# Why is the observable Universe larger than its age would suggest?

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 deals with this misconception as follows:

This reasoning would only make sense if the flat, static Minkowski spacetime conception under special relativity were correct. In the real Universe, spacetime is curved in a way that corresponds to the expansion of space, as evidenced by Hubble's 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"

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?

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).

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, a metric which includes curvature and expansion. This is something that the Minkowski metric lacks.

• 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. – GDav Nov 27 '13 at 12:22
• 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 Nov 27 '13 at 18:07
• @astromax +1 for the pretty formulas though. – iMerchant Feb 7 '18 at 2:21

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.

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.

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!

• 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 Oct 16 '19 at 17:29
• Also, why would you want to avoid "worrying about astrophysics and cosmology" on a site for astrophysics and cosmology? – pela Jan 27 at 10:33
• 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 Aug 10 at 17:27

I love Ned Wright's cosmology tutorial 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.)

• The link you give doesn't work for me. Is there a different one? – D. Halsey Aug 11 at 21:45
• @DHalsey The site was down; it's back up now. – benrg Aug 11 at 23:14