Starting with the Hydrogen atom as an example,
a dimensionless formulation of the Schrödinger equation for many electron atoms is presented. It is shown that the energy levels $E$ of such a system are only dependent on the fine structure constant $\alpha$, the atomic number $Z$, the speed of light $c$ in vacuum and elementary particle (rest) masses $m$ of electron and nucleus. If $\alpha$ and $c$ may be assumed to be constants of nature, then $E$ is only dependent on mass $m$. Introducing the Variable Mass hypothesis (VM) leads to formulas for the so-called intrinsic redshift. Ticks of Atomic clocks $\Delta t$ are shown to be inversely proportional to mass $m$.
Atomic clocks may seem the most accurate nowadays. Common timing in life, however, is still based upon seconds, minutes and hours, as parts of daylength, and days as parts of a year. This may be called the Orbital timeframe. By considering orbits of planets such as the earth, it is argued that ticks of Orbital clocks $\Delta T$ are inversely proportional to the square root of mass $m$.
If it is accepted that there are two clocks instead of one, then there is a discrepancy between the Atomic ticks $\Delta t$ and the Orbital ticks $\Delta T$, due to Variable Mass. A simple differential equation is derived that relates the two timeframes. For mass in Atomic time $t$, a formula can be derived that shows exponential growth. With help of the differential equation, a formula can be derived for mass in Orbital time $T$ as well, showing quadratic growth and
a beginning for $T=-A$, meaning that there is an age $A$.
Empirical evidence for both the quadratic and exponential growth formulas is found in the Shrinking Kilogram.
Quite unexpectedly, our version of the VM theory is consistent with the infamous C-decay. On the other hand, Variable Mass offers an explanation for the anomalous Earth's Rotation Retardation. At last, it is proved that Leap Seconds cannot be avoided with Atomic time and Orbital time, when combined with the Variable Mass hypothesis. And there is a very simple formula for calculating them.
CONTENTS
From e.g. Wikipedia we have that the Schrödinger equation for the Helium atom [4] may be written as $$ \left[-\frac{\hbar^2}{2m}\nabla_1^2-\frac{\hbar^2}{2m}\nabla_2^2 -\frac{2e^2}{4\pi\epsilon_0 |\vec{r}_1|}-\frac{2e^2}{4\pi\epsilon_0 |\vec{r}_2|} +\frac{e^2}{4\pi\epsilon_0 |\vec{r}_1-\vec{r}_2|}\right] \psi(\vec{r}_1,\vec{r}_2) = E\mbox{ }\psi(\vec{r}_1,\vec{r}_2) $$ Further meaning of symbols:
The idea of two clocks is found with several authors.
An exact matching with our own theory is not claimed.
| According to: | QUANTUM | GRAVITY |
| Barry Setterfield [33] | Atomic time | Orbital time |
| Arthur Milne | Dynamical time | Kinematic time |
| Fred Hoyle [34] | Minkowski frame | Einstein frame |
| Mike Helland [35] | Expanded time | Minkowski time |
The picture below and corresponding naming by Barry Setterfield [18].
The accompanying formulas relating Orbital clock ticks $\Delta T$ and Atomic clock ticks $\Delta t$ to Variable Mass $m$ have been discovered as the formulas (6) and (7) respectively.
$$
\frac{\Delta t}{\Delta t_0} = \frac{m_0}{m} \quad \mbox{and} \quad \frac{\Delta T}{\Delta T_0} = \sqrt{\frac{m_0}{m}}
$$
In order to be able to proceed, we must assume that Orbital time $T$ and Atomic time $t$ are synchronized at the here and now timestamp. This means human action for setting the clocks:
The Variable Mass Theory, according to Jayant Narlikar and Fred Hoyle, is about gravity only. This means that their timeframe is likely to be Orbital. In our notation: $t \rightarrow T$, replace $t$ by $T$ in the first place.
Another issue is their choice of the timestamp $t=T=0$. In the picture by Narlikar and Hoyle, time is set to zero at the timestamp where the "age" of the particle is zero as well as its mass. So let us clarify what the age is in our own formalism for $T=-A$.
$$
\frac{m}{m_0} = 0 \quad \Longleftrightarrow \quad 1+\frac{H}{2}(-A) = 0 \quad \Longleftrightarrow \\
A = \frac{2}{H} \quad \mbox{or} \quad \frac{1}{2}H = \frac{1}{A} \qquad (11)
$$
It follows that Narlikar's law can also be written as
$$
\frac{m}{m_0} = \left(1+\frac{T}{A}\right)^2
$$
And with the differential equations (8) the other way around:
$$
\frac{dt}{dT} = \sqrt{\frac{m_0}{m}} = \frac{1}{1+T/A} \quad \Longrightarrow \quad
t/A = \int dt/A = \int \frac{d(T/A)}{1+(T/A)} = \ln(1+T/A) \\ \Longrightarrow \quad e^{t/A} = 1+T/A
\quad \Longrightarrow \quad e^{H t} = \left(1+\frac{1}{2}H T\right)^2 = \frac{m}{m_0}
$$
where the condition ( $m=m_0 \Longleftrightarrow T=t=0$ ) has been taken into account.
Now take a look at the 1993 publication [19]
by Jayant Narlikar and Halton Arp. Then we find similar results and more. Starting with their formula (8) and converting everything into our own notation (e.g. $t_0\rightarrow A$):
$$
1+z = \frac{t_0^2}{\left[t_0-(r/c)\right]^2} = \frac{A^2}{\left[A-(r/c)\right]^2}
\quad \Longrightarrow \quad \frac{1}{1+z} = \left(1+\frac{T}{A}\right)^2 \quad \mbox{with} \quad r = -c.T
$$
where $r=$ distance. Is that correct within our version of the Variable Mass Theory? Yes it is, because of (5).
$$
H = \frac{2}{A} \quad \mbox{in} \quad \frac{1}{1+z} = \frac{m}{m_0} = \left(1+\frac{H}{2}T\right)^2
$$
Near equation (13) of the Narlikar-Arp paper we quote: expansion of equation (8) gives for small redshifts
$$
z = \frac{2r}{ct_0} , \quad \mbox{i.e. ,} \quad H_0 = \frac{2}{t_0}
$$
Equivalently in our version that would be:
$$
z = \frac{2r}{cA} \quad ; \quad H = \frac{2}{A}
$$
Thus ultimately revealing the meaning of the constant $H$. It is similar to the $H_0$ constant in
Hubble's law [20].
Interesting to notice that - in order to arrive at analogous results as in [19] -
General Relativity is not needed.
GR is effectively replaced by VM. GR has not become invalid, but it's rather redundant for the purpose. Or, as stated in [19]: Finally, the Euclidean, flat spacetime becomes a natural, primary reference frame in which cosmological processes are most simply described. The Variable Mass Theory is compatible with a static Euclidean universe, which is eternal in Atomic time and has moments of creation (i.e. one or more beginnings) in Orbital time.
8. Shrinking Kilogram
It has been found that $H=H_0=$ the Hubble constant. The benefits of this finding are questionable, because there exists a so-called Hubble tension [20]. Therefore $H_0$ is not so much a constant. It can even be conjectured that every other way of establishing a value for $H_0$ might give rise to different and incompatible outcomes. Therefore a better name for $H$ would be the Hubble parameter.
It has also been found that $A=2/H$, where $A$ is an Age in the Orbital timeframe. But the same $A$ cannot be an Age in Atomic time, as is obvious from the exponential increase in $t$, without a beginning. Summarizing (9), (10), (11):
$$
\frac{m}{m_0} = \left(1+\frac{H}{2}T\right)^2 = e^{H.t} \quad \mbox{with} \quad H = \frac{2}{A}
$$
In case there exists more than one moment of creation - say one for the universe as a whole ($A=28.8$ billion years) and one for the earth in particular ($A=4.9$ billion years) - then there is an issue that resembles the above discrepancies with the Hubble parameter. It seems to be wise to anyway not identify under all circumstances our $H$ with the common $H_0$.
The Hubble parameter is determined from e.g. redshift data, resulting in an estimate for the (gravitational) age of the universe. This sort of knowledge is of course absent if we want to determine the age of the earth. But it's still possible to define a sort of Hubble parameter for the Earth, by employing the above relationship: Age is twice the Hubble time. Let it be so by definition.
But if such is the case with the Earth, why not then with other objects / features in the universe, such as CMB, BAO, Cepheids, Red Giants, Quasars, Megamasers, Galaxies. They all may have different creation timestamps and hence different Hubble parameters.
That being said, we are going to present some direct empirical evidence of Variable Mass on an earthly scale.
From Shrinking Kilogram Bewilders Physicists [21] we quote:
A kilogram just isn't what it used to be. The 118-year-old cylinder that is the international prototype for the metric mass, kept tightly under lock and key outside Paris, is mysteriously losing weight - if ever so slightly.
Were slightly means 50 micrograms. More details are found
elsewhere [22].
According to VM we have, with meaning of the symbols as in previous writings
$$
\frac{m}{m_0} = \left[1+\frac{H}{2}T\right]^2 = \left[1+\frac{T}{A}\right]^2 \approx 1+2\frac{T}{A} \\
\mbox{Alternatively:} \quad \frac{m}{m_0} = e^{H.t} = e^{2t/A} \approx 1+2\frac{t}{A}
$$ $$
\frac{m-m_0}{m_0} \approx \frac{2 \times 118}{4.54 \times 10^9} \approx 52 \times 10^{-9}
$$
Thus $(m-m_0)$ is indeed approximately 50 micrograms when compared with $m_0 = 1\mbox{ }kg$.
Has the standard kilogram really been shrinking? No! Thanks to its splendid isolation it has remained constant all the time. But all matter around it has been suffering from Increasing Mass. Due to its interaction with cosmos [8] the latter has become heavier, with the amount as calculated.
9. C-decay
There is an issue with the theory in the paper by Arp and Narlikar, due to the fact that only one kind of timeflow is recognized, which is Orbital time, also called gravitational time by some. But with our version of the Variable Mass Theory there is also the Atomic timeframe. The (vacuum) speed of light in both timeframes is defined by a distance $dr$ divided by a time increment, in two ways.
$$
c_t = \frac{dr}{dt} \quad \mbox{and} \quad c_T = \frac{dr}{dT}
$$
If it is assumed that the lightspeed in Atomic time is an absolute constant ( i.e. $c_t=c_0$ ) then the lightspeed in Orbital time must be different. Using (8) and (10) we have
$$
c_T = \frac{dr}{dT} = \frac{dr}{dt}\frac{dt}{dT} \quad \mbox{with} \quad
\frac{dT}{dt} = \sqrt{\frac{m}{m_0}} = 1+\frac{H}{2}T \quad \Longrightarrow \quad c_T = \frac{c_0}{1+\frac{H}{2}T}
$$
Therefore the infamous C-decay [23] (take $T\le 0$) is - quite unexpectedly - compatible with (our version of) the Variable Mass Theory.
In the Arp-Narlikar paper a bit more complicated derivation shall be needed for reproducing their formula (13). Of minor importance, because the equivalent result can be proved much easier within the Atomic timeframe, using (5) and (9).
$$
1+z = e^{-H.(-t)} \approx 1+H.t \quad \Longrightarrow \quad z = H.t = H.\frac{r}{c_0} = \frac{2r}{c_0.A}
$$
A far more serious consequence of c-decay is the following though. For $mc^2$ we have conservation of mass-energy in Orbital time, which is the proper timeframe for gravitation.
$$
c_0 = c\sqrt{\frac{m}{m_0}} \quad \Longrightarrow \quad
\frac{m c^2}{m_0 c_0^2} = \frac{m}{m_0}/\frac{c_0^2}{c^2} = \frac{m}{m_0}/\left(\sqrt{\frac{m}{m_0}}\right)^2 = 1
\quad \Longrightarrow \quad mc^2 = m_0c_0^2
$$
Recall the dimensionless formula (2) for the energy levels of an arbitrary atom. And apply conservation of mass-energy in there.
$$
\left[-\frac{1}{2}\sum_{i=1}^Z\nabla^2_i(\vec{\rho}) - Z\alpha\sum_{i=1}^Z\frac{1}{|\vec{\rho_i}|}
+\alpha\sum_{i=1}^Z\sum_{j=i+1}^Z\frac{1}{|\vec{\rho_i}-\vec{\rho_j}|}
\right]\psi(\vec{\rho}_i) = \frac{E}{mc^2}\mbox{ }\psi(\vec{\rho}_i) = \frac{E}{m_0c_0^2}\mbox{ }\psi(\vec{\rho}_i)
$$
Due to Quantum Mechanics, $E$ remains an absolute constant in Orbital time. So there is no Intrinsic Redshift at all, at least not within the pure gravitational framework as designed by Halton Arp & Jayant Narlikar!
10. Milne's Formula
The basic formulas (9),(10),(11) in (our version of) the VM theory can not be repeated frequently enough.
$$
\frac{m}{m_0} = \left(1+\frac{1}{2}H.T\right)^2 = e^{H.t} \quad \mbox{with} \quad H = \frac{2}{A}
$$
where $m=$ mass, $H=$ Hubble parameter, $T=$ orbital time, $t=$ atomic time, $A=$ age. The result can also be written as follows.
$$
\sqrt{\frac{m}{m_0}} \quad : \quad 1+\frac{1}{2}H.T = 1+\frac{T}{A} = e^{(H/2)t} = e^{t/A}
$$
Herewith it's easy to determine $T$ as a function of $t$, and $t$ as a function of $T$.
$$
T = A\left(e^{t/A}-1\right) \quad \mbox{and} \quad t = A\mbox{ }\ln\left(1+\frac{T}{A}\right)
$$
Professor Edward Arthur Milne too has distinguished two kinds of time: ephemeral or dynamical time $\tau$, and absolute or kinematic time $t$, as repeated over and over again in three of his papers:
In a book [27] by Tom van Flandern [28] we find in a footnote on page 95 the following.
This formula in the Meta model can be derived with Milne's (modified) formula.
$$
\sqrt{\frac{1}{1+z}}=\sqrt{\frac{m}{m_0}} = 1+\frac{H}{2}T \quad \mbox{with} \quad A = \frac{2}{H}
$$ $$
d = c.(-t) = -c.A\mbox{ }\ln\left(1+\frac{T}{A}\right) = -c\frac{2}{H}\mbox{ }\ln\left(\sqrt{\frac{1}{1+z}}\right)
= -c\frac{2}{H}(-\frac{1}{2})\ln(1+z)
$$ $$
d = \frac{c}{H}\mbox{ }\ln(1+z) \qquad (12)
$$
However, the above derivation with help of Milne's formula is too much of a good thing. If we restrict attention to the Atomic timeframe only, then the proof is reduced to a one-liner by using (5) and (9) again.
$$
1+z=e^{-H.t} \quad \Longrightarrow \quad \frac{H}{c}(-ct) = \ln(1+z) \quad \Longrightarrow \quad d = \frac{c}{H}\mbox{ }\ln(1+z)
$$
It should be noticed that the logarithmic distance formula is not at all restricted to van Flandern's Meta theory. And there are Hubble tensions in $H$, making it possible that there are different intrinsic redshifts $z$ at the same distance $d$.
We continue with a quote from the book Origin of Inertia [29] by Amitabha Ghosh, on page 13: It is now an established fact that the spin motion of the Earth is gradually slowing down, and the magnitude of this secular retardation is about $6 \times 10^{-22}\mbox{ } rad/s^2$. That amount of information is already sufficient.
11. Earth's Rotation Retardation
Let's resume an equivalent of the formulas (9),(10),(11) with $A=$ age.
$$
1+T/A = e^{t/A} \quad \Longleftrightarrow \quad t = A\mbox{ }\ln(1+T/A)
$$
Calculate velocities $v$ and accelerations $a$ from displacements $s$ within the two time frames.
$$
\vec{v}_T = \frac{d\vec{s}}{dT} = \frac{d\vec{s}}{dt}\frac{dt}{dT} = \frac{\vec{v}_t}{1+T/A}
$$ $$
(1+T/A)\vec{v}_T = \vec{v}_t \qquad (13)
$$
It is noticed that results in 9. C-decay are a special case, namely for $\vec{v}_T = c_T$ and $\vec{v}_t = c_0$.
Patiently differentiating further with product rule and chain rule gives
$$
\vec{a}_T = \frac{d^2\vec{s}}{dT^2} = \frac{d}{dT}\left(\frac{d\vec{s}}{dt}\frac{dt}{dT}\right) = \frac{d}{dT}\left(\frac{d\vec{s}}{dt}\right)\frac{dt}{dT} + \frac{d\vec{s}}{dt}\frac{d}{dT}\left(\frac{dt}{dT}\right) =
$$ $$
= \frac{d^2\vec{s}}{dt^2}\left(\frac{dt}{dT}\right)^2 + \frac{d\vec{s}}{dt}\frac{d^2t}{dT^2}
$$
Hence
$$
\vec{a}_T = \vec{a}_t\left(\frac{1}{1+T/A}\right)^2 + \vec{v}_t\frac{-1/A}{(1+T/A)^2}
$$ $$
(1+T/A)^2\,\vec{a}_T = \vec{a}_t - \vec{v}_t/A \qquad (14)
$$
Under earthly circumstances here and now we have that $t=T=0$ and so
$$
\vec{v}_T = \vec{v}_t \\
\vec{a}_T = \vec{a}_t - \vec{v}_t/A
$$
Conclusion: velocities in the two time frames are the same, but accelerations are not.
According to our own theory we have from the above: $\vec{a}_T - \vec{a}_t = - \vec{v}_t/A$. Here $\vec{a}_T$ is the acceleration in Orbital time, $\vec{a}_t$ is the acceleration in atomic time, and $A$ is kind of an age. Due to the non-linear relationship between atomic time $t$ and orbital time $T$ a discrepancy between orbital and atomic accelerations shows up. Regardless of the question which of the accelerations is the "true" one, the difference of the two is a vector in the (opposite) direction of one and the same velocity. Furthermore, all accelerations and velocities are proportional with $R=$ radius of the earth = constant, as required by the implicit use of (7).
Now let the unit vector in tangential direction be defined by
$$
\vec{e}_v = \frac{\vec{v}_t}{|\vec{v}_t|} = \frac{\vec{v}_T}{|\vec{v}_T|}
$$
The velocity (component) at the surface of the earth is simply the (constant) radius of the earth times angular velocity $\omega$.
$$
\left(\vec{v}_t\cdot\vec{e}_v\right) = \left(\vec{v}_T\cdot\vec{e}_v\right) = R\,\omega
$$
The component of the orbital minus atomic acceleration is in tangential direction too and can be nothing else than the radius times a change of the same angular velocity.
$$
\left(\vec{a}_T-\vec{a}_t\cdot\vec{e}_v\right) = R\,\dot{\omega}
$$
It follows that the rotation retardation must be described by
$$
\left(\vec{a}_T-\vec{a}_t\cdot\vec{e}_v\right) = -\left(\vec{v}_t\cdot\vec{e}_v\right)/A
$$ $$
\dot{\omega} = - \omega/A \qquad (15)
$$
Calculation (with MAPLE) using the age of the earth gives
# Age of the earth
age := 4.543*10^9;
# Seconds in a year
year := 31556926;
# Seconds in a day
day := 60*60*24;
# Rotation of the earth
omega := 2*Pi/day;
# Retardation
slower := evalf(omega/(age*year));
We find that the retardation is about $5 \times 10^{-22} \mbox{ } rad/s^2$
12. Leap Seconds
There is a small but positive discrepancy between the Atomic daylength and the Orbital daylength, which is exemplified by the existence of Leap Seconds [30].
Empirical
data are obtained from the Wikipedia reference Δ T (timekeeping) [31]. Quote: The value of $\Delta T$ for the start of 1902 is approximately zero; for 2002 it is about 64 seconds. So the Earth's rotations over that century took about 64 seconds longer than would be required for days of atomic time.
Atomic clocks are running somewhat too fast. Therefore it is necessary to "stop" them once or twice in a year or so (source of picture = Wikipedia).
Let's see if we can reproduce. According to common kinematics we have, after a century of the earth's rotation, for its shortage of radians: $$ \Delta \theta = \frac{1}{2}\dot{\omega} (\mbox{century})^2 $$ It's a simple matter now to derive with help of (15) the number of leap seconds: $$ \mbox{leaps} = \frac{\Delta \theta}{2\pi}\times\mbox{day} = \Delta \theta \times \frac{\mbox{day}}{2\pi} $$ But we are a factor two off:
theta := slower/2*(100*year)^2;
leaps := evalf(theta/(2*Pi)*day);
leaps := 34.73137354
There is another way to look at the Leap Seconds problem, which is rather based upon the length of a year instead of the daylength. Let's go back to the original definition of Orbital time as defined by the orbit of the earth around the sun and the basic formulas (9),(10),(11).
$$
1+\frac{1}{2} H T = e^{(H/2)t} \approx 1 + \frac{1}{2}H t + \frac{1}{2}\left[\frac{1}{2}H t\right]^2
$$
Let both the Atomic and Orbital clock be synchronized as usual ($t=T=0$) and let them run free for a century, that is
$$
T = (\mbox{century})_T \quad \mbox{and} \quad t = (\mbox{century})_t
$$ $$
\left(1+\frac{1}{2} H T\right) - \left(1 + \frac{1}{2}H t\right) = \frac{1}{2} H(T-t)
\approx \frac{1}{2}H.\frac{1}{4}H t^2 \quad \Longrightarrow \\
(\mbox{century})_T - (\mbox{century})_t \approx \frac{1}{4}H (\mbox{century})^2
= \frac{1}{2}(\mbox{century})^2/\mbox{age}
$$
It is seen that the Atomic clock over years is again running faster than the Orbital clock. Let's calculate:
leaps := (100*year)^2/(2*age*year);
leaps := 34.73137354
Surprisingly, the two leaps are exactly the same. That's the reason why I have shown all the decimals as calculated by the Computer Algebra System.
It is obvious, however, that the two calculations correspond with two different phenomena. The first one describes that there is an additional time leap needed for a rotation of the earth to complete its periods over a day.
The second one describes how atomic clocks are running ahead of orbital clocks, defined e.g. by the earth orbiting around the sun in a year.
Thus the orbital clock is running faster over a year with the same amount
as calculated with daytime! Is that a coincidence? Mathematically speaking: no! It turns out that
the calculation for daytime is actually too much of a good thing. It can be simplified a lot, giving exactly the same outcome as for the time of a year.
$$
\mbox{leaps} = \frac{1}{2} \left[ \left(\frac{2\pi}{\mbox{day}}\right)/\mbox{age} \right] (\mbox{century})^2
\times \left(\frac{\mbox{day}}{2\pi}\right) = \frac{1}{2}(\mbox{century})^2/\mbox{age}
$$
For our leap second this would mean that the outcome $\approx 34.7$ should be taken twice instead of once. Thus improved calculation of the leap interval would reveal that it is about $2\times 34.7 \approx$ 69 seconds.
Remember that observation of the leap interval reveals that it is about 64 seconds.
At last it is clear that the "Age of the Earth" is relevant for the calculations and not the "Age of the Universe", whatever the latter might mean. So what we finally have is leaps $=2\times(\mbox{century})^2/(2\times \mbox{age})$. Unless we have been seduced by wishful thinking, the
conclusion is that leap seconds can be calculated by the following extremely simple formula.
$$
\mbox{leap seconds} = \frac{(\mbox{ seconds in a century })^2}{\mbox{age of the earth in seconds}} \qquad (16)
$$
Acknowledgements: I am grateful to Louis Marmet, Mike Helland and Francois Zinserling for their valuable
comments in the ACG discussions
forum [32].
Conflicts of Interest: The author declares no conflicts of interest.
Funding: This research received no external funding.
13. References
video: Variable Mass Theory: Intrinsic Red-Shift
video: The theory to REPLACE the Big Bang | Nonexpanding Ep. 8
FLAT SPACETIME COSMOLOGY: A UNIFIED FRAMEWORK FOR INTERGALACTIC REDSHIFTS
Jayant Narlikar and Halton Arp. The Astrophysical Journal 405: 51-56, 1993 March 1
First Published 2000 by Apeiron. 4405, rue St-Dominique, Montreal, Quebec H2W 2B2 Canada. ISBN 0-9683689-3-X