Cosmology Lecture 07

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The lower the temperature of the universe, the easier life gets: once the temperature of the universe drops below the rest energy of a particle, it is no longer often created in particle/antiparticle pairs.

Thermodynamics of a Fermi/Bose Gas

Phase Space is a 6-dimensional space of directions and linear momenta. It has volume:

${\displaystyle V={d^{3}xd^{3}p \over h^{3}}\,\!}$

The Distribution Function ${\displaystyle f({\hat {x}},{\hat {p}},t)}$ describes the number of particles in a particular ${\displaystyle {\hat {x}},{\hat {p}}}$ state at a given time. Thus, the number of particles in a phase space volume is:

${\displaystyle dN=f({\hat {x}},{\hat {p}},t){d^{3}xd^{3}p \over h}\cdot g\,\!}$

where ${\displaystyle g}$ is used to account for degeneracy. Reminder of Fermi/Bose statistics:

${\displaystyle f({\hat {x}},{\hat {p}},t)={1 \over e^{E-\mu \over kT}\pm 1}\,\!}$

Where ${\displaystyle \mu }$ is the chemical potential (usually 0 for us), + is for fermions, and - is for bosons. The chemical potential is usually 0 because we are talking about photons, and it can be shown that since photon number is not conserved, then ${\displaystyle \mu =0}$.

• # density in thermal equilibrium (integrating ${\displaystyle dN}$ above, dividing by volume, recall ${\displaystyle E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}}$):
${\displaystyle n={g \over h^{3}}\int {4\pi p^{2}dp \over e^{E(p) \over kT}\pm 1}\,\!}$
• Energy density (${\displaystyle dN}$, weighted by energy):
${\displaystyle u={g \over hA^{3}}\int _{0}^{\infty }{E(p){4\pi p^{2}dp \over e^{E(p) \over kT}\pm 1}}\,\!}$
• Entropy density ${\displaystyle s={S \over V}={1 \over V}{1 \over T}(U+PV-\mu N)}$. ${\displaystyle P}$ is pressure. For ${\displaystyle \mu =0}$:
${\displaystyle s={u+P \over T}\,\!}$

Without proving, we’ll state ${\displaystyle P=\langle {p^{2}c^{2} \over 3E}\rangle }$. The proof comes from kinetic theory. Thus:

${\displaystyle s={g \over h^{3}}\int _{o}^{\infty }{{4\pi p^{2}dp \over e^{E(p) \over kT}\pm 1}\left({E(p) \over T}+{p^{2}c^{2} \over 3ET}\right)}\,\!}$

So we have ways to calculate 3 very useful quantities for the early universe. We can take two useful limits of these quantities:

Ultra-relativistic (${\displaystyle kT\gg mc^{2}}$) Particles in the Early Universe

In this limit, particles are effectively massless (their energy is dominated by their motion, ${\displaystyle E={\sqrt {m^{2}c^{4}+p^{2}c^{2}}}\approx pc}$).

• "A." Bosons (1) The # density of relativistic bosons is:
{\displaystyle {\begin{aligned}n&={g \over h^{3}}\int _{0}^{\infty }{4\pi p^{2}dp \over e^{pc \over kT}\pm 1}\\&={4\pi g \over (2\pi )^{2}}\left({kT \over \hbar c}\right)^{3}\int _{0}^{\infty }{y^{2}dy \over e^{y}\pm 1}\\\end{aligned}}\,\!}

where ${\displaystyle y\equiv {pc \over kT}}$. This integral is calculable, and is the definition of the Reimann-Zeta function:

{\displaystyle {\begin{aligned}\zeta (n)&\equiv {1 \over \Gamma (n)}\int _{0}^{\infty }{y^{n-1}dy \over e^{y}-1}\\\int _{0}^{\infty }{y^{2}dy \over e^{y}-1}=\Gamma (3)\zeta (3)&\approx 2\cdot 1.202\\\int _{0}^{\infty }{y^{3}dy \over e^{y}-1}=\Gamma (4)\zeta (4)&={\pi ^{4} \over 15}\\\end{aligned}}\,\!}

So getting back to the # density of bosons:

${\displaystyle {n_{B}={g \over \pi ^{2}}\zeta (3)\left({kT \over \hbar c}\right)^{3}}\,\!}$

(2) The energy density of relativistic bosons is:

{\displaystyle {\begin{aligned}u&={g \over h^{3}}\int _{0}^{\infty }{pc{4\pi p^{2}dp \over e^{pc \over kT}\pm 1}}\\&={4\pi g \over (2\pi )^{3}}{(kT)^{4} \over (\hbar c)^{3}}\int _{0}^{\infty }{y^{3}dy \over e^{y}\pm 1}\\\end{aligned}}\,\!}
${\displaystyle {u_{B}={\pi ^{2} \over 30}g{(kT)^{4} \over (\hbar c)^{3}}}\,\!}$

For photons, ${\displaystyle g=2}$. We can calculate a flux related to their energy density:

${\displaystyle F={1 \over 4}u_{\gamma }c={\pi ^{2} \over 60}{K^{4} \over \hbar ^{3}c^{3}}T^{4}=\sigma =5.67\cdot 10^{-8}{W \over m^{2}K^{4}}\,\!}$

(3) The entropy density of relativistic bosons is:

${\displaystyle {s_{B}={u_{B}+P_{B} \over T}={4 \over 3}{u_{B} \over T}={2\pi ^{2} \over 45}\left({kT \over \hbar c}\right)^{3}K=3.602n_{B}K}\,\!}$

Recall that for radiation, the energy density ${\displaystyle \rho \propto a^{-3(1+w)}\propto a^{-4}}$. Since ${\displaystyle u\propto \rho \propto T^{4}\Rightarrow T\propto {1 \over a}}$. This is only true if ${\displaystyle g}$ is fixed.

• "B." Fermions: here’s a trick:
${\displaystyle {1 \over e^{y}+1}={1 \over e^{y}-1}-{2 \over e^{2y}-1}\,\!}$

This is an identity. (1) # density. Since ${\displaystyle y\equiv {pc \over kT}}$, then ${\displaystyle 2y={pc \over K({T \over 2})}}$, then using our identity:

{\displaystyle {\begin{aligned}n_{F}(T)&={g_{F} \over g_{B}}[n_{B}(T)-2n_{B}({T \over 2})]\\&={g_{F} \over g_{B}}n_{B}(T)[1-2({\frac {1}{2}})^{3}]\\\end{aligned}}\,\!}
${\displaystyle {n_{F}(T)={g_{F} \over g_{B}}{3 \over 4}n_{B}(T)}\,\!}$

(2) Energy density:

{\displaystyle {\begin{aligned}u_{F}(T)&={g_{F} \over g_{B}}[u_{B}(t)-2u_{B}({T \over 2})]\\&={g_{F} \over g_{B}}u_{B}(t)[1-2({\frac {1}{2}})^{4}]\\\end{aligned}}\,\!}
${\displaystyle {u_{F}(T)={g_{F} \over g_{B}}{7 \over 8}u_{B}(T)}\,\!}$

(3) Entropy density:

${\displaystyle s_{F}(T)={4 \over 3}{u_{F} \over T}\,\!}$
${\displaystyle {s_{F}(T)={g_{F} \over g_{B}}{7 \over 8}s_{B}(T)}\,\!}$

When doing computations like this for the universe, we’ll have to keep track of the ${\displaystyle g}$’s of each particle’s contribution. To aid us in doing this, we’ll define an effective degeneracy:

${\displaystyle {g^{*}=\sum _{i=bosons}{g_{i}}+{7 \over 8}\sum _{i=fermions}{g_{i}}}\,\!}$

Using this definition, the total energy density ${\displaystyle u_{tot}\propto g^{*}T^{4}}$.

Non-Relativistic (${\displaystyle kT\ll mc^{2}}$) Particles In the Early Universe

In this limit, a particle’s energy is dominated by its rest mass: ${\displaystyle E={\sqrt {m^{2}c^{4}+p^{2}c^{2}}}\approx mc^{2}(1+{\frac {1}{2}}{p^{2}c^{2} \over m^{2}c^{4}}+\dots )}$. Then in general, the denominator inside the integral over momentum simplifies:

${\displaystyle e^{E \over kT}\pm 1\approx e^{E \over kT}\approx e^{{mc^{2} \over kT}+{p^{2} \over 2kTm}}\,\!}$

So dropping the “${\displaystyle \pm 1}$” in the denominator:

{\displaystyle {\begin{aligned}n&\approx {g \over h^{3}}\int _{0}^{\infty }{4\pi p^{2}dpe^{-{mc^{2} \over kT}}e^{-{p^{2} \over 2mkT}}}\\&\approx {4\pi g \over h^{3}}e^{-{mc^{2} \over kT}}(2mkT)^{3 \over 2}\int _{0}^{\infty }{dy\cdot y^{2}e^{-y^{2}}}\\\end{aligned}}\,\!}
${\displaystyle {n\approx {g \over h^{3}}\left({mkT \over 2\pi }\right)^{3 \over 2}e^{-{mc^{2} \over kT}}}\,\!}$

This the the Maxwell-Boltzmann distribution if we put ${\displaystyle \mu }$, the chemical potential, back in the exponent with the energy (${\displaystyle mc^{2}}$). Thus, the # density is exponentially suppressed for ${\displaystyle kT\ll mc^{2}}$. For example, at ${\displaystyle kT\ll 1GeV\sim m_{proton}}$, if we are in thermal equilibrium, then the neutron-to-proton ratio is:

${\displaystyle {n_{n} \over n_{p}}\approx e^{-{(m_{n}-m_{p})c^{2} \over kT}}\approx e^{-1.293MeV \over kT}\,\!}$

This determines the Hydrogen/Helium ratio!