# The age of the universe

Many times I have read statements like, "the age of the universe is 14 billion years" . For example this wikipedia page Big Bang.

Now, my question is, which observers' are these time intervals? According to whom 14 billion years?

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The isotropic ones. –  MBN Mar 18 at 21:47
Basically, observers who are traveling with the galaxies. –  WillO Mar 18 at 22:15
@WillO But I figure that surely not all galaxies will travel equivalently, so this is ambiguous, isn't it? –  silvrfück Mar 18 at 22:18
@silvrfuck: That depends on your model of spacetime, but in the models people have in mind when they make these "age of the universe" calculations (e.g. FLRW), it is either unambiguous or close enough to unambiguous that we can pretend it is. –  WillO Mar 19 at 2:39

An observer with zero comoving velocity (i.e. zero peculiar velocity). Such an observer can be defined at every point in space. They will all see the same Universe, and the Universe will look the same in all directions ("isotropic").

Note that here I'm talking about an "idealized" Universe described by the FLRW metric:

$$\mathrm{d}s^2 = a^2(\tau)\left[\mathrm{d}\tau^2-\mathrm{d}\chi^2-f_K^2(\chi)(\mathrm{d}\theta^2 + \sin^2\theta\;\mathrm{d}\phi^2)\right]$$

where $a(\tau)$ is the "scale factor" and:

$$f_K(\chi) = \sin\chi\;\mathrm{if}\;(K=+1)$$ $$f_K(\chi) = \chi\;\mathrm{if}\;(K=0)$$ $$f_K(\chi) = \sinh\chi\;\mathrm{if}\;(K=-1)$$

and $\tau$ is the conformal time:

$$\tau(t)=\int_0^t \frac{cdt'}{a(t')}$$

The peculiar velocity is defined:

$$v_\mathrm{pec} = a(t)\dot{\chi}(t)$$

so the condition of zero peculiar velocity can be expressed:

$$\dot{\chi}(t) = 0\;\forall\; t$$

The "age of the Universe" of about $14\;\mathrm{Gyr}$ you frequently hear about is a good approximation for any observer whose peculiar velocity is non-relativistic at all times. In practice these are the only observers we're interested in, since peculiar velocities for any bulk object (like galaxies) tend to be non-relativistic. If you happened to be interested in the time experienced by a relativistic particle since the beginning of the Universe, it wouldn't be terribly hard to calculate.

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Does it follow from anything that the FLRW metric is a good approximation to the existing one? I mean not the currently observable chunk of it but the most probable extrapolation to the whole universe? –  მამუკა ჯიბლაძე Mar 19 at 6:58
@მამუკაჯიბლაძე The FLRW is general enough to be applicable to any homogeneous and isotropic universe. That these two properties hold on arbitrarily large scales is one of the fundamental assumptions of cosmology. This appears to hold observationally. Smaller deviations from homogeneity/isotropy can be treated perturbatively in the framework of the background metric. –  Kyle Mar 19 at 15:15
Then I don't quite understand - in what sense is it "idealized"? –  მამუკა ჯიბლაძე Mar 20 at 6:38
@მამუკაჯიბლაძე maybe that word was poorly chosen. What I meant by it is that homogeneity and isotropy must hold. So it makes sense to use the FLRW metric to talk about the age of the Universe as a whole, since we think it's homogeneous/isotropic on large scales. But locally, this assumption breaks down. This would introduce (small, I think) corrections to the time experienced by an observer at a given location. –  Kyle Mar 20 at 14:47
If you're confused by why we think the Universe should be "isotropic" and "homogeneous", it might sound more convincing in simpler terms. When averaging over large enough parts, we expect the structure of the Universe to be the same no matter where we are (homogeneous) and the same no matter what direction we look (isotropic). Now, these are clearly not perfectly true: if I look towards Andromeda... well, there's Andromeda, but if I look somewhere else nearby, I see past it. But! If I look at a big enough patch of sky, it looks about the same as another big enough patch of sky. –  Warrick Mar 22 at 8:57