a-cosmology-group/acg#292

@RedshiftDrift . Confession: before I went to sleep it looked better than it looks now :-( Must go back to the drawing board, I guess.

Confession: before I went to sleep it looked better than it looks now @HanDeBruijn, confession: this happens to me all the time…

But that’s better than stubbornly keeping alive a bad idea (like a zombie Big Bang model).

There is an error in the above solution procedure for the equations

$$ \begin{cases} \large \frac{du_1}{dt} = H_{TL}(u_2-u_1) \ \large \frac{du_2}{dt} = H_{VM}(u_1-u_2) \end{cases} $$

The first equation is partly solved by addition, yes, but it went wrong:

$$ \frac{d(u_1+u_2)}{dt} = 0 \quad \mbox{must be} \quad \frac{d\left[(u_1/H_{TL})+(u_2/H_{VM})\right]}{dt} = 0 $$

Resulting in

$$ H_{VM}u_1+H_{TL}u_2 = A \quad \mbox{instead of} \quad u_1+u_2 = A $$

still to be combined ith $H_0 = H_{VM}+H_{TL}$ and

$$ u_1-u_2 = B e^{-H_0 t} $$

Now multiply the latter with $(H_{TL},H_{VM})$ and perform further addition and subtraction with the former.

$$ u_1 = \left[A + B H_{TL} e^{-H_0 t}\right]/H_0 \quad ; \quad u_2 = \left[A - B H_{VM} e^{-H_0 t}\right]/H_0 $$

Written otherwise, with $u_\infty = u_1(\infty) = u_2(\infty) = A/H_0 =(H_{VM}u_1+H_{TL}u_2)/(H_{VM}+H_{TL}) =$ weighted mean, and replacing time $t$ by distance $x=ct$ with $c=$ lightspeed:

$$ u_1(x) = u_\infty + B \frac{H_{TL}}{H_0} e^{-(H_0/c)x} \quad ; \quad u_2(x) = u_\infty - B \frac{H_{VM}}{H_0} e^{-(H_0/c)x} $$

Fortunately, nothing becomes essentially different with the above preliminary sketch. But I’m not satisfied with the result for Variable Mass. Indeed an exponential function comes out but previous “established” theory suggests that associated energy density goes

$$ \mbox{like} \quad \frac{\rho c^2}{\rho_0 c^2} = e^{-(H_0/c)x} \quad \mbox{and not like} \quad \simeq \left(1-e^{-(H_0/c)x}\right) $$

where $\rho=$ mass density.

Update(s).

It’s easy to see that $u_\infty = 0$. Because VM starts with zero energy (density) at $t=-\infty$ and TL ends with zero energy (density) at $t=+\infty$ .

$$ u_{TL}(x) = + B \frac{H_{TL}}{H_0} e^{-(H_0/c)x} \quad ; \quad u_{VM}(x) = - B \frac{H_{VM}}{H_0} e^{-(H_0/c)x} $$

Only some magic is needed now to flip the minus in $(-B)$ into a plus. Unfortunately though: magic is not science.

A conclusion seems to be inevitable. Tired Light cannot be caused by Variable Mass; Variable Mass cannot be caused by Tired Light. If the two phenomena are really existent in Nature, then anyway they are independent. It is not excluded that they occur in combination (such as is the case with other redshifts) but that’s all. Let’s summarize in two pictures.

Tired Light:

Variable Mass:

Accompanying formulas.

$$ \large -A \le t \le 0 \quad ; \quad \frac{1}{1+z_{TL}} = e^{-H(t+A)} \quad ; \quad \frac{m_{VM}}{m_0} = e^{Ht} $$

where $A=$ age, $z=$ redshift, $H=$ Hubble parameter, $t=$ time, $m=$ mass, $0=$ here and now. Both resulting in an observed redshift $\large z_0 = e^{HA}-1 = e^{2.5}-1\approx 11$.

Sigh! No further mistakes hopefully ..

@HanDeBruijn when is time zero? Now … or ‘going back when’? Why does TL start at t=0 but VM ends at t=0?

Something else I’m thinking of if you apply variable mass, is the measurement of wavelength from a varied mass will differ. If an incoming wavelength is constant, then a varying mass will not measure it as constant. I think you should try graph that too but the curves would diverge away from each other? If you use frequency instead of wavelength, the curves could run together? (I have not spent enough time thinking about this last line so feel free to challenge)

The easy example to quote is in a gravity well where time is distorted, can be seen as the clock of the ‘measuring mass’ is slower, hence the incoming waves appear (relative to its own clock) faster in time, and wavelength measured appears bluer. i.e. Gravitational Blueshift. It would be easy to associate clocks with mass, since it acts inversely just like a musical instrument - e.g. long string low frequency. (long string ≈ low energy ≈ low mass in this analogy). Also same implied in the energy equations E=hf and E=mc^2.

when is time zero? Now … or ‘going back when’? Why does TL start at t=0 but VM ends at t=0?

Sorry, @Francois-Zinserling . The pictures were wrong. I have provided new ones in silent updates of the reply.

See above replies for pictures and symbols. It has been proposed more than once that Variable Mass and Tired Light can be combined as follows.

First we have Variable Mass, resulting in a far away redshift $z_{VM}$. From there, Tired Light is emitted with a redshift $z_{TL}$. If the Hubble parameters for both phenomena are the same ($H_{VM}=H_{TL}$), with dimensionless age $HA=1$, then the light arrives here and now with an observed redshift (+1)

$$ 1+z_0 = (1+z_{VM})(1+z_{TL}) = e^{H_{VM}A}.e^{H_{TL}A} = e^{(H_{VM}+H_{TL})A} = e^{1+1} \approx 7.4 $$

Is that right? No, it is plain wrong !! At least for STz. As has been stated correctly in a reply by @RedshiftDrift : >VM uses the hypothesis of varying mass, which contradicts the basic physics STz is built on. VM is not compatible with STz.

Indeed. The problem is in the Hubble parameter for STz (same argument for NTL too). We repeat :

$$ H_{STz} = \frac{2}{\pi}\frac{\alpha h^2}{m_e^2 c} n_e(STz) \qquad ; \qquad H_{NTL} = \frac{1}{\pi}\frac{\alpha h^2}{m_e^2 c} n_e(NTL) $$

Because the electron mass $m_e$ is with a Variable Mass now [ $m/m_0 = \exp(H_{VM}t)$ ] we get instead (for TL = STz here)

$$ H_{TL} = \frac{2}{\pi}\frac{\alpha h^2}{m_e^2 c} n_e(STz) \large \times e^{-2H_{VM}.t} $$

Leading to all kind of absurdities. The conclusion is inevitable.😥

VM is not compatible with STz / NTL . It is one or the other but not both.

VM is not compatible with STz / NTL . It is one or the other but not both.

I think you are giving up too early @HanDeBruijn… Where is your usual “checking the units”?

$m_e$ has dimensions of [ $kg$ ]. $h$ has dimensions of [ $kg \ m^2 \ s^{-1}$ ] $H_{TL}$ is a constant in VM, so the first equation is correct.

Come on !! Where then is the end of our VM purported “new physics”? Are we going to vary Planck’s constant $h$ too? Only because there is a $kg$ in its dimensionality? In that case, $G$ must be a vaying gravitational constant as well, giving that much of the Variable Mass becomes meaningless, because of the Standard gravitational parameter $GM$ in theory.

In that case, G must be a vaying gravitational constant as well

Yes. Or at least GM must be variable, as you already say. It is the simplest explanation that does away with DM. This solution is not ad-hoc (like MOND ) either, since the presence of ‘DM’ is predictable, hence DM can be cancelled by varying GM in cosmic scenarios, e.g. Galactic Disk, SMBH.

Little misunderstanding, @Francois-Zinserling . According to its dimensionality $G$ would vary like $\mbox{mass}^{-1}$ and $M$ of course like mass itself. This would mean that $GM$ as a whole is constant in time, thus effectively killing much of our beloved Variable Mass theory.

@HanDeBruijn Your VM is based on (as far as I recall - please correct me if I’m wrong):

• Low mass emits a signal e.g. hydrogen line
• High mass receives the signal and reads it as low relative to its own reference of a hydrogen line.

You would say because high mass has a higher internal clock, it perceives the incoming wave as redshifted. But who is driving the clock? It cannot be internal to the mass because clocks require energy. Just as do electric fields and gravity. So mass doesn’t have its own clock but requires (something like) an aether to ‘tick the clock’. But yes from the simple relation E = mc^2 = hf we can infer that a mass’ internal clock will speed up if the mass of the fundamental particles increased. And here the ‘more mass’ is not ‘more atoms’, but each nuclei and electron’s mass increased.

Then, if GM is considered as a measure of strength (in vicinity of a gravitational mass M), a stronger GM causes us to read an incoming wave as more energy, hence waves appear blueshifted. Just as we would experience in a gravity well and we know this is a real and observed and tested phenomenon. We could thus say GM is a measure of the local aether gradient.

NB to note in the above example of gravitational blueshift, the wave hasn’t changed, but the observer’s clock has changed. Is this also so in your method? In that case yes, your system is not TL (STz or otherwise) at all because your wave was never changed. It does not mean your idea is wrong, but yes it’s likely one or the other - unlikely both.

By extension, my hypothesis, a distant signal that appears redshifted locally, if we were to blame GM for the transition, must mean that ‘GM’ has weakened over time or distance. Since M is local (and also $f(local-aether)$!) it means G has weakened. I vote distance. This is in line with @ExpEarth ideas, although I still find the shell to be exotic and possibly even unnessecary. Still chewing on all that.

Here your increased (local) M would have had the same observation effect, so it’s not yet time to chuck out the baby.

The “heat exchanger” analogy simply describes energy exchange.

The cosmic system you describe moves energy inward and outward inside a closed volume.

It would be like using a heat exchanger with the two outputs connected to the two inputs (if I understand this correctly? As in diagram below?)

Not exactly the kind of thing engineers would like to see 😉 You will have to make your own diagram!

“Poor artist’s” version of a 1D ‘radial’ heat exchanger showing the energy flow in the shell universe (image from this site) The colour coding is misleading because “red” is hot but represents “blue”-shifted photons…!

Thanks for the useful links, Louis. I agree that engineers won’t like this heat exchanger universe. It’s doing no work - useless!

I will try to get a diagram together.

Look what you started @HanDeBruijn and @RedshiftDrift ! I think I will be modifying my gravity model as a result of this exercise.

Okay, it works something like this. The CMB photons that contact baryons in the shell are scattered toward the interior. The CMB photons are the thin coloured arrows. The recoil from the reemission gives the baryons a push towards the exterior (large white arrows). Similarly, when a reheated CMB photon passes through the universe and hits a shell baryon, it pushes the baryon towards the exterior (white arrow again). So both interactions tend to push the shell towards a larger radius. This also has the effect of pulling the connected filaments of spacetime in that direction.

Now the CMB photons are redshifted as they move inward and blueshifted as they move back towards the shell. In the redshift, they impart some of their momentum to the spacetime filaments, pushing them towards the centre (small black arrows). When the CMB photons head back to the shell, they are reheated by the spacetime filaments in those regions. This causes the filaments to lose a portion of their momentum that was in the direction of the CMB photon, resulting again in a net push towards the centre.

The end result is that the forces on the shell baryons and the spacetime filaments are balanced. In essence, hotter CMB photons lose energy to cooler CMB photons moving in the opposite direction. (On this point, I notice a slight correspondence with your model @RedshiftDrift , where you have wave energy being transferred to a wave moving the opposite way.)

This is a work in progress. Hey, I’m not expecting anyone to love this heat exchanger!

Continuing this discussion of @ExpEarth’s gravity model here.

Hydrogen is the most simple example of an atom. We could try to solve the Schrödinger equation for its energy levels. But it’s much easier to use instead good “old” quantum mechanics, especially the Bohr model of the hydrogen atom. The latter can be considered anyway as a crude prototype for all atoms. An electron with charge $e$ and mass $m$ describes a circular orbit with radius $r$ and velocity $v$ around a nucleus with the same (opposite) charge. Then with centripetal acceleration and Coulomb’s law we have

$$ m\frac{v^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2} $$

Quantum Mechanics is entered into the theory by assuming that the circumference of the orbit contains a whole number $n$ of Matter wave -lengths $\lambda$ of the electron, according to

$$ 2\pi r = n\lambda = n\frac{h}{p} = n\frac{h}{mv} \quad \Longrightarrow \quad v=\frac{n\hbar}{mr} $$

where $p=$ momentum, $\hbar=$ reduced Planck’s constant. And so for the radii $r_n$ of the orbits:

$$ r = \frac{e^2}{4\pi\epsilon_0 m v^2} = \frac{e^2 (mr)^2}{4\pi\epsilon_0 m (n\hbar)^2} $$

$$ 1 = \frac{e^2 mr}{4\pi\epsilon_0 n^2 \hbar^2} \quad \Longrightarrow \quad r_n = \frac{4\pi\epsilon_0 n^2 \hbar^2}{e^2 m} $$

$$ r_n = \frac{4\pi\epsilon_0\hbar c}{e^2}\frac{\hbar}{mc}n^2 = \frac{\hbar}{mc}\mbox{ }\frac{1}{\alpha}\mbox{ }n^2 $$

where $\hbar/(mc)$ is recognized as the Reduced Compton wavelength of the electron and $\alpha$ is the fine-structure constant. The (main) energy levels for hydrogen, at last, are found as follows.

$$ -E_n = \frac{1}{2}mv^2 = \frac{1}{2}\frac{e^2}{4\pi\epsilon_0 r_n} = \frac{1}{2}\frac{e^2}{4\pi\epsilon_0}\frac{e^2 m}{4\pi\epsilon_0 n^2 \hbar^2} = \frac{1}{2}\left(\frac{e^2}{4\pi\epsilon_0 \hbar c}\right)^2 mc^2 \frac{1}{n^2} = mc^2\mbox{ }\frac{\alpha^2}{2}\mbox{ }\frac{1}{n^2} $$

where $(mc^2)$ is recognized as mass-energy of the electron. All this is in concordance with general theory.

@PatrGT . A varying fine-structure constant $\alpha$ is incompatible with any version of the Variable Mass theory .

The following result has been found. The energy levels $E$ of a single atom are given in general by a factor times a dimensionless spectral function $S$ .

$$ E = mc^2.S(\alpha,Z\alpha,\vec{Q}) $$

where $c=$ lightspeed in vacuum, $Z=$ atomic number, $\alpha=$ fine-structure constant, $\vec{Q}=$ quantum numbers and last but not least, $m=$ (inertial) mass, to be more precise:

$$ m = \frac{M m_e}{M + m_e} = \frac{(M/m_e)}{(M/m_e)+1}m_e \approx m_e $$

with $M=$ mass of nucleus, $m_e=$ mass of electron and $M \gg m_e$. Relativistic effects and nuclear physics have not been taken into account. It is important to understand that $\alpha$ is a universal constant. If an electron jumps between orbits, that event is accompanied by an emitted or absorbed amount of electromagnetic energy, also called a photon $h\nu$ .

$$ h\nu = E_1 - E_2 = mc^2\left[S(\alpha,Z\alpha,\vec{Q_1})-S(\alpha,Z\alpha,\vec{Q_2})\right] $$

Now let us asssume there is increasing Variable Mass (VM theory). Quantities here and now are denoted with a subscript $0$ .

$$ h\nu_0 = m_0c^2\left[S(\alpha,Z\alpha,\vec{Q_1})-S(\alpha,Z\alpha,\vec{Q_2})\right] $$

where $\nu=$ frequency and $m_0 \gt m$ .

$$ \frac{h\nu}{h\nu_0} = \frac{mc^2}{m_0c^2} \quad \Longrightarrow \quad \frac{\nu}{\nu_0} = \frac{m}{m_0} $$

The lightspeed $c$ as well as the Planck constant $h$ are supposed to be invariant constants of nature throughout the cosmos. Wavelengths $\lambda$ are more convenient than frequencies. With $\nu=c/\lambda$ we have, at last

$$ \large \frac{\lambda}{\lambda_0} = \frac{m_0}{m} $$

Assuming that (elementary particle rest) mass $m$ in the past has been smaller than mass $m_0$ here and now, we see that all wavelengths $\lambda$ as received from the past are larger than wavelengths $\lambda_0$ here and now. This is called Cosmological Redshift. Expressed into common $z=$ redshift notation:

$$ z = \frac{\lambda-\lambda_0}{\lambda_0} = \frac{\lambda}{\lambda_0} - 1 \quad \Longrightarrow \quad \large 1+z = \frac{m_0}{m} $$

The fact that redshift here is caused by a change of mass inside atoms has given rise to the name intrinsic redshift.

Suppose that we are building an Atomic clock instead of observing galaxies. Then the frequency or rather the period / tick $\Delta t$ of the clock is a more relevant quantity than the wavelength. The two are intimately related though.

$$ \lambda = c.\Delta t \quad \Longrightarrow \quad \frac{\lambda}{\lambda_0} = \large \frac{\Delta t}{\Delta t_0} = \frac{m_0}{m} $$

Therefore, according to VM, atomic clock ticks have been larger / dilated in the past. Such is in concordance with the Time Dilated Past theory by Mike Helland.

Not so long ago, there was the good old definition of the Second as a Fraction of an ephemeris year, defined by a fixed number of seconds (31 556 926) in a year, together with certain restrictions on the “here and now”. What if we had still adopted this Orbital time as the “true” time nowadays, combined with our Variable Mass Theory ? Classical Orbital Mechanics is relevant in this context, especially the theory of Elliptical orbits. The Orbital period $\Delta T$ of a body traveling along an elliptic orbit can be computed as:

$$ \Delta T = 2\pi\sqrt{\frac{a^3}{\mu}} \quad \mbox{with} \quad \mu = GM $$

where $a=$ the orbit’s semi-major axis, $G=$ gravitational constant, $M=$ mass of the more massive (attracting) body. Let $M=$ mass of our sun and $a=$ distance between the sun and the earth, then $\Delta T$ cannot be else than a year. Very much the same outcome can be reproduced though by a far simpler model, assuming a circular orbit instead of an elliptical one. With (centripetal force = gravitational force) we have, for the angular velocity $\omega = 2\pi/\Delta T$ and a circle with radius $a$ :

$$ m \omega^2 a = \frac{GMm}{a^2} \quad \Longrightarrow \quad \Delta T = 2\pi \sqrt{\frac{a^3}{GM}} $$

Of interest to us is only the (increasing) mass $M$. We see that, in general, there is sort of a Time Delayed Past for Orbital clocks too, which is however different from the one for Atomic clocks, namely

$$ \frac{\Delta T}{\Delta T_0} = \sqrt{\frac{M_0}{M}} $$

Part below dismissed, thanks to @Francois-Zinserling

We are not finished yet. The following definition of the Second was internationally adopted too for more than a century. The second (symbol: s) is a unit of time derived from the division of the day first into 24 hours, then to 60 minutes, and finally to 60 seconds each (24 x 60 x 60 = 86400). This means that the rotation of the earth has been - and still is - relevant for timekeeping in the Orbital timeframe. Let’s investigate with help of Google AI. The moment of inertia of a solid sphere is given by the formula I = (2/5)MR^2, where M is the total mass of the sphere and R is its radius. This formula applies when the sphere is rotating about an axis passing through its center. Wikipedia informs us further about the kinetic energy $E_{rot}$ of such a rotation (→ day) , apart from our orbit around the sun (→ year) . With $m=$ mass of the Earth:

$$ E_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{5}mR^2\omega^2 \quad \Longrightarrow \quad \Delta T = 2\pi\sqrt{\frac{E_{rot}}{mR^2/5}} $$

Assuming conservation of kinetic energy $E_{rot}$ , despite changing Variable Mass, we thus find the same basic formula as before.

$$ \large \frac{\Delta T}{\Delta T_0} = \sqrt{\frac{m_0}{m}} $$

The result is that an Orbital second can be defined in two ways, by 86,400 seconds in a day or by 31,556,926 seconds in a year. That makes no difference for the Orbital timeframe, when combined eventually with Variable Mass. For the sake of clarity, I do not think that Atomic time as such is “better” than Orbital time. We have to consider both timeframes, Orbital and Atomic, in order to be able to proceed.

Just thinking out loud here: In the case of the central mass (e.g. sun), increasing the mass should increase the ‘gravity’, hence increase the orbital speed. There is no reason to think the orbital radius will change, hence ΔT will get smaller. This is intuitive and appears correct in your equation where ΔT is inversely proportional to sqrt(M). Orbital ΔT will decrease for an increasing m (the rest can be considered constants)

However, in the case of revolving earth, an increase in mass also must have an increase in radius, since $m=4/3*ρπr^3$, while taking density as constant. Granted, the earth radius only increases by the third root of an increasing m, but still cannot be ignored and r remains a variable. Your last equation should then have $m_0r_0^2/mr^2$ inside the square root, and the $r_0$ vs $r$ components do not cancel out. It would appear from this new equation that there is still a decrease in ΔT for an increasing m but not to the same scale as the orbital ΔT. There should be a measurable variance between the two scenarios.

As you reason, kinetic energy is conserved. Angular momentum must also be conserved in the case of earth rotation. There is no reason to believe that the increase in mass will come with its own exact added proportion of angular momentum. An increase in mass, and a subsequent increase in radius - given a fixed total angular momentum - must result in a slower rotation $ω$ as is also evident to balance your $E_{rot}$ equation. In this case $ω$ decreases, and ΔT increases and there is a disconnect between the scenarios.

A neat way to escape from this discussion is to simply admit that I am wrong and you are right. Proceeding in the way as proposed by me here would have lead to contradictions in the theory. VM now predicts a faster rotation of the earth while instead it must result in a slower rotation. As has been argued on Jan 30, 2024 , by myself ! Thank you very much for pointing this out, @Francois-Zinserling ! You are my guardian angel 🥇

I will not gloat on this because I sincerely admire your efforts (and also because I have yet to make my equations look as elegant as yours).

Here on the fringes of science we only have each other.

Asking about the unit of clock time, at least three definitions of the Second have been in use. And still are. We quote from Wikipedia.

• The second is derived from the division of the day first into 24 hours, then to 60 minutes, and lastly to 60 seconds each (24 × 60 × 60 = 86,400). The definition that is based on 1/86,400 of a rotation of the earth is still used by the Universal Time 1 (UT1) system.
• In 1956, the second was redefined in terms of a year relative to that epoch. The second was thus defined as “the fraction 1/31,556,925.9747 of the tropical year for 1900 January 0 at 12 hours ephemeris time”. This definition was adopted as part of the International System of Units in 1960.
• The current and formal definition in the International System of Units (SI) is more precise. The second […] is defined by taking the fixed numerical value of the caesium frequency, $\Delta \nu_{Cs}$ , the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, to be 91 92 631 770 when expressed in the unit $Hz$ , which is equal to $s^{-1}$.

The result is that an Orbital second can be defined in two ways, by 86,400 seconds in a day or by 31,556,926 seconds in a year. We have adopted only the latter definition for the Orbital timeframe, while the Atomic second is the same as is currently accepted as the “more precise” one. For the sake of clarity, personally I do not think that Atomic time as such is “better” than Orbital time. Anyway in our theory we have to consider both timeframes, Orbital and Atomic.

The picture below and corresponding naming is copyrighted so to speak by Barry Setterfield.

The accompanying formulas relating Orbital clock ticks $\Delta T$ and Atomic clock ticks $\Delta t$ to variable Mass $m$ have been heuristically discovered.

$$ \frac{\Delta t}{\Delta t_0} = \frac{m_0}{m} \quad \mbox{and} \quad \frac{\Delta T}{\Delta T_0} = \sqrt{\frac{m_0}{m}} $$

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 :

• Here and now is made the same for both clocks. Formally : $t=0$ if and only if $T=0$. In short: $t=T=0$ .
• The two clock ticks are made equal too for $t=T=0$ . Formally : $\Delta T_0 = \Delta t_0$ must be taken care of.

With these two precautions taken, we can derive two important formulas in our theory.

$$ \frac{\Delta T}{\Delta T_0} / \frac{\Delta t}{\Delta t_0} = \sqrt{\frac{m_0}{m}} / \frac{m_0}{m} = \sqrt{\frac{m}{m_0}} $$

Switching to an “exact” result by taking limits for infinitesimal small clock ticks ; obtaining a convenient analytical approximation in reality.

$$ \lim_{\Delta\to 0} \frac{\Delta T}{\Delta t} = \sqrt{\frac{m}{m_0}} \quad \Longrightarrow \large \quad \frac{dT}{dt} = \sqrt{\frac{m}{m_0}} \quad \mbox{or} \quad \frac{dt}{dT} = \sqrt{\frac{m_0}{m}} $$

Giving extremely simple differential equations for $T(t)$ or $t(T)$ , in principle. The boundary condition $T=t=0$ must be kept in mind.

We have found a formula for calculating the Variable (elementary particle rest) Mass as a function of Atomic time or as a function of Orbital time, provided that one of the two is known : $m(t)$ or $m(T)$ . Up to now we have no idea how to determine one of these. The Atomic time function seems to be the most natural one to start with. An obvious assumption shall be that the change of (elementary particle rest) mass $m$ is uniform. That is : it does not depend upon the origin $(0)$ of its own (proper) atomic time $t$. Then we have, for small and equal time ticks $\Delta t$ :

$$ \frac{m(\Delta t)}{m_0} = \frac{m(2\Delta t)}{m(\Delta t)} = \frac{m(3\Delta t)}{m(2\Delta t)} = \cdots $$

It is seen that

$$ \frac{m(\Delta t)^2}{m_0^2} = \frac{m(2\Delta t)}{m_0} \quad ; \quad \frac{m(3\Delta t)}{m_0} = \frac{m(\Delta t)}{m_0}\frac{m(2\Delta t)}{m_0} = \frac{m(\Delta t)^3}{m_0^3} $$

Quite in general, for a number of increments $=N$ :

$$ \frac{m(N\Delta t)}{m_0} = \left(\frac{m(\Delta t)}{m_0}\right)^N $$

where $N\Delta t = t$ is atomic time itself. And $\Delta t = (t/N)$ for the increments. We can write $m(\Delta t)/m_0$ as unity $(=m_0/m_0)$ plus its first derivative around $t=0$ .

$$ \frac{m(\Delta t)}{m_0} = 1 + \frac{d}{dt}\frac{m(t)}{m_0}(0).\Delta t = 1 + H\frac{t}{N} $$

where $H$ is some yet to be determined constant with the dimension of $t^{-1}$ or $Hz$ . Next step :

$$ \frac{m(N\Delta t)}{m_0} = \frac{m(t)}{m_0} = \left(1 + H\frac{t}{N}\right)^N $$

The result for $N\to\infty$ is a well known limit.

$$ \lim_{N\to\infty}\left(1 + \frac{Ht}{N}\right)^N = e^{H.t} \quad \Longrightarrow \quad \large \frac{m}{m_0} = e^{H.t} $$

Herewith the Atomic time function for Variable Mass has been found, apart from a constant named $H$ .

We have found a formula for calculating the (elementary particle rest) Variable Mass as a function of Atomic time or as a function of Orbital time, provided that one of the two is known : $m(t)$ or $m(T)$ . Meanwhile there is a result for $m(t)$ and thus it must be possible to calculate for $m(T)$ .

$$ \frac{dT}{dt} = \sqrt{\frac{m}{m_0}} \quad \mbox{and} \quad \frac{m}{m_0} = e^{H.t} $$

By solving the differential equation

$$ \frac{dT}{dt} = \sqrt{e^{H.t}} = e^{(H/2)t} \quad \Longrightarrow \quad T = \int e^{(H/2)t} dt = \frac{2}{H}e^{(H/2)t} + C $$

Use the boundary condition for $t=T=0$ :

$$ 0 = \frac{2}{H} + C \quad \Longrightarrow \quad C = - \frac{2}{H} $$

$$ T = \frac{2}{H}e^{(H/2)t} - \frac{2}{H} \quad \Longrightarrow \quad 1 + \frac{H}{2}T = e^{(H/2)t} $$

$$ \left(1 + \frac{H}{2}T\right)^2 = e^{H.t} $$

So we have found a second expression in (our version of) the Variable Mass Theory, namely (elementary particle rest) mass as a function of Orbital time.

$$ \large \frac{m}{m_0}(T) = \left(1+\frac{H}{2}T\right)^2 $$

Let’s evaluate the result with help of the video.

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 $T$ . 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 $T=-A$ .

$$ \frac{m}{m_0} = 0 \quad \Longleftrightarrow \quad 1+\frac{H}{2}(-A) = 0 \quad \Longleftrightarrow \quad A = \frac{2}{H} $$

Now take a look at the 1993 publication by Jayant Narlikar and Halton Arp. Then we find similar results, and more. Starting with 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

$$ H = \frac{2}{A} \quad \mbox{in} \quad \frac{1}{1+z} = \frac{m}{m_0} = \left(1+\frac{H}{2}T\right)^2 $$

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 our constant $H$ . It is precisely the constant in Hubble’s law : $H=H_0=$ the Hubble constant.

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. 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 a different and incompatible outcome. 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.

$$ \large \frac{m}{m_0} = \left(1+\frac{H}{2}T\right)^2 = e^{H.t} \quad \mbox{with} \quad H = \frac{2}{A} $$

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 ging to present some direct empirical evidence of Variable Mass on an earthly scale. From Shrinking Kilogram Bewilders Physicists 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 here. According to the Variable Mass Theory we have, with meaning of the symbols as in previous writings

$$ \frac{m}{m_0} = \left[1+\frac{H}{2}T\right]^2 = \left[1+\frac{T}{A}\right]^2 \approx 1+2\frac{T}{A} $$

$$ \frac{m-m_0}{m_0} \approx \frac{2 \times 118}{4.54 \times 10^9} \approx 52 \times 10^{-9} $$

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 the cosmos the latter has become heavier, with the amount as calculated.

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 of course Orbital time. But with our own 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.

$$ c_T = \frac{dr}{dT} = \frac{dr}{dt}\frac{dt}{dT} \quad \mbox{with} \quad \frac{dT}{dt} = \sqrt{\frac{m}{m_0}} = 1+\frac{H}{2}T \quad \Longrightarrow \quad c_T = \frac{c_0}{1+\frac{H}{2}T} $$

Therefore the infamous C-decay (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 formula (13). Of minor importance, because the equivalent result can be proved much easier within the Atomic timeframe.

$$ 1+z = e^{-H.(-t)} \approx 1+H.t \quad \Longrightarrow \quad z = H.t = H.\frac{r}{c_0} = \frac{2r}{c_0.A} $$

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.

$$ c_0 = c\sqrt{\frac{m}{m_0}} \quad \Longrightarrow \quad \frac{m c^2}{m_0 c_0^2} = \frac{m}{m_0}/\frac{c_0^2}{c^2} = \frac{m}{m_0}/\left(\sqrt{\frac{m}{m_0}}\right)^2 = 1 \quad \Longrightarrow \quad mc^2 = m_0c_0^2 $$

Now take a look at the dimensionless formula as derived for the energy levels of an arbitrary atom. And apply conservation of mass-energy in there.

$$ \left[-\frac{1}{2}\sum_{i=1}^Z\nabla^2_i(\vec{\rho}) - Z\alpha\sum_{i=1}^Z\frac{1}{|\vec{\rho_i}|} +\alpha\sum_{i=1}^Z\sum_{j=i+1}^Z\frac{1}{|\vec{\rho_i}-\vec{\rho_j}|} \right]\psi(\vec{\rho}_i) = \frac{E}{mc^2}\mbox{ }\psi(\vec{\rho}_i) = \frac{E}{m_0c_0^2}\mbox{ }\psi(\vec{\rho}_i) $$

The consequences are dramatic. Due to Quantum Mechanics, $E$ remains an absolute constant in Orbital time. There is no intrinsic redshift at all, within the pure gravitational framework as designed by Arp & Narlikar !!

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 of course Orbital time.

“Of course”? Why should this be ‘obvious’?

Actually, physicists who recognize only one kind of timeflow use atomic time.

The following basic formula in our Variable Mass Theory can not be repeated often 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:

• Kinematics, Dynamics, and the Scale of Time
• Kinematics, Dynamics, and the Scale of Time II
• Kinematics, Dynamics, and the Scale of Time III

The relationship between the timeframes $\tau$ and $t$ by Milne is (from the first reference) :

$$ \tau = t_0\mbox{ }\log\left(\frac{t}{t_0}\right)+t_0 \qquad \mbox{(A)} $$

Compare with our relationship between the timeframes $t$ and $T$ .

$$ t = A\mbox{ }\ln\left(1+\frac{T}{A}\right) $$

Like in the Arp & Narlikar paper, the origin of time is taken as $t_0$ being the present value of the age of the universe at ourselves. With other words: $t_0 = A$ . A next correction is renaming of variables and substitution: $\tau \rightarrow t+A$ and $t \rightarrow T+A$ , thus bringing the origins of atomic time $t$ and orbital time $T$ back to here and now. Having done all that, there still is a discrepancy between Milne’s theory and ours, coming from his physical interpretation of $\tau$ and $t$ , which seems to be exactly the other way around !! Anyway, a picture says more than a thousand words.

In the book Dark Matter, Missing Planets & New Comets by Tom van Flandern we find in the footnote on page 95 the following.

The formula in the Meta model can be derived with Milne’s (modified) formula.

$$ \sqrt{\frac{1}{1+z}}=\sqrt{\frac{m}{m_0}} = 1+\frac{H}{2}T \quad \mbox{with} \quad A = \frac{2}{H} $$

$$ d = c.(-t) = -c.A\mbox{ }\ln\left(1+\frac{T}{A}\right) = -c\frac{2}{H}\mbox{ }\ln\left(\sqrt{\frac{1}{1+z}}\right) = -c\frac{2}{H}(-\frac{1}{2})\ln(1+z) $$

$$ \large d = \frac{c}{H}\mbox{ }\ln(1+z) $$

However, the above derivation with help of Milne’s formula is too much of the good. If we restrict attention to the Atomic timeframe only, then the proof is reduced to a one-liner.

$$ 1+z=e^{-H.t} \quad \Longrightarrow \quad \frac{H}{c}(-ct) = \ln(1+z) \quad \Longrightarrow \quad d = \frac{c}{H}\mbox{ }\ln(1+z) $$

It is important to notice that the logarithmic distance formula is not at all restricted to van Flandern’s book.

There is another interesting consequence that follows from our formulas.

$$ 1+\frac{1}{2} H T = e^{(H/2)t} \approx 1 + \frac{1}{2}H t + \frac{1}{2}\left[\frac{1}{2}H t\right]^2 $$

where $H=$ Hubble parameter, $T=$ Orbital time, $t=$ Atomic time. Let both the Atomic and Orbital clock be synchronized as usual ($t=T=0$) and let them run free for a century, that is

$$ T = (\mbox{century})_T \quad \mbox{and} \quad t = (\mbox{century})_t $$

Then we see that the Atomic clock is anyway running faster than the Orbital clock:

$$ \frac{1}{2} H(T-t) = \frac{1}{2}H.\frac{1}{4}H t^2 \quad \Longrightarrow \quad \large (\mbox{century})_T - (\mbox{century})_t \approx \frac{1}{4}H (\mbox{century})^2 $$

Let’s further resume some of the formulas we started with; let $A=$ age.

$$ 1+T/A = e^{t/A} \quad \Longleftrightarrow \quad t = A\mbox{ }\ln(1+T/A) $$

Now calculate, for example, velocities $v$ and accelerations $a$ from displacements $s$ within the two time frames.

$$ \vec{v}_T = \frac{d\vec{s}}{dT} = \frac{d\vec{s}}{dt}\frac{dt}{dT} = \frac{\vec{v}_t}{1+T/A} $$

$$ (1+T/A)\vec{v}_T = \vec{v}_t $$

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

$$ \vec{a}_T = \vec{a}_t\left(\frac{1}{1+T/A}\right)^2 + \vec{v}_t\frac{-1/A}{(1+T/A)^2} $$

$$ (1+T/A)^2 \vec{a}_T = \vec{a}_t - \vec{v}_t/A $$

Under eartly circumstances here and now we have that $T=0$ 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.

Application

A quote from the book Origin of Inertia by Amitabha Ghosh: 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. According to our theory we have - as repeated from the above:

$$ \vec{a}_T = \vec{a}_t - \vec{v}_t/A $$

Here $\vec{a}_T$ is the “true” Newtonian acceleration (namely in orbital time), $\vec{a}_t$ is the acceleration as measured nowadays with atomic clocks and $A$ is an age. All accelerations are proportional with $R=$ radius of the earth. If there is a discrepancy $\Delta\omega$ between the orbital and atomic angular velocities, then it must being caused by the negative term, which is a vector in tangential direction slowing down the rotation. The calculation is a one-liner.

$$ -|\vec{v}_t |/A = -\omega R/A \quad \Longrightarrow \quad \dot{\omega} = -\left(\frac{2\pi}{\mbox{day}}\right)/\mbox{age} $$

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 $5 \times 10^{-22} \mbox{ } rad/s^2$

There is indeed a small but positive discrepancy between atomic time and orbital day time, which is exemplified by the existence of Leap Seconds. Atomic clocks will run too fast in the long run. Therefore it is necessary to “stop” them once or twice in a year or so (source of picture = Wikipedia).

Data are obtained from the following Wikipedia reference: Δ T (timekeeping) . 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. Let’s see if we can reproduce this. 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 to derive the number of leap seconds from this, but we are a factor of two off:

theta := slower/2*(100*year)^2;
leaps := evalf(theta*day/(2*Pi));

                         leaps := 34.73137354

As has been explained in the above, Atomic clocks are running faster than Orbital clocks, with an amount of $H(\mbox{century})^2/4$ . For the Hubble constant of the earth, the leap second as calculated by this effect has the same magnitude. Not at all by coincidence, because the former turns out to be another calculation for the latter.

$$ \mbox{leaps} = \frac{1}{4}H(\mbox{century})^2 = \frac{1}{2} \left[ \frac{1}{2} H \left( \frac{2\pi}{\mbox{day}} \right) \right] (\mbox{century})^2 \left(\frac{\mbox{day}}{2\pi}\right) $$

Remember, however, that observation of the leap interval reveals that it is about 64 seconds. It is now obvious, however, that the two calculations correspond with two different phenomena. The first one describes how atomic clocks are running ahead of orbital clocks, defined e.g. by the earth orbiting around the sun in a year. The second one describes that there is an additional time leap needed for a rotation of the earth to complete its periods over a day. For our leap second this means that the outcome $34.73137354$ should be taken twice instead of once. Thus improved calculation of the leap interval reveals that it is about $2\times 34.73137354 = 69.46274708 \approx$ 69 seconds. Furthermore 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 $H =(2/\mbox{Age})$ and leaps $=2\times (\mbox{century})^2 /(2\times \mbox{Age})$ . Conclusion :

$$ \mbox{leap seconds} = \frac{(\mbox{ seconds in a century })^2}{\mbox{age of the earth in seconds}} $$