I'm not quite sure how to ask this so that it can be answered in layman's terms, but I have lately seen, in several places, that with cosmological inflation, there was a point where the universe expanded faster than the speed of light.

In this explanation, the following is stated:

  • Gravity itself would have split off at the Planck time, 1e-43 of a second
  • The strong nuclear force by about 1e-35 of a second
  • Within about 1e-32 of a second, the scalar fields would have done their work, doubling the size of the Universe at least once every 1e-34 of a second (some versions of inflation suggest even more rapid expansion than this).

It goes on to say that this rapid expansion is enough to take a quantum fluctuation 1e-20 times smaller than a proton and inflate it to a sphere about 10 cm across in about 15 x 1e-33 seconds.

The conclusion is that this expansion occurred at faster than light speed.

My main question is about the role of time in all of this, and I'm only indirectly interested in this apparent violation of the cosmic speed limit.

I have watched at least one interview (and I can't find a link to it right now) where a well-known physicist sums up the meaning of time as a big unknown, not being intuitively "fundamental" the way space is. However, in the very least, and perhaps most abstract way, time is an important part to the mathematics when it comes to explaining how things work.

My own understanding of time is that it is a way of gauging change relative to some known standard that is changing. (And acknowledging that it can change, depending on one's frame of reference.) That standard might be the swinging of a pendulum or the oscillations of some subatomic particle.

But in the earliest part of the universe's lifetime, before and during inflation, what was the standard for measuring time? Why would it be acceptable to say that the time duration of the inflationary period was so short and thus violated an accepted principle. Couldn't an alternative view be that the inflationary period lasted longer and nothing travelled faster than the speed of light? How was it decided that the duration of time for the inflationary period was so short?


2 Answers 2


This is a common point of confusion, not only with regards to inflation, but any time an expanding universe comes up...

The "cosmic speed limit" as you call it says that no particle or signal can move through spacetime faster than the speed of light. Spacetime is a very specifically defined thing, described with a coordinate system. There is no restriction, in terms of speed, on what spacetime itself is allowed to do. Let me illustrate with an example.

Imagine a photon. Relativity tells us that it always travels at speed $c$ (exactly at the speed limit). Let's say the photon has a path 10 light years long to travel along (remember light years are a measure of distance, $1\mathrm{ly} =$ the distance travelled by a photon in 1 year). The photon leaves and travels for 5 years, covering a distance of 5 light years. Then very suddenly, the universe doubles in size! The photon continues on its journey. The 5 remaining light years to travel have doubled in size, so it travels 10 more years to cover the last 10 light years. The journey has lasted 15 years. But the photon is now 20 light years from its starting point. Naively, we might compute its speed as $v = 20\mathrm{ly}/15\mathrm{yr} = \frac{4}{3}c$, faster than the speed of light. But in reality, it was just moving at speed $c$ the whole time in a universe that expanded.

In a more realistic scenario, the universe doesn't "suddenly" double in size, it does it gradually, but conceptually the same thing happens... you just need to use integrals to work out the math.

As to the meaning of time, that's somewhat more philosophical. However, I'll point out that, at least in general relativity, time is on an equal footing with space. Spacetime is described by a mathematical object called a metric. One example of a metric looks like:

$$ds^2 = c^2dt^2-dx^2-dy^2-dz^2$$

$x,y,z$ are the spatial coordinates and $t$ is the time coordinate; $s$ is a sort of generalized measure of spacetime length. As you can see, other than the constant $c$ (which could be set to equal 1 with a clever choice of units, so it's really rather unimportant), and a negative sign, time and space are equivalent in this formalism. If you understand space, then time should also make sense, as it's simply related to space by your "cosmic speed limit".

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    $\begingroup$ While I'm pondering the information here, shouldn't that be c-squared, to keep the units consistent? (Maybe you're just checking to see if I'm paying attention :) ) $\endgroup$
    – Jim
    Mar 26, 2014 at 17:48
  • $\begingroup$ @Jim c is a constant, so there are no units attached to it. $\endgroup$ Mar 26, 2014 at 18:05
  • 2
    $\begingroup$ @Jim Ah yes, that should be a $c^2$. My mistake. $\endgroup$
    – Kyle Oman
    Mar 26, 2014 at 18:11
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    $\begingroup$ @PeterMichealLacey-Bordeaux constants most certainly have units (well, when it's relevant anyway). $c$ has units of length/time, and indeed is crucial to the dimensional validity of the metric. You can do all sorts of trickery to shuffle around the dimensions, re-express constants as dimensionless, set them equal to 1 and not worry about dimensions if you know what you're doing, and so on. But generically (some) constants do have units. $\endgroup$
    – Kyle Oman
    Mar 26, 2014 at 18:15
  • $\begingroup$ @Kyle Rather than taking the point of view of the photon, what about two objects separated by some distance (space) before inflation. (I don't know what those objects would be at this point in the timeline, but I want to assume they are real and not abstract, like a hypothetical comoving observer might be.) If space expands faster than the speed of light, wouldn't the distance between these objects separate at a speed faster than light? $\endgroup$
    – Jim
    Mar 26, 2014 at 23:03

The time used in describing the evolution of the universe is comoving time. This is the time that would be measured by a freely moving observer on their wristwatch (assuming the high temperatures didn't melt both the observer and the wristwatch :-).

Time is not a simple thing to define in general relativity, however we can always unambiguously define proper time. This is the time that would be measured by an observer who is floating freely i.e. not subject to any external forces. The proper time is an invariant and would have the same value for all observers.

The comoving coordinates are defined so that the time is the same as the proper time for these freely floating observers. When we say the universe is 13.8 billion years old (or whatever the current estimate is) we mean this is the time measured by these freely moving observers. You and I are approximately freely moving observers because the various peculiar motions due the the Earth moving round the Sun, Sun moving round the Milky Way etc are small compared to the speed of light. We are approximately at rest with respect to the cosmic microwave background.

Even though the expansion of the universe was extraordinarily rapid during inflation, any observers present would still have been freely moving and would still measure time in exactly the same way you and I measure time today. When we quote a figure for the duration of the inflationary phase it's this time that we are using.

Response to comments:

I think it's worth expanding on the discussion about the meaning of time in the comments. If we make a few simplifying assumptions about the early universe (principally that it was isotropic and homogenous) the early universe was described by the metric:

$$ ds^2 = -dt^2 + a^2(t) \left(dx^2 + dy^2 + dz^2 \right) $$

were $t$, $x$, $y$ and $z$ are comoving coordinates and $a(t)$ is the scale factor. It's important to emphasise that these coordinates are just one of the many choices we may have made so there is not necessarily any physical significance to them, though as you'll see they do have physical significance.

This is called the FLRW metric - I think strictly speaking the metric doesn't apply to the inflationary period, but all inflation did was change the scale factor $a(t)$ by a factor of $e^{50}$ or so in a short time. The form of the metric remains the same.

Note also that $t$ is just a coordinate like $x$ etc. It starts at zero and extends to infinity and is continuous in between so it is defined at all points in spacetime no matter how closely spaced they are.

Now, the definition of a comoving observer is that they don't move in space so $dx = 0$ etc. That means for a comoving observer the metric simplifies to:

$$ ds^2 = -dt^2 $$

The proper time is defined by $d\tau^2 = -ds^2$ so:

$$ d\tau = dt $$

And this immediately integrates to give:

$$ \tau = t + C $$

where $C$ is the constant of integration that we can set to zero by setting our watches appropriately.

The point of all this is that the proper time agrees with the elapsed time for a freely moving, i.e. comoving, observer. So the observer time is the same as the comoving time. This emerges from the metric and doesn't rely on the observer having some form of clock. The observer's time is just a coordinate like comoving time and takes all values from zero to (potentially) infinity.

  • $\begingroup$ The references are interesting. The definition of comoving time assumes the user has a clock. One thing I'm having difficulty with is what type of standard would be a useful (meaning "meaningful") clock during those early moments? $\endgroup$
    – Jim
    Mar 26, 2014 at 17:11
  • $\begingroup$ Time exists whether we measure it or not, and the definition of comoving time does not depend on having a clock to measure it any more than the definition of distance relies on us having a ruler to measure it. It might be interesting to consider what sort of clock could be used to measure such short timescales, but this is essentially just an engineering question. $\endgroup$ Mar 26, 2014 at 17:21
  • $\begingroup$ That seems to contradict the reference you gave, unless I'm not understanding: "The comoving time coordinate is the elapsed time since the Big Bang according to a clock of a comoving observer and is a measure of cosmological time." $\endgroup$
    – Jim
    Mar 26, 2014 at 17:26
  • $\begingroup$ That quote is strictly correct, but does imply the clock causes time and not the other way around. The phrase according to a clock of a comoving observer just means what the comoving clock measures agrees with the elapsed comoving time. An observer who wasn't comoving, e.g. an accelerating observer, would measure a different time so you could have an inflationary twin paradox. That doesn't mean the cosmological time changes if your're accelerating. The proper time is precisely defined and happens to match the comoving clock. $\endgroup$ Mar 26, 2014 at 17:42
  • $\begingroup$ @Jim: I've edited my answer to ramble on about what is meant by time. Hopefully this clears things up a bit. $\endgroup$ Mar 26, 2014 at 18:31

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