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Cosmic Background Radiation and Decoupling
------------------------------------------
Photons get scattered in a plasma (this is the
reason why we can't see through the sun). When
looking into space the further the distance the
further back in time we see and the higher the
temperature. Eventually, we get to point where
temperature is high enough for ionization (~3000K)
and the universe becomes opaque. This is called
the SURFACE OF LEAST SCATTERING. The reason why
we don't see this glow in the sky is that the
boundary is moving very fast and the waves are
redshifted to the point where their energy and
temperature are very low (~3K). Instead, what
we see is the COSMIC BACKGROUND RADIATION. This
is the radiation from the SLS.
Blackbody Radiation
-------------------
CMB radiation has a blackbody spectrum described
by the Planck Radiation Law.
I(ν,V) = (8πh/c^{3})(ν^{3}/e^{hν/kT} - 1) ... 1.
where I is the energy per unit volume (energy
density) per unit frequency.
The peak is at λ ~ hc/k_{B}T
The temperature and density variation of the
CMB (δT/T and δρ/ρ) is approximately equal to 10^{-5}.
Decoupling Temperature and Time
-------------------------------
Prior to the time of decouping, highly energetic
photons prevented electrons and protons from
forming neutral atoms of Hydrogen (and later
Helium). This can be represented by:
H + γ <--> p + e
As the universe cooled, the photons no longer
had enough energy to cause ionization, and
neutral atoms started to form. This caused
a rapid reduction of free electrons, and photons
no were no longer scattered. The result is that
the very first elements starred to form and, in
the process, the universe becomes transparent.
This process is referred to as DECOUPLING. The
question is at what temperature and time did this
occur?
From Plancks law, the photon number density is
found by dividing equation 1. by the energy, hν,
and integrating from 0 to ∞. Thus,
n_{γ} = (8π/c^{3})∫(ν^{2}/e^{hν/kT} - 1)dν
This is a tricky integral to solve so we shall
just state the result:
n_{γ} = 0.243(k_{B}T/hc)^{3} ... 2.
To get the total energy density we integrate
equation 1. from 0 to ∞.
I = (8πh/c^{3})∫(ν^{3}/e^{hν/kT} - 1)dν
This is another tricky integral that yields:
E_{Total} = (π^{2}/15)(k_{B}T)^{4}/(hc)^{3}
= 0.658(k_{B}T)^{4}/(hc)^{3} ... 3.
This can also be written as:
E_{Total} = αT^{4} = 4σ/c where α is the radiation
constant and σ is Stephan's constant.
Dividing equation 3. by equation 2. gives the
energy of a photon from the CMB:
E_{γ} ~ 2.7k_{B}T
We next consider the SAHA equation which describes
the degree of ionization of a plasma as a function
of the temperature, density, and ionization energies
of the atoms. The Saha equation follows directly
from the Boltzmann distribution.
(1 - X)/X = n_{p}(k_{B}Tm_{e}/2πh^{2})^{-3/2}exp(Q/k_{B}T)
Where X = n_{p}/(n_{p} + n_{H}) = n_{e}/(n_{p} + n_{H}) and satisfies
the condition of charge neutrality.
When X = 1 the gas is fully ionized. When
X = 0 the gas is neutral. We can also write:
n_{H} = (X - 1)n_{p}/X
Now define the baryon to photon ratio, η:
η = n_{B}/n_{γ} = n_{p}/Xn_{γ} (from n_{p}/[n_{γ}n_{p}/(n_{p} + n_{H})] = n_{B}/n_{γ})
∴ n_{p} = ηXn_{γ}
Substituting into equation 2. gives:
n_{p} = 0.243ηX(k_{B}T/hc)^{3}
According to present observational measurements
of the baryonic density, Ω_{m,0}, from the CMB, n_{γ}
is approximately equal to 1.7 x 10^{9}. The Saha
equation becomes:
(1 - X)/X = 0.243ηX(k_{B}T/hc)^{3}(k_{B}Tm_{e}/2πh^{2})^{-3/2}exp(Q/k_{B}T)
Which simplifies to:
(1 - X)/X^{2} ~ 3.84η(k_{B}T/m_{e}c^{2})^{3/2}exp(Q/k_{B}T)
If we assume X = 0.5 and Q for Hydrogen to be
13.6 eV (2.2 x 10^{-18} J) and solve for T we get
that the temperature at the time of decoupling
was about 3000K.
Now consider the equation for redshift:
z + 1 = T_{E}/T_{T} = a_{T}/a_{E} = λ_{T}/λ_{E} = ρ_{E}/ρ_{T} = t_{T}^{2/3}/t_{E}^{2/3}
where T = Today, E = Emitted.
Now, T_{T} = 2.7 K (the temperature of the CMB)
and T_{E} ~ 3000 K (the temperature at the time
of decoupling), so for the mass dominated
period when decoupling occurred, we can write:
z + 1 = T_{E}/T_{T} = 1100 = t_{T}^{2/3}t_{E}^{2/3}
∴ z + 1 = t_{T}/t_{E} = 1100^{3/2} ~ 37,000
∴ t_{E} ~ 365,000 years if the universe is about
13.5 x 10^{9} years old. Therefore, decoupling
occurred roughly 350,000 years after the Big
Bang.