From 'Distance measures (cosmology)' on Wikipedia:
Cosmologists commonly use the following measures for distances from the observer to an object at redshift $z$ along the line of sight:
- Comoving distance: $$d_C(z)=d_H\int_0^z\frac{dz'}{E(z')}.$$
- Transverse comoving distance: $$d_M(z)=\left\{\begin{array}{ll} \frac{d_H}{\sqrt{\Omega_k}}\sinh\left(\frac{\sqrt{\Omega_k}d_C(z)}{d_H}\right) & \Omega_k>0,\\ d_C(z) & \Omega_k=0,\\ \frac{d_H}{\sqrt{|\Omega_k|}}\sin\left(\frac{\sqrt{|\Omega_k|}d_C(z)}{d_H}\right) & \Omega_k<0. \end{array}\right.$$
- Angular diameter distance: $$d_A(z)=\frac{d_M(z)}{1+z}.$$
- Luminosity distance: $$d_L(z)=(1+z)d_M(z).$$
- Light-travel distance: $$d_T(z)=d_H\int_0^z\frac{dz'}{(1+z')E(z')}.$$
Note that the comoving distance is recovered from the transverse comoving distance by taking the limit $\Omega\to0$, such that the two distance measures are equivalent in a flat universe.
What do you do with the dH that comes up before the integration symbol under 'comoving distance' and 'light-travel distance'?
Not just here, but in general...
Also, the z is a constant in astronomy, so how do you take the derivatives of z in those same two equations? If the derivative of a constant is always 0? It doesn't make sense to me....
P.S. I posted something similar in Astronomy S.E., I hope that is okay... Here, I wanted to understand the general math aspect...
We can rewrite the equations with every "calculus $\mathrm d$" deitalicised, so$$d_C(z)=d_H\int_0^z\tfrac{\mathrm dz^\prime}{E(z^\prime)},\,d_T(z)=d_H\int_0^z\tfrac{\mathrm dz^\prime}{(1+z^\prime)E(z^\prime)}.$$The other equations don't contain any such $\mathrm d$, so I won't reproduce those. For all the difference it makes, $d_A,\,d_C,\,d_H,\,d_L,\,d_M,\,d_T$ might as well have been called $r_A,\,r_C,\,r_H,\,r_L,\,r_M,\,r_T$.
It helps to understand why these definite integrals start at $0$. The value of $a:=1+z$ is proportional to the width of the universe, so is a function of time. The usual convention is to take $a=1$, or equivalently $z=0$, in the present day. Then $d_C,\,d_T$ are distances from our present-day location to something observed by dint of light it emitted in the past: in particular, these distances are $0$ if $z=0$.
Of course, $z=0\implies d_C=d_T=0\implies d_A=d_L=d_M=0$. But $d_H$ is approximately proportional to $a$.