Essential reading is the first article in this series, which is about Atomic Time, Orbital Time and the Variable Mass Theory. It has been shown in this article that General Relativity (GR) can be effectively replaced by Variable Mass (VM) in a Static Euclidean Universe (SEU), which is eternal in Atomic time and has moments of creation (i.e. one or more beginnings) in Orbital time.
An obvious consequence is that the cosmological redshift is intrinsic: it depends on variable mass. Because that is where the VM theory has been designed for.
Another convenient consequence is that Newton's theory is good enough for describing gravity at a cosmic scale. It is demonstrated that the latter, when combined with the Variable Mass hypothesis, solves for (two versions of) well known Gravitational Paradoxes.
CONTENTS
Apart from physics, the whole idea of Minkowski space-time [16] is doubtful in another respect. Let us quote from an article by Norman (N J ) Wildberger,
titled Evolution versus Intelligent Design: a mathematician's view [17].
Going more out on a limb, I will call on modern neuro-psychology and Einstein's theories of relativity
to destabilize the familiar view of the universe as a dynamic place which is unfolding as we watch, along lines
that we can potentially understand. I will suggest that the world has already been created in its historical entirety.
We are just too low dimensional and internal to it to witness it across both space and time. This is consistent with
the idea of an omniscient deity, whose existence probably implies that everything is already known, and so any idea
of free will must be an illusion. Alright, so humanity is guilty of nothing. Because we couldn't
have done otherwise, even if we wanted to.
Or, as formulated by Doug Marett (2013) in his
Spacetime- Right or Wrong? [18]:
What space-time implies is that all events, past, present and future, are already written in the fabric of space.
Such a conclusion relegates all human activity to simply being automatons acting out the frames of a preordained film strip.
If we were to believe Einstein we would have to believe that there is no free will at all.
That in and of itself should be reason enough to reject the space-time concept without further consideration!
From Relativity Reexamined [19] at page 29: Einstein wanted to reduce all physics to pure geometry; he
thought that a conveniently curved space-time universe would provide an explanation for all physical laws from electromagnetism
to gravitation. This was his avowed aim and he worked toward this goal for half of his life. [ .. ]
But the goal was never reached. Einstein managed to interconnect curved geometry and gravitation in a brilliant way, but
his unitary theory, as he called it, was never achieved.
A chain reference in our key article [1] is the 1993 publication [32] by Jayant Narlikar [4] and Halton Arp [33]. From that publication we quote: Finally, the Euclidean, flat spacetime becomes a natural, primary reference frame in which cosmological processes are most simply described. Actually Narlikar has proved herewith that General Relativity is redundant on a cosmic scale.
The geometries in Einstein's General Relativity theory have been made "possible" by the discovery
of Non-Euclidean geometry [20] (Wikipedia) in the beginning of the 19th century. This has lead to the
(mis)understanding that non-Euclidean geometries might possibly be relevant to physics. However, the
the only thing it shows is that Euclid's Fifth's Postulate - the one of the parallel lines - cannot be
deduced from the other postulates; it's an independent axiom. Which as such is an important result indeed.
Non-Euclidean geometries can always be embedded in Euclidean geometry - there are many
models of non-Euclidean geometry [21] in common geometry.
In a posting at Mathematics Stack Exchange, titled
Non-Euclidean Geometrical Algebra for Real times Real? [22], it is argued that Euclid's Fifth's
Postulate is not only essential for Euclidean geometry, but it's also essential for elementary algebra.
I have yet to see any decent example contradicting this evidence: without the axiom of parallel lines
elementary algebra cannot even exist. However, Einstein's tensor calculus eventually rests on elementary
algebra. How then can space-time of General Relativity be non-Euclidean?
4. The Big Bang never happened
Let's proceed with another quote from the above mentioned article [17] by N J Wildberger:
Scientific types ought to be a bit more open-minded about things which border on the unknowable.
Dramatic denials of God or any kind of cosmic intelligence by biologists, esteemed as they may be in their own communities, seem to me to be almost as unjustifiable as the emphatic assertions of religious extremists,
or the overly confident descriptions of the first trillionth of a second by Big-Bangers. Namely according to the standard
Chronology of the universe [23] (Wikipedia).
Another quote comes from the book by Hans Jörg Fahr - Mit oder ohne Urknall [24] - and is copied / translated from page 33 without permission. It is noticed that page 314 contains literally the same piece of text, unless there is
something wrong with my memory. Or the message is so important that it cannot be repeated often enough.
Starts with a Bang [25]:
| Warum also drängt sich das Bild des Urknalls dennoch dem menschlichen Verstand mit so starker Suggestivität auf, förmlich wie eine unvermeidliche Vision? Natürlich haben Bilder einer Atombombenexplosion hierbei ihre ungeheuerliche Suggestivkraft entfaltet. Doch fragt man sich dann aber angesichts solcher Bilder: Entsteht hier eine Welt? Oder vergeht hier eher eine? Auch bei Bildern einer Supernova Explosion geht es einem kaum anders; auch hier glaubt man so etwas wie einen lokalen Weltbeginn wahrzunehmen, und man fragt sich: Könnte nicht vielleicht die Welt als ganze in Form einer globalen Meganova-Explosion, sozusagen wie aus einer gigantischen Wasserstoffbombe hervorgegangen sein? | So why is the image of the Big Bang so pressing upon the human mind with such strong suggestiveness, punctiliously like an inevitable vision? Of course images of an atomic bomb explosion have meanwhile unfold their tremendous suggestive power. Yet one is asking in the face of such images: is a world emerging here? Or does one rather perish? Also with images of a supernova explosion things go hardly different; here also one believes to perceive something like a local world beginning, and one wonders: Couldn't the world as a whole have been born in the form of a global meganova explosion, so to speak as if it had emerged from a gigantic hydrogen bomb? |
The glorification of violence is indeed Bankrupting Physics.
Think about particle accelerators for example, especially the Large Hadron Collider [26]. As if Depth of Truth is proportional to Heaviness of Bombardment. As if all this smashing of matter can
reveal anything else than space debris. The Embarrassing Nonsense of Particle Physicists - No, we do not need a New Collider [27].
It is clear, however, that modern physics is herewith just following a common trend in our nowadays "civilized" society,
where you can hardly watch a movie without a murder in it. Death seems to be more interesting than life. Standard (cosmological) models of nowadays physics say more about us humans than they reveal about the outside world.
But okay, let's forget about all this and go back to business. Is it true that no other than
Censored Papers Demolish the Big Bang Hypothesis [28]?
According to the Big Bang model, any Static Euclidean Universe is out of the question indeed. One of the most frequently
employed arguments to defend this stand is Olbers' Paradox (Wikipedia) [29] which is also known as the "dark night sky paradox".
We certainly have to deal with Olbers' Paradox in a SEU. However, it's not impossible that observational evidence is in favour of the latter. Particularly relevant in this respect is the information that has recently reached us via the James Webb Telescope [31].
5. The Cosmological Redshift
Cosmological Redshift [34] is the phenomenon that all wavelengths $\lambda$ as received from cosmic objects - in the past - are larger than corresponding wavelengths $\lambda_0$ here and now on earth. Expressed into common $z=$ redshift notation that is:
$$
z = \frac{\lambda-\lambda_0}{\lambda_0} = \frac{\lambda}{\lambda_0} - 1 \gt 0
$$
Take a look at the picture below, depicting a timeline in cosmos. Let (2) be our position on earth, (1) is the position of a far away civilization like ours, (0) is the position of a galaxy observed by observers on both habitable planets (1) and (2).

Let $\lambda=$ wavelength of light. Then it is mathematically obvious that:
$$
\frac{\lambda_1}{\lambda_0} \frac{\lambda_2}{\lambda_1} = \frac{\lambda_2}{\lambda_0}
$$
And therefore, by definition of any redshift as $z_{ab}=(\lambda_b-\lambda_a)/\lambda_a$ :
$$
1+z_{02} = (1+z_{01})(1+z_{12})
$$
It is reasonable to assume that the whole $\Delta t_{02}$ is the sum of the parts $\Delta t_{01}$ and $\Delta t_{12}$ and so
$$
1+z = F(\Delta t) \quad \mbox{where} \quad F(\Delta t_{02}) = F(\Delta t_{01}).F(\Delta t_{12}) = F(\Delta t_{01} + \Delta t_{12})
$$
This makes the function $F$ a special one, because it is seen now that the following property must hold for all $(t,\tau)$.
$$
\ln(F(t+\tau)) = \ln(F(t)) + \ln(F(\tau))
$$
We need additional mathematics, namely: Additive functions and measure theory [35]. From that reference it is inferred that the logarithm of our function $F$ is linear. And so the redshift itself (plus one) is an exponential function.
$$
\ln(F(t)) = \kappa\cdot t \quad \Longrightarrow \quad F(t) = 1+z = e^{\kappa\cdot t}
$$
where $\kappa$ is a constant that has to be determined from physics.
Essential reference is Atomic Time, Orbital Time and the Variable Mass Theory [1], which is to be considered as Part I of the current series on VM.
In there an alternative derivation of the above is found, namely in the section 6. Mass in Atomic Time. Taking a further look at the content of the paper, we have respectively the formulas (5), (9) and (11):
$$
1+z = \frac{m_0}{m} \quad ; \quad \frac{m}{m_0} = e^{H\cdot t} \quad ; \quad
A = \frac{2}{H} \quad \mbox{or} \quad \frac{1}{2}H = \frac{1}{A}
$$
where $z=$ intrinsic redshift [30], which is typical for the Variable Mass Theory.
Furthermore $m=$ (elementary particle rest) mass, $A=$ age (as measured in Orbital time).
From the previous it is easily concluded that our constant kappa is minus the familiar Hubble parameter:
$$
1+z = e^{\kappa\cdot t} = \frac{m_0}{m} = e^{-H\cdot t} \qquad (1)
$$
A further consequence is the logarithmic distance formula, which is formula (12) in our key reference [1]:
$$
d = \frac{c}{H}\mbox{ }\ln(1+z) \qquad (2)
$$
where $d=$ distance, $c=$ lightspeed in empty space. It should be noticed that there are Hubble tensions in $H$, making it possible that there are different intrinsic redshifts $z$ at the same distance $d$.
6. A Paradox of Gravitation
Key reference for this section is A Paradox of Newtonian Gravitation and Laplace's Solution [36] by
Amitabha Ghosh [37]
and Ujjal Dey.
Consider a test mass $\,m\,$ at position $(x,y,z)$ inside an infinitely large medium (ipse est Cosmos) with mean density $\,\rho_c\,$ and calculate the gravitational force $\,\vec{F}\,$ upon that test mass with Newton's third law, where $G\,$ is the Gravitational constant [38].
$$
\vec{F}(x,y,z) = G\cdot m \iiint \begin{bmatrix} x-\xi \\ y-\eta \\ z-\zeta \end{bmatrix}
\frac{\rho_c\,d\xi\,d\eta\,d\zeta}{\left[(x-\xi)^2 + (y-\eta)^2 + (z-\zeta)^2\right]^{3/2}}
$$
Introducing a Spherical coordinate system [39]:
$$
\left\{ \begin{array}{l}
\xi = r \sin(\theta) \cos(\phi) \\
\eta = r \sin(\theta) \sin(\phi) \\
\zeta = r \cos(\theta) \end{array} \right.
\qquad \mbox{where} \quad 0 \le \theta \le \pi \quad \mbox{and} \quad 0 \le \phi \le 2\pi
$$
Because of the expected spherical symmetry of the problem, we only have to consider a ray in the $\,z\,$ direction,
which means that $x=0$ , $y=0\,$. Also put $\,z=R\,$, then we have:
$$
d\xi\,d\eta\,d\zeta = r^2dr\sin(\theta)\,d\theta\,d\phi \\
(x-\xi)^2 + (y-\eta)^2 + (z-\zeta)^2 = \xi^2+\eta^2+\zeta^2 - 2 z \zeta + z^2 = r^2 - 2\,r R \cos(\theta) + R^2
$$
The integral then becomes, after suitable rearrangement:
$$
F(R) = G \cdot m \cdot \rho_c
\int_0^{2\pi} d\phi \int_0^\infty r^2 dr
\int_0^\pi \frac{\left[R - r\cos(\theta)\right] \sin(\theta) \, d\theta}
{\left[r^2 - 2\,r R \cos(\theta) + R^2\right]^{3/2}}
$$
Substitute $\,y=\cos(\theta)\,$ and find that $\,dy = \sin(\theta)\,d\theta\,$. Replace $\,y\,$ by $\,(-y)\,$.
$$
F(R) = G \cdot m \cdot \rho_c \cdot 2\pi \int_0^\infty r^2 dr \int_{+1}^{-1}\frac{R-ry}{(r^2-2\,ryR+R^2)^{3/2}} dy \\
F(R) = - G \cdot m \cdot \rho_c \cdot 2\pi \int_0^\infty r^2 \left[ \int_{-1}^{+1}\frac{R+ry}{(R^2+r^2+2\,ryR)^{3/2}} dy \right] dr
$$
At last, substitute $\,x = r/R\,$ to get
$$
F(R) = - 2\pi\,G\,\rho_c\,R\,m\int_0^\infty x^2 \left[\int_{-1}^{+1}\frac{xy+1}{(1+x^2+2\,xy)^{3/2}} dy \right] dx
$$
It should be noticed for later use that the distance between the test mass and an infinitesimal element in space
$\,= R\sqrt{1+x^2+2\,xy}\,$. Now calculate the inner integral, the one between square brackets.
int((x*y+1)/(1+x^2+2*x*y)^(3/2),y=-1..1); latex(%);The first line by MAPLE [40], the rest by hand. The outcome is a (Heaviside) step function. $$ -{\frac {-\sqrt { \left( x-1 \right) ^{2}}x-\sqrt { \left( x-1 \right) ^{2}}+\sqrt { \left( x+1 \right) ^{2}}x-\sqrt { \left( x+1 \right) ^{2}}}{\sqrt { \left( x+1 \right) ^{2}}\sqrt { \left( x-1 \right) ^{2}}}} \\ = - \left[-\frac{x}{|x+1|}-\frac{1}{|x+1|}+\frac{x}{|x-1|}-\frac{1}{|x-1|}\right] = \frac{x+1}{|x+1|}-\frac{x-1}{|x-1|} = 1-\frac{x-1}{|x-1|} \\ \\ \Longrightarrow \quad \int_{-1}^{+1} \frac{xy+1}{(1+x^2+2xy)^{3/2}}dy = \begin{cases} 2 \quad \mbox{for} \quad x \le 1 \\ 0 \quad \mbox{for} \quad x \ge 1 \end{cases} $$ It follows that $$ F(R) = - 2\pi\,G\,\rho_c\,R\,m\left[\int_0^1 2\,x^2\,dx + \int_1^\infty 0\,x^2\,dx\right] = - \frac{4}{3}\,\pi\,G\,\rho_c\,R\,m $$
![]() | The Paradox of Newtonian Gravitation is in the fact that, according to the above calculations, any test mass $\,m\,$ anywhere in the universe is being attracted (with a force $\,-F\,$) towards the center $C$ of a (spherical) coordinate system, which is also an arbitrary point in the universe. The paradox can be derived in a much simpler way by employing two well known results in gravity theory. Both are a consequence of Newton's Shell theorem [41]. The first result is that the force upon a test mass inside a hollow sphere is exactly zero. The second result is that the force upon a test mass outside a massive sphere is the same as the force towards a point at the center of the sphere with the total mass of the sphere concentrated in it. Therefore the total force is simply $\,F = - G\,m(\rho_c\cdot 4/3\pi R^3)/R^2 = -4/3\,\pi\,G\,\rho_c\,R\,m\,$ , quite in concordance with the more elaborate theory. |
g(x,alpha) := int(exp(-alpha*sqrt(1+x^2+2*x*y))*(x*y+1)/(1+x^2+2*x*y)^(3/2),y=-1..1);
F := int(g(x,alpha)*x^2,x=0..infinity);
simplify(%,assume=positive);
0
The problem has become the subject of a posting at the Mathematics Stack Exchange forum (MSE for short):
Prove or disprove the equality of these two integrals [43]. And indeed there is an analytical solution. Even better, it's entirely by hand: Brain on, computer off !
The gravitational force on a test mass in a universe with uniform density is zero everywhere. However, this is only the case with "assume=positive". The paradox pops up again as soon as $\,\alpha\,$ is exactly zero.
We can see that, at a positive infinitesimal distance from $\,\alpha = 0\,$, the integral suddenly changes from $2/3$ to $0$. Thus $\,\alpha = 0\,$ or $\,\alpha \gt 0\,$ makes a hell of a difference !!
# 1 megaParsec
Mpc := 3.08567758*10^22;
# Hubble parameter (2022-02-08)
H := 73.4*1000/Mpc;
# Speed of light
c := 299792458;
# Gamma
H/c;
-26
0.7934595873 10
It turns out to be very small indeed, approximately $\,8 \times 10^{-27}\,m^{-1}$. Compare with Laplace's own estimate as mentioned in [36]: $\,\Gamma \lt 7\times 10^{-18}\,m^{-1}$.Note. The initials L.M. in the name of the author of the awarded answer in [43] stand for Louis Marmet [44].
8. Seeliger's Paradox (NASA)
Key reference for this section is Seeliger's Gravitational Paradox and the Infinite Universe [45] by Leonardo Sarasúa.
A dimensions check of the formulas in the paper would have prevented us to revisit the maths. Fig. 1 is copied from the paper as it is.

List of symbols: $F=$ force, $G=$ gravitational constant, $m=$ test mass, $\rho=\rho_c=$ mean density of cosmos, $R=$ radius, $h=$ height, $r=\sqrt{R^2+h^2}=$ distance.
The plane in the above picture has thickness zero; but it must be $\,dh\,$ instead. Giving a slightly
but essentially different expression for the infinitesimal vertical component $dF$ of the force on the particle with mass $m\,$ as it is attracted by the plane.
$$
dF = \frac{G\,m(\rho_c\,2\pi R\,dR\,dh)}{\left(\sqrt{R^2+h^2}\right)^2}\frac{h}{\sqrt{R^2+h^2}} =
2\pi G m \rho_c\,h\,dh\frac{R\,dR}{(R^2+h^2)^{3/2}}
$$
The total force experienced by the test mass from all of the mass below it is
$$
F = 2\pi G m \rho_c \int_0^{\infty} h \left[ \int_0^{\infty} \frac{R\,dR}{(R^2+h^2)^{3/2}}\right] dh
$$
The integral between square brackets is, with $\,x=R/h\,$ and $\,u=x^2+1\,$:
$$
\int_0^{\infty} \frac{R\,dR}{(R^2+h^2)^{3/2}} = \frac{h^2}{h^3} \frac{1}{2} \int_{x=0}^{\infty} \frac{d(x^2+1)}{(x^2+1)^{3/2}}
= \frac{1}{h} \frac{1}{2} (-2) \left[-\frac{1}{\sqrt{u}}\right]_1^\infty = -\frac{1}{h}
$$
Giving for the total force as experienced by our test mass an infinite outcome.
$$
F = - 2\pi G m \rho_c \int_0^{\infty} dh = - \infty
$$
Minus another minus infinite outcome if the mass of the universe at the other side (above it) is taken into account.
Therefore the total force is undefined. This comprises the paradox.
However, let's now introduce the panacea as proposed by Laplace and Seeliger and observe what happens.
$$
F = 2\pi G m \rho_c \int_{-\infty}^{+\infty} h \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+h^2}}R\,dR}{(R^2+h^2)^{3/2}}\right] dh
$$
With :-(brainless)-: help of MAPLE [39] we see that the outcome is zero again,
as desired and required. Physical truth should not depend upon the way you look at it.
h*int(R*exp(-Gamma*sqrt(R^2+h^2))/(R^2+h^2)^(3/2),R=0..infinity);
f(h) := simplify(%,assume=positive);
int(f(h),h=-infinity..infinity);
0
It must be possible again to do this with Brain on, computer off.
Define $\,I\,$ as half the above integral:
$$
I = \int_0^{\infty} h \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+h^2}}R\,dR}{(R^2+h^2)^{3/2}}\right] dh \\
u = \sqrt{R^2+h^2} \quad ; \quad v = \frac{h}{\sqrt{R^2+h^2}} \quad ; \quad h = uv
\quad ; \quad du\,dv = \frac{\partial(u,v)}{\partial(R,h)}dR\,dh = \left| J \right| dR\,dh
$$
From the picture we see that $\,u=r\,$ and $\,v=\sin(\theta)\,$,
giving the integration bounds $\,0\le u \lt\infty\,$ and $\,0\le v \le 1\,$.
$$
\left| J \right| = \begin{vmatrix} \partial u/\partial R & \partial u/\partial h \\
\partial v/\partial R & \partial v/\partial h \end{vmatrix} =
\begin{vmatrix} R/\sqrt{R^2+h^2} & h/\sqrt{R^2+h^2} \\ -Rh/(R^2+h^2)^{3/2} & 1/\sqrt{R^2+h^2} - h^2/(R^2+h^2)^{3/2} \end{vmatrix}
= \frac{R}{R^2+h^2} \\
u^2 = R^2+h^2 \quad \Longrightarrow \quad R^2 = u^2-h^2 = u^2(1-v^2) \quad \Longrightarrow \quad R = u\sqrt{1-v^2} \quad \Longrightarrow \quad \left| J\right| = \frac{\sqrt{1-v^2}}{u} \\
I = \iint uv \cdot \frac{e^{-\Gamma\,u}\,u\sqrt{1-v^2}}{u^3} \left(\frac{u}{\sqrt{1-v^2}}\right)du\,dv =
\int_0^\infty e^{-\Gamma\,u}\,du \cdot \int_0^1 v\,dv = \frac{1}{2\Gamma}
$$
If the force components on the test particle from below and from above are considered separately,
then we observe that each of the corresponding integrals is equal to the same (extremely large) value.
So these finite outcomes neatly cancel each other out. Formally, with $\,g=-h\,$:
$$
F = 2\pi G m \rho_c \left\{ \int_{-\infty}^{0} h \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+h^2}}R\,dR}{(R^2+h^2)^{3/2}}\right] dh
+ \int_{0}^{\infty} h \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+h^2}}R\,dR}{(R^2+h^2)^{3/2}}\right] dh \right\} \\
F = 2\pi G m \rho_c \left\{ - \int_0^\infty g \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+g^2}}R\,dR}{(R^2+g^2)^{3/2}}\right] dg
+ \int_{0}^{\infty} h \left[ \int_0^{\infty} \frac{e^{-\Gamma\sqrt{R^2+h^2}}R\,dR}{(R^2+h^2)^{3/2}}\right] dh \right\} \\
F = 2\pi G m \rho_c \left\{ - \frac{1}{2\Gamma} + \frac{1}{2\Gamma} \right\} = 0 \qquad (3)
$$
The physical meaning of $\,1/\Gamma=c/H\,$ is the speed of light in vacuum times a Hubble time. Which in some circles is called the size of the Observable universe. Thus what we finally have is the force exerted on the test particle by half the universe below it, canceled by the force of half the universe above it. So to speak. We will return to this in another article of the series.