Derivation of the Hubble parameter Hubble's law
start friedmann equation:
h
2
≡
(
a
˙
a
)
2
=
8
π
g
3
ρ
−
k
c
2
a
2
+
Λ
c
2
3
,
{\displaystyle 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}},}
where
h
{\displaystyle h}
hubble parameter,
a
{\displaystyle a}
scale factor, g gravitational constant,
k
{\displaystyle k}
normalised spatial curvature of universe , equal −1, 0, or +1, ,
Λ
{\displaystyle \lambda }
cosmological constant.
matter-dominated universe (with cosmological constant)
if universe matter-dominated, mass density of universe
ρ
{\displaystyle \rho }
can taken include matter so
ρ
=
ρ
m
(
a
)
=
ρ
m
0
a
3
,
{\displaystyle \rho =\rho _{m}(a)={\frac {\rho _{m_{0}}}{a^{3}}},}
where
ρ
m
0
{\displaystyle \rho _{m_{0}}}
density of matter today. know nonrelativistic particles mass density decreases proportional inverse volume of universe, equation above must true. can define (see density parameter
Ω
m
{\displaystyle \omega _{m}}
)
ρ
c
=
3
h
2
8
π
g
;
{\displaystyle \rho _{c}={\frac {3h^{2}}{8\pi g}};}
Ω
m
≡
ρ
m
0
ρ
c
=
8
π
g
3
h
0
2
ρ
m
0
;
{\displaystyle \omega _{m}\equiv {\frac {\rho _{m_{0}}}{\rho _{c}}}={\frac {8\pi g}{3h_{0}^{2}}}\rho _{m_{0}};}
so
ρ
=
ρ
c
Ω
m
/
a
3
.
{\displaystyle \rho =\rho _{c}\omega _{m}/a^{3}.}
also, definition,
Ω
k
≡
−
k
c
2
(
a
0
h
0
)
2
{\displaystyle \omega _{k}\equiv {\frac {-kc^{2}}{(a_{0}h_{0})^{2}}}}
and
Ω
Λ
≡
Λ
c
2
3
h
0
2
,
{\displaystyle \omega _{\lambda }\equiv {\frac {\lambda c^{2}}{3h_{0}^{2}}},}
where subscript nought refers values today, ,
a
0
=
1
{\displaystyle a_{0}=1}
. substituting of friedmann equation @ start of section , replacing
a
{\displaystyle a}
a
=
1
/
(
1
+
z
)
{\displaystyle a=1/(1+z)}
gives
h
2
(
z
)
=
h
0
2
(
Ω
m
(
1
+
z
)
3
+
Ω
k
(
1
+
z
)
2
+
Ω
Λ
)
.
{\displaystyle h^{2}(z)=h_{0}^{2}\left(\omega _{m}(1+z)^{3}+\omega _{k}(1+z)^{2}+\omega _{\lambda }\right).}
matter- , dark energy-dominated universe
if universe both matter-dominated , dark energy- dominated, above equation hubble parameter function of equation of state of dark energy. now:
ρ
=
ρ
m
(
a
)
+
ρ
d
e
(
a
)
,
{\displaystyle \rho =\rho _{m}(a)+\rho _{de}(a),}
where
ρ
d
e
{\displaystyle \rho _{de}}
mass density of dark energy. definition, equation of state in cosmology
p
=
w
ρ
c
2
{\displaystyle p=w\rho c^{2}}
, , if substituted fluid equation, describes how mass density of universe evolves time, then
ρ
˙
+
3
a
˙
a
(
ρ
+
p
c
2
)
=
0
;
{\displaystyle {\dot {\rho }}+3{\frac {\dot {a}}{a}}\left(\rho +{\frac {p}{c^{2}}}\right)=0;}
d
ρ
ρ
=
−
3
d
a
a
(
1
+
w
)
.
{\displaystyle {\frac {d\rho }{\rho }}=-3{\frac {da}{a}}\left(1+w\right).}
if w constant, then
ln
ρ
=
−
3
(
1
+
w
)
ln
a
;
{\displaystyle \ln {\rho }=-3\left(1+w\right)\ln {a};}
ρ
=
a
−
3
(
1
+
w
)
.
{\displaystyle \rho =a^{-3\left(1+w\right)}.}
therefore, dark energy constant equation of state w,
ρ
d
e
(
a
)
=
ρ
d
e
0
a
−
3
(
1
+
w
)
{\displaystyle \rho _{de}(a)=\rho _{de0}a^{-3\left(1+w\right)}}
. if substituted friedman equation in similar way before, time set
k
=
0
{\displaystyle k=0}
, assumes spatially flat universe, (see shape of universe)
h
2
(
z
)
=
h
0
2
(
Ω
m
(
1
+
z
)
3
+
Ω
d
e
(
1
+
z
)
3
(
1
+
w
)
)
.
{\displaystyle h^{2}(z)=h_{0}^{2}\left(\omega _{m}(1+z)^{3}+\omega _{de}(1+z)^{3\left(1+w\right)}\right).}
if dark energy derives cosmological constant such introduced einstein, can shown
w
=
−
1
{\displaystyle w=-1}
. equation reduces last equation in matter-dominated universe section,
Ω
k
{\displaystyle \omega _{k}}
set zero. in case initial dark energy density
ρ
d
e
0
{\displaystyle \rho _{de0}}
given by
ρ
d
e
0
=
Λ
c
2
8
π
g
{\displaystyle \rho _{de0}={\frac {\lambda c^{2}}{8\pi g}}}
,
Ω
d
e
=
Ω
Λ
.
{\displaystyle \omega _{de}=\omega _{\lambda }.}
if dark energy not have constant equation-of-state w, then
ρ
d
e
(
a
)
=
ρ
d
e
0
e
−
3
∫
d
a
a
(
1
+
w
(
a
)
)
,
{\displaystyle \rho _{de}(a)=\rho _{de0}e^{-3\int {\frac {da}{a}}\left(1+w(a)\right)},}
and solve this,
w
(
a
)
{\displaystyle w(a)}
must parametrized, example if
w
(
a
)
=
w
0
+
w
a
(
1
−
a
)
{\displaystyle w(a)=w_{0}+w_{a}(1-a)}
, giving
h
2
(
z
)
=
h
0
2
(
Ω
m
a
−
3
+
Ω
d
e
a
−
3
(
1
+
w
0
+
w
a
)
e
−
3
w
a
(
1
−
a
)
)
.
{\displaystyle 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).}
other ingredients have been formulated recently.
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