# The Edge and the Age of the Universe: Where 13.8 Billion Years and 46.5 Billion Light-Years Come From

> How the reciprocal of the Hubble constant yields 13.8 billion years, why the observable universe reaches 46.5 billion light-years, and whether space itself has an edge.
> https://rikai.mugen-giken.com/en/physics/cosmology/age-and-size-of-the-universe

## 0. Key points

- The figure of 13.8 billion years for the age of the universe was not dug out of some stratum. It was **computed backwards from the rate of expansion**. The crudest estimate is a single division by the present expansion rate, that is, the reciprocal $1/H_0$ of the Hubble constant.
- Converting $H_0 = 67.4\ \mathrm{km/s/Mpc}$ into years gives $1/H_0 = 14.5$ billion years. The actual figure of 13.8 billion years is that number multiplied by $0.951$, and inside this $0.951$ sits the tug-of-war between matter and dark energy. We compute that coefficient by hand in this article.
- The radius of the observable universe is **not** 13.8 billion light-years but about 46.5 billion light-years. During the 13.8 billion years the light was in flight, the point it started from receded as well.
- Most of the objects whose light is reaching us now are, at this moment, receding from us faster than light. This does not contradict relativity.
- "The edge of the universe itself" is an entirely different question from "the edge of what we can see". Space may be infinite; and even if it is finite, it may have no edge. Observation tells us that it is flat out to a radius of at least about 230 billion light-years.

## 1. Motivation: does the universe have a birthday?

We live in an age where the question "how old is the universe?" gets the immediate answer "13.8 billion years". But how was that number measured? The universe has no growth rings, and no witness was present at the moment of its birth.

There is exactly one clue: the fact that **the universe is expanding**. In 1929 Edwin Hubble reported that the more distant a galaxy is, the faster it recedes — the recession speed $v$ is proportional to the distance $d$. That seems to make matters easy. If every galaxy has been receding at its present speed all along, then running the film backwards should bring them all together at a single point. That "rewind time" is an estimate of the age of the universe.

This naive calculation ran into trouble immediately. The constant of proportionality Hubble obtained was about $500\ \mathrm{km/s/Mpc}$, giving a rewind time of roughly **2 billion years**. Radiometric dating had already shown that rocks on Earth were at least 2 billion years old. The universe being younger than the Earth is a rather awkward conclusion.

The culprit behind this contradiction was "distance". Hubble measured distances to galaxies using Cepheid variables as his ruler, but the ruler was mis-calibrated. In 1952 Walter Baade realised that there are two distinct classes of Cepheid; distances more than doubled at a stroke, and the constant of proportionality dropped to less than half (on this mix-up of scales, see <Ref to="physics/cosmology/cosmic-distance-ladder#rem-metallicity" />). We leave to [How we know the distance to a star](/en/physics/cosmology/cosmic-distance-ladder) the story of how much labour goes into measuring distances (how the errors pile up as the rungs are spliced together is summarised in <Ref to="physics/cosmology/cosmic-distance-ladder#prop-ladder-additivity" />), but bear in mind that the age of the universe is, in the end, **the precision of distance measurement itself**.

And whenever the age comes up, the question "so where is the edge of the universe?" follows as a matter of course. That question has two entirely different meanings, and any discussion that conflates them gets lost. In this article we carry the age calculation through to the end, and then separate the two "edges" properly.

## 2. Preliminaries: tools for measuring an expanding universe

### 2.1. Scale factor and comoving distance

Picturing cosmic expansion as "galaxies flying apart through space" will almost certainly lead you astray. The correct picture is that **space itself stretches, and the galaxies, stuck to it, are carried apart**. The following definition turns that picture into formulas.

<Definition id="def-scale-factor" title="Scale factor and comoving distance">
Suppose the universe expands homogeneously (the same everywhere) and isotropically (the same in every direction). If the actual distance between two galaxies (their **proper distance**) $d(t)$ can be written, using a constant $\chi$ independent of time and a single function $a(t) > 0$ common to all pairs of galaxies, as

$$
d(t) = a(t)\,\chi
$$

then $a(t)$ is called the **scale factor** and $\chi$ the **comoving distance**. We normalise so that $a(t_0) = 1$ at the present time $t_0$. The comoving distance is therefore equal to the distance expressed in present-day units.
</Definition>

In short, $a(t)$ is the magnification of the photograph called the universe. At an epoch with $a = 0.5$, every distance was half of what it is now. The comoving distance $\chi$, on the other hand, is like a jersey number assigned to a galaxy: expansion does not change it.

### 2.2. The Hubble–Lemaître law

<Definition id="def-hubble" title="Hubble constant">
The rate of increase of the scale factor divided by $a$ itself,

$$
H(t) = \frac{1}{a(t)}\frac{da}{dt}
$$

is called the **Hubble constant** (more properly, the Hubble parameter). We write $H_0 = H(t_0)$ for its present value. $H$ has the dimensions of inverse time.
</Definition>

<Proposition id="prop-hubble-law" title="The Hubble–Lemaître law">
Suppose the universe expands homogeneously and isotropically in the sense of <Ref to="def-scale-factor" />. Then, as seen by any observer, the recession speed $v$ of a galaxy at distance $d$ satisfies

$$
v = H(t)\,d .
$$

The constant of proportionality $H(t)$ depends neither on the direction nor on which galaxy is chosen.
</Proposition>

<Proof of="prop-hubble-law">
By <Ref to="def-scale-factor" />, the proper distance for the pair of galaxies under consideration is $d(t) = a(t)\chi$, where $\chi$ is a constant independent of time. The recession speed is the rate of change of the proper distance, so differentiating with $\chi$ held constant,

$$
v = \frac{d}{dt}\bigl(a(t)\chi\bigr) = \frac{da}{dt}\,\chi .
$$

Substituting $\chi = d(t)/a(t)$ gives

$$
v = \frac{da}{dt}\cdot\frac{d(t)}{a(t)} = \left(\frac{1}{a}\frac{da}{dt}\right) d(t) = H(t)\,d(t)
$$

and the $H$ of <Ref to="def-hubble" /> appears. Since $a(t)$ was assumed to be a function common to all pairs of galaxies, $H$ depends neither on which galaxy is chosen nor on the direction.
</Proof>

What deserves attention in this proof is that **no particular centre was used anywhere**. Choosing any galaxy as the origin yields the same formula. Consequently, an inference such as "the more galaxies recede rapidly from you, the further you are from the centre of the universe" simply does not hold.

<Example id="ex-raisin" title="Checking with a raisin loaf">
Suppose three raisins A, B, C sit in a line in a lump of dough, with A to B measuring $1\ \mathrm{cm}$ and A to C measuring $3\ \mathrm{cm}$. Bake for an hour and let the dough swell to exactly twice its size, so that $a$ goes from $1$ to $2$ in one hour.

Seen from A, B goes from $1\ \mathrm{cm}$ to $2\ \mathrm{cm}$, hence recedes at $1\ \mathrm{cm/h}$, while C goes from $3\ \mathrm{cm}$ to $6\ \mathrm{cm}$, hence recedes at $3\ \mathrm{cm/h}$. Speed is proportional to distance, with constant of proportionality $1\ \mathrm{h^{-1}}$.

What about seen from B? From B, A goes from $1\ \mathrm{cm}$ to $2\ \mathrm{cm}$, so $1\ \mathrm{cm/h}$; from B, C goes from $2\ \mathrm{cm}$ to $4\ \mathrm{cm}$, so $2\ \mathrm{cm/h}$. The constant of proportionality is again $1\ \mathrm{h^{-1}}$, exactly as seen from A. Every one of the three raisins can claim to be the centre, and none of them is wrong. So much for the debate about the centre of the universe.
</Example>

### 2.3. The units of the Hubble constant

$H_0$ is quoted in the peculiar unit $\mathrm{km/s/Mpc}$ (kilometres per second per megaparsec). One megaparsec ($1\ \mathrm{Mpc}$) is $3.0857\times10^{19}\ \mathrm{km}$, or about 3.26 million light-years. Saying $H_0 = 67.4\ \mathrm{km/s/Mpc}$ means that the recession speed increases by $67.4\ \mathrm{km}$ per second for every $1\ \mathrm{Mpc}$ of separation. Since $\mathrm{km}$ and $\mathrm{Mpc}$ are both lengths, cancelling them leaves an inverse time. That cancellation is the star of the next section.

## 3. The age of the universe: from the reciprocal of the Hubble constant to 13.8 billion years

<Definition id="def-hubble-time" title="Hubble time">
The reciprocal of the present value of the Hubble constant,

$$
t_H = \frac{1}{H_0}
$$

is called the **Hubble time**. Likewise $c\,t_H = c/H_0$ is called the **Hubble radius**.
</Definition>

<Example id="ex-hubble-time" title="Converting the Hubble time into years">
We use the standard value $H_0 = 67.4\ \mathrm{km/s/Mpc}$ based on observations by the Planck satellite. Substituting $1\ \mathrm{Mpc} = 3.0857\times10^{19}\ \mathrm{km}$,

$$
H_0 = \frac{67.4\ \mathrm{km/s}}{3.0857\times10^{19}\ \mathrm{km}} = 2.184\times10^{-18}\ \mathrm{s^{-1}} .
$$

The length units cancel, leaving only "per second". Taking the reciprocal,

$$
t_H = \frac{1}{2.184\times10^{-18}\ \mathrm{s^{-1}}} = 4.578\times10^{17}\ \mathrm{s} .
$$

One year is $3.156\times10^{7}\ \mathrm{s}$, so

$$
t_H = \frac{4.578\times10^{17}}{3.156\times10^{7}}\ \mathrm{year} = 1.451\times10^{10}\ \mathrm{year} \approx 14.5\ \text{billion years} .
$$

The same number serves for the Hubble radius: $c\,t_H = 14.5$ billion light-years (since $c \times 1\ \text{year} = 1\ \text{light-year}$, the numerical value carries over unchanged).
</Example>

Fourteen and a half billion years. That is already remarkably close to our destination of 13.8 billion years. This is no accident, and we now see why, step by step.

### 3.1. If the expansion rate never changed, the age is exactly the Hubble time

<Proposition id="prop-constant-rate" title="Age of a uniformly expanding universe">
Suppose the scale factor grows in proportion to time, that is, $a(t) = kt$ for some constant $k > 0$. Then the time $t_1$ at which $a(t_1) = 0$ is $t_1 = 0$, and the time elapsed from then until the present $t_0$ equals

$$
t_0 = \frac{1}{H_0} .
$$
</Proposition>

<Proof of="prop-constant-rate">
Substitute $a(t) = kt$ into <Ref to="def-hubble" />. Since $da/dt = k$,

$$
H(t) = \frac{1}{a}\frac{da}{dt} = \frac{k}{kt} = \frac{1}{t} .
$$

Evaluating this at the present time $t_0$ gives $H_0 = 1/t_0$, that is, $t_0 = 1/H_0$. Moreover $a(t) = kt$ vanishes only at $t = 0$, so "the instant when all distances were $0$" is $t = 0$, and the time elapsed from then until now is exactly $t_0$.
</Proof>

The Hubble time <Ref to="def-hubble-time" /> is therefore **the age of the universe under the assumption that the expansion rate has never changed**. In the real universe the expansion rate has changed, so $1/H_0$ is no more than an estimate. By how much is it off?

### 3.2. The general formula for the age

<Proposition id="prop-age-integral" title="Integral expression for the age of the universe">
Suppose the scale factor $a(t)$ is a strictly increasing function of time and that there is a past time at which $a \to 0$. If $H$ can be expressed as a function $H(a)$ of the scale factor, then the time elapsed from $a = 0$ to the present ($a = 1$) is given by

$$
t_0 = \int_0^1 \frac{da}{a\,H(a)} .
$$
</Proposition>

<Proof of="prop-age-integral">
Solving the defining relation $H = (1/a)(da/dt)$ of <Ref to="def-hubble" /> for $dt$,

$$
\frac{da}{dt} = a H \quad\Longrightarrow\quad dt = \frac{da}{a H} .
$$

Since $a$ is a strictly increasing function of time, we may reparametrise $t$ by $a$. The time elapsed while $a$ runs from $0$ to $1$ is obtained by integrating both sides:

$$
t_0 = \int_{t(a=0)}^{t(a=1)} dt = \int_0^1 \frac{da}{a\,H(a)} .
$$

<Ref to="prop-constant-rate" /> is the special case of this formula in which $H(a) = H_0/a$ (for $a = kt$ we have $H = 1/t = k/a = H_0/a$); substituting it gives $\int_0^1 da/H_0 = 1/H_0$.
</Proof>

Once $H(a)$ is known, the age is obtained by integration alone. And $H(a)$ is supplied by the Friedmann equation, which follows from general relativity. We do not derive it here but simply use the result. For a spatially flat universe,

$$
H(a) = H_0\sqrt{\frac{\Omega_m}{a^3} + \frac{\Omega_r}{a^4} + \Omega_\Lambda}\,,
\qquad \Omega_m + \Omega_r + \Omega_\Lambda = 1 .
$$

Here $\Omega_m$ is the fraction of the present energy of the universe contributed by matter (ordinary matter plus dark matter), $\Omega_r$ the fraction contributed by radiation (light and light neutrinos), and $\Omega_\Lambda$ the fraction contributed by dark energy. The differing powers of $a$ have a reason. Matter is merely diluted by the volume $\propto a^3$, whereas light is on top of that stretched in wavelength by a factor $a$ and thereby loses energy, giving $a^4$. Dark energy is not diluted at all, hence a constant. Observations by the Planck satellite give

$$
\Omega_m = 0.315,\qquad \Omega_\Lambda = 0.685,\qquad \Omega_r \approx 9.2\times10^{-5} .
$$

All three are defined as ratios to the critical density (<Ref to="physics/cosmology/dark-matter-and-dark-energy#def-critical-density" />); for what they actually are, see [Dark matter and dark energy](/en/physics/cosmology/dark-matter-and-dark-energy).

<Example id="ex-eds" title="Forget dark energy and the universe comes out too young">
Until the mid-1990s many researchers regarded $\Omega_m = 1$, $\Omega_\Lambda = 0$ (a flat universe of matter alone) as standard. In that case $H(a) = H_0 a^{-3/2}$, so <Ref to="prop-age-integral" /> gives

$$
t_0 = \int_0^1 \frac{da}{a\cdot H_0 a^{-3/2}} = \frac{1}{H_0}\int_0^1 a^{1/2}\,da = \frac{1}{H_0}\left[\frac{2}{3}a^{3/2}\right]_0^1 = \frac{2}{3}\cdot\frac{1}{H_0} .
$$

Substituting the $14.5$ billion years of <Ref to="ex-hubble-time" />,

$$
t_0 = \frac{2}{3}\times 14.5\ \text{billion years} = 9.67\ \text{billion years} .
$$

Meanwhile the oldest stars in the globular clusters of the Milky Way are estimated to be around 13 billion years old. A universe of 9.7 billion years has no room for a 13-billion-year-old star. This was the "age problem" of the 1990s. The answer came in 1998: the expansion of the universe is not decelerating but **accelerating**.
</Example>

### 3.3. Getting 13.8 billion years

<Theorem id="thm-lcdm-age" title="Age of a flat universe of matter and dark energy">
Suppose space is flat, radiation is negligible ($\Omega_r = 0$), $\Omega_m + \Omega_\Lambda = 1$, and $0 < \Omega_\Lambda < 1$. Then the age of the universe is given by

$$
t_0 = \frac{2}{3\sqrt{\Omega_\Lambda}}\,
\ln\!\left(\frac{1+\sqrt{\Omega_\Lambda}}{\sqrt{\Omega_m}}\right)\cdot\frac{1}{H_0} .
$$
</Theorem>

<Proof of="thm-lcdm-age">
Substitute $H(a) = H_0\sqrt{\Omega_m a^{-3} + \Omega_\Lambda}$ into <Ref to="prop-age-integral" />:

$$
t_0 = \frac{1}{H_0}\int_0^1 \frac{da}{a\sqrt{\Omega_m a^{-3} + \Omega_\Lambda}} .
$$

Putting the contents of the square root over a common denominator gives $\Omega_m a^{-3} + \Omega_\Lambda = (\Omega_m + \Omega_\Lambda a^3)/a^3$, so pulling $a^{-3/2}$ out of the root and combining it with the leading $1/a$,

$$
t_0 = \frac{1}{H_0}\int_0^1 \frac{a^{1/2}\,da}{\sqrt{\Omega_m + \Omega_\Lambda a^3}} .
$$

Now substitute $u = a^{3/2}$. Then $du = \tfrac{3}{2}a^{1/2}\,da$, that is, $a^{1/2}\,da = \tfrac{2}{3}\,du$, and $a: 0 \to 1$ corresponds to $u: 0 \to 1$. Since also $a^3 = u^2$,

$$
t_0 = \frac{2}{3H_0}\int_0^1 \frac{du}{\sqrt{\Omega_m + \Omega_\Lambda u^2}} .
$$

This integral is evaluated by the formula $\displaystyle\int \frac{du}{\sqrt{A + Bu^2}} = \frac{1}{\sqrt{B}}\ln\!\left(u + \sqrt{u^2 + A/B}\right) + C$ (for $A, B > 0$). Indeed, differentiating the right-hand side with respect to $u$,

$$
\frac{1}{\sqrt{B}}\cdot\frac{1 + \dfrac{u}{\sqrt{u^2+A/B}}}{u+\sqrt{u^2+A/B}}
= \frac{1}{\sqrt{B}}\cdot\frac{1}{\sqrt{u^2+A/B}} = \frac{1}{\sqrt{A+Bu^2}}
$$

returns the integrand. Substituting $A = \Omega_m$ and $B = \Omega_\Lambda$,

$$
t_0 = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}
\left[\ln\!\left(u + \sqrt{u^2 + \frac{\Omega_m}{\Omega_\Lambda}}\right)\right]_0^1
= \frac{2}{3H_0\sqrt{\Omega_\Lambda}}
\ln\!\frac{1 + \sqrt{1 + \Omega_m/\Omega_\Lambda}}{\sqrt{\Omega_m/\Omega_\Lambda}} .
$$

Finally, multiplying numerator and denominator by $\sqrt{\Omega_\Lambda}$ and using the hypothesis $\Omega_m + \Omega_\Lambda = 1$, which makes $\sqrt{\Omega_\Lambda + \Omega_m} = 1$, we obtain

$$
t_0 = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\ln\frac{\sqrt{\Omega_\Lambda} + \sqrt{\Omega_\Lambda + \Omega_m}}{\sqrt{\Omega_m}}
= \frac{2}{3\sqrt{\Omega_\Lambda}}\ln\!\left(\frac{1+\sqrt{\Omega_\Lambda}}{\sqrt{\Omega_m}}\right)\frac{1}{H_0} .
$$
</Proof>

<Example id="ex-lcdm-number" title="Substituting the numbers to get 13.8 billion years">
Put $\Omega_m = 0.315$ and $\Omega_\Lambda = 0.685$ into <Ref to="thm-lcdm-age" />. First the square roots:

$$
\sqrt{\Omega_\Lambda} = \sqrt{0.685} = 0.8276,\qquad \sqrt{\Omega_m} = \sqrt{0.315} = 0.5612 .
$$

The argument of the logarithm is

$$
\frac{1 + 0.8276}{0.5612} = \frac{1.8276}{0.5612} = 3.256,\qquad \ln 3.256 = 1.1806 .
$$

The prefactor is $\dfrac{2}{3\times 0.8276} = 0.8054$, so

$$
t_0 = 0.8054 \times 1.1806 \times \frac{1}{H_0} = 0.9509\,\frac{1}{H_0} .
$$

Multiplying by the $1/H_0 = 14.51$ billion years of <Ref to="ex-hubble-time" />,

$$
t_0 = 0.9509 \times 14.51\ \text{billion years} = 13.80\ \text{billion years} .
$$

This agrees beautifully with the official Planck value of $13.797 \pm 0.023$ billion years. Neglecting radiation ($\Omega_r$) affects the result by less than $0.05\%$, in the direction of reducing the age by about $0.01$ billion years (we check this in <Ref to="ex-python-check" />).
</Example>

It is worth pausing over why the coefficient came out to be $0.9509$, so close to $1$. The universe decelerated in its first half (during the epoch when the gravity of matter dominated, the coefficient drops to $2/3$, as in <Ref to="ex-eds" />) and accelerated in its second half under dark energy. That deceleration and acceleration happen to cancel, returning us almost exactly to $1/H_0$, the age computed on the assumption of constant expansion. That is what $0.9509$ really is. We live in an era in which this coincidence holds.

<Remark id="rem-hubble-tension" title="The Hubble constant is still contested">
The age is inversely proportional to $H_0$, so a different value of $H_0$ changes the age directly. And the measurement of $H_0$ suffers from a discrepancy that is still unresolved. Inferred indirectly from the cosmic microwave background one gets $H_0 = 67.4 \pm 0.5\ \mathrm{km/s/Mpc}$, whereas direct measurement from nearby supernovae and Cepheids gives $H_0 = 73.0 \pm 1.0\ \mathrm{km/s/Mpc}$; the two disagree far beyond their error bars. This is the **Hubble tension**. The same discrepancy viewed from the side of the distance ladder is <Ref to="physics/cosmology/cosmic-distance-ladder#ex-hubble-tension" />. Adopting the latter value lowers the age to 12.7 billion years, leaving painfully little margin against the age of the oldest stars (<Ref to="exr-h0-73" />). Whether this mismatch is a systematic error in the measurements or a crack in the standard model is one of the hottest open problems today.
</Remark>

## 4. The edge of the observable universe: why 46.5 billion light-years?

If the universe is 13.8 billion years old, light can have travelled at most 13.8 billion light-years, so the radius of what we can see is 13.8 billion light-years — that is the tempting conclusion. Unfortunately it is wrong, because space kept stretching while the light was on its journey.

### 4.1. Definition and formula for the horizon

<Definition id="def-particle-horizon" title="Particle horizon (radius of the observable universe)">
Let $\chi_{\mathrm{hor}}$ be the largest comoving distance that light setting out at the beginning of the universe ($a = 0$) can have covered by the present time $t_0$. The corresponding proper distance today,

$$
d_{\mathrm{hor}}(t_0) = a(t_0)\,\chi_{\mathrm{hor}} = \chi_{\mathrm{hor}}
$$

is called the **particle horizon**, or the **radius of the observable universe**. From objects further away than this, not a single photon has yet reached us, even in principle.
</Definition>

<Proposition id="prop-horizon-formula" title="Integral expression for the particle horizon">
In a universe as in <Ref to="def-scale-factor" />, in which light travels locally at speed $c$, the particle horizon is given by

$$
d_{\mathrm{hor}}(t_0) = \int_0^{t_0} \frac{c\,dt}{a(t)} = \int_0^1 \frac{c\,da}{a^2 H(a)} .
$$
</Proposition>

<Proof of="prop-horizon-formula">
In an infinitesimal time $dt$, light covers a proper distance $c\,dt$. By <Ref to="def-scale-factor" />, proper and comoving distance are related by $d = a\chi$, so the increment of comoving distance corresponding to this motion is

$$
d\chi = \frac{c\,dt}{a(t)} .
$$

Summing from $t = 0$ to $t = t_0$, the total comoving distance the light has gained is $\chi_{\mathrm{hor}} = \int_0^{t_0} c\,dt/a(t)$. By the normalisation $a(t_0) = 1$, the proper distance today has the same value.

To change the variable of integration from $t$ to $a$, substitute the relation $dt = da/(aH)$ used in the proof of <Ref to="prop-age-integral" />:

$$
\int_0^{t_0}\frac{c\,dt}{a} = \int_0^1 \frac{c}{a}\cdot\frac{da}{aH(a)} = \int_0^1\frac{c\,da}{a^2H(a)} .
$$

Note the $1/a^2$ that has appeared in the integrand. That extra factor of $1/a$ encodes the effect that "space stretched by a factor $1/a$ after the light set out".
</Proof>

<Example id="ex-horizon-numbers" title="Computing the horizon">
**(1) A universe of matter alone** (the setting of <Ref to="ex-eds" />) has $a(t) = (t/t_0)^{2/3}$. Substituting into the left-hand form of <Ref to="prop-horizon-formula" />,

$$
d_{\mathrm{hor}} = \int_0^{t_0} c\left(\frac{t}{t_0}\right)^{-2/3} dt
= c\,t_0^{2/3}\left[3t^{1/3}\right]_0^{t_0} = 3c\,t_0 .
$$

That is exactly **three times** the naive expectation $c\,t_0$ based on the time the light was in flight.

**(2) In the real universe**, $H(a)$ is given by the Friedmann equation above, and the integral cannot be written in elementary functions, so we compute it numerically (<Ref to="ex-python-check" />). The result is

$$
d_{\mathrm{hor}}(t_0) \approx 46.1\ \text{billion light-years}
$$

whereas the value usually quoted in the literature is about **46.5 billion light-years**. The difference of about 1% comes from small differences in the treatment of radiation and neutrinos in the early universe. Either way, it is roughly $3.3$ times $c\,t_0 = 13.8$ billion light-years. In diameter that is about 93 billion light-years.
</Example>

<Figure caption="How the distances of cosmology compare. The thin line at the bottom is the distance, at the time of emission, of the place that emitted the cosmic microwave background now arriving.">
<svg viewBox="0 0 720 300" width="100%" role="img" aria-label="Horizontal bar chart comparing cosmological distances">
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    <text x="720" y="16" text-anchor="end" font-size="12">13.8 Gly</text>
    <rect x="0" y="24" width="142" height="14" fill="currentColor" opacity="0.35" />

    <text x="0" y="64">Hubble radius c/H₀ (where recession reaches c)</text>
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    <rect x="0" y="72" width="150" height="14" fill="currentColor" opacity="0.35" />

    <text x="0" y="112">Event horizon (limit for light emitted today)</text>
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    <rect x="0" y="120" width="172" height="14" fill="currentColor" opacity="0.35" />

    <text x="0" y="160">Source of the CMB: distance today</text>
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    <rect x="0" y="168" width="467" height="14" fill="currentColor" opacity="0.35" />

    <text x="0" y="208">Radius of the observable universe (particle horizon)</text>
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    <rect x="0" y="216" width="480" height="14" fill="var(--sl-color-accent)" />

    <text x="0" y="256">Source of the CMB: distance at emission</text>
    <text x="720" y="256" text-anchor="end" font-size="12">0.041 Gly</text>
    <rect x="0" y="264" width="3" height="14" fill="var(--sl-color-accent)" />
    <line x1="6" y1="271" x2="60" y2="271" stroke="currentColor" stroke-width="1" stroke-dasharray="3 3" />
    <text x="66" y="275" font-size="11">(41 million ly — thinner than a single line at this scale)</text>
  </g>
</svg>
</Figure>

The cosmic microwave background arriving now is light emitted when the universe was 380,000 years old, at about $a = 1/1100$ (for how that time is obtained, see <Ref to="physics/cosmology/big-bang-evidence#ex-recombination-redshift" />; for the whole picture, see [Was there really a Big Bang?](/en/physics/cosmology/big-bang-evidence)). At the moment of emission, the place that emitted this light was a mere **41 million light-years** from us. That distance was stretched by a factor of 1100 over 13.8 billion years and is now **45.2 billion light-years** — which is what the figure above shows. The edge of the universe is far away not because the light travelled for a long time, but because the road stretched while the light was on it.

### 4.2. Why we can see objects that recede faster than light

By <Ref to="prop-hubble-law" />, an object at distance $d$ recedes at $v = H_0 d$. The distance at which this $v$ equals $c$ is precisely the Hubble radius $c/H_0 = 14.5$ billion light-years. Yet the place that emitted the background radiation is now 45.2 billion light-years away, so its recession speed is

$$
v = H_0 \times 45.2\ \text{billion light-years} = \frac{45.2}{14.5}\,c = 3.1\,c .
$$

Light has reached us from a place receding at three times the speed of light.

<Aside type="note">
This does not contradict relativity. What special relativity forbids is that two objects at the same place pass each other faster than light. The recession speed of a distant galaxy is just the amount by which the intervening space has stretched, expressed as a difference of distances; nothing is passing anybody at superluminal speed. In fact every galaxy is at rest where it sits, and light passing that place always overtakes the galaxy at exactly the speed of light.
</Aside>

So how does light get to us from a place receding faster than light? The following example reproduces the mechanism exactly.

<Example id="ex-ant" title="An ant walking on a stretching rubber band">
An ant sits at the left end of a rubber band of length $L_0 = 1\ \mathrm{m}$. The band stretches uniformly along its whole length, with the right end receding at $k = 1\ \mathrm{m/s}$. The ant walks to the right at $v = 1\ \mathrm{cm/s} = 0.01\ \mathrm{m/s}$ relative to the band. The ant's speed is a hundredth of the stretching rate, and at first the gap to the right end only widens. Can the ant ever reach the right end?

Write the ant's position as $x(t)$ and the length of the band as $L(t) = L_0 + kt$, and track the **fraction covered** $s = x/L$. This $s$ increases only by the ant's own walking (stretching carries the ant and the right end by the same proportion and so leaves $s$ unchanged), giving

$$
\frac{ds}{dt} = \frac{v}{L(t)} = \frac{v}{L_0 + kt} .
$$

Integrating both sides from $0$ to $t$,

$$
s(t) = \frac{v}{k}\ln\!\left(1 + \frac{kt}{L_0}\right) .
$$

The logarithm is slow, but it grows without bound. Hence there is necessarily a finite time at which $s(t) = 1$, namely

$$
t = \frac{L_0}{k}\left(e^{k/v} - 1\right) = 1\times\left(e^{100}-1\right)\ \mathrm{s} \approx 2.7\times10^{43}\ \mathrm{s} .
$$

In years that is about $8.5\times10^{35}$, some $10^{26}$ times the age of the universe. **The ant does arrive.** It merely asks a certain fortitude of whoever is waiting.

Light in the universe is in the same position as this ant. Distant galaxies recede faster than light, but light keeps steadily gaining comoving distance, and in the early universe it had a far better deal than the ant, because the expansion was decelerating rapidly. That is why light emitted from a place fleeing at three times the speed of light has reached us.
</Example>

<Remark id="rem-horizon-personal" title="Everyone has their own horizon">
The particle horizon <Ref to="def-particle-horizon" /> is "the sphere of radius 46.5 billion light-years centred on the observer". Like the horizon seen from a ship, **each observer has a different sphere**. Inhabitants of a galaxy 10 billion light-years away also have an observable universe of radius 46.5 billion light-years, which overlaps ours substantially but does not coincide with it. They see regions we cannot. When we estimate how many civilisations the universe holds, only what lies inside this sphere can be counted (<Ref to="physics/cosmology/fermi-paradox#def-drake-equation" />, [How many aliens are there?](/en/physics/cosmology/fermi-paradox)). And the finiteness of the horizon is itself the reason the night sky is dark (<Ref to="physics/cosmology/olbers-paradox#prop-finite-age" />, [Why is the night sky dark?](/en/physics/cosmology/olbers-paradox)).
</Remark>

## 5. The edge of the universe itself: is there a boundary?

Every "edge" so far has been **the edge of what is visible**. There is no wall there. Galaxies exist beyond the horizon, and their inhabitants take themselves to be at the centre of the universe. But does space itself end somewhere?

<Figure caption="Two separate questions contained in the single phrase 'the edge of the universe'">
<Mermaid code={`flowchart TD
  Q["What is 'the edge of the universe'?"] --> A["(1) Edge of what we can see"]
  Q --> B["(2) Edge of space itself"]
  A --> A1["Radius about 46.5 billion ly<br/>a sphere centred on the observer"]
  A1 --> A2["Certainly exists<br/>but sits elsewhere for each person"]
  B --> B1["If space is infinite, there is no edge"]
  B --> B2["Even if finite, a 3-sphere or torus has no edge"]
  B --> B3["An edge would require an 'outside'"]
  B3 --> B4["No evidence for that has been found"]
  B1 --> C["Observation: curvature nearly zero<br/>flat out to at least 230 billion ly"]
  B2 --> C`} />
</Figure>

### 5.1. The option of being finite without an edge

"Finite implies bounded by an edge" is in fact an overreach of intuition. Consider the surface of the Earth. Its area is finite (about $5.1\times10^{8}\ \mathrm{km^2}$), yet no matter how far you walk there is no cliff. Two-dimensional creatures living on a sphere would say "our world is finite but has no edge".

The same thing happens in three dimensions. If space is a 3-sphere (the analogue of the surface of a ball seen from four dimensions), then its volume is finite and it has no edge. Fly straight ahead and you eventually come back to where you started. The same holds for a doughnut-shaped torus, which manages to be finite while keeping curvature $0$ — the structure in which a character leaving the right edge of a game screen reappears on the left.

"So is the universe sitting inside something?" is the natural next question, but nothing of the sort is needed. A sphere, or a 3-sphere, can be defined mathematically in its own right, with no ambient space. When we draw a sphere on paper we use a three-dimensional sheet, but that is a convenience of the artist, nothing more.

### 5.2. What observation says

Whether space is closed or infinite can be settled by measuring the **curvature**. In a universe with curvature the Friedmann equation acquires an extra term,

$$
H(a)^2 = H_0^2\left(\frac{\Omega_m}{a^3} + \Omega_\Lambda + \frac{\Omega_K}{a^2}\right)
$$

and the sign of $\Omega_K$ fixes the shape of space. If $\Omega_K < 0$ the universe is closed (finite), if $\Omega_K > 0$ it is open, and if $\Omega_K = 0$ it is flat. When $\Omega_K \neq 0$, the radius of curvature $R_c$ is given by

$$
R_c = \frac{c/H_0}{\sqrt{|\Omega_K|}} .
$$

Combining Planck satellite data with observations of the galaxy distribution gives

$$
\Omega_K = 0.0007 \pm 0.0019 .
$$

This is consistent with $0$ within the errors: the universe looks flat. From this we can say how large the universe must be even if it were closed.

<Proposition id="prop-curvature-radius" title="Lower bound on the radius of curvature">
If $|\Omega_K| \le 0.004$ (corresponding to the $2\sigma$ upper limit of the measurement above), then the radius of curvature of space satisfies

$$
R_c \ge 229\ \text{billion light-years} .
$$

This is about $4.9$ times the radius of the observable universe, or about 120 times its volume.
</Proposition>

<Proof of="prop-curvature-radius">
Since $R_c = (c/H_0)/\sqrt{|\Omega_K|}$ is a decreasing function of $|\Omega_K|$, the minimum of $R_c$ occurs when $|\Omega_K|$ takes its upper limit. By <Ref to="ex-hubble-time" />, $c/H_0 = 14.5$ billion light-years, and $\sqrt{0.004} = 0.0632$, so

$$
R_c \ge \frac{14.5\ \text{billion light-years}}{0.0632} = 229.4\ \text{billion light-years} .
$$

Compared with the $46.5$ billion light-years of <Ref to="ex-horizon-numbers" />, this is a factor $229.4/46.5 = 4.93$, and since volume goes as the cube of the radius, a factor $4.93^3 = 120$ in volume.
</Proof>

So the universe may well be finite, but even if it is, what we see is less than 1% of the whole. We are like ants in a small room, spending our lives without noticing that the Earth is round.

Moreover, finiteness can occur without curvature as well (as with a torus). In that case space "wraps around", so light from the same object arrives from another direction too, and **circles with identical patterns** should appear in two places on the map of the cosmic microwave background. This search for "circles in the sky" has been carried out on the WMAP and Planck maps, and nothing was found. Hence, even if the universe does wrap around, its circumference is at least larger than the observable universe.

<Remark id="rem-infinite-honestly" title="Honestly, we do not know">
Present observations cannot answer the question "is the universe infinite?". All we can measure is that the curvature is extremely close to $0$ within the visible region, and one cannot conclude from that that the whole is infinite — just as one cannot conclude that the Earth is a plane from the fact that the ground looks flat. Showing it is finite would require detecting curvature or "circles in the sky"; showing it is infinite is in principle even harder. Here, "we do not know" is the honest answer.
</Remark>

## 6. The future: the visible universe grows, the visible galaxies dwindle

We close with one more "edge". Whereas the particle horizon <Ref to="def-particle-horizon" /> was the limit of light arriving from the past, this one is the limit of light arriving in the future.

<Definition id="def-event-horizon" title="Event horizon">
Let $\chi_{\mathrm{eh}}$ be the largest comoving distance that light setting out at the present time $t_0$ can cover in the whole of the remaining time $t \to \infty$. The corresponding proper distance today,

$$
d_{\mathrm{eh}}(t_0) = \chi_{\mathrm{eh}} = \int_{t_0}^{\infty}\frac{c\,dt}{a(t)} = \int_1^{\infty}\frac{c\,da}{a^2 H(a)}
$$

is called the (cosmological) **event horizon**. Events happening today further away than this distance will never reach us, however long we wait.
</Definition>

<Proposition id="prop-event-horizon-finite" title="An accelerating universe has an event horizon">
In a flat universe with $\Omega_\Lambda > 0$, the integral in <Ref to="def-event-horizon" /> converges and satisfies

$$
d_{\mathrm{eh}}(t_0) \le \frac{c}{H_0\sqrt{\Omega_\Lambda}} .
$$

For $\Omega_m = 0.315$, $\Omega_\Lambda = 0.685$ and $H_0 = 67.4\ \mathrm{km/s/Mpc}$, this upper bound is 17.5 billion light-years.
</Proposition>

<Proof of="prop-event-horizon-finite">
In the Friedmann equation $H(a) = H_0\sqrt{\Omega_m a^{-3} + \Omega_\Lambda}$, the first term under the root is non-negative, so for every $a$

$$
H(a) \ge H_0\sqrt{\Omega_\Lambda} .
$$

The integrand is therefore bounded above by

$$
\frac{c}{a^2 H(a)} \le \frac{c}{a^2 H_0\sqrt{\Omega_\Lambda}}
$$

and hence

$$
d_{\mathrm{eh}}(t_0) \le \frac{c}{H_0\sqrt{\Omega_\Lambda}}\int_1^\infty \frac{da}{a^2}
= \frac{c}{H_0\sqrt{\Omega_\Lambda}}\left[-\frac{1}{a}\right]_1^\infty = \frac{c}{H_0\sqrt{\Omega_\Lambda}} .
$$

Numerically, from $c/H_0 = 14.5$ billion light-years in <Ref to="ex-hubble-time" /> and $\sqrt{0.685} = 0.8276$, we get $14.5/0.8276 = 17.5$ billion light-years.

Note that this argument does not apply to the particle-horizon integral of <Ref to="prop-horizon-formula" />. In the integral running into the past, $H$ diverges as $a \to 0$, and whether the integral converges is decided by how matter and radiation contribute. In fact, in an accelerating universe both horizons exist simultaneously.
</Proof>

Numerical integration gives a present event horizon of about **16.7 billion light-years**, slightly below the upper bound of 17.5 billion light-years derived above; in the distant future, when the universe becomes completely dark-energy dominated, it will approach that value.

A somewhat melancholy conclusion follows. Among the galaxies we see today, those whose present distance exceeds 16.7 billion light-years — in terms of redshift, objects with roughly $z > 1.8$ — will never deliver to Earth the light they emit **today**. Those galaxies remain visible, but what we see is their past, and as time goes on only ever older images will reach us, stretched, dimmed and reddened. If the acceleration continues, then in about 100 billion years every galaxy outside the Local Group will have disappeared from view, and the astronomers of that era will conclude that "the universe consists of our galaxy alone, and it is static". We happen to occupy one of the limited epochs in which the history of the universe can still be read.

<Aside type="tip">
That the event horizon of a black hole carries the same name is no coincidence. Both are boundaries beyond which information about events never reaches us again, and the mathematics is similar. The difference is that a black hole's horizon lies over there, whereas the cosmological horizon surrounds us. The definition of the horizon over there is in <Ref to="physics/cosmology/black-holes#def-event-horizon" />; for details, see [What lies beyond a black hole](/en/physics/cosmology/black-holes).
</Aside>

<Example id="ex-python-check" title="Doing the numerical integration yourself">
In both the integrals of <Ref to="prop-age-integral" /> and <Ref to="prop-horizon-formula" />, the integrand varies rapidly as $a \to 0$. We therefore substitute $a = u^2$ (so $da = 2u\,du$) to refine the sampling near the origin, and apply the trapezoidal rule.

```python
import numpy as np

H0 = 67.4                  # km/s/Mpc
Om, OL, Or = 0.315, 0.685, 9.2e-5
Mpc_km = 3.0857e19         # 1 Mpc [km]
yr_s = 3.1557e7            # 1 year [s]

tH = Mpc_km / H0 / yr_s    # Hubble time [years]
print(f"1/H0             = {tH/1e9:.2f} billion years")

def E(a):                  # H(a) / H0
    return np.sqrt(Om / a**3 + OL + Or / a**4)

u = np.linspace(1e-8, 1.0, 2_000_001)   # substitution a = u^2
a = u**2
age = np.trapezoid(2 * u / (a * E(a)), u) * tH
hor = np.trapezoid(2 * u / (a**2 * E(a)), u) * tH

print(f"age              = {age/1e9:.2f} billion years")
print(f"particle horizon = {hor/1e9:.1f} billion light-years")
```

The output is as follows.

```
1/H0             = 14.51 billion years
age              = 13.79 billion years
particle horizon = 46.1 billion light-years
```

The age is smaller than the hand computation of 13.80 billion years in <Ref to="ex-lcdm-number" /> by $0.01$ billion years, exactly the amount contributed by including radiation. The horizon of 46.1 billion light-years differs by about 1% from the commonly quoted 46.5 billion light-years. That difference stems from the value of $\Omega_r$, in particular from whether neutrinos are treated as radiation or as matter; taking $\Omega_r = 4.2\times10^{-5}$ (photons only) gives 46.4 billion light-years. Do not take "46.5 billion light-years" for a settled three-significant-figure value.
</Example>

## 7. Exercises

<Exercise id="exr-h0-73" difficulty="Easy">
Adopt the value $H_0 = 73.0\ \mathrm{km/s/Mpc}$ from direct measurement using supernovae and Cepheid variables. (1) Compute the Hubble time $1/H_0$ in years. (2) Multiply by the coefficient $0.9509$ of <Ref to="thm-lcdm-age" /> (with $\Omega_m = 0.315$, $\Omega_\Lambda = 0.685$) to obtain the age of the universe, and compare it with the age of the oldest globular clusters, 13 billion years.

<Solution>
**(1)** The procedure is that of <Ref to="ex-hubble-time" />. From $1\ \mathrm{Mpc} = 3.0857\times10^{19}\ \mathrm{km}$,

$$
H_0 = \frac{73.0}{3.0857\times10^{19}}\ \mathrm{s^{-1}} = 2.366\times10^{-18}\ \mathrm{s^{-1}},
$$

$$
\frac{1}{H_0} = 4.227\times10^{17}\ \mathrm{s} = \frac{4.227\times10^{17}}{3.156\times10^{7}}\ \text{year} = 1.339\times10^{10}\ \text{year} = 13.4\ \text{billion years} .
$$

Since $H_0$ is $1.083$ times $67.4$, the Hubble time can also be checked by division: $14.51/1.083 = 13.4$ billion years.

**(2)** $t_0 = 0.9509 \times 13.39\ \text{billion years} = 12.73\ \text{billion years}$. This falls below the roughly 13 billion years of the oldest globular clusters. Stellar age estimates carry uncertainties of about 500 million years, so this is not an immediate contradiction, but there is almost no margin left. This is one of the reasons the Hubble tension of <Ref to="rem-hubble-tension" /> is suspected of being more than mere measurement error.
</Solution>
</Exercise>

<Exercise id="exr-eds-horizon" difficulty="Standard">
Consider a flat universe of matter alone ($\Omega_m = 1$, $\Omega_\Lambda = \Omega_r = 0$). (1) Using the method of <Ref to="prop-age-integral" />, verify that the scale factor of this universe can be written $a(t) = (t/t_0)^{2/3}$. (2) Compute the particle horizon from the right-hand form of <Ref to="prop-horizon-formula" /> (the integral over $a$) and show that it equals $3c\,t_0$.

<Solution>
**(1)** For $\Omega_m = 1$ the Friedmann equation reads $H(a) = H_0 a^{-3/2}$. By <Ref to="def-hubble" />, $da/dt = aH = H_0 a^{-1/2}$, so separating variables,

$$
a^{1/2}\,da = H_0\,dt \quad\Longrightarrow\quad \frac{2}{3}a^{3/2} = H_0 t
$$

(using the condition $a = 0$ at $t = 0$). Hence $a = (3H_0t/2)^{2/3}$. The condition $a = 1$ at the present $t = t_0$ gives $3H_0t_0/2 = 1$, that is $t_0 = (2/3)/H_0$, in agreement with <Ref to="ex-eds" />. Putting this $H_0 = 2/(3t_0)$ back in yields $a(t) = (t/t_0)^{2/3}$.

**(2)** Substituting $H(a) = H_0a^{-3/2}$ into <Ref to="prop-horizon-formula" />,

$$
d_{\mathrm{hor}} = \int_0^1 \frac{c\,da}{a^2\cdot H_0a^{-3/2}}
= \frac{c}{H_0}\int_0^1 a^{-1/2}\,da
= \frac{c}{H_0}\Bigl[2a^{1/2}\Bigr]_0^1 = \frac{2c}{H_0} .
$$

The integrand diverges as $a \to 0$, but the divergence $a^{-1/2}$ is integrable, so the value is finite. Substituting $1/H_0 = (3/2)t_0$ from part (1),

$$
d_{\mathrm{hor}} = 2c\cdot\frac{3}{2}t_0 = 3c\,t_0
$$

in agreement with part (1) of <Ref to="ex-horizon-numbers" />.
</Solution>
</Exercise>

<Exercise id="exr-cmb" difficulty="Standard">
The cosmic microwave background is light emitted at the epoch $a = 1/1100$, and the place that emitted it is now at a distance of 45.2 billion light-years. (1) How far from us was that place at the instant the light was emitted? (2) Compute the present recession speed of that place using <Ref to="prop-hubble-law" /> and compare it with the speed of light. (3) Explain, using <Ref to="ex-ant" />, why the answer to (2) does not contradict special relativity.

<Solution>
**(1)** By <Ref to="def-scale-factor" />, the proper distance is $d(t) = a(t)\chi$ with the comoving distance $\chi$ unchanged. The distance today, at $a = 1$, is 45.2 billion light-years, so $\chi = 45.2$ billion light-years, and at the moment of emission $a = 1/1100$, so

$$
d = \frac{1}{1100}\times 45.2\ \text{billion light-years} = 0.0411\ \text{billion light-years} = 41.1\ \text{million light-years} .
$$

On astronomical scales that counts as "right next door". This light set out from 41 million light-years away and yet took 13.8 billion years to arrive, because space kept stretching and the road kept getting longer.

**(2)** Substituting $H_0 = 1/(14.5\ \text{billion years})$ and $d = 45.2$ billion light-years into <Ref to="prop-hubble-law" />,

$$
v = H_0 d = \frac{45.2\ \text{billion light-years}}{14.5\ \text{billion years}} = 3.12\ \text{light-years/year} = 3.1\,c .
$$

That is 3.1 times the speed of light.

**(3)** The situation is that of the ant on the rubber band. The right end of the band receded a hundred times faster than the ant walked, yet the ant steadily increased the **fraction** $s$ it had covered and arrived in finite time. In the universe likewise, light moves at exactly the speed of light at every point, and no object is being overtaken. The recession speed $3.1c$ is the amount by which the space between two separated points stretches, divided by distance; it is not the speed of anything passing anything else. Special relativity forbids only the latter.
</Solution>
</Exercise>

## References

- E. Hubble, "A relation between distance and radial velocity among extra-galactic nebulae", *Proceedings of the National Academy of Sciences* 15 (1929), 168–173. The original paper reporting the discovery of the expansion, giving about $500\ \mathrm{km/s/Mpc}$ for the constant of proportionality.
- Planck Collaboration, "Planck 2018 results. VI. Cosmological parameters", *Astronomy & Astrophysics* 641 (2020), A6. [arXiv:1807.06209](https://arxiv.org/abs/1807.06209) — the source of the values of $H_0$, $\Omega_m$, $\Omega_\Lambda$, $\Omega_K$ and the age of the universe used in this article.
- A. G. Riess et al., "A Comprehensive Measurement of the Local Value of the Hubble Constant with 1 km/s/Mpc Uncertainty from the Hubble Space Telescope and the SH0ES Team", *The Astrophysical Journal Letters* 934 (2022), L7. [arXiv:2112.04510](https://arxiv.org/abs/2112.04510) — one party to the Hubble tension.
- T. M. Davis and C. H. Lineweaver, "Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe", *Publications of the Astronomical Society of Australia* 21 (2004), 97–109. [arXiv:astro-ph/0310808](https://arxiv.org/abs/astro-ph/0310808) — the standard paper correcting misconceptions about horizons and superluminal recession. Its figures are exceptionally clear.
- Planck Collaboration, "Planck 2013 results. XXVI. Background geometry and topology of the Universe", *Astronomy & Astrophysics* 571 (2014), A26. [arXiv:1303.5086](https://arxiv.org/abs/1303.5086) — the results of the search for "circles in the sky".
- Takahiko Matsubara, *Gendai Uchūron: Jikū to Busshitsu no Kyōshinka*, University of Tokyo Press, 2010 (in Japanese) — Chapters 2 and 3 give a careful account of the Friedmann equation and of the various distances and horizons.