I would like to know the exact form of Hubble's law for the luminosity distance $d_L$, where $d_LH_0=f(z)$ in a flat FRW universe with energy density dominated by a component satisfying $\rho \sim a^{-n}$, where $a$ is the cosmic scale factor.
As I am trying to find the exact form, I don't want to use the expansion of the cosmic scale factor.
I think for distance you should use comoving distance, instead of luminosity distance, because on higher redshifts, luminosity distance doesn't work very well. Comoving distance is defined as: $$\chi(t)=\int^{t_0}_t\frac{dt'}{a(t')}=\int^{1}_{a(t)}\frac{da'}{a'^2H(a')}=\int^{z}_{0}\frac{dz'}{H(z')}$$ Comoving distance work well even on high redshifts. Also, comoving distance approx to $\chi\approx z/H_0$. But in the FRW flat metric, Hubble parameter equals to: $$H(t)=\bigg(\frac{\dot{a}}{a}\bigg)$$ So It's scale factor dependent. But also, evolution of the hubble parameter can be explained as: $$\frac{H^2(t)}{H_0^2}=\sum_{s=r,m,\nu,\Lambda}\Omega_s[a(t)]^{-3(1+\omega_s)}$$ Where $\omega_s=\frac{P_s}{\rho_s}$, $P$ is a pressure. It's so called equation of state, which was obtained from stress-energy tensor: $$T_{\mu\nu}=\begin{bmatrix} -\rho & 0 & 0 & 0\\ 0 & P & 0 & 0 \\ 0 & 0 & P & 0 \\ 0 & 0 & 0 & P \end{bmatrix} + \begin{bmatrix} -\rho_\Lambda& 0 & 0 & 0\\ 0 &-\rho_\Lambda& 0 & 0 \\ 0 & 0 & -\rho_\Lambda& 0 \\ 0 & 0 & 0 & -\rho_\Lambda \end{bmatrix} $$ From this tensor, $P_\Lambda=-\rho_\Lambda$, so $\omega_\Lambda=-1$. As well, for radiation $\omega_r=1/3$, for matter $\omega_m=0$. But this equation (Hubble parameter evolution) also dependent on scale factor, as can be seen.
