Entropy of the cosmological constant and the laws of thermodynamics?

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?

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

• Particle number is not a constant, neither in the universe nor in a simple chemistry experiment, so nothing that you are deriving here describes cosmology in any special way. The 2nd law of thermodynamics is the definition of temperature and it is only useful in systems near equilibrium in which temperature is actually a well defined quantity. The universe is not even close to being near equilibrium. The third law of thermodynamics is required to match the theory to observations. Whether it holds for the entire universe is not clear (and I have a hunch that it may not). – CuriousOne Feb 5 '16 at 14:52
• "nothing that you are deriving here describes cosmology in any special way" ... Can I have a reference to people already using this as the entropy of the cosmological constant? "The 2nd law of thermodynamics is the definition of temperature" ... I thought that it was the entropy always increases or remains constant? "(and I have a hunch that it may not)" Can you elaborate why you think so? – drewdles Feb 5 '16 at 17:08
• Also I thought this might be relevant for: "Particle number is not a constant, neither in the universe" ... A universe without the cosmological constant ... I doubt it: physics.stackexchange.com/questions/135983/… – drewdles Feb 5 '16 at 17:25
• @CuriousOne Also, temperature is only useful for systems in near equilibrium?! One can define temperature for any system which has entropy and energy $T=(\frac{\partial U}{\partial S})_{V,N}$ – drewdles Feb 5 '16 at 20:36
• I dont think I get your point ... energy is not constant and neither is $S$ ... In a situation when temperature is not defined what do you think a termometer will measure (also I feel we are digressing) – drewdles Feb 5 '16 at 21:24

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.