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?

  • 2
    $\begingroup$ The isotropic ones. $\endgroup$
    – MBN
    Commented Mar 18, 2014 at 21:47
  • $\begingroup$ Basically, observers who are traveling with the galaxies. $\endgroup$
    – WillO
    Commented Mar 18, 2014 at 22:15
  • 1
    $\begingroup$ @WillO But I figure that surely not all galaxies will travel equivalently, so this is ambiguous, isn't it? $\endgroup$
    – Yossarian
    Commented Mar 18, 2014 at 22:18
  • $\begingroup$ Question to clarify: It was written down as 14 billion years everywhere, but are those European or American billions? $\endgroup$
    – Someone
    Commented Jun 17, 2020 at 13:16
  • $\begingroup$ I'm unable to support this question because of the OP's previous acceptance of one of the two answers whose difficulties motivated my comments on the answers provided by Oman and Gibbs: It's possible that the universe, or an inflationary multiverse comprised of "local universes", is eternal to the past, with the latter version of those two possibilities being described in the reference I've cited under Gibbs' answer, and the former in the "Conformal cyclic cosmology" of Roger Penrose, a winner (albeit for reasoning not limited to that cosmological model) of 2020's Nobel Prize in physics. $\endgroup$
    – Edouard
    Commented May 1, 2021 at 3:49

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$ Commented Mar 19, 2014 at 6:58
  • 2
    $\begingroup$ @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. $\endgroup$
    – Kyle Oman
    Commented Mar 19, 2014 at 15:15
  • $\begingroup$ Then I don't quite understand - in what sense is it "idealized"? $\endgroup$ Commented Mar 20, 2014 at 6:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Kyle Oman
    Commented Mar 20, 2014 at 14:47
  • 2
    $\begingroup$ 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. $\endgroup$
    – Warrick
    Commented Mar 22, 2014 at 8:57

You can define the age of the universe roughly as the proper time for a hypothetical observer who is comoving with the galaxies and not too near a strongly gravitating object. This is imprecise because the galaxies are themselves moving around and the age would depend on exactly the worldline of the observer and how it moved to avoid heavy objects that dilate time etc.

This definition is good enough for cosmological measurements because the universe is roughly homogeneous, but if you want a very precise definition of the age of the universe at any given place and time which does not rely on the comoving flow then this is easily done. The age of an event can simply be defined as the longest possible proper time along any time-like worldline that starts at the big bang singularity and ends at the event of space and time. To maximise this proper time an observer must avoid gravitating objects and high velocities that would cause time dilation. This maximum is well defined provided the big bang is considered as a singularlty everywhere in the past of the observable universe and that there are no closed time-like curves that would spoil hyperbolicity. It avoids the assumption that the universe is homogeneous or modeled by a particular cosmology such as FLRW. Of course in the special case of FLRW the general definition is equivalent to the simple comoving time.

  • $\begingroup$ The assumption that the big bang was a past singularity is attacked by Nikodem Poplawski in 2010-2021 papers found by his name on the Arxiv website. $\endgroup$
    – Edouard
    Commented May 1, 2021 at 3:28

We use the Friedmann equations and EFE : $$\begin{cases} 3\frac{\dot{a}^2}{a^2}+3\frac{kc^2}{a^2}-\Lambda c^2=\frac{8\pi G}{c^2}\rho \qquad(1) \\[2ex] -2\frac{\ddot{a}}{a}-\frac{\dot{a}^2}{a^2}-\frac{kc^2}{a^2}+\Lambda c^2=\frac{8\pi G}{c^2}p \qquad(2)\\[2ex] R_{ij}-\frac{1}{2}Rg_{ij}=\frac{8\pi G}{c^4} T_{ij} \qquad(3) \end{cases}$$ If we toke $k=0; \Lambda\neq 0$ than our universe is flat and its expansion is accelerated; thus the EFE can be written : $$R_{ij}-\frac{1}{2}Rg_{ij}-\Lambda g_{ij}=\frac{8\pi G}{c^4}T_{ij}$$ Or it can be also written : $$R_{ij}-\frac{1}{2}Rg_{ij}=\frac{8\pi G}{c^4}T_{ij}+\Lambda g_{ij} \Leftrightarrow R_{ij}-\frac{1}{2}Rg_{ij}=\frac{8\pi G}{c^4}\Bigr(T_{ij} \frac{c^4\Lambda }{8\pi G}g_{ij}\Bigl)$$ We express the stress-energy tensor for vacuum : $$T_{ij}^{\mathbf{Vacuum}}=\frac{c^4\Lambda g_{ij}}{8\pi G}$$ Comparing it with perfect fluid: $$T_{ij}^{\mathbf{Matter}}=-p.g_{ij}+\Bigl(\frac{p}{c^2}+\rho_0\Bigr)u_i u_j$$ We can simulate vacuum as a fluid $$\begin{cases} \mathbf{Pressure}\ :\ p=-\frac{\Lambda c^4}{8\pi G} \\ \mathbf{Energy\ density}\:\ \rho=-p=\frac{\Lambda c^4}{8\pi G} \end{cases}$$ Adding cosmological parameters: $$\begin{cases} \Omega_v+\Omega_m=1 \\ 2q-1+3\Omega_v =0 \end{cases} \iff \begin{cases} \Omega_m =1-\Omega_v \\ \Omega_v =\frac{1-2q}{3} \end{cases}$$ Let $t=t_0$ and $\Lambda = \frac{3\Omega_{v_0}H_0^2}{c^2}$

$$\begin{cases} \Omega_{v_0}+\frac{8\pi G \rho_0}{3c^2 H_0^2}=1 \\[2ex] \Omega_{v_0}=\frac{\Lambda c^2}{3H_0^2}=\frac{1-2q_0}{3} \end{cases} \iff \begin{cases} 1-\Omega_{v_0}=\frac{8\pi G \rho_0}{3c^2 H_0^2} \iff (1-\Omega_{v_0})\frac{H_0^2}{c^2}=\frac{8\pi G \rho_0}{3c^2} \\[2ex] \Omega_{v_0}=\frac{1-2q_0}{3} \iff 1-\Omega_{v_0}=\frac{2}{3} (1+q_0) \end{cases}$$

Thus we obtain : $$\frac{8\pi G\rho_0}{3c^4}=\frac{2}{3} \frac{H_0^2}{c^2} (1+q_0)$$ In the first equation, we are having the following :

Recall $\ k=0, \Lambda \neq 0\ $: $$3\frac{\dot{a}^2}{a^2} -\Lambda c^2=\frac{8 \pi G}{c^2}\rho$$ We know that : $\rho a^3=\rho_0 a_0^3$; thus : $\rho=\frac{\rho_0 a_0^3}{a^3} $ $$\Rightarrow 3\frac{\dot{a}^2}{a^2} -\Lambda c^2=\frac{8 \pi G}{c^2}\frac{\rho_0 a_0^3}{a^3} \iff \dot{a}^2=\frac{8 \pi G}{3c^2}\frac{\rho_0 a_0^3}{a}+\frac{\Lambda c^2 a^2}{3} \iff da=\sqrt{\underbrace{\frac{8 \pi G\rho_0 a_0^3}{3c^2}}_{K}\frac{1}{a}+\underbrace{\frac{\Lambda c^2 }{3}}_{B}a^2} dt$$ Let $ \ K=\frac{8 \pi G\rho_0 a_0^3}{3c^2}\ $ and $\ B=\frac{\Lambda c^2}{3}\ $ : $$da=\sqrt{\frac{K}{a}} \sqrt{1+\frac{B}{K} a^3} dt \iff dt=\frac{da.a^{1/2}}{\sqrt{K}\sqrt{1+\frac{B}{K} a^3}}$$ Integrating, we get the following : $$\int dt=\int \frac{da.a^{1/2}}{\sqrt{K}\sqrt{1+{\frac{B}{K} a^3}}}$$ Let $x^2=\frac{B}{K} a^3\ $ thus : $$\begin{cases} 3a^2da=2\frac{K}{B} x dx \\ a^2=\bigl(\frac{K}{B}\bigr)^{2/3} x^{4/3} \\ a^{1/2}=\bigl(\frac{K}{B}\bigr)^{1/6}x^{1/3} \end{cases}$$ So (I'm going to skip math here !) : $$\int \frac{\frac{2}{3} \bigr(\frac{K}{B}\bigl)^{1/2}dx}{\sqrt{K}\sqrt{1+\underbrace{\frac{B}{K}a^3}_{x^2}}}=\int dt \iff \frac{2}{3B^{1/2}} \text{arcsh}(x)=t \qquad(\text{I'm Skiping math !}) $$ $$\fbox{$a^3=\frac{8\pi G \rho_0 a_0^3}{c^4\Lambda}\text{sh}^2\Bigr(\frac{c}{2}\sqrt{3\Lambda}t\Bigl)$} $$ Recall : $\Lambda = \frac{3\Omega_{v_0}H_0^2}{c^2}$ and $\frac{8\pi G\rho_0}{3c^4}=\frac{2}{3} \frac{H_0^2}{c^2} (1+q_0)$

$$\require{cancel} a^3=\frac{2H_0^2(1+q_0)a_0^3}{c^2\Lambda }\text{sh}^2\Bigr(\frac{c}{2}\sqrt{3\Lambda}t\Bigl) \iff a^3=\frac{2\cancel{H_0^2}(1+q_0)a_0^3 \cancel{c^2}}{3\cancel{c^2}\Omega_{v_0}\cancel{H_0^2}}\text{sh}^2\Biggr(\frac{\cancel{c}}{2}\sqrt{\frac{9\Omega_{v_0}H_0^2}{\cancel{c^2}}}t\Biggl)$$ Therefore: $$ a^3=\frac{2a_0^3(1+q_0)}{3\Omega_{v_0}}\text{sh}^2 \Bigr(\frac{3H_0}{2}\sqrt{\Omega_{v_0}}t\Bigl)$$ and now, let us calculate this $t$, well we are going to assume that $t=t_0$ and $a=a_0$: $$\begin{aligned}\require{cancel}\cancel{a_0^3}=\frac{2\cancel{a_0^3}(1+q_0)}{3\Omega_{v_0}}\text{sh}^2 \Bigr(\frac{3H_0}{2}\sqrt{\Omega_{v_0}}t_0\Bigl) \iff \frac{3\Omega_{v_0}}{2(1+q_0)}=\text{sh}^2 \Bigr(\frac{3H_0}{2}\sqrt{\Omega_{v_0}}t_0\Bigl)\\ \iff \frac{3H_0}{2}\sqrt{\Omega_{v_0}}t_0=\text{arcsh}\Biggr(\sqrt{\frac{3\Omega_{v_0}}{2(1+q_0)}}\Biggl) \\ \iff t_0=\frac{2}{3} \frac{1}{H_0\sqrt{\Omega_{v_0}}}\text{arcsh}\Biggr(\sqrt{\frac{3\Omega_{v_0}}{2(1+q_0)}}\Biggl) \end{aligned}$$ And here you go, the formula of our universe's age : $$\fbox{$t_0=\frac{2}{3} \frac{1}{H_0\sqrt{\Omega_{v_0}}}\text{arcsh}\Biggr(\sqrt{\frac{3\Omega_{v_0}}{2(1+q_0)}}\Biggl)$}$$ The numerical substitution : $$\begin{cases} H_0^{-1}\approx 14.56\times 10^9 \\ q_0\approx -0.5245 \\ \Omega_{v_0}\approx 0.683 \end{cases}$$ Thus : $$t_0=\frac{2}{3}\times 14.56\times 10^9 \frac{1}{\sqrt{0.683}}\text{arcsh}\Biggr(\sqrt{\frac{3\times 0.683}{2(1-0.5245)}}\Biggl)\approx 13.8\times 10^9\text{y}$$ I hope now that you understood my comment, it depends on the numerical values of the cosmological parameters.

And yeah one more thing, Sorry I skipped a lot of steps in the proof because of my laziness and it is long.

Good luck !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.