When one models the earlierearly universe it is most often assumed to be in thermal equilibrium i.e. the entropy is maximized and therefore constant $$dS=0$$
As far as I understand this can be explained as follows:
The change of entropy with Volume is given as: $$\frac{\partial S}{\partial V}=\frac{1}{T}\frac{\partial U}{\partial V}$$
therefore if we model a part of the early universe as a closed box of Volume $V_0$ with internal Energy $U_0$ we find $$\frac{\partial U}{\partial V}=0$$ and therefore $$\frac{\partial S}{\partial V}=0$$
i.e. as the Universe expands the entropy stays constant.
(One could also argue that a thermal equilibrium can be defined as the entropy being maximized and ignore the above...)
However, if we take a look at the Einstein equation including the cosmological constant $\Lambda$ $$G^{\mu\nu}=\frac{8\pi G}{c^4}T^{\mu\nu}-\Lambda g^{\mu\nu}$$ The cosmological constant is often interpreted as additional stress energy tensor i.e. we define the total tensor $$T^{'\mu\nu}=T^{\mu\nu}-\Lambda g^{\mu\nu}$$ with a new energy density $$T^{'00}=\rho-\Lambda$$ if we assume a perfect fluid with energy density $\rho$ Now the volume can be defined in terms of the scale paramter $a$ as $$V=V_0a^3$$ such that $\rho\propto a^{-3}$ i.e. $$\rho=\rho_0a^{-3}$$ $\Lambda$ however is just a constant. Therefore the overall energy $$U=\int dV\,\,T^{'00}$$ is no longer constant in $a$ and therefore $$dS\neq 0$$
Probably I got a mistake in this somewhere but I don't see it. I guess its either
- my assumption taking the cosmological constant into account is not valid or
2)I got some basics of thermodynamics wrong...
I would be glad if someone could show me where my mistake(s) is(are).