3
$\begingroup$

Let me preface by stating that I have no experience with General Relativity. I am working on a project for school that requires a little knowledge of it, so I am hoping to find some help. I do have experience with Special Relativity.

On to the question. I know that one can calculate the age of the Universe using the Lambda-CDM model. After making a few simplifying assumptions, one can find the relation

$$H\left ( a \right )=\frac{\dot{a}}{a}=H_{0}\sqrt{\frac{\Omega_{m}}{a^{3}}+\frac{\Omega_{rad}}{a^{4}}+\Omega_{\Lambda }}.$$

One can numerically integrate to find $t_{0}$, the age of the Universe.

$$t_{0}=\int_{0}^{1}\frac{da}{aH_{0}\sqrt{\frac{\Omega_{m}}{a^{3}}+\frac{\Omega_{rad}}{a^{4}}+\Omega_{\Lambda }}}$$

Now, if I am correct, when performing this calculation for the age, I was working in a co-variant coordinate system (the system that expands with the universe or the system of CMB). For my project, I want to calculate the age of the Universe in a different coordinate system. More specifically, I would like to calculate the age in a coordinate system that is not expanding with the Universe. I know from other articles on here that I cannot use Special Relativity, but I am unsure how to go about this. If someone could show me how to go about this, keeping in mind my knowledge on this subject is very limited, I would be appreciative.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

You can write $ds^2$ as $a^2$ times a static metric, introducing a new time coordinate viz. $d\eta=dt/a$. We say the full metric is conformal to the simpler one (this adjective refers to angle preservation upon the rescaling), so $\eta$ is called conformal time. The $\eta r\theta\phi$ coordinate system fits your criterion.

Note that $d\eta=da/(a^2 H)$; the conformal age of the universe, in the convention where $a=1$ today, is about $46.8$ Gigayears, which is why the Hubble zone is $93.8$ light Gigayears wide. Thus the conformal age is not the quantity you want! You should still integrate $da/(aH)$ instead, but write $a,\,H$ as functions of $\eta$ instead of $t$.

Improve this answer
$\endgroup$
15
  • $\begingroup$ So here is what I got: $$H\left ( a \right )=\frac{\dot{a}}{a\left ( t \right )}=\frac{1}{a}\frac{\mathrm{d} a}{\mathrm{d} t}=\frac{1}{a}\frac{\mathrm{d} a}{\mathrm{d} \eta }\frac{\mathrm{d} \eta }{\mathrm{d} t}=\frac{{a\left ( \eta \right )}'}{a\left ( \eta \right )^{2}}$$$$a\left ( t \right )=a\left ( \eta \right )$$ This does not quite seem correct. If it is not, what have I done wrong? $\endgroup$
    – Java Newbie
    Commented Apr 1, 2018 at 22:20
  • $\begingroup$ @JavaNewbie No, you are right; your result is equivalent to my own calculation. $\endgroup$
    – J.G.
    Commented Apr 1, 2018 at 22:57
  • $\begingroup$ Using the Friedmann equation $$\frac{\dot{a}^{2}+kc^{2}}{a^{2}}=\frac{8\pi G\rho +\Lambda c^{2}}{3}$$ and using $k=0$, I found $${a}'=\sqrt{\frac{8\pi G\rho+\Lambda c^{2}}{3}}a^{2}.$$ From the definitions from my last comment, $$t_{0}=\int_{0}^{1}\frac{a}{{a}'}\mathrm{d} a=\int_{0}^{1}\frac{1}{a}\sqrt{\frac{3}{8\pi G\rho+\Lambda c^{2}}}\mathrm{d}a.$$ Is this correct? Also, how do I then get $\rho$ in terms of $\Omega_{m}$, $\Omega_{rad}$, and $\Omega_{\Lambda }$. Finally, will the values of these parameters remain unchanged from the original ones? $\endgroup$
    – Java Newbie
    Commented Apr 2, 2018 at 1:30
  • $\begingroup$ On second glance, I found $$\eta _{0}=\int_{0}^{1}\frac{\mathrm{d} a}{H_{0}\sqrt{\Omega _{m}a+\Omega _{rad}+\Omega _{\Lambda }a^{4}}}\cong 46.3\:\mathrm{billion\:years.}$$ I kept all the parameters the same as before. I am a little confused as this is the quantity you said I didn't want. $\endgroup$
    – Java Newbie
    Commented Apr 2, 2018 at 2:28
  • $\begingroup$ @JavaNewbie You want the age of the. universe, which is the smaller value called $t_0$. $\endgroup$
    – J.G.
    Commented Apr 2, 2018 at 6:43
0
$\begingroup$

In short, the same way as before but assume $\Omega_{rad}$ is currently very small

$$\frac {\dot a}{a}=H_{0}\sqrt{\Omega_{m}a^{-3} + \Omega_{\Lambda}}$$

which has the solution

$$a(t)=(\Omega_{m}/\Omega_{\Lambda})^{1/3}\sinh^{2/3}(t/t_{\Lambda})$$

where $t_{\Lambda}=2/(3H_{0}\sqrt{\Omega_{\Lambda} })$

Set $a=1$ which gives you $t=t_{0}$ the current age of the Universe.

See the Lambda-CDM model in Wikipedia.

Improve this answer
$\endgroup$

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.