0
$\begingroup$

What is the maximum distance and the maximum time a photon has traveled which is observed by us today at the particle horizon in the LCDM cosmological model applying general relativity?

$\endgroup$
  • $\begingroup$ What do you mean by "applying GR" ? $\endgroup$ – Reign Oct 31 at 4:42
  • $\begingroup$ i.e., not using Special Relativity for calculations - which anyway should be evident for evaluations of cosmological models. $\endgroup$ – Rene Kail Oct 31 at 7:03
0
$\begingroup$

For Particle Horizon distance you can use the

$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)}$$ But you can write this equation in terms of z and (making limits from $\infty$ to $0$) to make calculations easier.

To calculate the conformal time part you have to divide above equation by c.

$$\eta = \int_{t_e}^{t_o}\frac{dt}{a(t)}$$

Edit: So we know that $$1 + z = a(t)^{-1}$$

hence

$$\frac{dz}{dt} = \frac{dz} {da} \frac{da} {dt}$$ $$\frac{dz}{dt} = -\frac{1}{a^2} \dot{a}$$ $$dz = -\frac{H}{a}dt = -H(1+z)dt$$ $$dt = -\frac{dz}{H(1+z)}$$

Since at $t_e$ corrsponds to $z = \infty$ and $t_0 = 0$ we have

$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)} = -c\int_{\infty}^{0}\frac{dz}{H}$$

Also we know that $$H = H_0\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}$$

Finally

$$ r_{horizon} = -\frac{c}{H_0}\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$

and

$$ \eta = -\frac{1}{H_0}\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$

For $\Omega_r = 9 \times 10^{-5}$, $\Omega_m = 0.31$ and $\Omega_{\Lambda}= 0.69$

we get

$$-\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}} = 3.200$$

So

$$r_{horizon} = 3.2 \times c/H_0 = 3.2 \times 4408.712 Mpc = 14107.88 Mpc = 46.013 Gly$$

$$ \eta = 3.2 \times 1/H_0 = 3.2 \times 14.38Gyr = 46.016 Gyr$$

So light has traveled 46.013 Gly since the Big Bang and the corresponding conformal time would be 46.013 Gyr. This time is not the age of the universe.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.