Problem 5.1.1: Chemical potential
Show that the photon gas in thermal equilibrium has zero chemical potential.
Chapter 5 · Section 1 of 6
Show that the photon gas in thermal equilibrium has zero chemical potential.
In the next 4 problems we consider some volume $V$, filled with black-body radiation of temperature $T$.
Find the number of photons with frequencies in the interval $[\omega ,\omega +d\omega]$.
Find the total number of photons.
What is this number for a gas oven at room temperature and at maximum heat?
What is the energy of photons with frequencies in the interval $[ \omega ,\omega +d\omega]$?
Calculate the free energy, entropy and total energy of black-body radiation.
Calculate the thermal capacity of black-body radiation.
Find the pressure of black-body radiation and construct its state equation.
Find the adiabatic equation for the black-body radiation.
Why CMB cannot be used to warm up food like in the microwave oven?
The relic radiation, or CMB, corresponds to the black-body radiation with temperature $T_{CMB}=2.725\:K.$ According to the main principle of thermodynamics, heat cannot transfer from a less heated body to more heated one, and thus the body (food in our case), which initially had temperature $T_0>T_{CMB},$ will emit more energy in the environment then absorb back, until the equilibrium installs with the CMB radiation at temperature $T_{CMB}.$
The binding energy of electron in the hydrogen atom equals to $13.6\ eV$. What is the temperature of the Planck distribution with this average photon energy?
The Planck distribution has maximum at frequency $\omega_m = 2.822 kT/\hbar.$ Then one obtains $ kT = 13.6/2.822 = 4.82\mbox{eV}$ and $T\approx 5.6 \cdot {10^4}\mbox{K}$.
Find in power-law cosmology (see Chapter 3) time dependence of the CMB temperature $T(t)$.
In power-law cosmology the scale factor $a(t)$and the CMB temperature $T(t)$ are related through the relation \[\frac{T_{0} }{T} =\frac{a}{a_{0} } =\left(\frac{t}{t_{0} } \right)^{\alpha } =\left(\frac{t}{t_{0} } \right)^{1/1+q} \] where $q$is the deceleration parameter. Consequently, \[T(t)=T_{0} \left(\frac{t}{t_{0} } \right)^{-\alpha } =T_{0} \left(\frac{t}{t_{0} } \right)^{-1/1+q} \]