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 $l_{H0} \sim 10^{28} \mbox{\it cm}$ 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 \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. \]