Cosmology Assignment 3, May 2008
Joachim Mortensen
14. maj 2008
Textbook: Barbara Ryden, An Introduction to Cosmology, 2003
4.5
Joachim Mortensen
Computation of an Equation-of-State Parameter
Wave-particle duality ⇒ particle with momentum has de Broglie wavelength λ = h/p. Expansion of the
Universe ⇒ λ ∝ a ⇒ λ = ka for some constant k, and thus p ∝ a−1 . The total energy of the gas of
particles (all of same mass m and momentum p) can be written as ε = nE where n is the number density
of particles, and E the energy per particle:
"
2 4
2 2 1/2
E = [m c + p c ]
2 4
= m c +
hc
k
2
1
a2
#1/2
(1)
To compute the equation-of-state parameter w of the gas of particles, I need the equation-of-state:
P = wε,
(2)
ȧ
ε̇ + 3 (ε + P ) = 0.
a
(3)
and the fluid equation:
Solving the latter for P :
1a
− ε.
(4)
3 ȧ
The last thing needed before inserting everything into the equation-of-state (solved for w), is to calculate
the time-derivative ε̇ of ε:
P = −ε̇
"
2 #1/2
hc
1
d
2 4
m c +
ε̇ = n
dt
k
a2
ε̇ =
n
h
2 m2 c4 +
hc 2 1
k
a2
(5)
(hc)2 −2ȧ
i1/2 k 2 a3 ,
(6)
giving:
ε̇ = h
−n(hc)2 ȧ
.
i1/2
hc 2 1
2
3
2
4
k a
m c + k
a2
(7)
Substituting into the equation-of-state, things gets a bit lumpy:
w=
−
w=
w=
−ε̇ 13 ȧa − ε
P
=
ε
ε
−n(hc)2 ȧ
i1/2
2
h
m2 c4 +( hc
k )
h
n m2 c4 +
1
a2
(8)
!
k2 a3
i1/2
hc 2 1
k
a2
1a
3 ȧ
−1
(hc)2 ȧa
(hc)2
i
−1
−
1
=
2 1
3 [k 2 m2 c4 a2 + (hc)2 ]
2 3
3 m2 c4 + hc
k
a2 k a ȧ
h
(9)
(10)
The two limits become:
lim w =
a→0
1
2
−1=−
3
3
(highly relativistic),
(11)
(nonrelativistic).
(12)
and
lim w = 0 − 1 = −1
a→∞
Not the expected values. I do not understand why?!
Cosmology Assignment 3, May 2008
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