Expresion of comoving distance

Question

I have a simple question :

How to prove the following relation :

The comoving distance to an object at redshift $z$ can be computed as

$$r(z)=\dfrac{c}{H_{0}} \int_{0}^{z} \dfrac{\mathrm{d} z}{E(z)}$$

from the relation :

$$r(t)=\int_{0}^{t} \dfrac{c\mathrm{d} t}{R(t)}$$

I tried to use with the definition : $1+z= \dfrac{R_{0}}{R(t)}$ but I can't conclude.

Any help is welcome.

UPDATE 1 : @Tosic's demonstration seems to be correct. But the factor $R_{0}$ is not disappearing. Indeed, If I do :

$$\dfrac{\text{d}(1+z)}{\text{d}t} = \dfrac{\text{d}z}{\text{d}t} = -\dfrac{H(t)}{R(t)}\,R_{0}$$

which implies :

$$\int_{0}^{z} \dfrac{c\text{d}z}{H(z)} = \int_{0}^{t}c\text{d}t\dfrac{R_{0}}{R(t)} = R_{0} \int_{0}^{t}\dfrac{c\text{d}t}{R(t)}$$

How to get rid of the factor $R_{0}$ ? Since if I multiply the comoving coordinate $r(t)$ by $R_{0}$, I get the cosmological horizon (the limit of observable universe if I integrate up to $z=1100$), don't I ?

Answer 1

According to this, the Hubble constant for redshift z is $H_0E(z)$. Meaning we need to prove that $$\int_{0}^{z_0}\frac{cdz}{H(z)} = \int_{0}^{t_0}\frac{cdt}{R(t)}$$ Take the first derivative of both sides of your second equation to obtain, by the chain rule, and the equation (this is the definition of the Hubble constant) $H = \frac{R(t)'}{R(t)}$ the following: $$\frac{dz}{dt} = -\frac{1}{R(t)^2}*R'(t) = -\frac{H(t)}{R(t)}$$ That the redshift is zero for the current time, and some value greater than zero for some time before the current time should explain the minus sign. After multiplying with $c$ and placing the small changes on both sides this becomes $$\frac{cdz}{H(z)}=-\frac{cdt}{R(t)}$$, which looks like it can be integrated to obtain what we need to prove (the integral from z to 0 is the one from 0 to t).
This proof is not very formal, but it's the best I can do, so I hope it is somewhat correct and that someone will give a more detailed answer if this is not good enough.