Variable Mass Theory and the
Cosmic Microwave Background

Han de Bruijn, retired Engineer,
Theoretical Physicist by education
E-mail: umumenu@gmail.com

This paper is the fourth one in a series of articles about the Variable Mass theory. Earlier publications in the VM series are: Atomic Time, Orbital Time and the Variable Mass Theory [1] and The Variable Mass Theory and Gravitational Paradoxes [2] and Variable Mass Theory and Observable Universe [3]. The content of the first three papers is supposed to be familiar to the reader.
Last but not least, essential reading is, of course, the Wikipedia page describing our current subject, which is the Cosmic Microwave Background [8].

CONTENTS

1. Gravitational Self Energy Field
2. Employing the Hubble Tension
3. CMB Temperature here and now
4. References

The gravitational field of a point mass and the electric field of a point charge are structurally similar. Each may be represented by a vector field in which the vectors are directed along radial lines emanating from the point, and the vector magnitudes decrease as the inverse square of the distance from the point. The electric field has a scalar energy density field associated with it. When the field vector at a point is $\vec{E}$, then the energy density at the same point is $$ u_E = \frac{1}{2}\epsilon_0(\vec{E}\cdot\vec{E}) = \frac{1}{2}\epsilon_0 E^2 $$ where $\epsilon_0$ is the permittivity of free space. Energy density is a measure of the energy stored in the field per unit volume of space.
Question: Is there an energy density associated with a Gravitational Field? [5].
Yes, there is! Let us use the similarity between the gravitational and electric fields to construct a gravitational energy density term. Start by noticing how the permittivity of free space enters into the expression for $E$ : $$ E = kQ/r^2 \quad \mbox{where} \quad k = 1/(4\pi\epsilon_0) $$ is a universal constant, $Q$ is the source charge, and $r$ is radial distance from the source. The gravitational field is given by an analogous expression $$ g = GM/r^2 $$ where $G$ is a universal too, namely the gravitational constant, $M$ is the source mass. The electrical field energy density may be written in terms of $k$ as $$ u_E = \frac{1}{2}\cdot 1/(4\pi k)\cdot E^2 = E^2/(8\pi k) $$ By analogy, a candidate gravitational energy density term may now be constructed, where the minus sign is according to [5]: $$ u_G = -g^2/(8\pi G) $$ We will use Newton's law of gravitation without modification, and incorporate the energy density term described above. $$ g = GM/r^2 \quad \mbox{so that} \quad u_G = - GM^2/(8\pi r^4) $$ Integrate the above to find the gravitational energy of the entire Observable universe [3]. $$ U_G = \int_{R_O}^\infty - \frac{GM_O^2}{8\pi r^4}4\pi r^2 dr = - \frac{GM_O^2}{2}\int_{R_O}^\infty\frac{dr}{r^2} = - \frac{GM_O^2}{2R_O} $$ Substitute values from the sections 7 and 8 in the same paper [3] for mass and radius of the Observable VM universe, the formulas (11) and (13) to be precise. $$ M_O = \frac{c^3W}{2GH} \quad \mbox{and} \quad R_O = W\frac{c}{H} $$ where $c=$ speed of light in empty space, $G=$ gravitational constant, $H=$ Hubble parameter, $W=$ dimensionless number = 3.383634283. And it appears that the total energy of the gravitational Self Energy Field is minus one quarter of the mass-energy of the VM Observable universe. $$ U_G = - \frac{G}{2}\left(\frac{c^3W}{2GH}\right)^2\frac{H}{W.c} = - \frac{c^5.W}{8GH} = - \frac{1}{4}M_Oc^2 \qquad (1) $$ It is important to notice that the dimensionless constant $W$ appears in all of these formulas only as a result of integration over a region ranging from $0$ to $W$. This means that the numerical value $W$ in principle can be replaced by anything else, namely by integrating over a different region in the universe.

2. Employing the Hubble Tension

Elsewhere [3], the following formulas for the Hubble parameter $H$ have been found. $$ H^2 = \chi\,8\pi\,G\,\rho_c \quad \mbox{with} \quad \begin{cases} \chi = w & (\mbox{d}) \\ \chi = 1/3 & (\mbox{c}) \\ \chi = 1/6 & (\mbox{b}) \\ \chi = 1/8 & (\mbox{a}) \end{cases} $$ where $G=$ gravitational constant, $\rho_c=$ critical mass density of cosmos, $w=$ 0.3883945571 = dimensionless constant. The cases (a) and (b) will be ignored in the current context, leaving us with the cases (c) and (d).

In the Variable Mass theory there is not a single value for the Hubble parameter, because $H=(2/A)$ where A is an "Age" and A can even be the age of the earth. Let us first define the Hubble parameter of the Observable universe [3] in VM: $$ H_O^2 = w.8\pi.G.\rho_c \quad \mbox{(Late universe) or (d)} $$ And a different Hubble parameter for the Cosmic Microwave Background [8] (with ΛCDM): $$ H_{CMB}^2 = \frac{1}{3}.8\pi.G.\rho_c \quad \mbox{(Early universe) or (c)} $$ Then it follows that $$ H_{CMB}^2 = \frac{1}{3w} H_O^2 \quad \Longrightarrow \quad \boxed{ H_{CMB} = \frac{1}{\sqrt{3w}} H_O } $$ Numerical values are obtained at hand of the following data.

'Hubble constant -- at 67.4 (+/- 0.5) km/sec/Mpc', NASA Science, July 22, 2018
'Hubble constant at 73.2 (+/- 1.3) km/sec/Mpc', The Astrophysical Journal Letters, February 2021

Heuristics in the above section is a confirmation that we must use the Early universe entry $\Leftarrow W$ in the above picture. With the critical mass density of the universe assumed constant, this corresponds with a smaller Hubble parameter $H=2/A$ and thus indeed with a larger age $A$. Using that Early universe entry with our integration procedure we find with (1) for the CMB: $$ U_{CMB} = -U_G = \frac{1}{4}M_Oc^2 \quad \mbox{with} \quad M_O = \frac{c^3W}{2GH} \qquad (3) $$ where $U=$ total energy, $M_O=$ mass of Observable universe, $c=$ speed of light in empty space, $G=$ gravitational constant, $H=$ Hubble parameter, $W=$ dimensionless constant = 2.096433787 (numerically), at the place where the Observability function [3] in VM is equal to $1/3$: $$ f_O(x) = \frac{2-e^{-x}(x^2+2x+2)}{x} = \frac{1}{3} \quad \Longrightarrow \quad x = 2.096433787 \Leftarrow W $$ As a second ingredient it is assumed that the CMB temperature, denoted as Θ, is proportional with cosmological redshift $(1+z)$. The redshift obeys the logarithmic formula $r=(c/H)\ln(1+z)$ - in TL (Tired Light theories) as well as in VM - and so $$ \Theta = \Theta_0 (1+z) = \Theta_0 e^{H/c.r} $$ where the subscript (0) denotes the CMB temperature at our place and $r=$ distance. The Stefan-Boltzmann law [7] expressed as a formula for radiation energy density is: $$ u_{CMB} = \frac{4}{c}\sigma \Theta^4 $$ where the constant of proportionality, $\sigma$, is called the Stefan-Boltzmann constant.
The total energy of the CMB is thus expressed by the following integral over a sphere with radius $R_O=W(c/H)$ of the Observable (VM) universe, taking $x=(H/c)r$. $$ U_{CMB} = \int_0^{R_O} \frac{4}{c}\sigma \left(\Theta_0 e^{(H/c)r}\right)^4 . 4\pi r^2 dr = \frac{4}{c}\sigma \mbox{ }\Theta_0^4 . 4\pi \left(\frac{c}{H}\right)^3 \int_0^W e^{4x}x^2dx \qquad (4) $$ The integral is calculated to be $$ II = \int_0^W e^{4x}x^2dx = e^{4W}\left(\frac{1}{4}W^2-\frac{1}{8}W+\frac{1}{32}\right)-\frac{1}{32} $$ The end-result is found by combining (3) and (4). $$ \Theta_0 = \left[\frac{M_Oc^2/4}{(4\sigma/c).4\pi.(c/H)^3.II}\right]^{1/4} $$ which should give the CMB temperature at our place, while adopting the Riess value for Hubble:

W := 2.096433787;
G := 6.67408*10^(-11);
Mpc := 3.08567758*10^22;
H := 73.4; # Riess
H := H*1000/Mpc;
c := 299792458;
sigma := 5.670374419*10^(-8);
M_O := c^3*W/(2*G*H);
II := exp(4*W)*(W^2/4-W/8+1/32)-1/32;
Theta_0 := (M_O*c^2/4/(4*sigma/c*4*Pi*(c/H)^3*II))^(1/4);
Theta_0 := evalf(Theta_0);