Difference between revisions of "Thermodynamical Properties of Elementary Particles"

The energy density equals to $$ \rho = \left\{ \begin{array}{c} \frac{\pi^2}{30}gT^4 \quad \; for\;bosons\\ \frac{7}{8}\frac{\pi^2}{30}gT^4 \quad for \;fermions \end{array} \right. $$ Here $g$ equals to the number of internal degrees of freedom for the particle. For photon (with spin $S=1$) $g=2$ due to absence of longitudinal polarization for the zero-mass particles. The number density is the following: \[ n = \left\{ \begin{array}{c} \frac{\zeta (3)}{\pi ^2 }gT^3 \quad \;for\;bosons \\ \frac{3} {4}\frac{\zeta (3)}{\pi ^2 }gT^3 \quad \; for\;fermions \end{array} \right. \] Here $\zeta (x)$ is the Riemann $\zeta$-function; $\zeta (3) \approx 1.202$.

\[ \frac{n_i}{n_\gamma} = \left\{ \begin{gathered} \frac{1}{2}g_i \quad for\; bosons\\ \frac{3}{8}g_i \quad for\; fermions\\ \end{gathered} \right.;\quad \frac{\rho_i}{\rho_\gamma} = \left\{ \begin{gathered} \frac{1}{2}g_i \quad for\; bosons\\ \frac{7}{16}g_i \quad for\; fermions\\ \end{gathered} \right. \] \[ \begin{gathered} \left(\frac{n_i }{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{\nu_{\mu ,e}} = \left( \frac{3}{8},\frac{7}{16} \right);\;\;\left( \frac{n_i}{n_\gamma}, \frac{\rho_i}{\rho_\gamma} \right)_{e^\pm} = \left( \frac{3}{4},\frac{7}{8} \right);\\ \left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_\pi = \left( \frac{1}{2},\frac{1}{2} \right);\;\;\left( \frac{n_i}{n_\gamma},\frac{\rho_i}{\rho_\gamma} \right)_{n,p} = \left( \frac{3}{4},\frac{7}{8} \right) \\ \end{gathered} \]

Average energy per particle reads$\left\langle \varepsilon \right\rangle = \rho /n$. For $T \gg m$ \[ \left\langle \varepsilon \right\rangle = \left\{ \begin{gathered} \frac{\pi ^2}{30\zeta (3)}T\quad \; for\; bosons\\ \frac{7\pi ^4}{180\zeta(3)}T\quad for\; fermions\\ \end{gathered} \right. \] In the non-relativistic limit $(T \ll m)$ one obtains the following: \[ \left\langle \varepsilon \right\rangle = m + \frac{3}{2}T \approx m \]

For quark one obtains the following: \[ g = 2(\, spin) \times 3(\, particle) \times 2(\, particles + \, antiparticle) = 12 \]

In the considered case one gets: \[ s = \frac{\rho + p}{T};\quad p = \frac{1}{3}\rho \] Using the results of problem of the present Chapter one obtains the following: \[ s = \frac{2\pi ^2}{45}g^* T^3, \] where $g^* = 1$ for bosons and $g^* = 7/8$ for fermions.

\[ \rho = \sum\limits_{i = bosons} g_i \frac{\pi^2}{30} T^4 + \sum\limits_{j = fermions} \frac{7}{8}g_j\frac{\pi^2}{30} T^4 \frac{\pi^2}{30}g^* T^4, \] where $g^* $ is the effective number of degrees of freedom: \[ g^* = \sum\limits_{i = bosons} g_i + \frac{7}{8}\sum\limits_{j = fermions}g_j. \]

\[ g^* = \sum\limits_{i = bosons} {g_i \left(\frac{T_i}{T} \right)^4 + \frac{7}{8}\sum\limits_{j = fermions}g_i \left( \frac{T_j}{T}\right)^4 \]

At temperature $1\, TeV$ all the Standard Model particles are relativistic (see table). Therefore \[ g^* = 28 + \frac{7}{8} \times 90 = 106.75. \]

At temperature $T \approx 0.2\, GeV$ quarks and gluons transform into colorless hadrons (protons and neutrons). Therefore \[ \Delta g = 16\,(gluons) + 12 \times 6\,(quarks) - 8\,(nucleons) = 80. \]

Represent the expression for energy density of relativistic particles in the following form: \[ \rho = \left( \sum\limits_{i = bosons} \frac{g_i}{2} + \sum\limits_{j = fermions} \frac{7}{16}g_j \right)\rho _\gamma. \] At the temperature under consideration the relativistic particles are presented by photons, electron, positrons and neutrinos. Therefore \[ \rho = \left( 1 + \frac{7} {4} + \frac{7}{4} \right)\rho _\gamma = \frac{9}{2}\rho _\gamma = \frac{9}{2}\alpha T^4, \] where the radiation constant \[\alpha = \frac{\pi^2}{15}\frac{k^4}{(\hbar c)^3} \approx 7.56 \cdot 10^{-16}\frac{ J}{ m^3K^4}.\]

Unlike the case considered in the previous problem, now only photons and neutrinos remain relativistic, because $10^8 K \sim 10^{ - 2} \mbox{\it MeV}$, and electron mass is $0.511 \mbox{\it MeV}$ is considerably greater. Therefore in the case under consideration one obtains the following: \[ \rho = \left( 1 + \frac{7}{4} \right)\rho _\gamma = \frac{11}{4}\rho _\gamma \Rightarrow \frac{\rho\left(T_2 = 10^{10} K \right)}{\rho \left( T_1 = 10^8 K \right)} = \frac{18} {11}\left(\frac{T_2}{T_1}\right)^4\simeq1.6\cdot10^8.\]

For a relativistic Fermi-gas the Fermi energy reads the following $E_F = \hbar c\left( 3\pi ^2 n \right)^{1/3} ;$ $\hbar c \simeq 197.3 \, MeV\cdot \, fm$, $n \simeq 300 \, cm^{-3}$, so $$E_F \simeq 0.4 \times 10^{-3} \, eV.$$