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At the beginning, right after the Big Bang, the universe was the size of a coin. One millionth of a second after the universe was the size of the Solar System (acording to, which is much much faster than speed of light. Can space expand with unlimited speed?

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Related: –  John Rennie Apr 9 '13 at 11:39
More related: –  Chris White May 17 '13 at 23:48
Possible duplicates: and links therein. –  Qmechanic May 22 '13 at 20:31

3 Answers 3

up vote 12 down vote accepted

Yes, the expansion of space itself is allowed to exceed the speed-of-light limit because the speed-of-light limit only applies to regions where special relativity – a description of the spacetime as a flat geometry – applies. In the context of cosmology, especially a very fast expansion, special relativity doesn't apply because the curvature of the spacetime is large and essential.

The expansion of space makes the relative speed between two places/galaxies scale like $v=Hd$ where $H$ is the Hubble constant and $d$ is the distance. When this $v$ exceeds $c$, it means that the two places/galaxies are "behind the horizons of one another" so they can't observer each other anytime soon. But they're still allowed to exist.

In quantum gravity i.e. string theory, there may exist limits on the acceleration of the expansion but the relevant maximum acceleration is extreme – Planckian – and doesn't invalidate any process we know, not even those in cosmic inflation.

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@Motl Then one can assume that sending messages between such galaxies is impossible, since they are receding from each other faster than the speed of light? How then they still affect each other by gravity, which has the speed of light? –  Force Apr 9 '13 at 13:46
Dear Jim, yes, as I said, the fact that the mutual speed exceeds the speed of light - although it's not really well-defined - means that they can't see each other or otherwise communicate. Otherwise, they also can't send signals to each other gravitationally - gravitational signals propagate by the speed of light, too. However, both faraway galaxies still feel the gravitational field - but in general relativity, gravity is given by the local curvature of the spacetime which is there regardless of what the other distant galaxy is doing right now. –  Luboš Motl Apr 9 '13 at 19:05
"the fact that the mutual speed exceeds the speed of light - although it's not really well-defined - means that they can't see each other or otherwise communicate" Yes, it's not uniquely defined, and that's the fundamental answer to the question. However, for the most common cosmological definition of the speed, the remainder of this sentence is false. See . –  Ben Crowell May 8 '13 at 12:22
Dear Ben, maybe I am not using the "most common definition of the speed" in a curved spacetime, but the statement of mine is surely correct for the "most natural definition of the speed" in a curved spacetime for any comparison with special relativity. –  Luboš Motl Jul 7 '13 at 5:57
Still, photons can travel between galaxies that are receding from each other faster than light already at the time of emission; see the Davis & Lineweaver paper linked by @Pulsar below. By the way, impressingly, Davis wrote this paper as part of her Ph.D. –  Thriveth Dec 13 '13 at 22:15

There are quite a few common misconceptions about the expansion of the universe, even among professional physicists. I will try to clarify a few of these issues; for more information, I highly recommend the article "Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe" from Tamara M. Davis and Charles H. Lineweaver.

I will assume a standard ΛCDM-model, with $$ \begin{align} H_0 &= 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} &= 9.24\times 10^{-5},\\ \Omega_{M,0} &= 0.315,\\ \Omega_{\Lambda,0} &= 0.685,\\ \Omega_{K,0} &= 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0} = 0. \end{align} $$

The expansion of the universe can be described by a scale factor $a(t)$, which can be thought of as the length of an imaginary ruler that expands along with the universe, relative to the present day, i.e. $a(t_0)=1$ where $t_0$ is the present age of the universe.

From the standard equations, one can derive the Hubble parameter $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}, $$ such that $H(1)=H_0$ is the Hubble constant. In a previous post, I showed that the age of the universe, as a function of $a$, is $$ t(a) = \frac{1}{H_0}\int_0^a\frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}}, $$ which can be numerically inverted to yield $a(t)$, and consequently $H(t)$. It also follows that the present age of the universe is $t_0=t(1)=13.8$ billion years.

Now, another consequence of the Big Bang models is Hubble's Law, $$ v_\text{rec}(t_\text{ob}) = H(t_\text{ob})\,D(t_\text{ob}), $$ describing the relation between the recession velocity $v_\text{rec}(t_\text{ob})$ of a light source and its proper distance $D(t_\text{ob})$, at a time $t_\text{ob}$. In fact, this follows immediately from the definition of $H(t_\text{ob})$, since $v_\text{rec}(t_\text{ob})$ is proportional to $\dot{a}$ and $D(t_\text{ob})$ is proportional to $a$.

However, it should be noted that this is a theoretical relation: neither $v_\text{rec}(t_\text{ob})$ nor $D(t_\text{ob})$ can be observed directly. The recession velocity is not a "true" velocity, in the sense that it is not an actual motion in a local inertial frame; clusters of galaxies are locally at rest. The distance between them increases as the universe expands, which can be expressed as $v_\text{rec}(t_\text{ob})$. Some cosmologists therefore prefer to think of $v_\text{rec}(t_\text{ob})$ as an apparent velocity, a theoretical quantity with little physical meaning.

A related quantity that is observable is the redshift of a light source, which is the cumulative increase in wavelength of the photons as they travel through the expanding space between source and observer. There is a simple relation between the scale factor and the redshift of a source, observed at a time $t_\text{ob}$: $$ 1 + z(t_\text{ob}) = \frac{a(t_\text{ob})}{a(t_\text{em})}, $$ such that the observed redshift of a photon immediately gives the time $t_\text{em}$ at which the photon was emitted.

The proper distance $D(t_\text{ob})$ of a source is also a theoretical quantity. It's an "instantaneous" distance, which can be thought of as the distance you would obtain with a (very long!) measuring tape if you were able to "stop" the expansion of the universe. It can however be derived from observable quantities, such as the luminosity distance or the angular diameter distance. The proper distance to a source, observed at time $t_\text{ob}$ with a redshift $z_\text{ob}$ is $$ D(z_\text{ob},t_\text{ob}) = a_\text{ob}\frac{c}{H_0}\int_{a_\text{ob}/(1+z_\text{ob})}^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}, $$ with $a_\text{ob} = a(t_\text{ob})$. The furthest objects that we theoretically can observe have infinite redshift; they mark the edge of the observable universe, also known as the particle horizon. Ignoring inflation, we get: $$ D_\text{ph}(t_\text{ob}) = a_\text{ob}\frac{c}{H_0}\int_0^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ In practice though, the furthest we can see is the CMB, which has a current redshift $z_\text{CMB}(t_0)\approx 1090$.

A source that has a recession velocity $v_\text{rec}(t_\text{ob})=c$ has a corresponding distance $$ D_\text{H}(t_\text{ob})=\frac{c}{H(t_\text{ob})}. $$ This is called the Hubble distance.

Almost there, just a few more quantities need to be defined. The photons that we observe at a time $t_\text{ob}$ have travelled on a null geodesic called the past light cone. It can be defined as the proper distance that a light source had at a time $t_\text{em}$ when it emitted the photons that we observe at $t_\text{ob}$: $$ D_\text{lc}(t_\text{em},t_\text{ob})= a_\text{em}\frac{c}{H_0}\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ There are two special cases: for $t_\text{ob}=t_0$ we have our present-day past light cone (i.e. the photons that we are observing right now), and for $t_\text{ob}=\infty$ we get the so-called cosmic event horizon: $$ D_\text{eh}(t_\text{em})= a_\text{em}\frac{c}{H_0}\int_{a_\text{em}}^\infty\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ For light emitted today, $t_\text{em}=t_0$, this has a special significance: if a source closer to us than $D_\text{eh}(t_0)$ emits photons today, then we will be able to observe those at some point in the future. In contrast, we will never observe photons emitted today by sources further than $D_\text{eh}(t_0)$.

One final definition: instead of proper distances, we can use co-moving distances. These are distances defined in a co-ordinate system that expands with the universe. In other words, the co-moving distance of a source that moves away from us along with the Hubble flow, remains constant. The relation between co-moving and proper distance is simply $$ D_c(t) = \frac{D(t)}{a(t)}, $$ so that both are the same at the present day $a(t_0)=1$. Thus $$ \begin{align} D_\text{c,ph}(t_\text{ob}) &= \frac{D_\text{ph}(t_\text{ob})}{a_\text{ob}},\\ D_\text{c,lc}(t_\text{em},t_\text{ob}) &= \frac{D_\text{lc}(t_\text{em},t_\text{ob})}{a_\text{em}},\\ D_\text{c,H}(t_\text{ob}) &= \frac{D_\text{H}(t_\text{ob})}{a_\text{ob}}. \end{align} $$ In fact, it would have been more convenient to start with co-moving distances instead of proper distances; in case you've been wondering where all the above integrals come from, those can be derived from the null geodesic of the FLRW metric: $$ 0 = c^2\text{d}t^2 - a^2(t)\text{d}\ell^2, $$ such that $$ \text{d}\ell = \frac{c\,\text{d}t}{a(t)} = \frac{c\,\text{d}a}{a\,\dot{a}} = \frac{c\,\text{d}a}{a^2\,H(a)}, $$ and $\text{d}\ell$ is the infinitesimal co-moving distance.

So, what can we do with all these tedious calculations? Well, we can draw a graph of the evolution of the expanding universe (after inflation). Inspired by a similar plot in the article from Davis & Lineweaver, I made the following diagram:enter image description here

This graph contains a lot of information. On the horizontal axis, we have the co-moving distance of light sources, in Gigalightyears (bottom) and the corresponding Gigaparsecs (top). The vertical axis shows the age of the universe (left) and the corresponding scale factor $a$ (right). The horizontal thick black line marks the current age of the universe (13.8 billion years). Co-moving sources have a constant co-moving distance, so that their world lines are vertical lines (the black dotted lines correspond with sources at 10, 20, 30, etc Gly). Of course, our own world line is the thick black vertical line, and we are currently situated at the intersection of the horizontal and vertical black line.

The yellow lines are null geodesics, i.e. the paths of photons. The scale of the time axis is such that these photon paths are straight lines at 45° angles. The orange line is our current past light cone. This is the cross-section of the universe that we currently observe: all the photons that we receive now have travelled on this path. The path extends to the orange dashed line, which is our future light cone. The particle horizon, i.e. the edge of our observable universe, is given by the blue line; note that this is also a null geodesic. The red line is our event horizon: photons emitted outside the event horizon will never reach us.

The purple dashed curves are distances corresponding with particular redshift values $z(t_\text{ob})$, in particular $z(t_\text{ob}) = 1, 3, 10, 50, 1000$. Finally, the green curves are lines of constant recession velocity, in particular $v_\text{rec}(t_\text{ob}) = c, 2c, 3c, 4c$. Of course, the curve $v_\text{rec}(t_\text{ob}) = c$ is nothing else than the Hubble distance.

What can we learn from all this? Quite a lot:

  • The current (co-moving) distance of the edge of the observable universe is 46.2 billion ly. Of course, the total universe can be much bigger, and is possibly infinite. The observable universe will keep expanding to a finite maximum co-moving distance at cosmic time $t = \infty$, which is 62.9 billion ly. We will never observe any source located beyond that distance.
  • Curves of constant recession velocity expand to a maximum co-moving distance, at $t_\text{acc} = 7.7$ billion years, and then converge again. This time $t_\text{acc}$, indicated by the horizontal black dashed line, is in fact the moment at which the expansion of the universe began to accelerate.
  • Curves of constant redshift also expand first, and converge when $t$ becomes very large. This means that a given source, which moves along a vertical line, will be observed with an infinite redshift when it enters the particle horizon, after which its redshift will decrease to a mimimum value, and finally increase again to infinity at $t = \infty$. In other words, every galaxy outside our local cluster will eventually be redshifted to infinity when the universe becomes very old. This is due to the dominance of dark energy at late cosmic times. Photons that we currently observe of sources at co-moving distances of 10, 20, 30 and 40 Gly have redshifts of 0.87, 2.63, 8.20 and 53.22 respectively.
  • The edge of the observable universe is receding from us with a recession velocity of more than 3 times the speed of light. $3.18c$, to be exact. In other words, we can observe sources that are moving away from us faster than the speed of light. Sources at co-moving distances of 10, 20, 30 and 40 Gly are receding from us at 0.69, 1.38, 2.06 and 2.75 times the speed of light, respectively.
  • Sources outside our particle horizon are moving away even faster. There is no a priori limit to the maximum recession velocity: it is proportional to the size of the total universe, which could be infinite.
  • The Hubble distance lies completely inside the event horizon. It will asymptotically approach the event horizon (as well as the curve of constant redshift 1) as $t$ goes to infinity. The current Hubble distance is 14.5 Gly (corresponding with $z=1.48$) , while the current distance to the event horizon is 16.7 Gly ($z=1.87$). Photons emitted today by sources that are located between these two distances will still reach us at some time in the future.
  • Although the difference between the Hubble distance and the event horizon today is rather small, this difference was much larger in the past. Consider for example the photons that we observe today, emitted by a source at a co-moving distance of 30 Gly. It emitted those photons at $t=0.62$ Gy, when the source was moving away from us at $3.5c$. The source continued its path along the vertical dotted line, while the photons moved on our past light cone. At $t=0.83, 1.64, 4.06$ Gy those photons passed regions that were moving away from us at $3c, 2c, c$ respectively. Along the way, those photons accumulated a total redshift of 53.22.

From all the above, it should be clear that the Hubble distance is not a horizon. I should stress again that all these calculations are only valid for the standard ΛCDM-model.

Apologies for the very lengthy post, but I hope it has clarified a few things.

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+1 for the Davis and Lineweaver reference - I can't cite that paper enough around here. Also awesome entire rest of the post! And a belated welcome to Physics Stackexchange! –  Chris White May 17 '13 at 23:46
@Chris White Thanks, much appreciated! –  Pulsar May 18 '13 at 0:00
+1, very nice answer. I, too, am a "Davis-Lineweaver apostle"; that paper really gets things cleared up. Nice figure! –  Thriveth Jun 26 '13 at 22:02
@Lubos Vacuous philosophical babbling??? All my calculations are correct and well known to cosmologists. I expected better from you. –  Pulsar Jul 7 '13 at 13:20
+1 What a beautiful explanation and also what a magnificent job with the plot: what a work of art, and so clear too. How long did that take to put together? I propose you think about putting it alongside the isometric embedding plot at… as it would really help clarify what the reader is looking at. Having the isometric embedding is pretty funky, but your plot can put a great deal more info on essentially the same plot, so they would go well together. –  WetSavannaAnimal aka Rod Vance Dec 17 '13 at 7:09

Your question is based on a fundamental misconception. You say:

At the beginning, right after the Big Bang, the universe was the size of a coin

but it's more accurate to say "the observable universe was the size of a coin" i.e. the 13.7 billion light year bit that we can currently see was at one time the same radius as a coin. The universe may well be infinite in size, and if so it has always been infinite in size right back to the moment of the Big Bang.

There is no point in the observable universe that is moving away from us at faster than the speed of light, but assuming the universe is infinite, or at least much bigger than the bit we can see, everything farther away from us than the edge of the observable universe is moving away from us faster than the speed of light. As Luboš says this doesn't violate relativity since it's space that's expanding not the objects themselves moving, and there is no limit to the expansion rate of space. In fact if there was a period of inflation immediately after the Big Bang, during this period space expanded at a rate that makes the speed of light look positively glacial.

If you're interested in a bit more detail about how we model the expansion of the universe search this site for "FLRW metric", or Google for it.

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"There is no point in the observable universe that is moving away from us at faster than the speed of light[...]" This is a common misconception. See Davis and Lineweaver . The answer by Pulsar is the only one that addresses the fundamental issue inherent the question, which is that GR doesn't have a well defined notion of the velocity of one object relative to some other distant object. (The Davis and Lineweaver uses one particular definition in the context of cosmology.) –  Ben Crowell May 8 '13 at 12:20

protected by Qmechanic Jan 25 '14 at 21:20

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