Editing Hubble's law (section)

== Derivation of the Hubble parameter ==
{{More citations needed section|date=March 2014}}

Start with the [[Friedmann equations|Friedmann equation]]:

<math display="block">H^2 \equiv \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho - \frac{kc^2}{a^2}+ \frac{\Lambda c^2}{3},</math>

where {{mvar|H}} is the Hubble parameter, {{mvar|a}} is the [[scale factor (universe)|scale factor]], {{mvar|G}} is the [[gravitational constant]], {{mvar|k}} is the normalised spatial curvature of the universe and equal to −1, 0, or 1, and {{math|Λ}} is the cosmological constant.

=== 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>

=== Matter- and dark energy-dominated universe ===

If the universe is both matter-dominated and dark energy-dominated, then the above equation for the Hubble parameter will also be a function of the [[equation of state (cosmology)|equation of state of dark energy]]. So now:

<math display="block">\rho = \rho_m (a)+\rho_{de}(a),</math>

where {{mvar|ρ{{sub|de}}}} is the mass density of the dark energy. By definition, an equation of state in cosmology is {{math|1= ''P'' = ''wρc''{{sup|2}}}}, and if this is substituted into the fluid equation, which describes how the mass density of the universe evolves with time, then

<math display="block">\begin{align}
\dot{\rho}+3\frac{\dot{a}}{a}\left(\rho+\frac{P}{c^2}\right)=0;\\
\frac{d\rho}{\rho}=-3\frac{da}{a}(1+w).
\end{align}</math>

If {{mvar|w}} is constant, then

<math display="block">\ln{\rho}=-3(1+w)\ln{a};</math>

implying:

<math display="block">\rho=a^{-3(1+w)}.</math>

Therefore, for dark energy with a constant equation of state {{mvar|w}}, {{nowrap|<math>\rho_{de}(a)= \rho_{de0}a^{-3(1+w)}</math>.}} If this is substituted into the Friedman equation in a similar way as before, but this time set {{math|1= ''k'' = 0}}, which assumes a spatially flat universe, then (see [[shape of the universe]])

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

If the dark energy derives from a cosmological constant such as that introduced by Einstein, it can be shown that {{math|1= ''w'' = −1}}. The equation then reduces to the last equation in the matter-dominated universe section, with {{math|Ω{{sub|''k''}}}} set to zero. In that case the initial dark energy density {{math|''ρ''{{sub|''de''0}}}} is given by<ref>{{cite book|last1=Carroll|first1=Sean|title=Spacetime and Geometry: An Introduction to General Relativity|edition=illustrated|date=2004|publisher=Addison-Wesley|location=San Francisco|isbn=978-0-8053-8732-2|page=328|url=https://books.google.com/books?id=1SKFQgAACAAJ}}</ref>

<math display="block">\begin{align}
\rho_{de0} &= \frac{\Lambda c^2}{8 \pi G} \,, \\
\Omega_{de} &=\Omega_{\Lambda}.
\end{align}</math>

If dark energy does not have a constant equation-of-state {{mvar|w}}, then

<math display="block">\rho_{de}(a)= \rho_{de0}e^{-3\int\frac{da}{a}\left(1+w(a)\right)},</math>

and to solve this, {{math|''w''(''a'')}} must be parametrized, for example if {{math|1= ''w''(''a'') = ''w''{{sub|0}} + ''w''{{sub|''a''}}(1−''a'')}}, giving<ref>{{cite journal |last1=Heneka |first1=C. |last2=Amendola |first2=L. |date=2018 |title=General modified gravity with 21cm intensity mapping: simulations and forecast |journal=[[Journal of Cosmology and Astroparticle Physics]] |volume=2018 |issue=10 |page=004 |doi=10.1088/1475-7516/2018/10/004 |arxiv=1805.03629 |bibcode=2018JCAP...10..004H |s2cid=119224326 }}</ref>

<math display="block">H^2(z)= H_0^2 \left( \Omega_m a^{-3} + \Omega_{de}a^{-3\left(1+w_0 +w_a \right)}e^{-3w_a(1-a)} \right).</math>