<p>In short: things can not move faster that light by theirselves, but they can move faster than light due to universal expansion. The more far away, the faster they go away.</p>

CC BY-SA 3.0
2013-11-24T19:59:26.340Z
2013-11-24T19:59:26.340Z
Envite
<p>I was just thinking about that and here is my layman's explanation. Imagine you're tracing two dots on a crumpled piece of paper, the dots are moving, but as they are moving, so is the paper getting ‘uncrumpled’, the actual distance between the dots will be more than the sum of distances they have travelled. </p>

CC BY-SA 3.0
2018-02-13T10:23:12.843Z
2018-02-14T01:55:26.773Z
2018-02-14T01:55:26.773Z
Hany
<p>I love <a href="http://www.astro.ucla.edu/%7Ewright/cosmo_01.htm" rel="noreferrer">Ned Wright's cosmology tutorial</a> and I highly recommend it, but that statement by him is at the least very misleading. Superluminal recession speeds plainly can't be related to spacetime curvature because they don't vanish in the limit of zero curvature (zero energy density or zero <span class="math-container">$G$</span>).</p>
<p>The real reason that distances can be larger than <span class="math-container">$c$</span> times the current cosmological time is that the clocks that we use to measure cosmological time are not at relative rest, like the clocks in inertial coordinate systems, but are moving radially away from each other, making cosmological coordinates more like polar coordinates. If we have a family of uniformly distributed clocks, and we define <span class="math-container">$t$</span> to be the reading on the nearest clock and <span class="math-container">$x$</span> to be (the number of clocks between that one and the origin) × (the separation between adjacent clocks when they both read the same time), then <span class="math-container">$Δx/Δt\le c$</span> is a true statement if those clocks are at relative rest, but not if they're moving outward from a common origin point. In the latter case, there turns out to be no upper limit on <span class="math-container">$Δx/Δt$</span>, even in special relativity.</p>
<p>In the special-relativistic case, you can think of this as being due to time dilation. If you look at two clocks with respect to inertial center-of-velocity coordinates, they move in opposite directions at some speed <span class="math-container">$v$</span>. After an inertial coordinate time <span class="math-container">$t$</span>, they're an inertial coordinate distance <span class="math-container">$2vt$</span> apart, but the elapsed time they've recorded is smaller than <span class="math-container">$t$</span> by a factor of <span class="math-container">$γ=1/\sqrt{1-v^2/c^2}$</span>. Since <span class="math-container">$γ{\to}\infty$</span> as <span class="math-container">$v{\to}c$</span>, the ratio of the coordinate distance to the elapsed times on the clocks also goes to infinity as <span class="math-container">$v{\to}c$</span>.</p>
<p>In special relativity, there's a tendency to think of inertial coordinate times as the &quot;real&quot; times and readings on clocks as somehow distorted by time dilation, but that's really just a human prejudice. The universe doesn't care about coordinate systems, and it only &quot;cares&quot; about reference frames if they're actually instantiated by physical objects. There are no naturally occurring inertial reference frames at large scales in the real world, but there is a naturally occurring radial reference frame, given by the averaged motion of matter on large scales, or by the crossing points of wavefronts from the cosmic microwave background. The most natural coordinate system for the universe – and the one actually used by cosmologists – is based on that naturally occurring frame, and, as Ned Wright said, when you define distances and times in that way, the distance/time ratio <span class="math-container">$c$</span> has no special significance.</p>
<p>(Actually, all three of Ned Wright's sentences are correct. The trouble is that when you take them together, they seem to imply that superluminal expansion is related to spacetime curvature, and that isn't correct.)</p>

CC BY-SA 4.0
2020-08-11T21:45:55.597Z
The link you give doesn't work for me. Is there a different one?
D. Halsey
CC BY-SA 4.0
2020-08-11T23:14:24.433Z
@DHalsey The site was down; it's back up now.
benrg
CC BY-SA 4.0
2020-08-10T17:24:46.010Z
2020-08-10T17:24:46.010Z
benrg
<p>The completely unscientific explanation...</p>

<p>Imagine the universe to be a balloon.
Two bodies start close to each other but on opposite surfaces. The expansion of the balloon takes them away from each other at equal speed and such a rate that the light from one at its starting point takes almost the entire history of the universe to reach the other. The distance between the two NOW is not twice the age of the universe - because you cannot travel "through" the balloon - but must instead go round the surface of the balloon... 13.8 * PI billion light years = 43 billion light years.</p>

<p>Not strictly correct, but at least avoids too much worrying about astrophysics and cosmology!</p>

CC BY-SA 4.0
2019-10-16T17:29:14.843Z
The fact that the ratio between the radius of the observable Universe in Glyr, and the age of the Universe in Gyr, is close to $\pi$, is purely coincidental, and is not even correct to 1 decimal (it's roughly 3.354). The ratio varies with time.
pela
CC BY-SA 4.0
2020-01-27T10:33:54.243Z
Also, why would you want to avoid "_worrying about astrophysics and cosmology_" on a site for astrophysics and cosmology?
pela
CC BY-SA 4.0
2020-08-10T17:27:02.420Z
I think this answer deserves more love. It captures an important feature of the real answer, which is that distance/time is larger than you expect because the distances and times are defined by (a curved-spacetime analogue of) polar coordinates, not Cartesian coordinates.
benrg
CC BY-SA 4.0
2022-10-26T03:12:54.157Z
Observable universe diameter is around 93 billion light-years so that's not enough https://en.wikipedia.org/wiki/Observable_universe
tuxayo
CC BY-SA 3.0
2018-04-30T20:25:04.467Z
2018-04-30T20:25:04.467Z
mark
<p>The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static. </p>

<p>Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time. </p>

<p>A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is <em>currently</em> made of:</p>

<p>$$ H(a) = H_{0} \sqrt{\frac{\Omega_{m,0}}{a^{3}} + \frac{\Omega_{\gamma,0}}{a^{4}} + \frac{\Omega_{k,0}}{a^{2}} + \Omega_{\Lambda,0}} $$
where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a <strong>constant</strong> regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured <strong>today</strong>.</p>

<p>The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment. </p>

<p>Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:</p>

<p>$$H(a) = H_{0} \frac{\sqrt{\Omega_{\gamma,0}}}{a^{2}}$$</p>

<p>At around:</p>

<p>$$ \frac{\Omega_{m,0}}{a^{3}} = \frac{\Omega_{\gamma,0}}{a^{4}} $$
$$ a = \frac{\Omega_{\gamma,0}}{\Omega_{m,0}}  $$
we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an <strong>acceleration</strong> of the coordinates of universe - something which has been confirmed (see: <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/" rel="noreferrer">2011 Nobel Prize in Physics</a>).</p>

<p>So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:</p>

<p>$$ D_{h} = \frac{c}{H_{0}} \int_{0}^{z_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega_{m,0}(1+z)^{3} + \Omega_{\Lambda}}} $$</p>

<p>where the emission redshift $z_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the <a href="http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric" rel="noreferrer">Friedmann–Lemaître–Robertson–Walker metric</a>, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.</p>

CC BY-SA 3.0
2018-02-07T02:21:54.100Z
@astromax +1 for the pretty formulas though.
iMerchant
CC BY-SA 3.0
2013-11-27T12:22:03.433Z
Thank you for such a detailed and considered answer. You might've overlooked the "layman" element of the question - at least, the mathematics goes a long way over _my_ head - but I appreciate that there's probably a limit to how much a layman _can_ understand about such things.
GDVS
CC BY-SA 3.0
2013-11-27T18:07:53.190Z
Hmm - my apologies. I thought this would be a digestible chunk of cosmology. The real point I wanted to make is that it's an integral rather than a simple product between the age of the universe and the speed of light. Because different things act differently with expansion, you get "phases" that the universe goes through. The rate of expansion changes depending on which phase it happens to be in. Feel free to keep posting questions - I (and others) would be happy to try to make things as understandable as possible.
astromax
CC BY-SA 3.0
2013-11-24T17:19:24.797Z
2016-10-29T04:22:19.237Z
2016-10-29T04:22:19.237Z
astromax
<p>The age of the Universe is estimated at 13.8 billion years, and current theory states nothing can exceed the speed of light, which can lead to the incorrect conclusion that the universe can't have a radius of more than 13.8 billion light years.</p>

<p><a href="https://en.wikipedia.org/wiki/Observable_universe">Wikipedia</a> deals with this misconception as follows:</p>

<blockquote>
  <p>This reasoning would only make sense if the flat, static <a href="https://en.wikipedia.org/wiki/Minkowski_space">Minkowski spacetime</a> conception under special relativity were correct. In the real Universe, <a href="https://en.wikipedia.org/wiki/Spacetime">spacetime</a> is curved in a way that corresponds to the <a href="https://en.wikipedia.org/wiki/Metric_expansion_of_space">expansion of space</a>, as evidenced by <a href="https://en.wikipedia.org/wiki/Hubble%27s_law">Hubble's law</a>. Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance. → <em>Ned Wright, <a href="http://www.astro.ucla.edu/~wright/Dltt_is_Dumb.html">"Why the Light Travel Time Distance should not be used in Press Releases"</a></em></p>
</blockquote>

<p>That doesn't clear the matter up for me, and having no science or maths background beyond high school, further reading into Hubble's law isn't helping much either.</p>

<p>One layman's explanation I've seen offers explanation that the Universe itself isn't bound by the same laws as things <em>within</em> it. That would make sense – insofar as these things can – but the above quote (<em>"Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance"</em>) seems more general than that.</p>

<p>Can anyone offer (or direct me to) a good layman's explanation?</p>

CC BY-SA 3.0
2017-10-16T14:20:22.400Z
A duplicate of this question: [How can the universe be expanding faster than speed of light?](https://astronomy.stackexchange.com/q/23105/698)
Farhan
CC BY-SA 3.0
2014-04-13T17:21:58.943Z
you might want to look at the comments in this question? http://astronomy.stackexchange.com/q/2150/1227
chris
<universe><distances><space-time>
Why is the observable Universe larger than its age would suggest?
GDVS
# Why is the observable Universe larger than its age would suggest?

(Asked by: username_0 on 2013-11-24)

The age of the Universe is estimated at 13.8 billion years, and current theory states nothing can exceed the speed of light, which can lead to the incorrect conclusion that the universe can't have a radius of more than 13.8 billion light years.

[Wikipedia](https://en.wikipedia.org/wiki/Observable_universe) deals with this misconception as follows:

> This reasoning would only make sense if the flat, static [Minkowski spacetime](https://en.wikipedia.org/wiki/Minkowski_space) conception under special relativity were correct. In the real Universe, [spacetime](https://en.wikipedia.org/wiki/Spacetime) is curved in a way that corresponds to the [expansion of space](https://en.wikipedia.org/wiki/Metric_expansion_of_space), as evidenced by [Hubble's law](https://en.wikipedia.org/wiki/Hubble%27s_law). Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance. → *Ned Wright, ["Why the Light Travel Time Distance should not be used in Press Releases"](http://www.astro.ucla.edu/~wright/Dltt_is_Dumb.html)*

That doesn't clear the matter up for me, and having no science or maths background beyond high school, further reading into Hubble's law isn't helping much either.

One layman's explanation I've seen offers explanation that the Universe itself isn't bound by the same laws as things *within* it. That would make sense – insofar as these things can – but the above quote (*"Distances obtained as the speed of light multiplied by a cosmological time interval have no direct physical significance"*) seems more general than that.

Can anyone offer (or direct me to) a good layman's explanation?

**Tags:** universe, distances, space-time
> **Comments:**
> **username_2**: you might want to look at the comments in this question? http://astronomy.stackexchange.com/q/2150/1227  
> **username_1**: A duplicate of this question: [How can the universe be expanding faster than speed of light?](https://astronomy.stackexchange.com/q/23105/698)  

---

## 5 answers

### Answer by username_10 on 2013-11-24. Score: 24 | Accepted answer

The easiest explanation for why the maximum distance one can see is not simply the product of the speed of light with the age of the universe is because the universe is non-static.

Different things (i.e. matter vs. dark energy) have different effects on the coordinates of the universe, and their influence can change with time.

A good starting point in all of this is to analyze the Hubble parameter, which gives us the Hubble constant at any point in the past or in the future given that we can measure what the universe is *currently* made of:

$$ H(a) = H\_{0} \sqrt{\frac{\Omega\_{m,0}}{a^{3}} + \frac{\Omega\_{\gamma,0}}{a^{4}} + \frac{\Omega\_{k,0}}{a^{2}} + \Omega\_{\Lambda,0}} $$
where the subscripts $m$, $\gamma$, $k$, and $\Lambda$ on $\Omega$ refer to the density parameters of matter (dark and baryonic), radiation (photons, and other relativistic particles), curvature (this only comes into play if the universe globally deviates from being spatially flat; evidence indicates that it is consistent with being flat), and lastly dark energy (which as you'll notice remains a **constant** regardless of how the dynamics of the universe play out). I should also point out that the $0$ subscript notation means as measured **today**.

The $a$ in the above Hubble parameter is called the scale factor, which is equal to 1 today and zero at the beginning of the universe. Why do the various components scale differently with $a$? Well, it all depends upon what happens when you increase the size of a box containing the stuff inside. If you have a kilogram of matter inside of a cube 1 meter on a side, and you increase each side to 2 meters, what happens to the density of matter inside of this new cube? It decreases by a factor of 8 (or $2^{3}$). For radiation, you get a similar decrease of $a^{3}$ in number density of particles within it, and also an additional factor of $a$ because of the stretching of its wavelength with the size of the box, giving us $a^{4}$. The density of dark energy remains constant in this same type of thought experiment.

Because different components act differently as the coordinates of the universe change, there are corresponding eras in the universe's history where each component dominates the overall dynamics. It's quite simple to figure out, too. At small scale factor (very early on), the most important component was radiation. The Hubble parameter early on could be very closely approximated by the following expression:

$$H(a) = H\_{0} \frac{\sqrt{\Omega\_{\gamma,0}}}{a^{2}}$$

At around:

$$ \frac{\Omega\_{m,0}}{a^{3}} = \frac{\Omega\_{\gamma,0}}{a^{4}} $$
$$ a = \frac{\Omega\_{\gamma,0}}{\Omega\_{m,0}} $$
we have matter-radiation equality, and from this point onward we now have matter dominating the dynamics of the universe. This can be done once more for matter-dark energy, in which one would find that we are now living in the dark energy dominated phase of the universe. One prediction of living in a phase like this is an **acceleration** of the coordinates of universe - something which has been confirmed (see: [2011 Nobel Prize in Physics](http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/)).

So you see, it would a bit more complicating to find the distance to the cosmological horizon than just multiplying the speed of light by the age of the universe. In fact, if you'd like to find this distance (formally known as the comoving distance to the cosmic horizon), you would have to perform the following integral:

$$ D\_{h} = \frac{c}{H\_{0}} \int\_{0}^{z\_{e}} \frac{\mathrm{d}z}{\sqrt{\Omega\_{m,0}(1+z)^{3} + \Omega\_{\Lambda}}} $$

where the emission redshift $z\_{e}$ is usually taken to be $\sim 1100$, the surface of last scatter. It turns out this is the true horizon we have as observers. Curvature is usually set to zero since our most successful model indicates a flat (or very nearly flat) universe, and radiation is unimportant here since it dominates at a higher redshift. I would also like to point out that this relationship is derived from the [Friedmann–Lemaître–Robertson–Walker metric](http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric), a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.

> **Comments:**
> **username_0**: Thank you for such a detailed and considered answer. You might've overlooked the "layman" element of the question - at least, the mathematics goes a long way over _my_ head - but I appreciate that there's probably a limit to how much a layman _can_ understand about such things.  
> **username_10**: Hmm - my apologies. I thought this would be a digestible chunk of cosmology. The real point I wanted to make is that it's an integral rather than a simple product between the age of the universe and the speed of light. Because different things act differently with expansion, you get "phases" that the universe goes through. The rate of expansion changes depending on which phase it happens to be in. Feel free to keep posting questions - I (and others) would be happy to try to make things as understandable as possible.  
> **username_11**: @astromax +1 for the pretty formulas though.  

---

### Answer by username_3 on 2013-11-24. Score: 12

In short: things can not move faster that light by theirselves, but they can move faster than light due to universal expansion. The more far away, the faster they go away.


---

### Answer by username_4 on 2018-02-13. Score: 9

I was just thinking about that and here is my layman's explanation. Imagine you're tracing two dots on a crumpled piece of paper, the dots are moving, but as they are moving, so is the paper getting ‘uncrumpled’, the actual distance between the dots will be more than the sum of distances they have travelled.


---

### Answer by username_5 on 2020-08-10. Score: 8

I love [Ned Wright's cosmology tutorial](http://www.astro.ucla.edu/%7Ewright/cosmo_01.htm) and I highly recommend it, but that statement by him is at the least very misleading. Superluminal recession speeds plainly can't be related to spacetime curvature because they don't vanish in the limit of zero curvature (zero energy density or zero $G$).

The real reason that distances can be larger than $c$ times the current cosmological time is that the clocks that we use to measure cosmological time are not at relative rest, like the clocks in inertial coordinate systems, but are moving radially away from each other, making cosmological coordinates more like polar coordinates. If we have a family of uniformly distributed clocks, and we define $t$ to be the reading on the nearest clock and $x$ to be (the number of clocks between that one and the origin) × (the separation between adjacent clocks when they both read the same time), then $Δx/Δt\le c$ is a true statement if those clocks are at relative rest, but not if they're moving outward from a common origin point. In the latter case, there turns out to be no upper limit on $Δx/Δt$, even in special relativity.

In the special-relativistic case, you can think of this as being due to time dilation. If you look at two clocks with respect to inertial center-of-velocity coordinates, they move in opposite directions at some speed $v$. After an inertial coordinate time $t$, they're an inertial coordinate distance $2vt$ apart, but the elapsed time they've recorded is smaller than $t$ by a factor of $γ=1/\sqrt{1-v^2/c^2}$. Since $γ{\to}\infty$ as $v{\to}c$, the ratio of the coordinate distance to the elapsed times on the clocks also goes to infinity as $v{\to}c$.

In special relativity, there's a tendency to think of inertial coordinate times as the "real" times and readings on clocks as somehow distorted by time dilation, but that's really just a human prejudice. The universe doesn't care about coordinate systems, and it only "cares" about reference frames if they're actually instantiated by physical objects. There are no naturally occurring inertial reference frames at large scales in the real world, but there is a naturally occurring radial reference frame, given by the averaged motion of matter on large scales, or by the crossing points of wavefronts from the cosmic microwave background. The most natural coordinate system for the universe – and the one actually used by cosmologists – is based on that naturally occurring frame, and, as Ned Wright said, when you define distances and times in that way, the distance/time ratio $c$ has no special significance.

(Actually, all three of Ned Wright's sentences are correct. The trouble is that when you take them together, they seem to imply that superluminal expansion is related to spacetime curvature, and that isn't correct.)

> **Comments:**
> **username_6**: The link you give doesn't work for me. Is there a different one?  
> **username_5**: @DHalsey The site was down; it's back up now.  

---

### Answer by username_7 on 2018-04-30. Score: 4

The completely unscientific explanation...

Imagine the universe to be a balloon.
Two bodies start close to each other but on opposite surfaces. The expansion of the balloon takes them away from each other at equal speed and such a rate that the light from one at its starting point takes almost the entire history of the universe to reach the other. The distance between the two NOW is not twice the age of the universe - because you cannot travel "through" the balloon - but must instead go round the surface of the balloon... 13.8 \* PI billion light years = 43 billion light years.

Not strictly correct, but at least avoids too much worrying about astrophysics and cosmology!

> **Comments:**
> **username_8**: The fact that the ratio between the radius of the observable Universe in Glyr, and the age of the Universe in Gyr, is close to $\pi$, is purely coincidental, and is not even correct to 1 decimal (it's roughly 3.354). The ratio varies with time.  
> **username_8**: Also, why would you want to avoid "_worrying about astrophysics and cosmology_" on a site for astrophysics and cosmology?  
> **username_5**: I think this answer deserves more love. It captures an important feature of the real answer, which is that distance/time is larger than you expect because the distances and times are defined by (a curved-spacetime analogue of) polar coordinates, not Cartesian coordinates.  
> **username_9**: Observable universe diameter is around 93 billion light-years so that's not enough https://en.wikipedia.org/wiki/Observable_universe  

---

