Difference between revisions of "Entropy of Expanding Universe"

\[ dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V} \right)_T = \frac{\rho + p} {T} \Rightarrow S = \frac{\rho + p}{T}V + f(T).\] As entropy is proportional to volume, then $f(T) = 0.$ For the photon gas $p = \rho/3$, therefore $$ s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha {T^3}. $$

$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$

$$ p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}. $$

$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.

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 $T_0 \approx 2.725K$ \[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]

Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{cm}$ and therefore \[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]

It is because the heat flow is absent in the homogeneous and isotropic Universe.

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 \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\]

\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]

By definition, \[ p = - \left( \frac{\partial E}{\partial V} \right)_S. \] If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\]

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 \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. \]

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 \[\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\]

\[\begin{array}{l} {\frac{dp}{dT} =\frac{1}{T} \left(\rho +p\right)\to \frac{d\rho }{dT} =\frac{1+w}{w} \frac{\rho }{T} ,} \\ {\frac{w}{w+1} \frac{d\rho }{\rho } =\frac{dT}{T} ,} \\ {T\propto \rho ^{\frac{w}{w+1} } } \end{array}\]

\[T\propto \rho ^{\frac{w}{w+1} } ,\quad \rho \propto a^{-3(1+w)} ,\to T\propto a^{-3w} \]

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.

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.