# Cosmology Lecture 07

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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:

$V={d^{3}xd^{3}p \over h^{3}}\,\!$ The Distribution Function $f({\hat {x}},{\hat {p}},t)$ describes the number of particles in a particular ${\hat {x}},{\hat {p}}$ state at a given time. Thus, the number of particles in a phase space volume is:

$dN=f({\hat {x}},{\hat {p}},t){d^{3}xd^{3}p \over h}\cdot g\,\!$ where $g$ is used to account for degeneracy. Reminder of Fermi/Bose statistics:

$f({\hat {x}},{\hat {p}},t)={1 \over e^{E-\mu \over kT}\pm 1}\,\!$ Where $\mu$ 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 $\mu =0$ .

• # density in thermal equilibrium (integrating $dN$ above, dividing by volume, recall $E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}$ ):
$n={g \over h^{3}}\int {4\pi p^{2}dp \over e^{E(p) \over kT}\pm 1}\,\!$ • Energy density ($dN$ , weighted by energy):
$u={g \over hA^{3}}\int _{0}^{\infty }{E(p){4\pi p^{2}dp \over e^{E(p) \over kT}\pm 1}}\,\!$ • Entropy density $s={S \over V}={1 \over V}{1 \over T}(U+PV-\mu N)$ . $P$ is pressure. For $\mu =0$ :
$s={u+P \over T}\,\!$ Without proving, we’ll state $P=\langle {p^{2}c^{2} \over 3E}\rangle$ . The proof comes from kinetic theory. Thus:

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

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

• "A." Bosons (1) The # density of relativistic bosons is:
{\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 $y\equiv {pc \over kT}$ . This integral is calculable, and is the definition of the Reimann-Zeta function:

{\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:

${n_{B}={g \over \pi ^{2}}\zeta (3)\left({kT \over \hbar c}\right)^{3}}\,\!$ (2) The energy density of relativistic bosons is:

{\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}}\,\! ${u_{B}={\pi ^{2} \over 30}g{(kT)^{4} \over (\hbar c)^{3}}}\,\!$ For photons, $g=2$ . We can calculate a flux related to their energy density:

$F={1 \over 4}u_{\gamma }c={\pi ^{2} \over 60}{K^{4} \over \hbar ^{3}c^{3}}T^{4}=\sigma =5.67\cdot 10^{-8}{W \over m^{2}K^{4}}\,\!$ (3) The entropy density of relativistic bosons is:

${s_{B}={u_{B}+P_{B} \over T}={4 \over 3}{u_{B} \over T}={2\pi ^{2} \over 45}\left({kT \over \hbar c}\right)^{3}K=3.602n_{B}K}\,\!$ Recall that for radiation, the energy density $\rho \propto a^{-3(1+w)}\propto a^{-4}$ . Since $u\propto \rho \propto T^{4}\Rightarrow T\propto {1 \over a}$ . This is only true if $g$ is fixed.

• "B." Fermions: here’s a trick:
${1 \over e^{y}+1}={1 \over e^{y}-1}-{2 \over e^{2y}-1}\,\!$ This is an identity. (1) # density. Since $y\equiv {pc \over kT}$ , then $2y={pc \over K({T \over 2})}$ , then using our identity:

{\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}}\,\! ${n_{F}(T)={g_{F} \over g_{B}}{3 \over 4}n_{B}(T)}\,\!$ (2) Energy density:

{\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}}\,\! ${u_{F}(T)={g_{F} \over g_{B}}{7 \over 8}u_{B}(T)}\,\!$ (3) Entropy density:

$s_{F}(T)={4 \over 3}{u_{F} \over T}\,\!$ ${s_{F}(T)={g_{F} \over g_{B}}{7 \over 8}s_{B}(T)}\,\!$ When doing computations like this for the universe, we’ll have to keep track of the $g$ ’s of each particle’s contribution. To aid us in doing this, we’ll define an effective degeneracy:

${g^{*}=\sum _{i=bosons}{g_{i}}+{7 \over 8}\sum _{i=fermions}{g_{i}}}\,\!$ Using this definition, the total energy density $u_{tot}\propto g^{*}T^{4}$ .

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

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

$e^{E \over kT}\pm 1\approx e^{E \over kT}\approx e^{{mc^{2} \over kT}+{p^{2} \over 2kTm}}\,\!$ So dropping the “$\pm 1$ ” in the denominator:

{\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}}\,\! ${n\approx {g \over h^{3}}\left({mkT \over 2\pi }\right)^{3 \over 2}e^{-{mc^{2} \over kT}}}\,\!$ This the the Maxwell-Boltzmann distribution if we put $\mu$ , the chemical potential, back in the exponent with the energy ($mc^{2}$ ). Thus, the # density is exponentially suppressed for $kT\ll mc^{2}$ . For example, at $kT\ll 1GeV\sim m_{proton}$ , if we are in thermal equilibrium, then the neutron-to-proton ratio is:

${n_{n} \over n_{p}}\approx e^{-{(m_{n}-m_{p})c^{2} \over kT}}\approx e^{-1.293MeV \over kT}\,\!$ This determines the Hydrogen/Helium ratio!