The total energy of the universe, U, is conserved in a dark energy hypothesis [1] .

$U=U_{\lambda }+\left[M\left(dm\right)+M\left(b\right)\right]c^2$

The first term on the right is the dark energy and the second is the energies of dark matter and baryonic matter, which are in the ratio of five to one in the current epoch, λ = 7/10. The argument of the preceding paper is that there is a dark entropy that increases continuously without limit during cosmic evolution, S_λ = U_λ/T. The Helmholtz free energy, F (A in the chemist’s notation) is just a Legendre transformation of U in which the thermal variable is changed from S to T: F = U – TS. Hence, the Helmholtz free energy is given by

$F=\left[M\left(dm\right)+M\left(b\right)\right]c^2$

The baryonic mass is a constant so the focus will be on dark matter, which in the DEH is continuously converted into dark energy governed by the dimensionless representations of dark energy and dark matter,

$\lambda +\chi \left(dm\right)=0.95$

and the corresponding evolutionary rule:

$d\lambda =-d\chi \left(dm\right)>0$

From now on, F will mean F(dm), although for simplicity the qualifying symbol of dm will be omitted. Evidently,

$F=N\langle F\rangle$

where N is the number of quanta and $\langle F\rangle$ is the average value of one quantum:

$\langle F\rangle =m{c}^2$

The first task is to estimate the mass, m.

1) Presentation of numbers, equations, and terminology used in the subsequent analysis.

2) Estimation by the theory of fluctuations of the mass, m, of one quantum.

3) Presentation and discussion of the theory of fluctuations, which is due to Born and Einstein.

4) Presentation and discussion of equations of state for dark matter, its degeneracy, and the evolution implied by the equations.

5) Summary and conclusions.

Useful numbers, equations, and terminology

Given the current ratios of cosmological energies, dark energy: dark matter: baryonic matter: 70:25:5, and given the formalism of the DEH, the mass of dark matter is

$M=\frac{0.25\Gamma }{\kappa c^2}=8.45\times {10}^{49}\text{kg}$

where Γ = 6.306 × 10²⁴ m, which is the total energy of the universe expressed as a length, and κ is the Einstein gravitation constant whose units are length/energy. With a scale factor of a = 1.37 × 10²⁶ m, its density is

$\rho =3.29\times {10}^{-29}\text{kg}\cdot {\text{m}}^{-3}$

Because the density is so low, there will be little error in treating the dark matter gas as ideal, an assertion that will be confirmed later. (McQuarrie, Ch.3)

The temperature regime of interest is low as the following Table 1 illustrates.

Terminology is standard. An ideal gas has no interparticle forces. A classical

gas has no quantum effects. Quantum effects occur in an ideal Bose-Einstein gas, for example, because of the symmetry of the N-particle wave function, which means that the bosons behave cooperatively rather than acting individually.

Estimation of a dark particle mass

The derivation of the formula for $\langle F\rangle$ by the method of fluctuations; the accompanying discussion is given in the section following this one. Starting with the result leads directly to the mass.

$\frac{\langle F\rangle }{k_{B}T}=\frac{m{c}^2}{k_{B}T}=\frac{\beta \mu }{\exp \left(\beta \mu \right)-1}=\frac{\ln \left(n{\Lambda }_B^3\right)}{n{\Lambda }_B^3-1}$ (1)

In Equation (1), β = 1/k_BT as usual, μ is the chemical potential, n is the number density, and Λ_B is the de Broglie thermal wavelength:

${\Lambda }_B=\frac{h}{\sqrt{2\pi m{k}_{B}T}}$ (2)

Estimation by two different methods leads to similar results.

Method one. Since T is small, β is large from which exp(βμ) > 1, in which case

$\frac{m{c}^2}{k_{B}T}=\frac{\ln \left(n{\Lambda }_B^3\right)}{n{\Lambda }_B^3}$

The ratio ln(x)/x has a maximum of 1/e at x = e, leading to

$\frac{m{c}^2}{k_{B}T}=\frac{1}{e}$(3)

Choosing, say, the temperature of the present epoch leads to m = 1 × 10⁻⁴⁰ kg and mc² = 0.0864 meV.

Method two. Since T → 0 let μ → 0 in which case

$\frac{m{c}^2}{k_{B}T}=\frac{\beta \mu }{\exp \left(\beta \mu \right)-1}\to 1$

Speculating that μ = 0 at T = 1.71 K leads to m = 2.63 × 10⁻⁴⁰ kg and mc² = 0.147 meV.

Both methods give a mass in the same range, which is that of particles called axions by particle physicists, although the determination of mass here doesn’t show that the particles are axions. The point is that this is the business of making estimates, that given Equation (1) and the low temperature data, there’s no unique way of ma estimates, but however it’s done, they will all come out in this range.

A generic energy U can fluctuate around its mean value $\langle U\rangle$. The mean square fluctuation about the mean, also called the variance, is given by a well-known formula from statistics, that it is the mean of the square minus the square of the mean:

${\sigma }_U^2=\langle U^2\rangle -{\langle U\rangle }^2$

The variance is also given by

${\sigma }_U^2=-\frac{\text{d}\langle U\rangle }{\text{d}\beta }$

which is a relationship easy to establish.

The thermodynamic quantity of interest is the Helmholtz free energy:

$F=-pV+\mu N$

which will be modeled as an ideal classical gas.

$\langle F\rangle =\frac{F}{N}=-k_{B}T+\mu =-\frac{1}{\beta }+\mu$

Suppose that the variance in F is due only to the fluctuation in temperature. Then

${\sigma }_F^2=\frac{1}{{\beta }^2}={\langle F\rangle }^2$

On the other hand, suppose that the fluctuation is due only to the second term in which the number of particles fluctuates at constant μ. The average value F is $\langle F\rangle =\mu \langle N\rangle$, and the variance of F would be due to the variance of N:

${\sigma }_F^2={\mu }^2{\sigma }_N^2$

But curiously the variance of N is just $\langle N\rangle$. (McQuarrie, Ch. 3) Hence,

${\sigma }_F^2=\mu \langle F\rangle$

Now suppose that the two fluctuations occur independently of one another, which seems reasonable given that one is a temperature fluctuation and the other a number fluctuation. In this case the variance will be the sum of the two, and one has that

${\sigma }_F^2=-\frac{\text{d}\langle F\rangle }{\text{d}\beta }={\langle F\rangle }^2+\mu \langle F\rangle$

This is an example of a Bernoulli differential equation that can be integrated by setting $\langle F\rangle =1/z$. Given the solution, g(z), with the integration constant set at unity, and solving for $\langle F\rangle$ gives Equation (1).

The method is due to Max Born [2] who writes that it’s Einstein’s, but it may be the case that Born developed it from Einstein’s discussions of fluctuations [3] . Born/Einstein’s application is to the blackbody problem. The differential equation is that above but with $\langle F\rangle$ replaced by $\langle E\rangle$, the average energy of a radiation oscillator. There are two independent fluctuations in the radiation field, one due to the long wavelength, high temperature limit, that of Rayleigh-Jeans, which gives the term in ${\langle F\rangle }^2$, and the other due to the short wavelength, low temperature limit of Wien, which gives the other term with μ replaced by an oscillator energy E₀. Planck had derived the blackbody radiation law by interpolating between these two extremes using entropy as the interpolation tool. The Born/Einstein method gives

$\langle E\rangle =\frac{h\nu }{\exp \left(\beta h\nu \right)-1}$

Planck and Einstein found themselves at odds over the meaning of this formula. Einstein argued that the electromagnetic radiation was quantized, that hν represented a bundle of radiation, a light quantum, whereas Planck held that light was a wave phenomenon, that it was not quantized, and that hν was simply an interpolation parameter. It was many years before Planck came around to Einstein’s point of view [4] .

The denominator of Equation (1) appears in Bose-Einstein statistics. Does it imply that dark matter is a boson? Dark matter may be a boson, but that shouldn’t be inferred from Equation (1), which is the solution of a differential equation that has been generated without any input from BE statistics.

However, the hypothesis now will be that the dark matter quantum is a boson. The reason is simple, that dark energy is spinless, and if dark matter carried a spin, there would be a requirement to explain why it wasn’t conserved. A spinless particle has to be a boson.

These are Bose-Einstein equations obtained by contour integration as shown in the appendices. They are given for the thermodynamic limit V → ∞.

$\frac{p}{n{k}_{B}T}=2i{\left[\ln \varphi \right]}^{1/2}=\frac{2}{{\pi }^{1/2}}n{\Lambda }_B^3$ (4a)

Equating the first and third terms is an example of an ideal, non-classical gas. Equating the second and third, and squaring gives (4b).

$\varphi =\exp \left[-\frac{{\left(n{\Lambda }_B^3\right)}^2}{\pi }\right]=\exp \left[-\frac{1}{\pi }{\left(\frac{T_C}{T}\right)}^3\right]$(4b)

The second equality in Equation (4b) comes from combining the critical temperature for transition to a Bose-Einstein condensate

$T_C=\frac{h^2{n}^{2/3}}{2\pi m{k}_B}$ (5)

with the de Broglie thermal wavelength, Equation (2). The symbol φ is for the

fugacity (aka absolute activity) of the BE gas and is bounded by 0 < φ < 1 as required ( Figure 1 ).

(For fugacity McQuarrie uses the symbol λ, but that’s dark energy in a DEH, whereas Annett uses the symbol z, but that’s a complex variable in this paper. [5] ) A boson may be found either in the condensate or gas phase, so their fractions are

$f_c+f_g=1$

The identification $f_g=\varphi$ is natural, giving

$f_c=1-\varphi$(6)

Equations (4b) and (6) show that the gas and condensate will coexist only if the ratio T_c/T is of order unity.

Contour integration gives the Helmholtz free energy as

$F=-N{k}_{B}T\ln \left(\varphi \right)$(7)

Since the chemical potential is

$\mu =\frac{\partial F}{\partial N}$

it follows that Equation (6) become

$f_c=1-\exp \left(-\beta \mu \right)$(8)

Numerical illustrations for the current epoch

For illustrative work assume that at this epoch μ = mc²: then T = 2.7255 K, m = 1.54 × 10⁻⁴⁰ kg, mc² = 1.38 × 10⁻²³ J and φ = 0.693. There is now enough information to generate numbers from the equations of state (4a-b).

$\ln \left(\frac{1}{\varphi }\right)=\frac{1}{\pi }{\left(\frac{T_c}{T}\right)}^3$

gives T_c = 2.86 K, which in turn gives the number density from Equation (5): n = 2.56 × 10⁷ m⁻³. By Equation (2)

${\Lambda }_B^3=4.19\times {10}^{-8}{\text{m}}^3$

The numerical value of the ideal, non-classical equation of state is p/nk_BT = 1.21; the deviation from unity is entirely due to quantum effects.

Dark matter is in a state of phase equilibrium

$dm\left(c\right)=dm\left(g\right)$

so the pressure calculated from the equations of state should be interpreted as a vapor pressure. By Equation (A.3) in the thermodynamic limit V → ∞ and with γ = π,

$p=\frac{2}{{\pi }^{1/2}}{\left(\frac{T_c}{T}\right)}^3\frac{k_{B}T}{{\Lambda }_B^3}=1.30\frac{k_{B}T}{{\Lambda }_B^3}=1.17\times {10}^{-15}\text{Pa}$

The Clapeyron equation is

$\frac{\text{d}p}{\text{d}T}=\frac{\Delta S}{\Delta V}=1.30\frac{k_B}{{\Lambda }_B^3}=4.28\times {10}^{-16}\text{Pa}\cdot {\text{K}}^{-1}$

ΔV is the volume difference between the gas and condensed phases, which is essentially that of the gas phase: ΔV = a³ = 2.57 × 10⁷⁸ m³. Hence the entropy of the phase transition is

$\Delta S=1.10\times {10}^{63}\text{J}\cdot {\text{K}}^{-1}$

which is that of a first-order phase change in the Ehrenfest classification.

It is instructive to calculate the free energy in two ways.

1) The scale factor at this epoch is a = 1.37 × 10²⁶ m (DEH II). Then the number of quanta is N = na³ = 6.58 × 10⁸⁵. Equation (7) gives F = 9.12 × 10⁶² J.

2) F = Mc² = 7.59 × 10⁶⁶ J using the mass at the beginning of this article.

The first calculation is based on BE statistics, so the gas is ideal and non-classical. The second implicitly assumes that the gas is ideal and classical. This is seen by calculating the number of quanta from ii) by N = M/m = 5.51 × 10⁸⁹—there is the factor of 10⁴ by which the two values of F differ. The BE calculation is preferred, which is a conclusion supported by the consideration of degeneracy that follows.

The density of states is

$\Phi =\frac{1}{{\Lambda }_B^3}\sqrt{\frac{6}{\pi }}=3.30\times {10}^7\text{states}{\text{m}}^{-3}$

With the number density given above $\frac{\Phi }{n}=1.29$ states/boson meaning that on average each state is singly occupied. This is in contrast to an ideal, classical gas of diatomic nitrogen, a boson, at 300 K and 1 bar where $\frac{\Phi }{n}={10}^6$ states/boson.

Almost all of the states are unoccupied. Degeneracy means proximity, that the quanta are close together, which is the environment in which quantum effects thrive.

A comparison of dark matter to ⁴He is useful. In superfluid He(II) the interatomic distance is d = 0.265 nm. The de Broglie thermal wavelength calculated at 2.17 K is Λ_B = 0.592 nm, giving Λ_B/d = 2.23. Annett writes of this relationship, “So we can expect that quantum mechanical effects are always important for liquid ⁴He.” (P. 24)

With the above number density for dark matter and with Chandrasekhar’s formula [6] for the nearest neighbor distance in a random distribution of gas particles, d = 0.552n^−1/3 = 1.87 mm. But Λ_B = 3.47 mm giving Λ_B/d = 1.86, so the inference is that quantum effects are important in dark matter.

Another indicator of degeneracy/quantum effects is the wave character of the boson quantum: with m = 1.54 × 10⁻⁴⁰ kg and T = 2.7255 K, the de Broglie thermal wavelength and Compton wavelength are comparable in size and in the range of a household ruler:

${\Lambda }_B=3.47\text{mm}$ and ${\lambda }_c=\frac{\hslash }{mc}=2.29\text{mm}$

In weak degeneracy, deviations from ideal behavior can be expressed in a virial expansion such as appears in the study of real gases (McQuarrie):

$\frac{p}{n{k}_{B}T}=1-\frac{n{\Lambda }_B^3}{2^{5/2}}+\cdots$

Given that

$\frac{n{\Lambda }_B^3}{2^{5/2}}=0.189$

is rather large for a “mere” correction, the notion that this BE gas is weakly degenerate is marginal.

One. Dark energy stands to dark matter as entropy stands to Helmholtz free energy.

Two. The mass of the dark matter quantum is independent of any particular numerical value of the number of bosons or boson density.

Three. A candidate particle for dark matter is a degenerate, spinless boson with mass in the neighborhood of 10⁻⁴⁰ kg/0.1 meV.

To obtain Equation (4a), take the thermodynamic limit $V\to \infty$ and ask

$\frac{p}{n{k}_{B}T}=?$

There are two ways to proceed.

1) Eliminate ${\Lambda }_B^3$ from between (A.2) and (A.3). Set γ = π/2 and ν = 0.

2) Eliminate ln(φ) from between them. Set γ = π and ν = π/2

To obtain Equation (7) in the thermodynamic limit, divide (A.4) by (A.2), set ω = π and ν = 0. The signs require management.