Editing Hubble's law (section)

=== Matter-dominated universe (with a cosmological constant) ===

If the universe is [[Matter-dominated era|matter-dominated]], then the mass density of the universe {{mvar|ρ}} can be taken to include just matter so

<math display="block">\rho = \rho_m(a) = \frac{\rho_{m_{0}}}{a^3},</math>

where {{math|''ρ''{{sub|''m''{{sub|0}}}}}} is the density of matter today. From the Friedmann equation and thermodynamic principles we know for non-relativistic particles that their mass density decreases proportional to the inverse volume of the universe, so the equation above must be true. We can also define (see [[density parameter]] for {{math|Ω{{sub|''m''}}}})

<math display="block">\begin{align}
\rho_c &= \frac{3 H_0^2}{8 \pi G}; \\
\Omega_m &\equiv \frac{\rho_{m_{0}}}{\rho_c} = \frac{8 \pi G}{3 H_0^2}\rho_{m_{0}};
\end{align}</math>

therefore:

<math display="block">\rho=\frac{\rho_c \Omega_m}{a^3}.</math>

Also, by definition,
<math display="block">\begin{align}
\Omega_k &\equiv \frac{-kc^2}{(a_0H_0)^2} \\
\Omega_{\Lambda} &\equiv \frac{\Lambda c^2}{3H_0^2},
\end{align}</math>

where the subscript {{math|0}} refers to the values today, and {{math|1= ''a''{{sub|0}} = 1}}. Substituting all of this into the Friedmann equation at the start of this section and replacing {{mvar|a}} with {{math|1= ''a'' = 1/(1+''z'')}} gives

<math display="block">H^2(z)= H_0^2 \left( \Omega_m (1+z)^{3} + \Omega_k (1+z)^{2} + \Omega_{\Lambda} \right).</math>