Entropy of the cosmological constant and the laws of thermodynamics?

Question

Convention

The convention being used is:

• $ A_{C} = $ The classical variable

Premise

Consider the following toy-model universe:

A universe with a positive cosmological constant.

Basic Assumptions

• We know in extreme cases of inflation particles (in high number)are formed due to Hawking's radiation. Hence, $\mathrm dN_C \neq 0$

• We also that Noether's theorem assumes space-time to be a constant background. This is not true here: $\mathrm dU_C \neq 0$

• As space-time is expanding: $\mathrm dV_c \neq 0$

Statistical Mechanics

Let the number of micro-states of the particles created (not virtual particles) be $\Omega = \frac{N!}{\prod n_j}$. Where $n_j$ is the particle in the $j$'th state and energy $(\epsilon_c)_j$

Hence,

$$ \ln(\Omega) = \ln(N!) - \sum_j \ln(n_j!)$$ $$ \implies \ln(\Omega) = N \ln N - N - \left(\sum_j n_j \ln n_j - n_j\right)$$ $$ \implies \ln(\Omega) = N \ln N - \left(\sum_j n_j \ln n_j \right)$$ $$ \implies \mathrm d (\ln(\Omega)) = (\ln N)\mathrm dN - \sum_j \ln n_j \mathrm d(n_j) $$

We note for $\Lambda =0 \implies \mathrm dN = 0$ :

$$ \implies \mathrm d (\ln(\Omega) = - \sum_j \ln n_j \mathrm d(n_j) = \mathrm d(\frac{S_c}{k_b}) = \frac{\mathrm dU_c - P_c \mathrm dV_c}{T}$$

Going back to $\Lambda \neq 0$

$$ \implies \mathrm d (\ln(\Omega) = \ln N \mathrm dN + \mathrm dU_c - P_c \mathrm dV_c $$

Rewriting in terms of $S$

$$ \implies \mathrm dS = \underbrace{k_b \ln N \mathrm dN}_{\mathrm dS_\Lambda} + \mathrm d(S_c) $$

Defining $ k_b \ln N \mathrm dN $ as the entropy of the cosmological constant $S_\Lambda$

$$ \implies \mathrm dS = \mathrm d S_\Lambda + \mathrm d S_c$$

Writing everything explicitly:

$$ T\mathrm dS = k_b T \ln N_c \mathrm dN_c + \mathrm dU_c - P_c \mathrm dV_c $$

Questions

I would have loved to ask about a whole series of questions but for now:

• Is the above correct? Has anyone already thought about this? If so a reference would be welcome :)

• Is the physical interpretation of $S_\Lambda$ justified?

$$ d(S_\Lambda) = k_b \ln N \mathrm dN $$

• How does one justify the $2$'nd law of thermodynamics? Or can one justify a postive cosmological constant by saying the second law of thermodynamics is true? (below $\dot{S}$ is the time derivative of $S$ )

$$ \dot{S} = \dot{S_\Lambda} + \dot{S_c} > 0$$

• How does one show the third law of thermodynamics for the cosmological constant? Or the opposite: Can one justify the small cosmological constant via low temperature?

$$ \lim_{T \to 0} S_{\Lambda} = 0 $$

Answer 1

After a long discussion with "curiousone" I would like to like to share the relevant points of our discussion (hopefully I will do them justice) and some extra bits I added after thinking it over

First Law of thermodynamics

While the equation

$$TdS = k_b T \ln N dN + dU -PdV $$

is quite general to any system where particle number is not conserved. We can make it more specific to our system by:

$$ \frac{dN}{dt} = \text{number of particles created via hawking radiation}$$

We also note that this assumes that the first law of thermodynamics is correct (the conservation of energy) which may not be the case as Noether's theorem does not hold.

Second Law of thermodynamics

Hence we stick to the below statement of entropy:

$$ dS = dS_{\Lambda} + dS_c$$

The physical justification of $dS_{\Lambda} = k_b \ln N dN$

The second law of thermodyamics is trivially true then:

$$ dS_{\Lambda} = k_b \ln N dN \implies \dot{S}= k_b \dot{ N } \ln N >0 $$

as $\dot{N}>0$ as particles are created and $\ln N>0$ as we operate in the limit of $N \to \infty$

Third law of thermodynamics

Nernst-Simon statement: The entropy of a system at absolute zero temperature either vanishes or becomes independent of the intensive thermodynamic parameters, e.g. pressure, magnetic field, electric potential, etc.

Unattainability statement: To bring a system to absolute zero temperature involves an infinite number of processes or steps.

The Nerst-Simon statement is not true in quantum gravity for example black holes (http://www.scholarpedia.org/article/Bekenstein-Hawking_entropy#Status_of_the_third_law_of_black_hole_thermodynamics)

The unattainability statement is probably true (in my opinion) but what would it phyiscally imply I'm still uncertain.