# Is there any method to know the real age of the universe? [duplicate]

Well, I was wondering about the real age of our universe, I found that it's estimated to be $$13.8\times 10^9$$ years.

Is it an approximation, or laws behind this age?

• of course there is physical, but this is not a good question, there's some models, we call it the universe and we use friedmann equations to differentiate between these models. I'll answer your question.
– user257151
Apr 2 '20 at 16:48
• You mean that using physics we can get the exact age of the universe ?
– user257533
Apr 2 '20 at 17:58
• Not the exact age, because "The age formula" has some constants and their values are approximated, thus the result isn't exact and it is approximated too, any way I have an answer typed out for you.
– user257151
Apr 2 '20 at 18:00
• If you manage to date Mother Nature long enough, she might tell you her birthday. Just don’t forget it! Apr 2 '20 at 19:00
• Does this answer your question? The age of the universe Apr 2 '20 at 19:09

As I said there's a mathematical laws behind this approximation.

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.

I hope you have understand that there is a laws behind this approximation and this is a way to compute our universe's age. Good luck !

Of course it is approximation, as any almost any measurement beyond simple counting involves approximation, but there is quite a bit of evidence based on observation fitted to known laws.

The main evidence springs from fitting observations of cosmological parameters, Hubble's constant $$H_0$$, density $$\Omega$$, curvature $$\Omega_k$$, and dark energy $$\Omega_\Lambda$$ to Friedmann models. This is done both from supernova data and from analysis of the cosmic microwave background, with broad, but not perfect agreement (recent measurements seem to suggest a small level of disagreement, but not enough that anyone is seriously concerned).

There is also supporting evidence, as it is possible to measure the age of some of the oldest stars from their constitution.

• Back when I was a student, we had "Journal Club", where students presented new research outside their field over lunch. I presented a paper back when the Universe was 14B y/o, and Willy Fowler ripped me a new one, and delved into Big Bang Nucleosynthesis arguments that it could not be older the 13.8. Since I had a bachelors and he had a Nobel Prize (for...Big Bang nucleosynthesis), I did not argue. Moral: don't forget BBN. (and stellar nucleosynthesis).
– JEB
Apr 2 '20 at 18:51
• Fowler's Nobel prize was actually for stellar nucleosynthesis, although it was Hoyle who did almost all the work. Apr 2 '20 at 18:57
• Yes, I just learned of the B2FH controversy. Nevertheless, he knew his BBN too, and apparently knew the age was closer to 13.8B when everyone else insisted it was 14B or older. I wish I could repeat his argument, but it was way over my head, and 30 years ago.
– JEB
Apr 2 '20 at 19:01