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Consider the following statement:

Hadron Epoch, from $10^{-6}$ seconds to $1$ second: The temperature of the universe cools to about a trillion degrees, cool enough to allow quarks to combine to form hadrons (like protons and neutrons).

What does it mean to say "from $10^{-6}$ seconds to $1$ second"?

How is time being measured?

One particle might feel just $10^{-20}\ \mathrm s$ having passed and another could feel $10^{-10}\ \mathrm s$ having passed.

Is saying "1 second after the big bang" a meaningful statement?

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    $\begingroup$ I have always wondered about this myself. When all of space-time was condensed into such a tiny volume, it seems that the passage of time would have no real meaning (same with black hole singularities). I suppose I am actually talking about a time frame earlier than the one mentioned above. $\endgroup$ Commented Mar 3, 2017 at 16:08
  • $\begingroup$ The implausibility of "condensation", usually associated with "cooling", into such a tiny and white-hot volume as conventional (field-based) inflation's "bubble from nowhere", is what's driven one of the 2 branches of mainstream physics into "future only" inflation: In the "past- and future-eternal" branch (whose models include the "Conformal cyclic cosmology" described by 2020 Nobel winner Penrose in his Nobel speech [that's freely available on Youtube]), gravity drives similar inflation thru the late (not early) stage of each iteration of a single universe, allowing a low-entropy beginning. $\endgroup$
    – Edouard
    Commented Feb 27, 2022 at 10:29

4 Answers 4

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We know that time passes differently for different observers, and the question is how can a time be given without telling which frame it is in. The answer is that there's a preferred reference frame in cosmology, the comoving frame, because of the fact that there's matter and radiation in it.

Intuitively, the special frame is the one that's "static" with respect to this matter and radiation content. More precisely, it is the one in which all observers that see an isotropic universe are static. Time measured in this system is called comoving time. The time from the beginning of the universe is usually given in this way, as a comoving time.


To get some intuition about the comoving frame one might consider the comoving observers, the ones that see isotropy and therefore have constant comoving coordinates. A comoving observer is such that when it looks around and adds the motion of the objects it sees zero net motion.

For example, we can look at the cosmic microwave background and detect some variation in the redshift depending on the direction. It's caused by Doppler effect and it means that we have some velocity relative to the comoving frame. On the other hand, a comoving observer sees the same redshift in any direction.

Another example: we can choose to measure the distances and velocities of galaxies. By Hubble's law, we expect the velocity to be proportional to the distance. If we find a deviation from this behavior, we know that the galaxy is moving with respect to the comoving frame, and thus has a peculiar velocity (we also have a peculiar velocity). If all galaxies had constant comoving coordinates, we would see perfect agreement with Hubble's law: the relative motions of galaxies would be due only to the expansion of the universe.

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    $\begingroup$ No. All comoving observers see the redshift. Otherwise, as @coconut described it, they'd detect some redshift in some direction' (quotes are words from coconut), and maybe a blueshift in another. Comoving Kobservers see the same redshift in all directions. Physical distances between comoving observers keep on increasing due to the expansion and thus we see the redshifts. What coconut described was no different than the Doppler shift due to a peculiar velocity. Comoving means at a fixed coordinate in the coordinate frame drawn in the expanding 3D balloon that is the universe. Yes, it is tricky. $\endgroup$
    – Bob Bee
    Commented Feb 26, 2017 at 21:53
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    $\begingroup$ What if this fixed coordinate is located very close to a very massive object where time passes much slower? Isn't this time measurement very prone to the mass distribution? $\endgroup$ Commented Feb 26, 2017 at 22:45
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    $\begingroup$ Am I right in guessing that the comoving frame is also the frame in which the time measured from the Big Bang is maximized? That is, an observer in any other frame will measure the same or a smaller time than an observer in the co-moving frame? $\endgroup$ Commented Feb 27, 2017 at 1:44
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    $\begingroup$ So, since this is new to me, does that mean that there is such a thing as "universal rest", and a $(0,0,0)$ point in the universe? $\endgroup$
    – Bobson
    Commented Feb 27, 2017 at 5:20
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    $\begingroup$ @Bobson There's some notion of rest relative to the content the universe has, but still there's no special point $\endgroup$
    – coconut
    Commented Feb 27, 2017 at 6:51
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While not at all obvious, it turns out that our best models of cosmology suggest that there exists a special frame of reference in which the distribution of the entire universe's matter and energy appears extremely uniform on very large (i.e. cosmological) scales. When we talk about the age of the universe, we always mean the age as viewed in this special frame. You are completely correct that particles that are moving very quickly with respect to this special frame will measure a very different age of the universe.

(Note that the existence of this special frame is completely compatible with special and general relativity, which say that the laws of physics themselves look the same in any inertial reference frame, not that the distribution of matter does.)

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    $\begingroup$ Is there any reason to believe that 'homogeneous' reference frame existed "10^-6" seconds after the big bang like it does today..? $\endgroup$
    – user541686
    Commented Feb 27, 2017 at 8:46
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    $\begingroup$ @Mehrdad If in fact the current spacetime metric is very homogeneous (i.e. well described by the FRW metric), then by symmetry, that homogeneity must be maintained as you run Einstein's equations backwards. In fact, it turns out that perturbations to that homogeneity tend to be amplified under forward time-evolution, so any small inhomogeneities (i.e. above the quantum scale) that long ago would have resulted in a very "rough" spacetime today, which we don't see. So if anything, spacetime was probably even smoother back then. (Although cosmic inflation complicates things a bit.) $\endgroup$
    – tparker
    Commented Feb 27, 2017 at 9:00
  • $\begingroup$ It was smoothed out by the cosmic inflation to a great degree, leaving only the quantum fluctuations, amplified. It's one of the values of cosmic inflation $\endgroup$
    – Bob Bee
    Commented Feb 28, 2017 at 0:29
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    $\begingroup$ @BobBee Well, I would say it "may have been" smoothed out by cosmic inflation - while inflation would indeed nicely explain the smoothness, it's far from a confirmed theory. $\endgroup$
    – tparker
    Commented Feb 28, 2017 at 7:33
  • $\begingroup$ Right. It is the only theory we have that has some backing and explains smoothing. After inflation it was smooth, and the lambda cmd model predicts the rest. The size of overdensities were set then, and evolution could be modeled after with semi known physics. And is $\endgroup$
    – Bob Bee
    Commented Feb 28, 2017 at 20:45
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Other answers have correctly pointed out that there's a special time measure, called cosmological time, that is singled out by the symmetries of the expanding matter, and times "since the big bang" are intervals of cosmological time.

No answer so far has mentioned that, probably, nothing special happened at t=0. That is, there probably wasn't a big bang, or any other unusual event, 1 second before the time that we call "1 second after the big bang."

As you go farther back in time, the energy density of the universe gets higher. The higher the density, the less idea we have of how physics works. If you assume that nothing fundamentally new happens at arbitrarily high density, you can extrapolate backwards to an infinite density (the big bang singularity) at a finite time in the past. Times "since the big bang" are measured from that time.

But we have no good reason to believe that this model is accurate, and at least one good reason to think it isn't, called the horizon problem: we know (from the CMBR) that the early universe was in extremely good thermal equilibrium, and in this model there isn't nearly enough time since the alleged big bang for the universe to equilibrate.

So there is a lot of interest in models that avoid this problem. The most famous are the inflationary models, where there's a period of approximately exponential growth in the scale factor that then transforms into a precursor of the well-understood high-density era.

"Exponential growth" sounds fast, and is usually described as fast, but it's actually slower than the growth of the model it replaces. Here's a picture:

This just joins an exponential curve on the left (for inflation) to a square-root curve on the right (for radiation-dominated expansion). In reality, inflation isn't exactly exponential, and the transition, called reheating, doesn't just happen at a point, but this gets the general idea across.

The dotted line is what the scale factor would be if the nothing-new-happens theory were correct. The time where it hits the x axis is the time from which times "since the big bang" are measured. But in the real model, represented by the solid line, nothing special happens at t=0. It's just an undistinguished moment in the late inflationary epoch, shortly before the end of it.

So why measure times according to a model that's probably wrong? Because it's more useful to be consistent than right. "After the big bang" would be an inconvenient measure if it kept jumping around according to the latest ideas about the very early universe. All of these models agree at later times (times when the density is low enough that the physics is well understood) so we may as well use the simplest model as our measurement standard when talking about well-understood times.

Besides, most alternative models can't predict the time since the big bang, or don't even have a big bang, so they wouldn't work as replacements for the after-big-bang zero point anyway.

The after-big-bang terminology does confuse people. For example when it's said that inflation ended $10^{-32}$ seconds after the big bang, people often assume that it could have lasted at most $10^{-32}$ seconds. In fact, though, it has to last at least 100 times longer than that in order to solve the horizon problem*, and in most (all?) inflationary models it lasts many orders of magnitude longer than that minimum.

(*Because it's usually said that inflation needs to last for at least 60 e-folds to solve the horizon problem, and the interval between 0 and $t_R$ in this diagram is ½ e-fold.)

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  • $\begingroup$ Great answer. Models without a big bang, or with a BB that's only "local", (i.e., "past- and future-eternal" models) include Aguirre & Gratton's "Steady-state eternal inflation", Nikodem Poplawski's "Cosmology with torsion", 2020 Nobel Winner Roger Penrose's "Conformal cyclic cosmology", and (most recently) Aguirre & Deutsch's "State-to-state cosmology". The Borde-Guth-Vilenkin Theorem (written before most of the aforementioned models) included, in the last footnote in its last (2003's) revision, an exemption for the 1st of Aguirre's models that I've mentioned; the others are more recent. $\endgroup$
    – Edouard
    Commented Aug 12, 2021 at 17:26
  • $\begingroup$ I wonder whether the OP is aware that the "accepted answer" can be changed from the one originally accepted: I was on the site for a few years when I found out that that is indeed the case! $\endgroup$
    – Edouard
    Commented Feb 27, 2022 at 10:55
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Adding a little more argument about co-moving, we know that momentum is conserved, even in relativity, so we can talk about the center of mass frame, the frame where the total momentum is zero. This is the preferred frame.

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    $\begingroup$ In General Relativity momentum, that is 3D momentum, is only conserved if spacetime is homogeneous. It is in the frame where you see the spatial distributions as homogeneous, and isotropic, that is the comoving frame. 3D momentum is not always conserved in General Relativity, so not a good starting point. The geometrical symmetries are the best starting points $\endgroup$
    – Bob Bee
    Commented Feb 27, 2017 at 2:46

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