The age of the universe

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 !