Wednesday, 22 July 2015

cosmology - Redshift and conformal time


What is the relationship between redshift and conformal time ?


For example in a paper i found:


taking $z_e = 3234 $ at the time of radiaton-matter equality yields the conformal time $\frac{\eta_e}{\eta_0} = 0.007$ and taking $z_E=0.39$ at matter-$\Lambda$ equality yields $\frac{\eta_E}{\eta_0}=0.894$ and setting redshift at decoupling $z_d=1089$ yields $\frac{\eta_d}{\eta_0} = 0.0195$ where $\eta_0$ is the present decoupling time.



Further some cosmological parameters are given as : $\Omega_r = 8.36 \times 10^{-5}, \Omega_m = \Omega_b + \Omega_{dm} = 0.044 +0.226, \Omega_\Lambda = 0.73, H_0=0.72$


Now how can i calculate all those $\eta_e, \eta_E, \eta_d, \eta_0$ from given redshift values and/or above parameters ? I searched whole of my text books trying to find an explicit relation for conformal time $\eta$ but all i got was $ \mathrm{d}t=a(t)\mathrm{d}\eta$. Any help would be very helpful.



Answer



As you state, conformal time is defined as $$ \eta(t) = \int_0^t\frac{\text{d}t'}{a(t')}. $$ Using $$ \dot{a} = \frac{\text{d}a}{\text{d}t}, $$ this can be written in the form $$ \eta(a) = \int_0^a\frac{\text{d}a}{a\dot{a}} = \int_0^a\frac{\text{d}a}{a^2H(a)}, $$ with $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}. $$ The scale factor $a$ is related to the redshift as $$ 1 + z = \frac{1}{a}, $$ so that $$ \eta(z) = \frac{1}{H_0}\int_0^{1/(1+z)}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ Basically, the conformal time is equal to the distance of the particle horizon, divided by $c$ (see this post for more info). $\eta_0$ refers to the current conformal age of the universe.




edit


I just checked the values that you posted with my own cosmology calculator. I get $$ \eta_0 = 45.93\;\text{Gigayears} $$ for the current conformal age of the universe, and $$ \begin{align} z_e &= 3234,& a_e &= 0.000309,& t_e &= 5.54\times 10^{-5}\;\text{Gy},&\eta_e &= 0.3804\;\text{Gy},\\ z_E &= 0.39,& a_E &= 0.719,& t_E &= 9.359\;\text{Gy},& \eta_E &= 41.08\;\text{Gy},\\ z_d &= 1089,& a_d &= 0000917,& t_d &= 0.00037\;\text{Gy},& \eta_d &= 0.911\;\text{Gy}, \end{align} $$ so that $$ \frac{\eta_e}{\eta_0} = 0.00828,\quad \frac{\eta_E}{\eta_0} = 0.894,\quad \frac{\eta_d}{\eta_0} = 0.0198. $$ So my results are almost the same, but there's a small discrepancy. Apparently there's a small numerical error somewhere.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...