a-cosmology-group/acg#276

The Scirp journals are not a bad choice (this one, Journal of Modern Physics, and a few more). They have page charges for authors. On one hand this is okay, since they are free access, but it also means they lean towards publishing. There is at least some peer review (I have reviewed a few articles for JMP). I agree they publish a lot of interesting articles.

Thanks for the headsup on this one @LifeIsStrange I am on VIXRA but need a sponsor for ARXIV. I am not looking for a sponsor though. They’ve lost the plot if they think they’re keeping the rifraff out.

I submitted to JAMP on the same day you posted above and was sent a review within the first week, with a conditional acceptance. They give very tight schedules for replies and follow-ups but I guess it’s to try and make it into the next volume. In the end I didn’t make it for the next volume. Not sure how they would react to a lengthy rewrite. (which they almost asked for but were ok with my minor revisions)

I would agree with @ExpEarth that their review may be a little lightweight but for this subject a heavyweight was not going to let it pass.

It was published right at the end of November so from first submit to publish took almost 2 months which is not that long, but also mercifully short.

The Mechanics of Electrostatic Attraction and Repulsion, a Speculative Conceptual Analysis DOI: 10.4236/jamp.2024.1211234

Congrats on getting your paper published, Francois! You are opening a whole new field of study here and there could be many good results to follow. It makes no difference at this stage whether the peer review is stringent or not. The most important thing is to have your ideas open access for everyone to see.

Congrats! Sorry Francois I have not yet read your paper (my maths skills are lacking though), but push gravity is an excellent research direction! So is the related concept of mock gravity:

https://www.semanticscholar.org/paper/Galaxy-formation-by-mock-gravity-Hogan-White/622065965bb3945ce2b31b8d2be2229cc94dd9ca https://arxiv.org/abs/1802.05648 https://www.semanticscholar.org/paper/Attractive-force-on-atoms-due-to-blackbody-Haslinger-Jaffe/328e4988c2e8b515d3d90476697a6c30e9d63f90

nice digression: https://www.semanticscholar.org/paper/Gravitational-Dispersion-Forces-and-Gravity-Pinto/1cfa3fd2187c23c59e4149b3dc9e9d74435c04bb

@LifeIsStrange I was accused of adding too many images, so you can just read the pictures :). Apologies - no offense intended! With all the links coming from you I’m sure you are underselling your math skills.

I am familiar with the mock gravity concept and in specific Hogan’s (and others’) work. These are based on absorption models, which was also our prior attempts at defining push-gravity. Fatio and Le Sage models historically also require absorption but this turned out to be the biggest obstacle, as shown by Maxwell, La Place, Poincare, Feynman and others that the absorbed energy would e.g. ‘blow up’ earth in an instant. These calculations are correct, unfortunately. Even more unfortunate is that the energy requirement for any current model of gravity would ‘blow up the earth’, from Newton through Einstein’s GR and of course MOND too. In a Newtonian ‘gravity comes out of mass’, the mass would radiate away all its self-energy in an instant. and GR would also ‘blow things up’, but it claims space energy is zero so there’s a conundrum as to how space exactly ‘curves’ or ‘tells matter how to move’. Even so, GR is very compatible with push-gravity.

So where the energy of gravity comes from is wholesale ignored in all current models.

Unless the energy mode can be transformed, and this has not been shown before.

In this work I believe I have shown the possibility of interactions with a photon aether, where photon pairs are initially ‘invisible’ (hence can penetrate matter or go right through without touching sides) and the fraction that does interact is transformed into electric fields, which is what we know as keeping atoms together and is not some new unknown ‘field’. This field dissipates through matter without generating heat. Thereby solving the push-gravity energy problem.

the mass would radiate away all its self-energy in an instant

Just curious, @Francois-Zinserling : how do you define / calculate the self-energy of a mass?

Dear @HanDeBruijn I’m not sure if that is a loaded question so here goes …

My obvious choice is from $E=mc^2$. Or $E=γmc^2$ more specific in a fixed reference frame. And by radiate away as above I imply that a (Newtonian) model that expects gravity to emanate from within an object must somehow provide the energy to maintain that action. Since we don’t observe every object’s mass as rapidly depleting, we can rule out that gravity is driven by self-energy, or else we must provide a mechanism that ‘manufactures’ energy within a mass.

GR gravity does not rely on self-energy, and neither does pushgravity. GR’s ‘space’ is pushgravity’s ‘aether’.

Slightly different note: I found where Brown tries to (naively) calculate the energy required for Le Sage gravity. About 2/3 down the page $E=10^{30}J/s$ Le Sage’s Shadows

In your second link the author claims:

It’s obvious from the picture that the slope of the tangent line at any point of the transversal wave must be greater than minus one.

It might seem obvious in that picture, but not at all in e.g. this first one ..

, which means A < λ/2π does not hold for all waves, or else A must get smaller and smaller for high frequency waves, which then seem to belie the equation E = hf. Not that I think an electromagnetic wave’s energy sits in the amplitude, but the author’s reasoning seems flawed.

Thanks for the feedback, Matt.

My lack of article writing experience is clearly showing. Others have commented on the abstract too. I will also remove section3 since it adds no value.

Section4 is the main event with fig16 and fig 18 visualising a push-type mechanism for both attraction and repulsion.

The (my) photons and pairs are visualised throughout from fig5 onward, but the 0E + 0B pair is shown in fig8. This is the one that should penetrate right through matter without any electromagnetic effects.

I did not realise Arto et. al. also proposed filaments. Although I do not think I will be easily swayed, please send a relevant link if you have one available. I am still not convinced by the filaments that could be infintiely long and with ~(10^80)! <- that’s a faculty sign - connections between them. I suppose photon pairs, being bosons, can manifest in many ways. Perhaps if I understand it better.

As for the size of photons, there doesn’t seem to be much consensus. Many QM students propose it as a plane wave that collapses at one point at time of measuring. This does not explain how a photon (plane wave) navigates through the cosmos without getting ‘measured’ upon its very first collision with any object. It would mean we should not see objects if there are other obstructions in the path of the plane, leaving not much to be seen. And if every collision is a measurement, then the wave is not a single entity … Also, I fail to see how a plane wave can impose a circular momentum RHC or LHC at the point of measure. Sometimes QM make things too complicated even for themselves.

Most studies propose a photon having a ‘cylinder-like’ spatial form. There are several proponents for a ‘wavelength long’ photon, but then a photon (with a +/- sine wave) cannot form e.g. a pure (+) electric field around a charged particle. So if the ‘fields’ are set up by photons (real photons), then there needs to be +E and -E photons. I’ve read quite a few papers on this, each with their own arguments. Liu’s half-wave length photons fit the scene very well for me at least.

Not to worry, if the ideas are good the writing can come after! Your work is quite thought provoking. I’m still trying to get a full grasp of it.

Arto does not use the term filaments, but his meaning is pretty clear. They think the photons have to be in long trains in order to confer structure to the vacuum. In the paper of theirs that you cite, Grahn et al, see Figure 3 in Part D:

I think Arto might agree with you that the filaments do not connect to particles. He just has them constituting the vacuum (or spacetime). It’s my work with the Hubble luminosity that convinces me that they must be attached to particles. It’s possibly not as improbable as it might seem, if you think of the filaments themselves being organized into bigger filaments and perhaps ultimately in cosmic filaments. I think if you followed that route, your task could possibly be a lot easier, but please don’t let me divert you from what you’re already doing there.

One small point. I notice you cite Kevin Brown. He was a major nuisance at the time PG came out. He totally degraded the Wikipedia article that some of us worked on. He insisted the theory was of historical interest only. He also had this weird infatuation with Fatio and almost insinuated that Le Sage plagiarized him. In fact Fatio was a minor character who never published on it. He did mention his PG hypothesis to Newton, and Newton had some similar ideas (also mentioned in PG), but it was Le Sage who is really our hero. He’s the one who was later rediscovered by Kelvin and others.

Can the gravitational constant G be calculated theoretically? From the answer on Mathematics Stack Exchange by Riad, one with zero (!) upvotes: I just came to know that Schrodinger in a 1925 (German paper) stated that G ≈ c^2 Ru/Mu wit no formula; Ru and Mu are the radius and mass of the universe. This then relates $G$ with the mass and radius of the observed universe and the constant speed of light. With other words, for the Observable universe according to (our version of) the Variable Mass Theory (VMT), with Mass and Radius denoted as $=M_O$ and $=R_O$ :

$$ R_O = W\frac{c}{H} \quad ; \quad M_O = \frac{W}{2}\frac{c^3}{G.H} \quad \Longrightarrow \quad G = \frac{1}{2} c^2 \frac{R_O}{M_O} $$

where $W=$ pure number (dropped), $c=$ speed of light, $H=$ Hubble parameter, $G=$ gravitational constant. But okay, not everybody likes the VMT, so let’s do it the classical way. I suppose most ACG members will agree with the following.

$$ R_O = \frac{c}{H} \quad ; \quad M_O = \frac{4\pi}{3}R_O^3\mbox{ }\rho_c $$

where the critical mass density $\rho_c$ according to ΛCDM is obtained via $H^2 = (8\pi/3)\rho_c G$ and so

$$ M_O = \frac{4\pi}{3}\left(\frac{c}{H}\right)^3\frac{3}{8\pi}\frac{H^2}{G} = \frac{c^2}{2}\frac{c}{H}\frac{1}{G} $$

It follows that the end-result is exactly the same as the one obtained with the VMT.

$$ \frac{M_O}{R_O} = \frac{c^2}{2G} \quad \Longrightarrow \quad G = \frac{1}{2} c^2 \frac{R_O}{M_O} $$

But the result is of little or no value if there are no independent means to determine the Mass and Radius of the observed universe.

Note. Choosing the ΛCDM critical density as the mean density has as a consequence that even the “classical” SEU is a Black Hole universe: $R_O$ is precisely the Schwarzschild radius.

$$ G = \frac{1}{2} c^2 \frac{R_O}{M_O} \quad \Longrightarrow \quad R_O = \frac{2GM_O}{c^2} $$

Hi Han, In 2020, I have proposed a way to determine the exact mass and radius of the observed universe. See my paper here: https://www.scirp.org/journal/paperinformation?paperid=103476 I am not sure it is the correct way but it would explain the empirical Weinberg formula (see paragraph 12 of the paper). As you will see (if you read my paper), I am also a proponent of a variable mass with time (see paragraph 5 of the paper) but instead of increasing with time like in your theory, the electron mass decreases with time. Regards, Patrick

Can the gravitational constant G be calculated theoretically? @ExpEarth . Let’s take a look at this paper by Matthew R. Edwards (April 2022): Optical Gravity in a Graviton Spacetime. Jump to the section

4. Optical Gravity with Gravitons

On page 10/11: For $\rho$ equal to the critical density $\rho_c = 3H^2/8\pi G$ [ .. ] On page 13: From this it can be seen that the term for the gravitational constant is

$$ G = \frac{H_0^2}{4\pi\rho} \qquad (18) $$

For $\rho = \rho_c = 5 \times 10^{−30}gm\mbox{ }cm^{−3}$ and $H_0$ as before, we find $G = 7 times 10^{−8}cm^3gm^{−1}s^{−2}$, which is very close to its actual value of $6.67 \times 10^{−8}cm^3gm^{−1}s^{−2}$. Thus, for $\rho \approx \rho_c$, the model is able to recover Newton’s law. The above approach is similar to versions of Le Sage’s theory of gravity in which a cosmic field of electromagnetic radiation gives rise to the force [ .. ]

Leaving Matt’s eloquent proze aside, let’s focus on the mathematics.

$$ G = \frac{H_0^2}{4\pi \rho} = \frac{H_0^2}{4\pi \rho_c} = \frac{H_0^2.8\pi G}{4\pi.3H_0^2} = \frac{2}{3} G $$

From this it is concluded to

$$ 1 = \frac{2}{3} \quad \mbox{or} \quad 1 = 0 $$

You would have avoided this highly trivial paradox, @ExpEarth, if you had taken serious your own advice on page 7 of the paper: Taking $\rho = 5\times 10^{−30} gm\mbox{ }cm^{−3}$, _which is about equal to the critical density $\rho_c = 3H^2/8\pi G$, [ .. ]

$$ \rho = \frac{H_0^2}{4\pi G} = \frac{2}{3}\frac{3H^2}{8\pi G} \quad \mbox{or} \quad \rho = \frac{2}{3}\rho_c \lt \rho_c $$

Google teaches us the following. The actual average density of matter in the universe, including dark matter, is about 0.3 times the critical density. When dark energy is also factored in, the total average density matches the critical density, which allows the universe to continue expanding at its current rate.

Yeah sure, but ACG doesn’t believe in dark matter. Thus for us there is no objection against taking the mass density of the universe lower than the critical density. Then there is still a vicious circle in the paper, but no contradiction anymore: $1=1$ . Here is another example of someone who tried to calculate for the Gravitational constant, and received analogous criticism from my side.

Nobody can calculate the value of the Gravitational Constant theoretically

In reply to your 1st question:

François, isn’t that last sentence of yours not just lip service to established “science” ??

I don’t think so. Since I am proposing a gravity that is driven by discrete photons, it is already quantised. Now it’s about getting the math right, and for that 4-D tensors will be required, sadly not my forté.

Aslo, as for Schrodinger’s $G ≈ c^2 Ru/Mu$

Since the R and M are in the same sphere, this reduces to .. $G ≈ 3/4πρ$ * $c^2 /Ru^2$ with the intention to show that $G ∝ 1/Ru^2$ which loosely relates back to @budrap00 representation of an ESW.

Agreed that G cannot be calculated theoretically since from my understanding it is not a constant. These guys have written some very readable papers on the difficulties to measurements G

29. C. Rothleitner and S. Schlamminger, (2017), Measurements of the Newtonian constant of gravitation, G, website: DOI 29

30. Junfei Wu et al, (2019), Progress in Precise Measurements of the Gravitational Constant, website: 30

31. Stephan Schlamminger, (date), Why is it so difficult to measure the gravitational constant?, website: 31

It may be concluded from a previous posting that adopting the “necessary” ΛCDM critical value for the mean mass density of the universe is disputable at least. In ACG contexts the Static Euclidean Universe (“SEU” as Eric Lerner says) is simply “flat” by definition. Adopting ΛCDM values, meant for an expanding universe instead, is likely to have undesirable consequences. This will demonstrated again by a subject in the same @ExpEarth paper as before. The main lesson to be learned is: be cautious if you consider borrowing anything from standard ΛCDM cosmology !!

On page 10 of Matt’s paper: [ .. ], where $\alpha=H_0/c$ [ .. ]

It should be noticed that an exponential decay $\exp(-\alpha.r)$ is attached to (IMHO non-existing) gravitons by Matthew Edwards and others (e.g. Amitabha Ghosh). In contrast, the VMT is attaching exponential decay to variable mass. The difference between the two approaches is subtle, because Gravitational constant and gravitational Mass are often combined into the Standard gravitational parameter, which makes it difficult to always distinguish between the two. That being said, let’s proceed with the paper.

If matter, chiefly in the form of galaxies, is considered to be distributed approximately evenly in space, $E_g$ can then be evaluated by integrating the gravitational potential energy of concentric shells of matter of thickness $dr$ centred about $m$, each of constant density $\rho$, out to infinite distance. We then have

$$ E_g = - \int_0^\infty \frac{4\pi G \rho m r^2 e^{-\alpha r}}{r}dr = - 4\pi G \rho m \int_0^\infty r e^{-\alpha r} dr \qquad (11) $$

Substituting from above for $\alpha$, we then have

$$ E_g = \frac{- 4\pi G \rho m c^2}{H_0^2} \qquad (12) $$

Whoa! Not so fast! With $\xi = H_0/c.r$ we have first

$$ \int_0^\infty r e^{-H_0/c.r} dr = \left(\frac{c}{H_0}\right)^2 \int_0^\infty \xi e^{-\xi}d\xi $$

$$ \frac{d}{d\xi}\left[-e^{-\xi}(\xi+1)\right] = e^{-\xi}(\xi+1) - e^{-\xi} = e^{-\xi}\xi $$

$$ \int_0^\infty \xi e^{-\xi}d\xi = \left[-e^{-\xi}(\xi+1)\right]_0^\infty = 1 $$

No errors found, so let’s proceed. For $\rho$ equal to the critical density $\rho_c = 3H^2/8\pi G$ we find

$$ E_g = - mc^2 \qquad (13) $$

Really? Let’s calculate again (silently assuming that $H=H_0$).

$$ E_g = \frac{- 4\pi G \rho_c m c^2}{H_0^2} = \frac{- 4\pi G m c^2}{H_0^2}\frac{3H_0^2}{8\pi G} = - \frac{3}{2} mc^2 $$

This trouble too would have been avoided with the same choice for $\rho$ as before.

$$ \rho = \frac{2}{3} \rho_c $$

Can it be concluded that the mean mass density of the universe, everywhere in Matt’s Optical Gravity model, should be $(2/3)$ times the ΛCDM critical mass density?

It’s interesting to compare the above derivation with an analogous (maybe more correct?) one in the book by Amitabha Ghosh: Origin of Inertia. I’ve taken the freedom to add more details to Amitabha’s theory in this page at my website. The final outcome may be summarized as follows. For $H_0^2 = (3/4)\pi\rho G$ the potential energy of a particle with mass $m$ is exactly equal to $-mc^2$, which implies that the total energy content of the universe is nil. So far so good, but what’s more interesting for us now is the relationship between mean mass density and critical mass density. It is given by

$$ H_0^2 = \frac{3\pi}{4}\rho G = \frac{3\pi}{8}\rho_c G \quad \Longrightarrow \quad \rho = \frac{1}{2}\rho_c $$

Up to now we have three expressions, and even two more, for the mean mass density as it is related to the ΛCDM critical density of the observable universe:

$$ \rho = \left(\frac{3}{8} \mbox{ or } \frac{1}{2} \mbox{ or } \frac{2}{3} \mbox{ or } 1 \mbox{ or } 3w \right) \rho_c $$

where $w = 0.3883945571$ (VMT), hence $3w \gt 1$.

Can we speak about a Mean_Mass_Density_of_the_Universe tension?

Update.

With the Variable Mass Theory we find a value of the mean mass density of the universe which is greater than the “critical” ΛCDM value. It is suggested herewith that it may be better to replace the latter by the former. The “critical” density is no longer critical for a Static Euclidean Universe. This means that the critical density must be re-defined and we should divide everything by $3w$ , resulting in

$$ \rho = \left(0.32 \lor 0.43 \lor 0.57 \lor 0.86 \lor 1 \right) \rho_c $$

@PatrGT . From the Introduction: >the gravitational coupling constant $\alpha_G$ acts as a scale factor between the quantum realm and the cosmos and decreases with cosmic time.

Yes, we have been going through this in Dimensions check at my website.

the Hubble parameter will always equal the inverse of the age of the Universe and that the Universe density will always equal the Universe critical density.

This is in complete agreement with (my version of) the Variable Mass Theory.

It is predicted that the mass and size of particles decrease with cosmic time. The Newtonian gravitational constant G and the Planck constant h are also predicted to decrease with cosmic time.

Just curious. Is there any Intrinsic Redshift predicted by your theory?

this change in scale remains unnoticed as all our surroundings, including our units of measurement [ .. ]

Agreement as far as the quote ends.

From Support by Hoyle expressions for the Planck units of length and mass are copied.

$$ l_P=\sqrt{\hbar G/c^3} \quad ; \quad m_P=\sqrt{\hbar c/G} $$

From your paper we have

$$ R_h = \frac{2\mu l_P}{\sqrt{\alpha_G^3}} \quad ; \quad M_u = \frac{\mu m_P}{\sqrt{\alpha_G^3}} $$

Proceed as follows.

$$ \frac{R_h}{M_U} = \frac{2 l_P}{M_U} = 2\sqrt{\hbar G/c^3}\sqrt{G/(\hbar c} = (2/c^2) G $$

By vicious circle reasoning then the desired result follows.

$$ G = \frac{1}{2}c^2\frac{R_h}{M_U} $$

I still have to read the rest of your paper. Sorry for the short and rather negative response so far.

Hi Han, I do believe there is intrinsic redshift predicted by my model but I still need to do some more work on it to propose something coherent. Why “ vicious circle“ ? The equation G=C^2Rh/2Mu simply follows from my first 3 equations (see paragraph 4 of my paper), there is nothing strange with that, it is just a normal deduction if we assume that my first 3 equations are correct. You were asking if one could calculate the value of the gravitational constant, I suggest that one can using equation 36 (or 37) of my paper, the main uncertainty would be on the age of the Universe. Thank you for your feedback.

Hi Patrick, >it is just a normal deduction if we assume that my first 3 equations are correct.

There is not an issue with being correct. The issue is that you have found just another proof for a triviality, namely $1=1$. And whatever complicated that proof apears to be, the result is nothing new.

The equation G=C^2Rh/2Mu simply follows from my first 3 equations

Unless I’m missing something obvious, that is not true: we need the additional definitions for Planck length and Planck mass. Nothing about $G$ follows from $R_h/M_U = 2l_P/m_P$ without those definitions.

Edit. >if one could calculate the value of the gravitational constant, I suggest that one can using equation 36 [ .. ]

Yes indeed, I don’t understand exactly why, but it works (apart from some Hubble tension).

m_p := 1.67262192*10^(-27);
m_e := 9.1093837*10^(-31);
mu := m_p/m_e;
G := 6.6743*10^(-11);
c := 299792458;
h := 6.62607015*10^(-34);
Mpc := 3.08567758*10^(22);
H_0 := 70*1000/Mpc;
G := H_0*2*(h/(2*Pi))^2/(mu^2*m_e^3*c);
G := evalf(%);
G := 6.6743*10^(-11);

Giving $G := 0.6604265832\times 10^{-10} \approx 0.66743 \times 10^{-10} m^3 kg^{-1} s^{-2}$ .

Hi Han, I am not sure I understand you correctly. Have you seen somewhere else an equation linking the age of the Universe to the electron Compton time ? (equation 1). Have you seen anywhere else an equation linking the Hubble radius to the electron Compton wavelength ? (equation 2) and have you seen anywhere else an equation linking the mass of the Universe to the electron mass ? (equation 3). If my equations are correct, it would be quite revolutionnary. The fact that by simply grouping these 3 equations I find that G=C^2Rh/2Mu and it seems to agree with yours and other people’s model is a hint that my equations are correct, it is not a triviality. The additional definitions for Planck length and Planck mass are known by any scientist, just look them up in Wikipedia (no need for “support from Hoyle”)

@Francois-Zinserling . >Now it’s about getting the math right, and for that 4-D tensors will be required, sadly not my forté.

Something similar is said by @ExpEarth in thread 256, when talking about his Shell Universe:

To do that properly would have required a deep dive into the Einstein field equations of liquid or gas spherical masses, something that I would have to seriously retool for.

Speaking for myself, it’s with General Relativity as with French: I can read it, but I cannot write it; I can understand more or less what they are saying, but I cannot properly respond to it.

GR has been invented more than a century ago (1915). There has been plenty of time to instruct people about it. And yet almost nobody speaks the language fluently. How would that come?

@PatrGT . Did you read the Edit in my previous reply? You are quite right: with equation 36 in your paper people can indeed calculate for the Gravitational constant. >I do believe there is intrinsic redshift predicted by my model but I still need to do some more work on it to propose something coherent.

It’s not difficult I think, because only the Rydberg constant (in energy units) has to be considered.

$$ Ry = m c^2\frac{\alpha^2}{2} \quad \mbox{with} \quad \alpha = \mbox{fine-structure constant} $$

I don’t know (so quickly) how the quantities composing Ry (and α) are varying with cosmic time in your theory. Otherwise I could have done the work myself.

@HanDeBruijn I believe that the fine structure constant is varying with time and that will be explained in my next paper but it is not ready yet. It is not as easy as you might think. BTW, why do you think that only the Rydberg constant has to be considered ?

@PatrGT . >I believe that the fine structure constant is varying with time [ .. ] It is not as easy as you might think.

Oh well, let us assume that energy $mc^2$ is conserved and that $\alpha$ is linear with (VMT orbital) time $T$ as a first approximation.

$$ \frac{\alpha}{\alpha_0} = 1+\frac{T}{A} \quad \mbox{with} \quad A = \mbox{Age} $$

where the subscript $0$ denotes here and now $T=0$ . Then we have for the Rydberg constant

$$ Ry = mc^2\frac{1}{2}\alpha_0^2\left(1+\frac{T}{A}\right)^2 \quad \mbox{with} \quad T \gt -A $$

having the same effect on Intrinsic Redshift as Narlikar’s law with the VMT.

BTW, why do you think that only the Rydberg constant has to be considered ?

I thought there is a relevant link to Intrinsic Redshift already in one of my earlier replies to you.

@HanDeBruijn you assume that energy is conserved but in your theory (and my model) mass vary with time. I don’t understand.

@PatrGT .

you assume that energy is conserved but in your theory (and my model) mass vary with time.

In my theory energy is only conserved in the so-called orbital timeframe, not in the atomic timeframe. For the reason that the speed of light is a variable too in the former frame. The end-result is that $mc^2=m_0c_0^2$ where the mass and lightspeed are variables on the left hand side.

I don’t understand.

Upon further investigation of your theory I see that the Ansatz as proposed by me is indeed not compatible with it. Let’s repeat

$$ Ry = m_e c^2\frac{\alpha^2}{2} \quad \mbox{with} \quad \alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} $$

In your theory we have in (16) : $m_e \sim t^{-2}$, $c=$ constant and in (19) : $h \sim t^{-3}$. There is no time dependence defined in your paper for the elementary charge $e$ and the electric permittivity of free space $\epsilon_0$. If the latter two may be supposed constant then we have

$$ Ry \sim t^{-2}\left(t^{-3}\right)^{-2} \sim t^4 $$

Translated back to VMT that’s the square of Narlikar’s law.

@PatrGT . The following picture is from A large numbers coincidence by @sahil5d .

$G=$ Gravitational constant , $c=$ Speed of light , $h=$ Planck’s constant , $H=$ Hubble parameter , $m_p=$ Proton mass , $m_e=$ Electron mass . If I am not mistaken, this is the complete list of all the independent variables in your paper. Another way to write the above equality is

$$ \frac{h}{m_e.c.2\pi} \times \frac{H}{c} = \frac{H\hbar}{m_ec^2} = \frac{2G}{c^2}.m_p \times \frac{m_p.c.\pi/2}{h} =\frac{G m_p^2}{2 \hbar c} $$

From which follows the formula that has been used for calculating the Gravitational constant theoretically.

$$ \frac{H\hbar}{m_ec^2} = \frac{G m_p^2}{2\hbar c} \quad \Longrightarrow \quad H = H_0 = \frac{m_p^2 m_e G c^2}{2\hbar^2c} = \frac{\mu^2 m_e^3 G c}{2\hbar^2} \qquad (36) $$

It is noticed that the large numbers coincidence is one equation with six (un)knowns. This means that only one (un)known can be calculated from the others. All the rest must be deemed as circular reasoning. Our choice is: calculate for the Hubble parameter $H_0$ with (36). Then equation (1) in your paper can safely be replaced by (the upside down version of) another equation, namely

$$ t_0 = 1/H_0 \qquad (4) $$

Continuing.

$$ R_h = c.t_0 \qquad (2) $$

Your universe is exactly like discussed in a previous reply. Because the ΛCDM expression for the (critical) density is the very same

$$ \rho_u = \frac{3H_0^2}{8\pi G} \qquad (7) $$

As a consequence, formula (3) can safely be replaced by

$$ R_h = \frac{2G M_u}{c^2} \quad \Longrightarrow \quad M_u = \frac{R_hc^2}{2G} \qquad (5) $$

Indeed, with

$$ V_h = \frac{4\pi R_h^3}{4} \qquad (6) $$

The above is (part of) an order of the formulas as I would prefer. Again, Patrick, something being wrong is not the issue here.

@HanDeBruijn I am not sure I understand the point you are trying to make here … Sahil’s coincidence is not a coincidence if the starting point is my 3 equations. Using my equations, you can just simply reduce Sahil’s coincidence to a perfect equality. What Sahil’s found by number crunching is in fact just a rearrangement of my equatilons. I think that you are missing the point here when you say you prefer to start with the equation Rh=2GMu/c^2. If you start with that you will never get to my equations 1,2 and 3 that are linking atomic values to cosmological values. You will just stay with an equation between cosmological values with no real mean to find their exact values. Do you see what I mean ?

@PatrGT

Do you see what I mean ?

Yes, I see what you mean. I shall no longer try to “improve” on the subject in this way. Have a good day and thanks for the hint how to do a theoretical calculation for the Gravitational constant.

@HanDeBruijn thank you for your comments and for taking the time to read my paper. When I write up about the variation of alpha and e, I will let you know. It is always good to have someone else’s feedback.

@PatrGT . >about the variation of alpha

Be cautious, Patrick! There is an astonishing thread 205 by @sahil5d , where it is claimed that $\alpha$ is a pure number. This would mean that it is independent of its physical constituents.

@HanDeBruijn thank you for the info, I will have a look