Problem 6.3.1
Transform the energy conditions for the flat Universe to conditions for the entropy density (see [1])
solution
Chapter 6
Section 3 of 8
Section 3 of 8
Chapter 6 · Section 3 of 8
Entropy of Expanding Universe
Problem 6.3.2
Find the entropy density for the photon gas.
solution
\[ dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V} \right)_T = \frac{\rho + p} {T} \Rightarrow S = \frac{\rho + p}{T}V + f(T).\] As entropy is proportional to volume, then $f(T) = 0.$ For the photon gas $p = \rho/3$, therefore $$ s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha {T^3}. $$
Problem 6.3.3
Use the result of the previous problem to derive an alternative proof of the fact that $aT=const$ in the adiabatically expanding Universe.
solution
$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$
Problem 6.3.4
Find the adiabatic index for the CMB.
solution
$$ p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}. $$
Problem 6.3.5
Show that the ratio of CMB entropy density to the baryon number density $s_\gamma/n_b$ remains constant during the expansion of the Universe.
solution
$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.
Problem 6.3.6
Estimate the current entropy density of the Universe.
solution
The current entropy of the Universe is determined by the CMB photons. A rough estimate neglects the contribution of other particles, then one obtains \[s = 2\frac{2\pi ^2}{45}T_0^3.\] For $T_0 \approx 2.725K$ \[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]
Problem 6.3.7
Estimate the entropy of the observable part of the Universe.
solution
Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{cm}$ and therefore \[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]
Problem 6.3.7b
Why is the expansion of the Universe described by the Friedman equations adiabatic?
solution
It is because the heat flow is absent in the homogeneous and isotropic Universe.
Problem 6.3.8
Show that entropy is conserved during the expansion of the Universe described by the Friedman equations.
solution
The first law of thermodynamics \[ dE + pdV = TdS \] can be transformed into the form \[ a^3 \left[\frac{d\rho}{dt} + 3H(\rho + p)\right] =T\frac{dS}{dt}.\] Then with the conservation equation \[\frac{d\rho}{dt} + 3H(\rho + p) = 0,\] which follows from the Friedmann equation, one obtains \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\]
Problem 6.3.9
Show that the entropy density behaves as $s\propto a^{-3}$.
solution
\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]
Problem 6.3.10
Using only thermodynamical considerations, show that if the energy density of some component is $\rho=const$ than the state equation for that component reads $p=-\rho$.
solution
By definition, \[ p = - \left( \frac{\partial E}{\partial V} \right)_S. \] If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\]
Problem 6.3.11
Show that the product $aT$ is an approximate invariant in the Universe dominated by relativistic particles.
solution
Consider a volume $V$, filled by photons and let it expands with the same rate as the whole Universe, i.e. $V \propto a^3 (t)$. For radiation $\rho = \alpha T^4 $, $p = \rho/3.$ The first law of thermodynamics applied to the expanding Universe with no heat flow reads \[ \frac{dE}{dt} = - p\frac{dV}{dt}. \] As $E = \rho V$, then \[ \frac{1}{T}\frac{dT}{dt} = - \frac{1}{3V}\frac{dV}{dt},\] or equivalently \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. \]
Problem 6.3.12
Show that for a system of particles in thermal equilibrium, \[\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\]
solution
The entropy of a system of particles in thermal equilibrium, is a function $S=S(V,T)$ such that \[dS=\frac{1}{T} \left(dE+pdV\right)=\frac{1}{T} \left[d(\rho V)+pdV\right]=\frac{1}{T} \left[V\frac{d\rho }{dT} dT+(p+\rho )dV\right]\] Hence \[\frac{\partial S}{\partial V} =\frac{1}{T} \left(\rho +p\right),\quad \frac{\partial S}{\partial V} =\frac{\partial S}{\partial T} =\frac{V}{T} \frac{d\rho }{dT} \] These derivatives must satisfy \[\frac{\partial }{\partial T} \left[\frac{1}{T} \left(\rho +p\right)\right]=\frac{\partial }{\partial V} \left(\frac{V}{T} \frac{d\rho }{dT} \right)\] Finally obtain \[\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\]
Problem 6.3.12b
Show that for the substance with the equation $p=w\rho$ \[T\propto \rho ^{\frac{w}{w+1} } \]
solution
\[\begin{array}{l} {\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\to \frac{d\rho }{dT} =\frac{1+w}{w} \frac{\rho }{T} ,} \\ {\frac{w}{w+1} \frac{d\rho }{\rho } =\frac{dT}{T} ,} \\ {T\propto \rho ^{\frac{w}{w+1} } } \end{array}\]
Problem 6.3.13
Show that for the substance with the equation $p=w\rho $ \[T\propto a^{-3w} \]
solution
\[T\propto \rho ^{\frac{w}{w+1} } ,\quad \rho \propto a^{-3(1+w)} ,\to T\propto a^{-3w} \]
Problem 6.3.14
Obtain a generalization of Stefan-Boltzmann law for cosmological fluid, described by generalized polytropic equation of state $p=w\rho +k\rho ^{1+1/n} $.
solution
For the polytropic equation of state the thermodynamical equation \[\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\] becomes \[\left(w+k(1+1/n)\rho ^{1/n} \right)\frac{d\rho }{dt} =\frac{1}{T} \left(w+1+k\rho ^{1/n} \right)\rho \] This equation can be integrated into \[T=T_{*} \left[1\pm \left(\rho /\rho _{*} \right)^{1/n} \right]^{\left(w+n+1\right)/\left(w+1\right)} \left(\rho /\rho _{*} \right)^{w/\left(w+1\right)} \] where $T_{*} $is a constant of integration, the upper sign corresponds to$k>0$, and the lower sign corresponds to$k<0$. This relation can be viewed as a generalization of Stefan-Boltzmann law.
Problem 6.3.15
Show that the velocity of sound for cosmological fluid described by generalized polytropic equation of state $p=w\rho +k\rho ^{1+1/n} $ vanishes at the point where the temperature is extremum.
solution
For the generalized polytropic equation of state the velocity of sound is given by \[\begin{array}{l} {c_{s}^{2} =\frac{dp}{d\rho } =w\pm \left(w+1\right)\frac{n+1}{n} \left(\frac{\rho }{\rho _{*} } \right)^{1/n} ,} \\ {\rho _{*} =\left[\frac{w+1}{\left|k\right|} \right]^{n} } \end{array}\] The upper sign corresponds to$k>0$, and the lower sign corresponds to$k<0$. As we have seen in the previous problem \[T=T_{*} \left[1\pm \left(\rho /\rho _{*} \right)^{1/n} \right]^{\left(w+n+1\right)/\left(w+1\right)} \left(\rho /\rho _{*} \right)^{w/\left(w+1\right)} \] The extremum of temperature (when it exists) is located at \[\rho _{e} =\rho _{*} \left[\mp \frac{wn}{\left(w+1\right)(n+1)} \right]^{n} \] It is easy to see that the velocity of sound vanishes at the point where the temperature is extremum.