However, other sets of coordinates will give you FLRW universes with the spatial curvature $k=-1,0,1$ which are not static with respect to the cosmological time coordinate:

$$d s^2 = - d t_{-1}^2 + \alpha^2\sinh^2(t_{-1}/\alpha) (d\rho^2 +\sinh^2\rho d\Omega^2)$$ $$d s^2 = - d t_{0}^2 + \exp(2t_{0}/\alpha) (dx^2 +dy^2+dz^2)$$ $$d s^2 = - d t_{1}^2 + \alpha^2\cosh^2(t_{1}/\alpha) (d\chi^2 +\sin^2\chi d\Omega^2)$$

In all these coordinates, the $t_{-1,0,1}=const.$ hypersurfaces represent congruences of observers on a constant-curvature space-like hypersurface in spacetime and $t$ the proper time of the observer at a given point of the congruence.

If we take an intersection of the $R=\alpha$ horizon with the $t_{-1,0,+1}$, we get a 2D space-like surface we might like to call the "formal horizon". The formal horizon indeed does expand or shrink depending on the curvature slicing we choose and the epoch the congruence is in (note that there is a "bounce" epoch at $t_{\pm 1}=0$ for the curved congruences), but this is only an effect in certain coordinates, with respect to a certain congruence of observers. In the expansion epoch we are in, the formal horizon certainly shrinks, but from the point of another set of moderately natural coordinates, it will be asymptotically constant.

(As to your comments, remember that thermodynamics require a specification of a time coordinate. The cosmological $t$ is the proper time of the cosmic fluid so it is good for determining the short-range thermodynamical properties of the fluid itself, but one must be very careful in thermodynamical arguments about other quantities.)

The entropy of the universe only grows if it is a closed system. You are likely mixing two different concepts: the radius of the entire universe, that keep growing and the cosmological horizon of a particular observer, which does not define any physical border, just a causal/observable one. The entropy inside the observable universe (and its radius) does not grow, and actually, by the holographic principle, the entropy gets actually reduced. This is because of the reduction of mass and energy and of available microstates as the radius diminishes. The extreme case will be in the future if the big rip happens, every fundamental particle and every black hole will become isolated from the rest of the universe, so each disconnected part of the universe will in fact have very low entropy.