Difference between revisions of "Entropy of Expanding Universe"
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">\[ |
− | \[ | + | |
dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V} | dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V} | ||
\right)_T = \frac{\rho + p} {T} \Rightarrow | \right)_T = \frac{\rho + p} {T} \Rightarrow | ||
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s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha | s = \frac{S}{V} = \frac{4}{3}\frac{\rho}{T} = \frac{4}{3}\alpha | ||
{T^3}. | {T^3}. | ||
− | $$ | + | $$</p> |
− | + | ||
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$</p> |
− | $$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$ | + | |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">$$ |
− | $$ | + | |
p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}. | p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}. | ||
− | $$ | + | $$</p> |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.</p> |
− | + | ||
− | $$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe. | + | |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">The current entropy of the Universe is determined by the CMB photons. A rough estimate neglects the contribution of other particles, then one obtains |
− | + | ||
− | The current entropy of the Universe is determined by the CMB photons. A rough estimate neglects the contribution of other particles, then one obtains | + | |
\[s = 2\frac{2\pi ^2}{45}T_0^3.\] | \[s = 2\frac{2\pi ^2}{45}T_0^3.\] | ||
For $T_0 \approx 2.725K$ | For $T_0 \approx 2.725K$ | ||
− | \[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\] | + | \[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]</p> |
− | </p> | + | |
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<div style="width:100%;" class="NavContent"> | <div style="width:100%;" class="NavContent"> | ||
− | <p style="text-align: left;"> | + | <p style="text-align: left;">Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{\it cm}$ and therefore |
− | Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{\it cm}$ and therefore | + | \[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]</p> |
− | \[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\] | + | |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">It is because the heat flow is absent in the homogeneous and isotropic Universe.</p> |
− | It is because the heat flow is absent in the homogeneous and isotropic Universe. | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">The first law of thermodynamics |
− | The first law of thermodynamics | + | |
\[ | \[ | ||
dE + pdV = TdS | dE + pdV = TdS | ||
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a^3 \left[\frac{d\rho}{dt} + 3H(\rho + p)\right] =T\frac{dS}{dt}.\] | a^3 \left[\frac{d\rho}{dt} + 3H(\rho + p)\right] =T\frac{dS}{dt}.\] | ||
Then with the conservation equation | Then with the conservation equation | ||
− | \[\frac{d\rho}{dt} + 3H(\rho + p) = 0,\] which follows from the Friedmann equation, one obtains \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\] | + | \[\frac{d\rho}{dt} + 3H(\rho + p) = 0,\] which follows from the Friedmann equation, one obtains \[ \frac{dS}{dt} = 0 \Rightarrow S = const.\]</p> |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]</p> |
− | \[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\] | + | |
− | </p> | + | |
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− | <p style="text-align: left;"> | + | <p style="text-align: left;">By definition, |
− | By definition, | + | |
\[ | \[ | ||
p = - \left( \frac{\partial E}{\partial V} \right)_S. | p = - \left( \frac{\partial E}{\partial V} \right)_S. | ||
\] | \] | ||
− | If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\] | + | If $\rho = const$, then \[\frac{\partial E}{\partial V} = \rho\Rightarrow p = - \rho.\]</p> |
− | </p> | + | |
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<div style="width:100%;" class="NavContent"> | <div style="width:100%;" class="NavContent"> | ||
− | <p style="text-align: left;"> | + | <p style="text-align: left;">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 |
− | 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}. | \frac{dE}{dt} = - p\frac{dV}{dt}. | ||
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\[ | \[ | ||
\frac{1}{T}\frac{dT}{dt} = - \frac{1}{3V}\frac{dV}{dt},\] or equivalently \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. | \frac{1}{T}\frac{dT}{dt} = - \frac{1}{3V}\frac{dV}{dt},\] or equivalently \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. | ||
− | \] | + | \]</p> |
− | </p> | + | |
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Revision as of 22:17, 1 October 2012
Problem 1
Transform the energy conditions for the flat Universe to conditions for the entropy density (see [1])
Problem 2
Find the entropy density for the photon gas.
\[ dS = \frac{dE}{T} + \frac{pdV}{T} = \frac{V}{T}\frac{d\rho}{dT}dT + \frac{\rho+ p}{T}dV \Rightarrow\] \[ \left( \frac{\partial S}{\partial V} \right)_T = \frac{\rho + p} {T} \Rightarrow S = \frac{\rho + p}{T}V + f(T).\] 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 3
Use the result of the previous problem to derive an alternative proof of the fact that $aT=const$ in the adiabatically expanding Universe.
$$S = \frac{4}{3}\alpha T^3V = const;\quad V \propto a^3 \Rightarrow aT = const.$$
Problem 4
Find the adiabatic index for the CMB.
$$ p \propto \rho \propto a^{ - 4};\quad V \propto {a^3} \Rightarrow pV^\gamma = const,\quad \gamma = \frac{4}{3}. $$
Problem 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.
$$s \propto T^3 \propto a^{ - 3};\quad n_b \propto a^{ - 3}.$$ Therefore their ratio remains constant during the expansion of Universe.
Problem 6
Estimate the current entropy density of the 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 $T_0 \approx 2.725K$ \[s \approx 1.5 \cdot 10^3 \, cm^{-3}.\]
Problem 7
Estimate the entropy of the observable part of the Universe.
Size of the observable part of the Universe is $l_{H0} \sim 10^{28} \mbox{\it cm}$ and therefore \[S = \frac{4\pi}{3}sl_{H0}^3 \sim 10^{88}.\]
Problem 7
Why is the expansion of the Universe described by the Friedman equations adiabatic?
It is because the heat flow is absent in the homogeneous and isotropic Universe.
Problem 8
Show that entropy is conserved during the expansion of the Universe described by the Friedman equations.
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 9
Show that the entropy density behaves as $s\propto a^{-3}$.
\[S = const \Rightarrow sV = const \Rightarrow sV_0 a^3 = const \Rightarrow sa^3 = const \Rightarrow s \propto a^{ - 3}.\]
Problem 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 $p=-\rho$.
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 11
Show that the product $aT$ is an approximate invariant in the Universe dominated by relativistic particles.
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 \[ \frac{d}{dt}(\ln T) = - \frac{d}{dt}(\ln a)\Rightarrow aT = const. \]