# Maximum distance traveled by a photon up to the present time

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?

• What do you mean by "applying GR" ? – Reign Oct 31 '19 at 4:42
• i.e., not using Special Relativity for calculations - which anyway should be evident for evaluations of cosmological models. – Rene Kail Oct 31 '19 at 7:03

## 1 Answer

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.