Faster Than The Speed of Light?

Back to the office after starting out early to make the long journey back to Brighton from Cardiff, all of which went smoothly for a change. I’ve managed to clear some of the jobs waiting for me on my return from the Christmas holidays so thought I’d take my lunch break and write a quick blog post. I hasten to add, however, that the title isn’t connected in any way with the speed of this morning’s train, which never at any point threatened causality.

What spurred me on to write this piece was an exchange on Twitter, featuring the inestimable Sean Carroll who delights in getting people to suggest physics for him to explain in fewer than three tweets. It’s a tough job sometimes, but he usually does it brilliantly. Anyway, the third of his tweets about the size of the (observable universe), and my rather pedantic reply to it, both posted on New Year’s Day, were as follows:

I thought I’d take the opportunity to explain in a little bit more detail how and why it can be that the size of the observable universe is significantly larger than what one naively imagine, i.e. (the speed of light) ×(time elapsed since the Big Bang) = ct, for short. I’ve been asked about this before but never really had the time to respond.

Let’s start with some basic cosmological concepts which, though very familar, lead to some quite surprising conclusions.  First of all, consider the Hubble law, which I will write in the form


It’s not sufficiently widely appreciated that for a suitable definition of the recession velocity v and distance R, this expression is exact for any velocity, even one much greater than the speed of light! This doesn’t violate any principle of relativity as long as one is careful with the definition.

Let’s start with time. The assumption of the Cosmological Principle, that the Universe is homogeneous and isotropic on large scales, furnishes a preferred time coordinate, usually called cosmoloogical proper time, or cosmic time, defined in such a way that observers in different locations can set their clocks according to the local density of matter. This allows us to slice the four-dimensional space-time of the Universe into three spatial dimensions of one dimension of time in a particularly elegant way.

The geometry of space-time can now be expressed in terms of the Robertson-Walker metric. To avoid unnecessary complications, and because it seems to be how are Universe is, as far as we can tell, I’ll restrict myself to the case where the spatial sections are flat (ie they have Euclidean geometry). This the metric is:

ds^{2}=c^{2}dt^{2} - a^{2}(t) \left[ d{r}^2 + r^{2}d\Omega^{2} \right]

Where s is a four-dimensional interval t is cosmological proper time as defined above, r is a radial coordinate and \Omega defines angular position (the observer is assumed to be at the origin). The function a(t) is called the cosmic scale factor, and it describes the time-evolution of the spatial part of the metric; the coordinate r of an object moving with the cosmic expansion does not change with time, but the proper distance of such an object evolves according to


The name “proper” here relates to the fact that this definition of distance corresponds to an interval defined instantaneously (ie one with dt=0). We can’t actually measure such intervals; the best we can do is measure things using signals of some sort, but the notion is very useful in keeping the equations simple and it is perfectly well-defined as long as you stay aware of what it does and does not mean. The other thing we need to know is that the Big Bang is supposed to have happened at dt=0 at which point a(t)=0 too.


If we now define the proper velocity of an object comoving with the expansion of the Universe to be

v=\frac{dR}{dt}=\left(\frac{da}{dt} \right)r = \left(\frac{\dot{a}}{a}\right) R = HR

This is the form of the Hubble law that applies for any velocity and any distance. That does not mean, however, that one can work out the redshift of a source by plugging this velocity into the usual Doppler formula, for reasons that I hope will become obvious.

The specific case ds=0 is what we need here, as that describes the path of a light ray (null geodesic); if we only follow light rays travelling radially towards or away from the origin, the former being of greatest relevance to observational cosmology, then we can set d\Omega=0 too and find:

dr =\frac{cdt}{a(t)}

Now to the nub of it. How do we define the size of the observable universe? The best way to answer this is in terms of the particle horizon which, in a nutshell, is defined so that a particle on the particle horizon at the present cosmic time is the most distant object that an observer at the origin can ever have received a light signal from in the entire history of the Universe. The horizon in Robertson-Walker geometry will be a sphere, centred on the origin, with some coordinate radius. The radius of this horizon will increase in time, in a manner that can be calculated by integrating the previous expression from t=0 to t=t_0, the current age of the Universe:

r_p(t_0)=\int_{0}^{t_0} \frac{cdt}{a(t)}.

For any old cosmological model this has to be integrated by solving for the denominator as a function of time using the Friedmann equations, usually numerically. However, there is a special case we can do trivially which demonstrates all the salient points. The matter-dominated Einstein- de Sitter model is flat and has the solution

a(t)\propto t^{2/3}

so that

\frac{a(t)}{a(t_0)} = \left(\frac{t}{t_0}\right)^{2/3}

Plugging this into the integral and using the above definitions we find that in this model the present proper distance of an object on our particle horizon is

R_p = 3ct_{0}


By the way, some cosmologists prefer to use a different definition of the horizon, called the Hubble sphere. This is the sphere on which objects are moving away from the observer according to the Hubble law at exactly the velocity of light. For the Einstein-de Sitter cosmology the Hubble parameter is easily found

H(t)=\frac{2}{3t} \rightarrow R_{c}= \frac{3}{2} ct_{0}.

Notice that velocities in this model are always decaying, so in it the expansion is not accelerating but decelerating, hence my comment on Twitter above. The apparent paradox therefore has nothing to do with acceleration, although the particle horizon does get a bit bigger in models with, e.g., a cosmological constant in which the expansion accelerates at late times. In the current standard cosmological model the radius of the particle horizon is about 46 billion light years for an age of 13.7 billion years, which is just 10% larger than in the Einstein de Sitter case.

There is no real contradiction with relativity here because the structure of the metric encodes all the requirements of causality. It is true that there are objects moving away from the origin at proper velocities faster than that of light, but we can’t make instantaneous measurements of cosmological distances; what we observe is their redshifted light. In other words we can’t make measurements of intervals with dt=0 we have to use light rays, which follow paths with ds=0, i.e. we have to make observations down our past light cone. Nevertheless, there are superluminal velocities, in the sense I have defined them above, in standard cosmological models. Indeed, these velocities all diverge at t =0. Blame it all on the singularity!

This figure made by Mark Whittle (University of Virginia) shows our past light cone in the present standard cosmological model:


If you were expectin the past light cone to look triangular in cross-section then you’re probably thinking of Minkowski space, or a representation involving coordinates chosen to resemble Minkowski space. Cosmological If you look at the left hand side of the figure, you will find the world lines of particles moving with the cosmic expansion labelled by their present proper distance which is obtained by extrapolating the dotted lines until they intersect a line parallel to the x-axis running through “Here & Now”.  Where we actually see these objects is not at their present proper distance but at the point in space-time where their world line intersects the past light cone.  You will see that an object on the particle horizon intersected our past light cone right at the bottom of the figure.

So why does the light cone look so peculiar? Well, I think the simplest way to explain it is to say that while the spatial sections in this model are flat (Euclidean) the four-dimensional geometry is most definitely curved. You can think of the bending of light rays shown in the figure as a kind of gravitational lensing effect due to all the matter in the Universe. I’d say that the fact that the particle horizon has a radius larger than ct is not because of acceleration but the curvature of space-time, an assertion consistent with the fact that the only familiar world model in which this effect does not occur is the (empty) purely kinemetic Milne cosmology, which is based entirely on special relativity.



21 Responses to “Faster Than The Speed of Light?”

  1. “but in Matter-dominated Einstein-de Sitter universe radius of particle horizon =3ct > ct but always decelerating”

    Not only that, but the Einstein-de Sitter universe does not have any non-Euclidean geometry. Actually, both the acceleration and the non-Euclidean geometry are red herrings.

    Folks interested in this stuff should check out Harrison’s textbook, Rindler’s classic paper on horizons, Stabell and Refsdal’s excellent 1966 paper on classification of cosmological models, and papers by Davis and Lineweaver. Fellow commentator Cusp (who also has his own blog) has also pointed out some confusion on this topic in the literature.

  2. It’s been a while since someone first pointed out that there is no royal road to geometry, and of course not everything can be meaningfully tweeted. As Feynman said to a journalist, “Listen, buddy, if I could explain it in three minutes it wouldn’t be worth a Nobel Prize”. 😐

  3. That Sean Carroll of all people should make that mistake! Reminded me of Appendix B of Davis and Lineweaver: [Annoyingly, they incorrectly refer to these recession rates as velocities too ;-)]. I saw a really awful example of this sort of thing via Twitter recently: (Myth 3).

    • telescoper Says:

      I’ve never seen that cosmic myths article before. It just replaces one myth with a worse one!

    • There is nothing wrong with referring to a recession rate as a velocity in cosmology as long as you don’t invoke the Doppler formula to calculate velocity from redshift.

      There are various distances defined in cosmology, and in general they change with time, but no such change is in general given by any Doppler formula.

      On the other hand, not all velocities are changes in distance. Suppose I observe a double star an interpret the change in the spectral shift with time as a velocity, due to the motion in the double-star system, due to the motion of the Earth, etc. I don’t know of anyone who wouldn’t interpret this as a velocity, though there is no distance one can point to which changes as a function of time. Along these lines, Bunn and Hogg showed that one can interpret cosmological redshifts as a Doppler shift, but this is more trouble than it is worth and in any case one cannot plug the redshift of an object into a Doppler formula and get the correct recession velocity.

      Edward Harrison wrote a fine ApJ paper about this, and named some big names of some people who got things wrong, at least in their explanations of what they were doing.

      • “There is nothing wrong with referring to a recession rate as a velocity in cosmology […]”

        Well the mathematical and conceptual foundations of cosmology are pretty clear and precise about what a velocity is and in that context a recession rate just isn’t one and I can’t see any utility in trying to interpret it as one. I know physicists typically don’t like to be pedantic about that sort of thing but with all the confusion and the big names getting things wrong…

      • There is certainly the velocity v = DH, where v is a velocity, D the proper distance, and H the Hubble constant. Why should this be called a recession rate and not a velocity. Yes, in general the velocity is not the derivative of any distance with respect to any time, and neither the velocity nor the distance are “directly observable”, but it still makes sense. Saying that recession rates are not velocities can also be confusing.

      • telescoper Says:

        As I explained in the post, this v is the rate of change of proper distance with cosmological proper time. It seems perverse to me not to call it a velocity.

      • I agree. In my comment above, it should read “not the derivative of any directly observable distance”. Of course, what is directly observable and what is calculated is to some extent a matter of definition.

      • Okay, I guess it’s just me then but I really don’t see how it can be confusing – or perverse – to maintain the distinction so that a velocity is always a genuine (local, SR) vector.

  4. “I’ve complained about it (and Myth 8)”

    Why myth 8 in particular?

  5. John Peacock Says:


    Not sure I agree with what you say on the Milne model. If you define the current distance as current scale factor times comoving distance, then this exceeds the proper distance to the particle horizon, c t_0, by a factor of int_0^1 dt / a [where t is normalised to 1 today, as is the scale factor a(t)]. In these terms, the Milne model is a=t, so the distance to the particle horizon is infinitely larger than c t_0.

    The reason for this paradoxical result is that the Milne model is negatively curved, with the transverse part involving a distance (c/H_0) sinh r, as opposed to the radial comoving distance (c/H_0) r. In this model, SR applies. Since this doesn’t affect transverse lengths, we see that the “true” distance to an object really is (c/H_0) sinh r, so the particle horizon lies at r = infinity. (c/H_0) r then diverges.

    • telescoper Says:

      Actually the Milne model is flat…it’s purely Minkowski, but that depends on how you define the time coordinate!

      But you are right. What I should have said is in a model in which the metric were exactly of the Minkowski form…

      • Phillip Helbig Says:

        It depends on what one means by “Milne model”. Milne’s idea in his kinematic relativity was to do cosmology with special but without general relativity. As such, he used Minkowski space, which is of course flat. However, the FRW model which is observationally equivalent (as regards the relationship between apparent magnitude and redshift, say) has negative curvature.

      • telescoper Says:

        These are identical. The only difference is how you define the spatial sections.

  6. John Duffield Says:

    “I’d say that the fact that the particle horizon has a radius larger than ct is not because of acceleration but the curvature of space-time”

    I’d say it’s because of “compound interest”. Neil Cornish used that analogy here. The curved spacetime of the expanding universe isn’t much like the curved spacetime of a gravitational field. For the former your measurements change over time. For the latter your measurements change with location. But what is interesting is that the expanding universe is a bit like pulling away from a black hole. I like the frozen-star black hole myself. I’m sure it’s right, because of this: “Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields”. But IMHO Jo Magueijo got his VSL back to front. What was the speed of light 13.8 billion years ago? Zero.

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