# How to Change Coordinate Systems in General Relativity

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.

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 $$t$$.

• 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? Commented Apr 1, 2018 at 22:20
• @JavaNewbie No, you are right; your result is equivalent to my own calculation.
– J.G.
Commented Apr 1, 2018 at 22:57
• 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? Commented Apr 2, 2018 at 1:30
• 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. Commented Apr 2, 2018 at 2:28
• @JavaNewbie You want the age of the. universe, which is the smaller value called $t_0$.
– J.G.
Commented Apr 2, 2018 at 6:43

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.