I have thus far been mostly a spectator in ACG, having had not much background to cosmology. Some of you may know me as a fierce proponent of pushgravity, which necessitates the existence of an aether. In an earlier work I derived Newton’s equation from this mechanism (as many others have done before me). My own take-out was this:
(A) $G=µc^2/4π$
, with µ as an absorption coefficient, but from first principle it applies to any transformation that follows the Beer-Lambert law of exponential decay. From this a value for µ can be calculated to be $9.33 * 10^{-27}$ m/kg but it implies that there may be a variable aspect to ‘G’ which is material related. We cannot account for this with current methods since all gravitational equations include $M*G$ which could make variations in G look like variations in M.
I went back to some of the earlier assumptions and approximations made and think that I can throw in a more elaborate equation into the mix, which could now make me a ‘participant’ of ACG :). For ease of writing I will only show how ‘G’ is changed and not try redo Newton again. I also imply this ‘variable G’ can be directly substituted into GR and it would probably eliminate both the need for dark matter and the cosmological constant. .. Also, G should be a tensor in this context so I am giving the much simplified version here:
(B) $G=(1-e^{-μρx} )/μρx*μc^2/4π$
From eq(B), eq(A) is easily derived by simplifying with the binomial expansion. The variables μρx include the original μ, ρ for material density (typically related to μ) and x being the linear path of decay (e.g. diameter of earth, sol, football, etc) But if we don’t simplify, and analyse the ‘G’ equation in (B), a few interesting conclusions follow:
- Earth-sized masses and smaller where ρ typical <10 and x up to $10^6$ m In this weak limit, for μρx << 1 eq(B) reduces to eq(A) and gravity appears ‘linear’.
- Sun sized masses where ρ typical <10 and x in the order of $10^9$ m Exponential decay is markedly higher and will appear to give an anomaly when masses from 1 and 2 interact, e.g. SOL vs Mercury.
- Neutron star: For very large ρ~ $10^{17}$ and smaller than sol diameter x~ $10^4$ m the exponential becomes markedly noticeable (but not reaching zero within this limit of x)
- Black hole: An extension of the above (3) but where opacity reaches maximum (aether transparency severely reduced, but probably still not zero)
- Galactic Disk: For a small ρ (<<1 e.g. disk), and ρ decreasing exponentially with respect to r from the centre, and a large x~ $10^{17}$ m and from a naive $v^2 = GM/r$ equation we get a velocity curve that looks very much like the disk mass curve (excl centre mass), which flattens out and appears remarkably like galactic rotation curves.
- And here is my ACG entry: Universe: Here ρ<<1 even much smaller than (5) but for an extreme x~ $10^{27}$ m gives a tiny attenuation of the flux over such a distance. Looks linear because it is so small but it is really exponential. Observed as a weakening of the origin flux and mistaken as velocity of the observed object. Because of $e^{-x}$ effect, far-away observations are pronounced, compared to (relatively) nearby. ‘Very’ nearby attenuation effects seem negligible due to ρ<<1, but not, as in (2), (3) and (4) where there is more mass present.
There is more to derive from this, e.g. that masses of sol objects or larger are overstated because G is really bigger through those objects. But for now I would hope for some feedback on the above.
Arguments of exponents must be dimensionless. However, when doing the check one gets
$$ \mu \rho x = \frac{m}{kg}.\frac{kg}{m^3}.m = m^{-1} \ne 1 \qquad \mbox{???} $$
@Francois-Zinserling Francois, I see some other problems other than what Han points out. Maybe I missed something, but it seems you could possibly be mixing two different theories, Le Sage’s (‘pushing gravity’) and Quirinio Majorana’s model.
In a PG model, Newton’s constant G contains terms for the energy density of the particles, waves or whatever that impart momentum and energy to masses, as well as a term for the absorption coefficient of mass to these particles or waves.
In Majorana’s model, he postulates that masses emit a gravitational flux of some kind that is absorbed by other bodies. The flux term is automatically contained within G in a sense.
I can’t see what kind of particle or wave is causing gravity in your description. So it does not seem like a PG model but more like Majorana’s model.
Some history on Majorana’s work which connects with cosmology too is in this Apeiron paper by Assis: https://www.ifi.unicamp.br/~assis/Apeiron-V13-p3-11(1992).pdf
Hi, Han The ‘p’ in the exponent represents a linear density as typically seen in xray equations. I’ll elaborate when I can get behind a pc but units are in kg/m^2
@ExpEarth definitely not a Majorana model although I revere the original experiment where he showed proof of absorption. Majorana himself was unsure of his results and interpreted it as ‘an absorption of gravity’, which lead to the malicious review of Russell. It is my opinion that Russell’s review led to Majorana doing his follow-up experiment on a shoestring budget which gave inconclusive results. Fortunately many, including Maartins in your book were supportive of Majorana’s results. In my opinion Majorana showed an absorption of Flux and his findings agree with what I would expect.
My $e^{-x}$ component stems from a Fatio/Le Sage absorption/transformation model and has not been ‘stuck onto’ Newton.
I find myself flat in bed due to an unfortunate incident that ironically also included gravity. I don’t have @HanDeBruijn skills of equations here and also working upside down with my phone but I will follow with a post how I got to ^^ that equation.
@Francois-Zinserling . >I find myself flat in bed due to an unfortunate incident that ironically also included gravity.
A bit too much of pull gravity, I suppose? That happening is not quite uncommon at our age. Wish you good recovery without too many unpleasant after-effects.
Thank you for the well-wishes. I could swear I was pushed down! 😁
We start with the basic Fatio/Le Sage pushgravity mechanism but we replace Fatio’s kinetic corpuscles with Annila’s photon pairs. Thereby eliminating all of the ‘Le Sage objections’ while attaining optical depth.
As per Rueda and Haisch we acknowledge $E=mc^2$ where E represents the strength of the aether that interacts with the object. Also derived by myself from relativistic Doppler equations. Since we will propose the same aether to cause gravity there is no distinction between inertial or gravitational mass.
We define $I_0$ as the primordial flux, homogenous, but we will only analyze in 1 dimension for simplicity. Also no velocities and vectors for the same reason.
For typical radiation absorption we have the equation: $I = I_0 e^{-μρx}$ With μ as linear absorption coefficient in m/kg With ρ as linear density in kg/m^2 With x in meters
Which in a symmetric aether leads a potential anywhere on the surface of a sphere as: $I(abs) = I_0 - I$ Which vector is everywhere pointing into the mass.
For μρx <<1 Simplifying with a binomial expansion we find: $I(abs) = I_0μρx$ Remembering that this is linear absorption we can conclude for volume: $I(abs.vol) = I_0μρxyz$ With ρ now in kg/m^3 $= I_0uM$ (I’ve done this for a sphere to same result)
Using the technique of a Gaussian sphere we find the field strength at a distance r $φ(r) = (I_0 μ M)/4πr^2$ Which we can already recognize as $GM/r^2$
And it remains to show as per Rueda and Haisch $I_0 = c^2$
Then finally $φ = g = μc^2M/4πr^2$
This reveals for the Newtonian limit, where μρx <<1 $G = μc^2/4π$
All this to show that my original equation (B) was not an ad-hoc addition to Newton but derived from 1st principles, and Newton is only one of the solutions where upx <<1.
Maybe now points 2,3,4… would make sense to someone.
@Francois-Zinserling writes:
So I’d rather think that’s a surface density, but okay.
It follows that anywhere on the surface of a sphere (I think)
$$ I(abs)=I_0-I=I_0(1-e^{-\mu\rho x})\approx I_0\mu \rho x $$
Then you suddenly define
$$ I(abs.vol) = I_0\mu\rho xyz $$
with dimensions
$$ \left(\frac{m}{s}\right)^2.\frac{m}{kg}.\frac{kg}{m^3}.m^3 = \left(\frac{m}{s}\right)^2.m $$
So the dimensionality of $I$ has changed into [length] times the dimensionality of $I_0$. Is that correct?
@Francois-Zinserling
This is better! I’m glad you are linking back to the Annila photon pairs model. Hopefully you will eventually connect to the interesting paper you did where you linked these photon pairs to electrostatic repulsion/attraction.
But how exactly are you eliminating the ‘Le Sage objections’? If the photon pairs are being absorbed by masses, would they not steadily heat them up? There are ways to get around that problem, but I don’t want to be constantly going back to my own model.
This little part seems quite amazing. I’m not sure about this using different angles for the photons and getting the term for mass!
You still have your photon flux term Io ending up outside the gravitational constant. As well, the u term should be squared. When you consider the two-body case, the attenuation of the flux in the first body creates the ‘shadow’ falling on the second body. But only a fraction of the missing corpuscles/waves represented in that shadow would have been absorbed by the second body. Lastly, by putting p in G, you have the force depending on a body’s density. (You might like to compare your term with Radzievskii’s in PG, noting that he uses the rho term for the gravitational radiation, not mass density).
@HanDeBruijn I have a similar confusion with the term ‘linear density’. It sometimes gets defined as $kg/m$ and sometimes $kg/m^2$, depending on context. Also sometimes called area density for the second but even that description is confusing for its usage. Beam types seem to define the distinction but in a descriptive way the second one might be shown as $kg/m/m$. Most x-ray calculations are done with the second which seemed apt for this radiation model.
For the first it’s simple you would multiply by distance and get the absorption in kg. But then if you want an absorption into a volume you end with $kg.m^2$. To keep units intact there will always be some manipulation. For the second you would multiply by an area to get kg, even though it is already called an ‘area density’. If you are now more confused then I have succeeded in my objective. 😁
In my later jump from linear to volume you may say i took a bit of a leap. The xyz for a cube seems oversimple but i have derived similar for a sphere to get to ‘M’. My basis is that if linear density applies to distance then I need to exchange that to volume density for volume … this all to get to the mass term we see in Newton so that it can be a usable equation with practical solutions. In reality all this needs tensors but then there are only 5 people on the planet that would read it.
As for the last, units of $I_0$ should be $m^2/s^2$ but *m in final units seems correct.
@ExpEarth I blame the curse of large fingers on a small phone screen while lying down for some of your confusion. I am now wielding a mouse and have gone back to correct all the φ,μ,ρ and π which is hopefully a bit more clear.
I will at some point propose the photon-pair transformation as the substitute for absorption. I am also considering @RedshiftDrift ‘photon through electron’ solution if it can be adapted to ‘photon pair through hadron’ because the solution does require a bit of absorption, but he didn’t want to bite last time I asked. 😁 For me now the Le Sage energy catastrophe is merely a problem waiting for a solution.
Your last comment is a bit unclear which I hope was (only) caused by my poor representation of variables. $I_0=c^2$ is inside G.
The term ‘shadow gravity’ is a big misnomer and is linked to (or cause of) the apparent ‘Le Sage speed of gravity problem’. True, he had maybe also set out on that path but you see in my derivation above of ‘g’ there is only the central mass with its surrounding gravitational field. It’s already gravity and only 1 object is involved. And we know that for any comparatively small approaching masses ‘g’ will be the acceleration, thus no knowledge of second (small) mass needed, and thereby concluded that no linked-shadow effect is needed. The small mass is merely falling into the field - that is already there - and in real-time updated at c. For large masses we don’t use ‘g’ because it gets a bit more complex. I would still not call it shadow though, merely an overlap of 2 gravitational fields. This approach should make it easier to analyse 3-body problems too.
This is a secondary effect and I think this may be what Majorana detected. I am leaning toward exploring this more. With μ calculated (back-calc from G) as $9.33*10^{−27}$ m/kg it indicates a very very small fraction of $I_0$ interacts with mass which is why it is so hard to recognize shielding, except on galactic scales. Effectively when μρx <<1 is no longer true and the Newton approximation falls apart
Tensors are not the problem. They are routinely used in classical parts of physics such as chrystallography and solid mechanics. General Relativity is different in that its four-dimensional non-existent space-time is curved not into a fifth dimension but into .. nothingness !! I am routinely downvoting anything that smells like a manifold.
I just read this thread and found it difficult to follow because of typesetting.
Markdown is ok for a post, but it would be much better if you could write and post a .pdf document (when you no longer need your bed for gravitational support) with everything in one place and proper equation numbering!
Also, I find equations such as $G = (1-e^{-\mu\rho x})/\mu \rho x * \mu c^2/4\pi$ unclear, it’s better to write this as
$$G = \frac{1-e^{-\mu\rho x}}{\mu \rho x} \times \frac{\mu c^2}{4\pi}$$
(assuming that’s what you mean!?)
This is confusing indeed. Better would be to stick to a 3D interpretation of these parameters, where μ as an absorption coefficient in m^2/kg (a cross section per kg of material) and ρ as a density in kg/m^3. This makes physical sense first, and then
$$I = I_0 e^{− μ ρ x}$$
is easy to interpret.
The reason I “don’t bite” :wink: is that the STz effect does not produce any absorption of photons! “Absorption” implies fewer particles are left after the process. However, if you take 10000000 photons and let them interact with one electron for 1Gyr, you will end up with the same 10000000 photons and 1 electron. What STz does is to decrease the energy of these photons and transfer it to the electron, without photon absorption.
I don’t quite understand how your model eliminates ‘Le Sage speed of gravity problem’ and ‘shadow effects’ (or not)! I think it is necessary to have absorption of ‘gravity particles’ to obtain a momentum transfer that gives a push! (STz produces random kicks but no net push on electrons… another reason for me not to bite!)
I would need to read about the physics first, before looking at equations that only quantify a mysterious process.
@RedshiftDrift . Repeating your version of the main formula by @Francois-Zinserling, with [ ] denoting dimensionality.
$$ G = \frac{1-e^{-\mu\rho x}}{\mu\rho x}\times\frac{\mu c^2}{4\pi} \quad \mbox{with dimensions} \quad \frac{m^3}{s^2kg} =1\times \frac{[\mu]m^2}{s^2} $$
$$ \Longrightarrow \quad [\mu] = \frac{m}{kg} \quad \Longrightarrow \quad [\mu\rho x] = \frac{m}{kg}\mbox{ }\frac{kg}{m^3}\mbox{ }m \ne 1 \quad \mbox{} $$
Nice try, but not quite, Louis ! Though you have suggested a remedy yourself: >$\mu$ as an absorption coefficient in m^2/kg (a cross section per kg of material)
Then it might be reasonable to replace the length $x$ by an area $A$ instead, giving
$$ G = \frac{1-e^{-\mu\rho A}}{\mu\rho A}\times\frac{\mu c^2}{4\pi} \quad \mbox{with} $$
$$ [\mu\rho A] = \frac{m}{kg}\mbox{ }\frac{kg}{m^3}\mbox{ }m^2 = 1 \quad \mbox{and} \quad \left[\frac{\mu c^2}{4\pi}\right] = \frac{m}{kg}\mbox{ }\left(\frac{m}{s}\right)^2 = \frac{m^3}{s^2kg} = [G] $$
That is mathematically consistent, but >I would need to read about the physics first, before looking at equations that only quantify a mysterious process.
Thank you @HanDeBruijn and @ExpEarth and @RedshiftDrift for your continued engagement.
Han, as per last the units of μ are still m/kg. It seems accepted units for μ are either $1/m$ or $m^2/kg$ depending on the application (with or without ‘ρ’ in the exponent). Both are interchangeably called linear density. If the coefficient is μρx then it means linear - because of the ‘x’ term. μ units must change for a different choice of ρ. If the coefficient contains an $x^2$ term then it is an area, agreed, but then the beam is also automatically redefined as an ‘area of beams’.
I think it is still a problem because of lack of clarity and maybe because I have only seen rare usage of $m/kg$. Either way if I use $m^2/kg$ as shown to get the right units for ‘G’ that still leaves me with a problem of a missing dimension to get ‘M’ in the Newtonian limit where μρx <<1. So I don’t think it was an error to multiply by ‘YZ’ (area) but I must work at making it crystal clear and correct.
There is also the possibility that I was hasty in assigning $I_0=c^2$ and that there may be another coefficient in there that would balance the units out. I’m re-reading Rueda and Haish and other papers on ‘inertial mass from aether’ to see how I need to make an adjustment. The ‘other’ μ could be hidden inside the original which doesn’t change the outcome, but if it only couples to $I_0$ it gets more complicated.
The solution works out well too if I just use the $1/m$ in the exponent as μx but then I lose some basis for my arguments in the original post, e.g. neutron star example.
I find great value in these discussions since the end result must be ‘correct’ in all ways of physics, but what’s been said so far has not made me doubt my original post.
Yes, Louis, I can only drool over yours and Hans’ equations. One day when I grow up, I want to write equations like that. :) Jokes aside. I mean by absorption the absorption of ‘some’ momentum energy. Going by the mechanism of ‘photon pair transformation to static electric fields’, there is no loss of photons or pairs. But I do need some energy (momentum) lost in the process to enact gravity. Your quantum STz provides that solution. If photon pairs (of very specific energy) can behave similar with all subatomic particles - as your photons do for electrons. Your STz also provides this solution without scatter.
@Francois-Zinserling . >what’s been said so far has not made me doubt my original post.
Don’t want to sound arrogant, but if I were you, I would start to doubt the original post.🤔
I have met these guys before: Haisch, Rueda, Puthoff.👎
Not doubting anything … fixing some unclarities and writing a pdf where I am able to do equations :) I hope I can eventually unlock it for you too - the aether and pushgravity I mean.
Puthoff maybe 👎. He took some lone dubious explorations. His work is of value, but his outcomes are lacking somewhat. But read Gravity and the Quantum Vacuum Inertia Hypothesis A solid work based on proper electrodynamics.
(edit: I see from your link above you have included it on your scolding page. I think you should read it again and not just discard it ‘because it claims too much’. When you are mining for something, you will get nothing … nothing … nothing. Until you hit the vein and then treasures will just keep falling out. <- This is my analogy of the dynamic aether aka ZPE aka ZPF aka Quantum Vacuum.)
Can’t resist … Replace $x \rightarrow xyz$ and so $\mu\rho xyz = \mu M$ where $M=$ mass, as proposed by @Francois-Zinserling . Then
$$ G = \frac{1-e^{-\mu\rho x}}{\mu\rho x}\times\frac{\mu c^2}{4\pi} \mbox{ }\rightarrow\mbox{ } G = \frac{1-e^{-\mu M}}{\mu M}\times\frac{\mu c^2}{4\pi} $$
Work out the prefactor (the math before $\times$).
$$ \frac{1-e^{-\mu M}}{\mu M} = \large \frac{1 - \left[1 + (-\mu M) + \frac{1}{2!}(-\mu M)^2 +\frac{1}{3!}(-\mu M)^3 + \frac{1}{4!}(-\mu M)^4 \cdots\right]}{\mu M} $$
$$ = 1 - \frac{1}{2}\mu M + \frac{1}{6}(\mu M)^2 - \frac{1}{24}(\mu M)^3 \cdots $$
Dimensionality of $\mu M$ is m/kg.kg=m and so, with a number g in meters m :
$$ \left[\frac{1-e^{-\mu M}}{\mu M}\right] \approx 1 - \frac{1}{2}g\mbox{ }m + \frac{1}{6}(g\mbox{ }m)^2 - \frac{1}{24}(g\mbox{ }m)^3 \mbox{= 1 - length + area - volume for } g \ne 0 $$
Which is nonsense. The only possible escape is $g=0$ , simply giving:
$$ G = \frac{\mu c^2}{4\pi} $$
But herewith the whole theory with exponential decays is gone !! One cannot sin against mathematics and get away with great words.
Sometimes the police @HanDeBruijn have to be policed ….
$$ \left[\frac{1-e^{-\mu M}}{\mu M}\right] \approx 1$$
On the other hand I may have been too revering of Rueda and Haisch. The mountains of tensor transformations do not appeal to me and smacks of trying to keep your eye on the cups to guess which one hides the pea. their derivation of Newton reads a bit like a comic book …
They should not have diluted their paper with all this other stuff. $m_i = m_g$ in a ZPF would have made a great paper on its own.
Agreed !! But it still escapes me how your units of μM are dimensionless. Please provide us with the final word on this.
Indeed. And I feel very much comfortable with your smart comments above !!
As you probably kow, there has been written a whole book on the subject of (inertial) mass by Amitabha Ghosh with title Origin of Inertia. That book has become one of my classics, though I do not agree with most of its contents (e.g. the common issue of varying $G$ - gravitons - versus Varying $M$, where varying ($GM$) is the “observable” quantity).
@HanDeBruijn the final word is out, much improved, I think …. With a fair bit of rework, thanks to your tenacity on units, and a re-read of Rueda and Haisch … and Louis’ insistence on a pdf. The results seem somewhat promising.
The short answer is:
$$G = \frac{1-e^{-\mu_ρ\rho x}}{\mu_ρ\rho x} \times \frac{\mu c^3}{4\pi ℎ} in [\frac{m^3}{kg s^2}]$$
with the 𝜇 in $[m^2]$ on the right a new constant (newmieuw!) incorporating other values from R&H and results of the derivation.
And the long answer is at: Vixra.org 2108.0123v3
including an elaboration with graphs on the original post of this thread.
@Francois-Zinserling I am at the bottom of page five of viXra:2108.0123. Equation (1) looks like a quantum superposition, but is not expressed in the standard notation of quantum mechanics. Should Eq. (1) be written as:
$$ |\psi_1 \rangle = |R_1 \rangle + |R_2 \rangle = 0, \quad \quad \quad \quad \quad \quad \quad \quad (L1) $$
describing a wavefunction in the Dirac “ket” notation $| \Psi_1 \rangle$?
Or does your notation “ $| \Psi )$ “ (with a round bracket) describe a wavefunction using a different quantum mechanical notation that is not Dirac’s notation?
Oh crap I blame the Word equation editor. I need to fix that!
@Francois-Zinserling Ok, I will read this as Dirac’s notation. I read on p. 5:
“Consider 2 right-hand polarised photons, where R1+R2 are not seen as two photons in tandem but are parallel and both existing within the same space and time, with condition:
$$ |\psi_1 \rangle = |R_1\rangle + | R_2 \rangle = 0 \quad \quad \quad \quad \quad (1)’’ $$
The statement explicitly says that both photons “exist within the same space and time”. We can check that by calculating the probability of finding the ‘photon-pair’ by multiplying the wavefunction $|\psi_1 \rangle$ by its Hermitian conjugate $\langle \psi_1 |$. From $|\psi_1 \rangle = 0$ and $0^* = 0$, we have $\langle \psi_1 | = 0$ and the probability of finding the photon-pair anywhere is
$$ \langle \psi_1 |\psi_1\rangle = 0. \quad \quad \quad \quad \quad (L1) $$
That is, there is zero probability of finding the photon-pair, and therefore it does not exist!
Quantum mechanics requires the wavefunction to be normalized to 1 so that the object we are talking about has a probability $\langle \Psi | \Psi \rangle =1$ of existing.
I can’t read further until the wavefunction is normalized properly to a probability of existence equal to 1.
The problem with your Eq. (1) (and your article in JAMP) is that it does not describe two particles, but instead two properties of a single particle that cancel each other (indeed, $|R_1\rangle = - |R_2 \rangle$).
Two particles would be described by the tensor product
$$ |\Psi_1 \rangle = |R_1\rangle \otimes |R_2\rangle, \quad \quad \quad \quad \quad (L2) $$
which can be normalized to $\langle \Psi_1 |\Psi_1\rangle=1$.
Thank you for that @RedshiftDrift . My QED seems marginally better than my GitHub equation skills, resulting in that gibberish. As a side, a warning to all, it appears JAMP has gone over to the dark side, which would partly explain the shallow review I got, and the result of ^^ that getting ‘published’. I see now how broken it is. Maybe I should retract and redo.
Since I desperately need you to read the rest of the paper, I’ve gone to read Grahn, Annila and Kolehmainen again, but again to realise that they define inertia as an emergent statistical phenomenon, not a fundamental force mediated by particle exchange in the usual QFT sense. Although they do define the photon pairs as ‘the carriers of inertia’, they do not propose a quantum formalism for interaction with objects of mass. Since I need the photon-pairs, but I also need interaction, I differ here from Annila and argue that interaction (and transformation) does happen, based on Rueda and Haisch’s interaction fraction η << 1 of ‘objects of mass’ with vacuum energy. I discarded my illegal QED attempt and created a new wording for the photon-pairs:
|R⟩ and |L⟩ represent single-photon states with |R⟩ as right-hand circular (RHC) and |L⟩ as left-hand circular (LHC) polarization (respectively spin +1 and -1 along propagation). Using |RR⟩ as shorthand for the two-photon state |R⟩₁ ⊗ |R⟩₂, meaning both photon 1 and photon 2 are in the RHC state. |RR⟩ has a definite spin of ⟨Ŝz⟩ = +2ħ. Similarly, |LL⟩ means both are LHC and |LL⟩ has a definite spin of ⟨Ŝz⟩ = -2ħ.
⟨R∣R⟩=1, similarly ⟨L∣L⟩=1, meaning both functions are normalised.
These states form the ‘dark’ of space through the superposition of |RR⟩ and |LL⟩ - the photon pairs - which have no measurable classical electromagnetic field because the oppositely rotating fields of the |RR⟩ and |LL⟩ components interfere destructively on average, and with a net spin=0 the pairs appear to be undetectable by electromagnetic and QED standards.
Equation for the superposition modified to:
$$|s⟩= \frac{1}{√2}(|RR⟩+|LL⟩) with ⟨s|S ̂_z |s⟩=0 and ⟨s|E ̂^+ |s⟩=0$$
@Francois-Zinserling
I would be a little cautious about applying all the rules of quantum physics to your aether model. Keep in mind that the filaments of photons being exchanged between particles could form the basic substructure of spacetime, the vacuum, the aether or whatever you want to call it. It’s quite possible that the ‘rules’ of QED, etc emerge from this substructure.
So, yes, the point by @RedshiftDrift that the photons must apply to two (or more) photons should be granted, but I would not necessarily try to make your system obey every point of electromagnetic theory as it is now understood. We can infer properties of it later based on other observations. Once a full system is described, then it could be used to make predictions.
And you shouldn’t feel shy about your paper in JAMP. It’s surely a stepping stone to better ones to come!
@Francois-Zinserling the state you define above makes sense and seems to define photon-pairs without spin and without electric field. Having browsed at Annila’s papers I suspect the magnetic field is also zero.
Which leads us to the next problem. Momentum is proportional to the Poynting vector $\vec P$ = $\vec E$ $\times$ $\vec B$. Since $\vec E$ and $\vec B$ are both zero for state $|s\rangle$, the state has no momentum $\vec p$ !
As per Newton, the force is $\vec F$ = d $\vec p$ /d $t$, and since there is no momentum there cannot be any force to get pushgravity to work.
I haven’t read the rest of your viXra:2108.0123 paper, but a hint from you explaining how you address this problem would help.
@ExpEarth if different “‘rules’ of QED, etc emerge from this substructure,” then you have to explain how that does not change the already observed rules of electrodynamics I am using above. Those rules cannot be changed too much as this would invalidate Maxwell’s equations, Einstein’s photoelectric effect, Schrödinger’s quantum mechanics, and even for momentum: Newton’s equations, all of which have been tested to high degree of accuracy.
It is not acceptable that pushgravity should destroy Newtonian mechanics and derived theories in order to explain Newtonian gravitation!
@RedshiftDrift , yes the B field is also zero. The photon pairs are completely dark to all electromagnetic effects, including momentum and spin. (edit) $E^2$ and $B^2$ are not zero but are not detectable.
However, Rueda and Haisch have shown a fraction η<<1 to interact with their ZPF, which η I take for granted, and for ZPF I substitute the sea of discrete Annila photon-pairs. This interaction causes the wavefunction to go ‘bright’ into either |RR⟩ or |LL⟩ state with a 50% probability for either. This η accounts for the ‘mass’ part from R&H where $ηρ_0V_0=mc^2$. (left side referring to vacuum energy)
Here @ExpEarth intuition pays off because there is a limit to the meaning of |RR⟩ in QED. For instance, it does not discern the phase of the photons which may be 180’ out, meaning the pair, and I take some license here |R,+⟩ ⊗ |R,-⟩ is still E and B dark in classic terms, except now it has a bright spin=+2. The proposed interaction will thus be a spin-spin interaction, a local vertex, not navigated by the particle cloud.
I intend to write this out when I redo the JAMP article, not for this gravity paper. The logic is still sound although the JAMP formalism is dodgy.
If the sea of photon-pairs is isotropic to the mass, then there is no net momentum transferred. However, if there is an asymmetry as there would be in the vicinity around any mass where $I_{in} > I_{out}$ then there is a net momentum transfer to a nearby particle.
A sensible alternative is : E=mc2 via Mach’s Principle (à la Ghosh)
Happy new year!
@Francois-Zinserling just to make sure I understand:
Are you saying that $\langle B \rangle = 0$ (the magnetic field is zero on average) but $\langle B^2 \rangle$ is nonzero and small, characterized by $\eta$, but big enough to produce an interaction in a sea of Annila photon-pairs? (This applies to $E$ as well.)
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You write
but you can write $|\Psi\rangle = | R \rangle \otimes | R \rangle \equiv |RR\rangle$ and impose
$$ \left \langle RR \left| \hat E^2 \right| RR \right\rangle \ne 0. $$
QED does not tell us how to build such a state (how do you physically generate such photon-pairs?), but if it exists QED can describe it.
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When you write
Which interaction, the spin-spin interaction? In that case the spins of photon pairs can interact, but that interaction does not carry linear momentum, only angular momentum.
There is still something missing to produce the pushgravity force.
@RedshiftDrift the photon pairs force an introduction of a hidden variable - well not so hidden because it is just a phase state - but it forces us to recognise the phase state as a parameter which is not standard in QED. This may sound as if I’m inventing stuff but the Annila pairs have to be written with an EM+ or EM- phase identifier for each photon as: $$|s⟩=1/√2(|R_+ R_-⟩+|L_+ L_-⟩) ,⟨s|S ̂_z |s⟩=0 ,⟨s|E ̂^+ |s⟩=0 $$ for gravity and electromagnetism to ensue. Just writing as $$|s⟩=1/√2(|RR⟩+|LL⟩)$$ still means they can be dark but there are limited interpretations there-of. Introducing them as $R_+R_-$ was a direct hit on the pinata for me. And the pinata just keeps on giving.
These pairs are electromagnetically dark but at a cross-section of ~ $L_p^2$ there is a defined and predictable rate of collisions. By the way I did not set out to find a Planck scale gravity - the results of deriving ‘G’ told me so! I was expecting a high freq>>> gamma rays but didn’t even dream Planck scale until it was thrown back at me.
While they may be dark they still carry the energy of 2 Planck photons and $E^2=E_p^2$ each, but is not measurable in any way in this state. Upon collision though, 1 photon of the pair will couple to the particle as per Rueda and Haisch - scaled by Compton wavelength - scaling $η=λ_p/λ_c$ revealing the particle’s mass as $mc^2=ηℏω_P$. For about $1t_P$ there is a momentum transfer [edit, yes, angular momentum] [edit2, yes spin-spin because the pairs are spin $^+_-$ 2.] along the direction of the photon - allowed by Heisenberg. At a rate of ~ $1/t_p$ in an isotropic flux of dark pairs these net momentum transfers average to zero and the particle does not gain momentum. However - the particle now depletes the dark pairs (transforming them to what will become electrostatic fields - but that’s the next paper), but another particle sees this depletion at $1/4πr^2$ and for that particle the dark photon flux is no longer isotropic - resulting in a net momentum transfer toward the first particle. <- there’s the push-gravity
@Francois-Zinserling I am trying to follow, but you are inventing stuff that is not standard QM and not Newtonian physics. I am not familiar with Rueda’s and Haisch’s QM. Let me know where my understanding differs from yours.
$$ S_z = \langle R | \hat S | R \rangle = +1 \quad \quad \quad {\rm and} \quad \quad \quad S_z = \langle L | \hat S | L \rangle = -1 $$
that is, $+1$ for “Right” and $-1$ for “Left” polarization.
$$ S_{z1} = \langle RR | \hat S_{z1} | RR \rangle = +1, \quad S_{z1} = \langle LR | \hat S_{z1} | LR \rangle = -1, \quad {\rm etc.} $$
$$ S_{z2} = \langle RR | \hat S_{z2} | RR \rangle = +1, \quad S_{z2} = \langle RL | \hat S_{z2} | RL \rangle = -1, \quad {\rm etc.} $$
The total spin is
$$ \langle RR | \hat S_{z1} + \hat S_{z2} | RR \rangle = +2, \quad \langle LL | \hat S_{z1} + \hat S_{z2} | LL \rangle = -2. $$
$$ |s’\rangle = \frac{1}{\sqrt 2} ( | RR \rangle + | LL \rangle ) $$
which does have a total spin $\langle s’ | \hat S_ z | s’ \rangle = 0$ as can be verified from
$$ \frac{1}{\sqrt 2} \Big( \langle RR | + \langle LL | \Big) \big[ S_{z1} + S_{z2} \big] \frac{1}{\sqrt 2} \Big( | RR \rangle + | LL \rangle \Big) = 0. $$
$$ |s\rangle = \frac{1}{\sqrt 2} \Big( | R+R- \rangle + | L+L- \rangle \Big) $$
Do I have this right so far?
$$ \hat E(r) = E_\omega \vec \epsilon \Big[ \hat a e^{i k\cdot r} - \hat a^\dagger e^{-i k \cdot r} \Big] $$
where $\hat a |n\rangle = \sqrt n |n-1\rangle$ is the photon destruction operator, $E_\omega = [\hbar \omega/(2\epsilon_0 L^3)]^{1/2}$ is the field of one photon of frequency $\omega$ in a volume $L^3$, and the field is linearly polarized along $\vec \epsilon$. (Adding the orthogonal polarization would produce a circular polarization, but I’ll ignore the spin for the moment.)
For one photon the electric field is $\langle 1| \hat E |1\rangle = E_\omega$, as expected.
However, for a Annila pair (not $|2\rangle$ but $|1, 1\rangle$) you want
$$ \langle 11| \hat E |11\rangle = E_\omega + E_\omega = 0 $$
$\ $
Here is where I am lost since $2E_\omega >0$, how do you handle that?
Both types of momentum, “angular” and “linear” are conserved in isolated systems, but angular momentum does not directly transfer linear momentum in the direction of a particle’s motion. This is incompatible with QM and Newtonian dynamics, and therefore accepting this means rejecting most of physics. Unless I don’t understand what you are saying, I cannot accept anything past this statement. Please explain.
@RedshiftDrift thank you for your patience and your continued engagement.
If we can agree that the Annila pairs are ‘dark’ then you may ignore these R- artifacts for now, and base it on the outcome, that the only way to get the dark pair out of superposition is a ‘direct collision’ with the particle which results in a spin-spin interaction as a local vertex. No EM interaction because up to this point the pair was dark. Then onto your momentum concern. This may be a representation problem from my side or an interpretation from your side but this must be a near to exact equivalent of your $ST_z$ interactions between photons and electrons. Difference being this is a spin2 pair that fall out of superposition and cannot alter the particle’s spin, so it escalates to angular momentum (Angular momentum transfer is orbital, not spin) $$L=r×Δp$$ where you may set r to Planck length, or the length of the cosmos, because for an interaction time of ~ $1t_p$ it will appear macroscopically linear. When the interaction concludes, the momentum Δp is retrieved by the pair and the particle is returned to its initial state. Momentum is most definitely conserved.
All seems ok, but for a finite time ~ $1t_p$ the particle had some momentum. In an isotropic flux these average to zero.
@Francois-Zinserling
The discussion has taken rather complex turns and I’m hesitant to jump in. However, at this point I am not really clear at what you mean when you talk about Annila’s photon pairs. For comparison here is a figure from one of his papers.
and here is the caption to that figure:
To me it seems he is describing the vacuum as being composed of ‘rays’ of photons. You can see in the figure he refers to unpaired and paired photons as discrete elements within these rays. I have referred to them as filaments of photons connecting all masses and Arto has said he does not disagree with that characterization. I’m not sure if you have incorporated this filamentary structure of the vacuum into your model or if you have the paired photons as being isolated pairs propagating in all directions.
The difference in my view is critical, as one can easily develop a gravity model using the idea of photon filaments connecting masses. I have talked about it in some papers. There is more than one way of doing it. For example, if the photon pairs are connecting all masses, then when two masses approach each other the wavelengths of their connecting pairs must decrease. This requires an energy input. Conversely, the wavelengths of the other filaments the two particles share with more distance masses must all increase. The energy required for particles to approach each other could thus come from the remote connections.
With the isolated pairs that you seem to be thinking of, however, I can’t get a sense of how gravity comes out of it. I don’t think Haisch, Rueda and Puthoff can be of much help here. Their ideas were inventions based on the zeropoint fields, itself a questionable invention lacking a basis in fact.
It seemed in your JAMP paper that you had the photons arranged in filaments, but have we lost that now? This is just my intuition about the problem, but I don’t think we can get to gravity without it. Without a clear gravity mechanism, I don’t see the point of all this discussion of QED and Annila pairs, but I likewise appreciate the efforts of @RedshiftDrift to get to the bottom of it. The photon pairs are the key to gravity and cosmology.
@ExpEarth thank you for engaging. Yes, I have seen that image of Annila. A ‘ray’ is just a stream of photons so we may both be analysing the same thing from different perspectives. My pairs act very local, so our approaches are literally lightyears apart but might converge on the same principle. Not Grahn, Annila nor Kolehmainen seem to have developed mathematics or formalism of the ‘dark pairs’ any further, so it may well be that I have departed from their original idea. Their interpretations are mostly statistical analysis based on entropy. I couldn’t find ‘gravity’ in there except for an ‘expectation’. The QED formalism is mostly my own and has improved slightly with my QED knowledge and interactions from e.g. @RedshiftDrift . Still, dark pairs are theirs so I will continue referencing them until they tell me to stop :)
You are right that Rueda and Haisch did not gain much traction with their ZPF either, but for me their vacuum equation for mass being revealed as a coupling effect to a ‘ZPF’ field has proven crucial.
Combining these two concepts and introducing ‘ZPF’ not as a collection of modes, but monochromatic, has led me to this latest gravity paper. The derivation is heuristic, although based on sound principles in my opinion.
It was when the derivation revealed that ‘G’ happens at the Planck scale with a cross-section of ~ $L_p^2$ that the pinata started giving, and it has not stopped. I did not force it toward ~ $L_p$ ; I just insisted on ‘monochromatic’ and next thing it just fell out and stared at me.
That JAMP paper now looks like it needs a lot of TLC but even there I was working with individual photons. Real, not virtual, and not really rays unless you consider the cumulative effects. I will soon present my paper on electrostatics as a next step to unifying mass/gravity and charge/electrostatics in a dark-photon aether. From what fell out the pinata so far, I can already confirm that the dark-photon aether must be real, or else I will soon be labelled as the next-best fantasy writer after Tolkien :)
Thanks @Francois-Zinserling for the helpful clarifications. I can now see that we are on quite different paths towards gravity which nonetheless share some common features.
One thing that puzzles me about your method of free ‘dark pairs’ is why they should be in that format. I can understand how photons traveling between particles could have standing wave interference patterns and how this could also be the basis for quantum entanglement. One thing you might want to consider is the extension of Annila’s pairs to describe the structures of all particles. He basically says all matter is made of photons. The smallest one made of photons is the neutrino and perhaps neutrinos could serve in the same way as your single ‘dark pairs’. Just guessing here.
One other cautionary note I would add. You seem to be betting a lot on the Planck values - Planck length, Planck time, etc. I have not published on it, but I am convinced these values are all incorrect. They were developed using incorrect assumptions. Will publish soon on this I hope.
I will take a back seat now and wish you the best in developing your gravity and electromagnetic models.
The point of this discussion is that QED provides a formal language to discuss photon-pairs and make quantifiable predictions on their behaviour. Without a formal language, it’s like trying to discuss the works of Shakespeare without understanding English. $$ϵ_R= \frac{1}{\sqrt 2} (\hat x +i \hat y) $$ clearly describes a circularly polarized wave, but “photons arranged in filaments” has no formal meaning: how are they ‘arranged’? The expression $\exp[{\rm i} ($ $\vec k$ $\cdot$ $\vec r$ $-$ $\omega t)]$ describes a moving wave in space and time, “photon pairs connecting all masses; when two masses approach each other the wavelengths of their connecting pairs must decrease” doesn’t tell anything: how does the emitted photon know the distance to the other mass that will be absorbing it later?
@Francois-Zinserling
This is where I cannot follow anymore. The electric field $E_\omega = [ \hbar \omega / ( 2\epsilon_0 L^3 ) ]^{1/2}$ must be a positive number since it is an amplitude that corresponds to a number of photons.
Indeed, it is incompatible with standard QED since the number of photons cannot be negative or imaginary. In QED, $n \in \mathbb N$. This means QED quantum operators, such as $\hat a |n\rangle = \sqrt n |n-1\rangle$, are not valid in your framework. Since $\hat a$ tells me what a photon is, I don’t know the formalism you are using and I don’t understand what you mean by a “photon”.
Do you have a description of the ‘Zinserling formalism’ I could read so I can understand how you build ‘photon-pairs’ with the “R+”?
It seems it is both a “representation problem from your side” and “an interpretation from my side” :smile: This is however not at all “near to exact equivalent” since the two photons in a photon-pair travel in the same direction, while photons must travel in opposite directions for an STz interaction. (Hence the negative sign in the exponent of the second term of $$\hat E(r) = E_\omega \vec \epsilon \Big[ \hat a e^{i k\cdot r} - \hat a^\dagger e^{-i k \cdot r} \Big]$$.)
The quantity $r$ must represent something involved in the electromagnetic interaction, e.g. the radius of the atom for hydrogen. Which one is it, r is small or large? What does it represent in your model?
When I published a semi-classical description of the STz mechanism in 2009, I had one useful comment from someone who asked: “from STz follow many interesting properties, but is the fundamental mechanism something you invent or is it known from quantum optics?” I was able to work out the properties of STz from standard QED and published that in 2024. So STz is not something I invent, it is a physical effect that follows from known physics, and when I use the word “photon” it has the standard meaning understood by physicists.
I am asking you the same thing: from the Zinserling-pushgravity mechanism follow many interesting properties (that fall out of the “piñata”), and the same applies to the Edwards-Shell-Universe, but is it something you invent or is it known from standard physics?
@Francois-Zinserling and @ExpEarth, from the above discussion it seems you use different physics to derive all those interesting properties. It would be better to publish the framework before studying the properties, otherwise nobody will ever understand what you are doing.
Perhaps a discussion of these two questions would help:
About the properties of a photon: Which experiment can I do to know if there was a photon in a certain volume? Or, which experiment tells me that I have detected one?
Electromagnetic interactions: one gram of lead in a volume of 1cm x 1cm x 1cm is very opaque, it absorbs light very efficiently! But one gram of molecular hydrogen ($H_2$) in a volume of 1cm x 1cm x 1cm is extremely transparent. Why is it that photons interact very differently with these two substances, but photon-pairs interact to produce the same gravity on them?
@RedshiftDrift writes: >I am not familiar with Rueda’s and Haisch’s QM. Let me know where my understanding differs from yours.
Actually I am a bit familiar with Rueda’s and Haisch’s QM. Though I can’t find my favorite reference(s) anymore: it’s a long time ago that I’ve been involved with Young Earth Creationism (YEC). The following gives an impression: Questions about John Hartnett’s book, Starlight, Time and the New Physics. It’s important to understand that Rueda’s and Haisch’s QM is not standard QED. Search for “ SED “ in the page by Barry Setterfield and you wil find that heir mechanism is not QED but Stochastic ElectroDynamics. One thing I can remember is that John Hartnett has not been amused by SED.
@Francois-Zinserling writes: >You are right that Rueda and Haisch did not gain much traction with their ZPF either,
And their SED instead of QED.
Update
Why didn’t you mention, @Francois-Zinserling ? As SED is abundant in the reference as recommended by you in this reply. >But read Gravity and the Quantum Vacuum Inertia Hypothesis. A solid work based on proper electrodynamics.
Nope, based on SED is not “on proper electrodynamics”. In common discourse it is QED.
I don’t have anything like a full framework for my model, but let me try a simple analogy to assist visualization.
In QED they say that when a photon is absorbed by an atom the photon is destroyed and its energy and momentum are transferred to other parts of the atom. A new photon is created when one is emitted from the excited atom.
Now this can be expressed in a different way. When a photon is absorbed by an atom, we can also imagine it to retain its photonic character even though its photon properties can no longer be detected. The absorbed photon might now be called a ‘dark photon’. It could be analogous in the photon filament model to a segment of paired photons (see the figures in my most recent post). You could likewise say the photon has been ‘destroyed’ in this case, whereas actually it is again only hidden. The paired-photon segments can then release ordinary photons, in analogy with the photon re-emission in QED.
The network of photon filaments in my model might thus be likened to the internal photon ‘filaments’ which we might infer to exist in atoms. Here the standing wave patterns afford a simple way of visualizing this internal structure.
So just as QED has little to say about the absorbed photons in atoms, having characterized them as ‘destroyed’, so might it have little to say about the paired photon filaments of spacetime in my model.
The utility of this approach is that the spacetime photonic network allows for photon energy to be absorbed and re-emitted and this property can be linked to gravity and cosmology. Returning to the Le Sage ‘pushing gravity’, the absorption of energy notably occurs in the space between masses not inside them. This makes life so much easier, since the heating effect would not cause masses to be incinerated with all this absorption.
@RedshiftDrift says:
This is true!
It looks like we are talking a about different ‘E’. TLDR version: “An R₋ photon has a field 𝐸−(𝑥,𝑡)=𝑠𝑖𝑛(𝑘𝑥−𝜔𝑡); an R₊ photon has 𝐸+(𝑥,𝑡)=𝑠𝑖𝑛(𝑘𝑥−𝜔𝑡−𝜋). A dark pair is the superposition 𝐸++𝐸− which vanishes identically, rendering it electromagnetically dark, except for a visible spin component.”
Figure showing two components of the dark pair. RHC/LHC not shown.
I did not think the inhouse EM charge notation would create such an upheaval, but has now been frowned upon from many sources. It is not an invention intended to overthrow all of QED but merely to relate to a property that arises from the Planck interactions that I cannot find described by standard QED. Apart from the ‘EM charge label’, all else is known physics.
I have made an Addendum which will be added to Gravity paper and Electrostatics paper. Please tell me if this clarifies the choice of notation: pdf version here: EM charge identifier.pdf
@Francois-Zinserling . >rendering it electromagnetically dark, except for a visible spin component.
Ah no! How can electromagnetically dark still have a visible spin component? Zero amplitude = no wave = no photons = no spin components, I would think.
Thanks @HanDeBruijn for the link to Barry Setterfield on John Hartnett’s book. I spent 2014 in Australia working with John Hartnett, but never heard of Haisch, Rueda, or Puthoff. However I’ve heard about Carmeli relativity and Hartnett’s five dimensions x, y, z, t1, and t2. I believe that John has now abandoned that idea of ‘physical time t1’ and ‘biblical time t2’, and no longer works on a Carmeli cosmology.
@ExpEarth
Well then, the problem was already solved by Haisch, Rueda, and Puthoff, according to Setterfield…
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Thanks for the simple analogy to assist visualization. But your answer does not describe any experiment or measurement procedure do detect entities that you call by various names. That only raises more questions:
You are describing entities with different words, but that doesn’t tell me what they are.
The photon is destroyed and becomes hidden in a ‘dark photon’, right?
QED says a lot more than that!
All these are calculated predictions that have been verified experimentally on numerous types of atoms, molecules, and ions interaction with various types of light.
What does the paired photon filaments of spacetime in your model say about the nine quantitative properties given above by QED? (Questions 14 through 22)
This contradicts the absorption by atoms described above where the photon was absorbed by the atom! 24. How do you know that? Just because it makes life easier? 25. How does the photon know to be absorbed in the empty space between masses? 26. If the energy goes to an empty space, what gets heated? How can the temperature of that ‘space’ be measured to verify that it has absorbed the energy? 27. If the momentum goes to the same place, in the empty space, how does a mass located elsewhere know how to acquire the momentum?
From the two questions I had, there are now 27 questions to answer and a contradiction to resolve… And the original questions have not been answered.
Science is more than just ‘descriptive’, it also needs a ‘predictive’ component, something that will give a number to measure and the possibility of verification or falsification. It’s fine to describe a flower and say “it’s a rose that reflects light of wavelength longer than 620 nm”, but this has no predictive power. You describe gravity as “photon filaments absorbed in the space between masses and emitted”, but this statement has no predictive power. Worse, it cannot be falsified since no experiment is described to verify the statement. For this model to be scientific, it needs to make quantitative predictions that can be experimentally measured, and verified or falsified.
The question you can ask yourself is: Which observations would make me abandon the Shell Universe model?
Indeed, I am talking about the QED ’ $E_\omega$ ’, while you describe two classical waves with opposite $\vec E$.
There is a positron now? I thought we were talking about electrons and photons! In normal matter there are very few positrons compared to the number of electrons; how could they contribute anything?
The QED electric field
$$ \hat E(r) = E_\omega \vec \epsilon \Big[ \hat a e^{i k\cdot r} - \hat a^\dagger e^{-i k \cdot r} \Big] $$
cannot describe the field you need. Even if everything else is known physics, this problem is a show stopper because the ‘EM charge label’ (a new quantum number?) has implications on known physics.
I’ll have a look at it later, but I can’t put much effort into this unless the new physics is described clearly. You must show that the ‘EM charge label’ can be incorporated into electrodynamics without breaking any already-verified predictions.
Hartnett’s ‘physical time t1’ and ‘biblical time t2’ have enjoyed a revival in Atomic Time, Orbital Time and the Variable Mass Theory. Especially take notice of the scheme below.
Atomic time loosely corresponds with ‘physical time t1’. Orbital time loosely corresponds with ‘biblical time t2’. An exact matching is not claimed. The Young Earth in YEC is replaced in VM by the Old Earth, with age as in mainstream geology.
With the Variable Mass theory as published at least three predictions on an earthly scale can be made:
It is noticed that three such meager experiments have been enough for acceptance of General Relativity as a mainstream theory.
Hi, @HanDeBruijn In superposition - totally dark. Meaning it will not react to any E or B fields in space. But of the individual pairs in superposition, one will fall out of superposition when it ‘collides’ with a point particle. In other words, if it comes in range of a spin-spin interaction which is a local vertex. From my gravity paper it was revealed - unforced - that the interaction area is $L_p^2$ (Planck area), meaning the dark pair has a very large probability of missing everything. Except there are ~ $10^{104}$ dark pairs per $m^3$ which means there’s a 100% probability there will be a dark-photon interaction with every particle at every instant (Planck time). Then, when it falls out of superposition, (whichever pair is selected with matching spin) the spin2 is bright and partakes in the interaction but E and B still dark until after the interaction.
@RedshiftDrift thou doth protest too much! I do appreciate your comments and criticism though - it has steered me to self-correct in a few instances.
Yes, there’s a positron. Has been since Dirac. :) If ‘you guys’ hadn’t invented fractional charged quarks we would have found the missing antimatter by now. My hypothesis covers all fundamental particles of integer charge and wouldn’t make it out the gate if it didn’t. If I claim the electron (all negative charged particles) interacts only with a negative photon phase in a Planck length, then it leaves the positive phase for the other half of the neutrally charged universe. Waves also travel so a (+) phase now becomes a (-) phase after a π-shift in space or time, so we won’t ‘use up’ all of the (-) phases. All I’m doing is labelling it in its phase at an instant of measurement. Not sure why that is considered QED blasphemy! :(
Agreed, and it’s at the heart of my (our problem). Hence the labels. I could maybe have done the gravity paper without that but not what follows. So the labels must stay, or be improved.
Technically yes, practically no because with wavelengths >> $λ_P$ which we usually deal with in physics the wave-phase is indiscernible, hence apparently no need for QED to label it thus far. It ‘may’ imply that electrons will only receive waves from electrons but that hasn’t been fully explored. In our matter vs (missing) antimatter universe though all the waves we measure are from electrons to electrons so again a reason why no label required yet. Waves are waves and phases shift, but when a particle has a clock that ‘measures a wave’ at every instant $t_P$ then it can discern a (+) from a (-) wave-phase. Too much to fully explore so my answer will be: No QED broken yet unless QED has something to say about Planck photons then I might be challenging that, or more likely not upheaving anything. It’s a perfect falsifier though. If it can be shown to break physics, I’ll have to fix it or pack it up.
A general remark at first. As an engineer, it escapes me completely why we would need still another theory to “explain” gravity, while we have already two machineneries doing the job: good old Newtonian gravity (for sure) and General Relativity. Another remark second. Let’s repeat the table from my previous reply.
The two time frames (Atomic time $t$ and Orbital time $T$) are related by
$$ 1+\frac{T}{A} = e^{t/A} \quad \mbox{with} \quad A=\frac{2}{H} $$
where $A=$ age (in Orbital time) and $H=$ Hubble parameter. In contrast with TL, there is no way in VM to determine $A$ and therefore $H$ from first principles. With the Variable Mass Theory, there is a sharp distinction between QUANTUM and GRAVITY, which is expressed by the existence of the two separate timeframes $t$ and $T$. With other words: Quantum Gravity does not exist and never will. This already makes any attempt to explain gravity by “dark photon pairs” a futile undertaking. Still apart from the question if it would not have been easier to identify these imaginary particle-pairs with single gravitons, being equally hypothetical, but popular in some (other) circles.
@RedshiftDrift
First off, thanks for going through my model in such detail!
At a fundamental level, science still does not have a good grasp on what a photon really is. We can see its effects in some cases but that doesn’t help us get the full picture. We don’t even know if it’s really a wave, a particle or both. So I tend to cast a pretty wide net and not to get too hung up on the precise definitions of things. Some of my models disagree with each other. This approach will often put me 180 degrees away from you on these questions and can only cause you a lot of confusion. That said, let’s see if I can address some of your concerns.
For Q1 and Q2 I think I have described the situation adequately enough. All I’m saying is that after the photon is absorbed by an atom it remains photon-like but that its photon properties are now hidden from view. In that sense it could be called a dark photon, but that term has different meanings and I should be more careful in using it here.
For Q3 you will recall my gravity model based on the passage of light through a spacetime optical medium: https://link.springer.com/article/10.1007/s10509-014-1864-4 In that model, it’s the momentum that photons lose to the envelope of optical medium around masses that gives rise to gravity. I link it to the Hubble redshift.
Keep in mind the Arto Annila idea that all matter is made up of photons and the vacuum (or spacetime) is as well. The reason why I say the photons are arranged in filaments is that that this linearity is compatible with the Hubble luminosity (what I am now calling the Lambda luminosity). This is defined by L = -UH0, where U is the gravitational potential energy of the system. If we suppose (a) that all masses are exchanging photons or virtual photons with each other, with these exchanges linked to some form of quantum gravity and (b) that the photons exchanged between two masses are equal in total energy to their mutual gravitational potential, the simplest situation possible is that the exchanged photons either describe lines between the masses or form filaments of photons between the masses.
Here we get into another feature of my work, which is expressed in this paper: https://www.sciencedirect.com/science/article/pii/S0030402622004284 The redshifted photons of the spacetime background intercepted by masses would have a range of frequencies. But my conjecture is that masses reprocess all these photons of various frequencies into smooth waves of uniform frequency. It is this uniformity which allows the emitted waves to form interference patterns and a spacetime lattice, which is important for producing the gravitational force in my model.
A dark photon would be the double-stranded photon, the actual photons the single-stranded.
By released, I just mean that the photon transforms from a hidden, double-stranded ‘dark photon’ to a detectable, single-stranded actual photon.
Did I already answer this question?
Already answered I hope.
Located within the atom in the interactions between charges. Could also be called virtual photons.
No, see answer to Q5. The standing waves are produced by the reprocessing of background photons by masses.
Q 14-22 did not appear with those numbers but I can see what they refer to. Here I should back up a little and clarify. The two situations – absorption of a photon by an atom and the absorption of photon energy by spacetime filaments – are NOT entirely analogous. As indicated in the answer to Q3 the absorption of photon energy by spacetime is a gradual redshift, not a total absorption of the photon as in an atom. Nonetheless the properties of the absorbed portion of redshifted photons in spacetime filaments would imbue the filaments with the collective properties of all the redshifted photons.
I hope I have explained this already. It’s not absorption of entire photons. In my papers linked to above, the magnitudes come out right.
For Q24, just a reminder that this is all a model (or models)! For Q25, as mentioned above for Q3, the absorption arises with refraction of photons in the spacetime optical material. Now for the heating of spacetime this is finally where we have to go to my shell universe model. You will recall the figure below, where the spacetime heating effect is linked to CMB temperature:https://scx2.b-cdn.net/gfx/news/hires/2024/do-we-live-in-a-shell-1.jpg
For Q27, the energy lost by photons to the spacetime filaments with the refractive Hubble redshift disrupts the spacetime lattice and can then exert forces on the masses attached to those filaments.
For me the model I describe has plenty of predictive power and is leading me to new predictions daily. Experimental proof of the Lambda luminosity is what is needed most. Some sort of laboratory study. Actually, you are the one possibly best suited to prove that it is right! I can only make more predictions. As to what would cause me to abandon the Shell Universe, it would be some proof that the Lambda luminosity does not occur or some study showing that the basic shell structure is not the actual one.
Great @ExpEarth, let me know which experiment will test “Lambda Luminosity”, and I will let you know if I can do that experiment.
@RedshiftDrift Sorry for the late reply. Going over the numbers again, it seems to be impossible to detect the very small Lambda luminosity of a lab-sized object, even using your magical optical powers!
The Le Sage speed of gravity problem has two components to it
I think the ‘shadow gravity’ was a path Browne took us on, and it was based on an incorrect premise that masses had to somehow make ‘visible’ contact and gravity would only ensue after there was a mutual shadow cast on each other. It may also have been an incorrect original premise by Fatio/Le Sage but I am not sure. This shadow would have required a ‘back and forth’ which again leads to ‘speed-of-gravity’ problems - which are not observed. However, as you can see from my derivation, it only takes one mass to establish ‘g’ so no shadow needed. IMO shadow is an effect which reveals secondary shielding, but it is going to be hard to find as Majorana found out. So if g=f(c) then there is a possible solution, except of course relativistic effects also need to be included. Shielding was termed as ‘it’s going to be another problem for Le Sage’ but again IMO since the entire model is based on shielding (of aether), there is no problem to be found here.
Here is an extensive rework of the earlier paper, addressing @RedshiftDrift concerns about QED formalism and refining the document based on comments from ACG and others. Manuscript now under consideration at AIP Advances and already been rejected by 3 others. Addressing anomalies
It is still a Le Sage push-gravity in a dark-pair-photon (Annila et. al) aether. No, it is not deterministic with e.g. an electron probability of ~ $$10^{-23}$$ per Planck time of ‘jittering’ I can safely say that nothing is written in stone.
The ease with which the Coulomb derivation follows (shortened as appendix) from the gravity derivation hints that this is no longer just a toy model. In short, it now shows the Planck units and measures are not just abstracts as we’ve been taught all our lives, but the Planckian aether promises to be the foundation of all physics. Ongoing work already covers a mechanistic Heisenberg, Pauli exclusion, magnetism, PHOTON!, although some of these are still a bit speculative.
Mouse click on Link: Addressing anomalies … gives:
DOI Not Found
10.13140/RG.2.2.17069.24809l
Muffins … Take two
also editing the above, thanks Han.
Sorry, @Francois-Zinserling , I still cannot see anything else than a circular definition between μ and G in your main formula.
$$ G = \frac{\mu c^3}{4\pi \hbar} \left( = \frac{\mu c^3}{2 h} \right) $$
Yeah, sure, why not invoke General Relativity too, as if that would help (provided that you master both).
@HanDeBruijn μ is defined in eq.17 from an origin in eq.7. Neither refer to ‘G’. A component appears that looks and acts like ‘G’ in Newton. μ is in there in eq.19, not $$L_P$$
The Planck-length relation to G is a known definition, and it sets $$L_P$$ in terms of G. I am not claiming to invent this but refer to it as a known entity.
From this relation it is clear that $$μ ∝ L_P^2$$ Thus I have derived G in terms of $$L_P$$. It’s not the same as the definition and brings a lot more to the table.
Yes, but in equation (22) you calculate μ from known constants, and among these is also G. How can you say that this is not circular? So I stay with my point that nobody can calculate for the gravitational constant.
Your concern is valid. Let me argue my case.
As a mathematician you know the basic rule that if $$A=B$$ and $$B=C$$ then $$A=C$$. This is not circular.
So the mathematic relation here is then if $$G=μc^3/4πℏ$$ (derived) and $$G=L_P^2c^3/ℏ$$ (known Planck relation) then $$μ = 4πL_P^2$$. Not quite the same as above, but allowed.
The link between G and $$L_P$$ is not new. I did not derive this relation; my earlier statement is thus over-ambitious and incorrect. I know of the relation but I always saw it as a mere abstract. The numeric value for μ (back-calculated from G) is not significant and here merely illustrative.
Admittedly, the $$μ ∝ L_P^2$$ relation is also back-calculated from G but it now gives new meaning to μ as a ‘cross section’, not just a mere consolidated constant. In the original ‘aether flux’ definition (eq.7) that later led to the consolidated constant μ, there is a proposed wavelength λ for a monochromatic aether. If $$μ = 4πL_P^2$$, the conclusion must be that $$λ ∝ L_P$$. Very significant!
In my push-gravity search for an aether-photon solution, I had always expected to find a very high frequency (short wavelength). High, even beyond extreme gamma. My argument was that if x-rays can mostly go through material, and gamma goes even deeper, then a much-much higher frequency ‘might’ go through everything but without the destructive effects of x-rays or gamma rays. It would just transfer momentum, thus ‘push’. I did not anticipate the Planck length as a wavelength but it immediately made perfect sense when it fell out.
If you look up Planck units then you will see that Planck length, Planck mass, Planck time and Planck temperature are definitions, ipse est not derived. G is in every one of these definitions. Such in contrast to your $G=\mu c^3/4\pi\hbar$ formula which is indeed derived. >The numeric value for μ (back-calculated from G) is not significant and here merely illustrative.
Denied. I find it very significant. And I think it is illustrative for the fact that your theory may be a beautiful construct of the mind (still a matter of taste) but it lacks predictive power.
Jumping on my hobby horse. This formula in VM might turn out to be wrong in the end but it certainly has predictive power.
$$ \mbox{leap seconds} = \frac{(\mbox{ seconds in a century })^2}{\mbox{age of the earth in seconds}} $$
@Francois-Zinserling
It seems you have improved the paper quite a lot, but I’m afraid I still disagree with the units you are using for G, which leads me to think your derivation is incorrect. They seem to be way more complicated than they need to be. Basically, in a Le Sage-type model, including one based on radiation as the pushing agent, G should equal the energy density of the field times the mass absorption coefficient squared divided by 4 pi. You have too many terms in your expression. In PG Martins did a one-dimensional Le Sage model. Perhaps you can use that to see how your model can be simplified. Also Radzievskii in PG develops gravity using radiation specifically.
There are some other points where our approaches differ, but where there is no clear right or wrong. On these we have to go with our gut feeling. But I share skepticism with Han that the Planck units are relevant, except for the Planck power. And Haisch, Rueda and Puthoff are also not a factor. At one point I thought they were and wrote papers using the ZPF, but there is no evidence for these fields and a massive CMB can provide just enough energy without giving way too much.
On your style, it is unconventional to split the intro into sections as you have done. Better to shorten the intro into a few pages.
The paper also does not develop the narrative you need to engage the reader. Most readers are not thrilled about Le Sage and so you will lose that whole crowd at line one. Better to define the problem you are trying to solve more generally and suggest a Le Sage approach to address it. Actually, you could go soft on Le Sage, just using the ideas to develop a new version. Your ideas about Grahn et al’s dark photons are key. We differ on how the Grahn-Annila ‘dark photons’ are organized.
Again on the intro, you start by blaming general relativity for dark matter and dark energy and using that as basis for new theory. Remember that Einstein favoured a static universe initially. It was the Hubble redshift that led to bad interpretations of GR and cosmology. It would be better to reframe your paper as solving particular aspects of GR that have created confusion in the past. As well all know, there are many! My main recommendation is not to address the Le Sage type readership mainly. We are a nefarious bunch and hard to convince! Better to go to a wider audience who might not have fixed views on such things.
Thank you for the continued comments, I fully appreciate the efforts to engage.
@ExpEarth the Newtonian equation and even ‘G’ has been derived by various models, countless times. There are even Newtonian derivations from entropy and even time dilation. They bring nothing new though than that a derivation is possible, so that alone can no longer score points. Paul Stowe has also done a lot of work recently with a radiation model.
I am not sure what concern you have with units. G units are very specific. Derivations, where energy densities were used, are in my opinion shortcuts that do not define ‘what energy’, just like GR does not say ‘what space’. The extra units in my derivation define this exact, right down to the wavelength of the ZPF. No, not the ZPF of so many because most of those broadbands of photons must produce heat and leave us looking for the band that does not, possibly causing gravity. And if there is a narrow band in there strong enough to cause gravity, then the rest of it is strong enough to roast us all every second. The ZPF is maybe not the word for my ‘aether’ but it had to be defined as monochromatic, lest we would see gravity strong over here and weak over there. No chance that a broadband ZPF of ‘leftover photons’ can be homogenous.
These derivations bring nothing new though. I might be totally sold on push-gravity but someone else is still looking for solutions in an ‘instant action at a distance’, strings, or curved space (except this space is not allowed to have properties). It will not help to try and cater to those crowds and sell them softly on Le Sage. Anyone who has given this serious thought must know that no other gravity model can practically deliver the observations. Maybe it’s the engineer in me that wants to understand and not just ‘shut up and calculate’. GR can work as a Le Sage, if the ‘space-time’ just had more properties than the ‘nothingness that can be curved’. So we need an aether, just as GR does.
I am happy with the equation I have for Annila’s dark pairs. It functions fully in my model and even maps directly to the creation and annihilation of particle pairs (triplets and singlets). This is the difference with my model … the bonuses are falling out like someone has hit a critical strike on the pinata, and not just gravity related. AIP is actively looking for someone to do a formal review. At least a review - no matter how critical - can be used to strengthen the paper.
Right now I do not know what to do with all of this if I cannot even get one paper accepted. The barrage of rejections and rebuttals is disheartening.
@Francois-Zinserling . >even ‘G’ has been derived by various models, countless times.
Oh yeah? Then show me such a derivation, without closed loops please.
Sure. Other people are happy with dark matter and dark energy.
Speaking for myself, I have given up trying to publish papers in journals. For a number of reasons: (1) it is too expensive for an individual, costs exceed the benefits. (2) they try to impose inferior standards, such as math formatting using Microsoft (instead of TeX) (3) still apart from rejections and rebuttals that are disheartening
Last but not least (asking Google AI). As of recent estimates, for example viXra hosts over 20,000 preprints and publications across its various open-access categories. The largest share of submissions falls under Relativity and Cosmology, Quantum Physics, Astrophysics, and Quantum Gravity and String Theory. And so my publication goes like in Raiders of the Lost Ark: In a vast warehouse, the Ark is crated up and stored among countless other crates.
Vixra is useful to claim ownership but to post there and hope someone sees it is like throwing a ball into the tardis. More chance to get ‘some response’ from ResearchGate or Academia if you’re prepared to mix it up with the other die-hards in the chat columns.
If it walks like G and quacks like G, it’s G.
This below is a basic derivation but contains some more recent work. From my references. There are many others in there, including Matt’s book.
see p53.
Sorry, François. Maybe I am blind for it, but I cannot see sort of a basic derivation on page 53 of [9]. For me, it’s like chickens clucking a lot, but not laying eggs. Or maybe there exist different opinions on what a derivation is, which I hope not.
@HanDeBruijn if you’re expecting a number to fall out you’re likely to be disappointed. No physical constant with a dimension (like c, G, ħ, e, etc.) has a value that is “derived” without measurement. You can derive e.g. c from other constants but then those need to be measured. So if G is ‘derived’ in terms of other constants (or variables), it’s not circular reasoning. If that is true, pretty much all of physics is built on circular reasoning.
It seems everyone got stuck in my paper at ‘G=….’ though. What I’m saying is that ‘deriving Newton’ is not profound and has been done many times before. I have studied push-gravity since ~2010 and have probably seen 50 variations of deriving Newton. What I did different in my ‘Extended Newton’:
Now my point is if none of this can get any of you ‘anti-establishment’ crowd (I include myself of course) excited, what chance do I stand with a formal journal?
@Francois-Zinserling . >No physical constant with a dimension (like c, G, ħ, e, etc.) has a value that is “derived” without measurement.
Okay. Same example again:
$$ \mbox{leap seconds} = \frac{(\mbox{ seconds in a century })^2}{\mbox{age of the earth in seconds}} $$
Asking Google AI and using common sense: (1) The Earth is estimated to be 4.54 billion years old, with a margin of error of 50 million years. (2) There are 31,536,000 seconds in a standard 365-day year and 31,622,400 seconds in a 366-day leap year. (3) There are 100 times the number in (2) seconds in a century. (4) a century might see anywhere from 30 to 60 leap seconds.
It is important to realize that the above calculation is totally independent of the observation (4). Now do the calculation and check if the leap seconds as calculated are in concordance with observation. >It seems everyone got stuck in my paper at ‘G=….’ though.
Yes, of course they do.
I’m not sure what you’re trying to say. The universe does not care about our day lengths or years, and our errors of actually getting that right. It’s because we try to fit an integer number of seconds into a real value that we have to keep adjusting every few years. It’s not a clock error. We will find our ‘second’ directly proportional to anyone else’s clock though.
that is a measurement.
@Francois-Zinserling . Another attempt to make my point. The Theory of Heat Radiation by Max Planck is a truly impressive book. Here is part of a page from it.
Obviously Planck didn’t know, when he wrote this page, that
$$ \sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90} $$
Giving for the energy density $u$ of black radiation the final outcome
$$ u = aT^4 \quad \mbox{with} \quad a = \frac{8\pi^5}{15}\frac{k^4}{(hc)^3} $$
where $T=$ absolute temperature, $k=$ Boltzmann constant, $h=$ Planck constant, $c=$ speed of light. For a black body (an ideal surface that absorbs and emits all radiation), the emissive power (E or P/A) is:
$$ E = \sigma T^4 $$
For real-world objects, which are not perfect emitters, the formula includes emissivity ($\epsilon$):
$$ E = \epsilon\sigma T^4 $$
As an engineer, I have done many such calculations on Multimode Heat Transfer in Sodium Pumps (: ask AI). The constants $a$ and $\sigma$ are related. See Stefan-Boltzmann law.
$$ a = \frac{4\sigma}{c} \quad \Longrightarrow \quad \sigma = \frac{2\pi^5k^4}{15c^2h^3} $$
It is noticed that $(k,h,c)$ and $\sigma$ are constants that can be determined independently of each other. Which makes that the above is a physics theory as it should be.
Yes, that is indeed beautifiul.
Let me try and make my point again …
Let A = BC (let’s call it a definition) through some analysis/derivation we find that A = XC Conclusion: X=B. what is so difficult about this?
You say it’s circular. It is not. I think you’re also stuck on $$L_P$$ as an abstract and cannot find the sense in this.
Translated into your theory:
$$ G = \frac{\mu c^3}{4\pi \hbar} \qquad A=BC $$
$$ G = \frac{4\pi L_P^2 c^3}{4\pi \hbar} \qquad A=XC $$
$$ \mu = 4\pi L_P^2 \qquad B=X $$
All of these equalities are identities. This means that there exist no independent measurements for $G$ , $\mu$ and $L_P$. And so these equalities are completely uninteresting from an empirical point of view.
I kind of get your drift but you are wrong. G is measured. G was around before $$L_P$$ and was not ’defined in $$L_P$$ terms. $$L_P$$ is a later definition; its definition relates it to ‘G’. Hitherto $$L_P$$ appeared to be an abstract definition of ‘some length’ only.
I defined a wavelength in equation 7. Monochromatic but of unknown magnitude. There are no ad-hoc additions, the derivation proceeds heuristic. There was no mention of G or $$L_P$$ at that time, only an expectation to find something that looks like G, and the hope of learning something new. The derivation continues to derive a component
$$ \frac{\mu c^3}{4\pi \hbar} \qquad$$
that looks and acts like G. Therefore G is now derived from ‘other’ parameters, including the expectation of a wavelength mentioned above. The derivation can end here but it brings nothing new and tells us nothing more about the postulated aether.
Then one of these ‘other’ parameters, the wavelength, ~equates to the Planck length. This confirms that gravity acts on particles at the Planck scale. Further affirmation with electrostatics, as shown in the supplement.
By borrowing $$L_P^2$$ from its definition which includes G, this now affirms that $$L_P$$ is not a mere abstract, and also gives a mechanistic description for the action of G. (and m, and e)
Your theory is good philosophy at best, but IMHO it is not physics as it ought to be. Everyone wants to see the numbers come out. Thus: be my hero and calculate for G !!
Another extremely simple example of an effective calculation is in my section on the Shrinking Kilogram. But so far, nobody has shown any serious interest in our VM work, despite of any evidence. Why bother.
By the way, your theory reminds me of Entropic gravity by Erik Verlinde. Perhaps it’s a good idea to take a look.