[General] Compton and de Broglie wavelengththe "error"

[Viv Robinson]

Hi Andre, Richard and All,

This discussion has gone on for many exchanges that have been based upon theoretical considerations. Surely it is time let experiment and observation determine the correctness or otherwise of the respective viewpoints. 

This is an issue that deals with relativistic mechanics, which has been around for over 100 years. I don’t suggest that the Hyperphysics website is the last word on physics, At http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html, it does cover this topic. It shows clearly Richard’s presentation that  
E^2 = p^2 c^2 + (mo c^2)^2  and not as you suggest.

This is at the core of all particle accelerators, particularly those using curved trajectories. Particle accelerators have been designed 
to get protons and heavy nuclei to very near the speed of light. The trajectory required to get high energy particles around those accelerators and collide them is quite precise. Accelerators do have an ability for minor adjustments to be made. They would not be sufficient to make the adjustments required if particle trajectories were based upon your equation 
E^2 = (pc)^2 + (m c^2)^2 

IMHO, experiment is the final arbiter of any theory. André it could help your case if you were to get confirmation from particle accelerator users that they did indeed use your equation and not that used in Hyperphysics, in standard text books and as indicated by Richard. 

Sincerely,

Vivian Robinson 



On 13 December 2017 at 2:42:43 PM, André Michaud (srp2 at srpinc.org) wrote:

Hi Richard,

Ok, I will try to explain as clearly as possible, but for my explanation to make any sense, you will need to get hold of Abraham's analysis of Kaufmann's data and verify for yourself the calculations on top of also analyzing the Marmet derivation and the analysis I carried out afterwards (see below for links to the papers).

There is not a single equation in all of my derivations that I have not calculated values with to confirm. I even often put the figures directly in my papers where I thought readers should do these calculations themselves to make certain, so they would not have to laboriously go look for examples on their own.

I used two calculators, the simple scientific Casio fx-991 that has about 40 standard constants preprogrammed, that makes most calculations easy:

https://www.amazon.com/Casio-fx-991MS-Scientific-Calculator-Display/dp/B0000VILI2

and a Ti-89 Titanium for the more complex cases:

https://www.amazon.com/Texas-Instruments-TI-89-Titanium-Calculator/dp/B000KUIKGQ/ref=sr_1_3?s=electronics&ie=UTF8&qid=1513133901&sr=1-3&keywords=ti-89+titanium+graphing+calculator

Kaufmann was an experimentalist. He did not interpret his data. He simply collected it, gave his opinion and published his findings.

Max Abraham analyzed it and found that the longitudinal non-linear energy growth curve obeyed the gamma factor established by Voigt, and that the transverse non-linear mass increment growth curve obeyed the gamma factor growth curve applied to half the rest mass of the electron.

Poincare studied the results and confirmed.

I studied the results and agree to Max Abraham conclusions.

What is more, Marmet`s derivation demonstrates that the magnetic field of an accelerating electron increases with velocity according to the transverse energy/mass growth curve established by Max Abraham.

That`s all.

Certainly you can see that the (mo c2 )2 element of SR equation E2 = p2 c2 + (mo c2 )2 is invariant, which goes counter the Kaufmann data regarding the transversemass increase.

And that the (mc2 )2 element of electromagnetic equation E2 = p2 c2 + (mc2 )2 where m=ymoc2 varies with velocity. This element is meant to correspond to the transversally measurable mass of the Kaufmann experiment and of Marmet's derivation.

If you correlate the increase in mass of element (mc2 )2 element of the electromagnetic equation with Marmet's derivation that confirms that the magnetic field of the electron in motion increases with the velocity, you will be able to verify that the increase in magnetic mass corresponds to this factor and to the figures collected by Kaufmann.

From Marmet's conclusion, it can be seen that the rest magnetic mass of the electron corresponds to exactly half the invariant rest mass of the electron, ad that this part of the mass is the one that seems to grow transversally according to the gamma factor. Actually, finer analysis reveals that this transverse mass increment belongs to the carrying energy of the electron. This mass increament is measurable longitudinally as well as transversally, just like the invariant rest mass of the electron.

I proposed the electromagnetic energy-momentum version simply so that people could see the difference with respect to the SR version, but in reality, I think that this SR energy-momentum version (and even the electromagnetic version) arein my view just unrequired complicated hair-splitting.

I admit that I am now getting a little mixed up into all these half of this, half of that. But the real derivations with correct forms from the Marmet derivation are in these 2 papers:

http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/2257

and http://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197

And the Marmet paper is here:

http://www.newtonphysics.on.ca/magnetic/index.html

If you can make abstraction of this antiquated notion ofkinetic energy converting into so called "potential energy" that was axiomatically drilled in to all of us, for the duration of your analysis, this would help a lot in understanding these derivations. If you try to think of energy as always being "active kinetic energy", this wouldhelp a lot also.

Best Regards


---
André Michaud
GSJournal admin
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On Tue, 12 Dec 2017 18:48:22 -0800, Richard Gauthier wrote:

Hello André,
You wrote:"The SR energy-momentum equation as formulated is simply wrong, and demonstrably so with any electron deflection experiment".
Could you pleaseexplain in a concise way why you say that any electron deflection experiment would demonstrate that thestandard relativistic energy-momentum equation E^2 = p^2 c^2 + (mo c^2 )^2 wrong, where E= gamma mo c^2 and p=gamma mo v? This would be really helpful. Thanks.
Richard
 
On Dec 11, 2017, at 4:19 PM, André Michaud <srp2 at srpinc.org> wrote:
 
Hi Richard,

My comments in red below

Best Regards
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André Michaud
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On Sun, 10 Dec 2017 14:53:20 -0800, Richard Gauthier wrote:

Hello André,
Thank you for yourquick reply. One quick question. You wrote:
Not exactly. I say that the first two gammas are incorporated into the new energy E and into the (mov definition of the p term) so that the equation above becomes my reformed relativistic energy-momentum equation.

But in the first line of equations in your Appendix B you define relativistic momentum as : p=γmov (5.65) and you usethis equation to obtain yourequation (B2).  

Yes. but in the line immediately following line (5.65), note that I mathematically separate the gamma factor from the ratio p/mov, so the mathematical group "mov" is henceforth separated from the gamma factor, and is by very definition the Newtonian mov of the Newtonian p(Newton)=mov definition.


Then you combinedequation (B2)with the relativistic energy equation (B1) to get the equation above (B5) ,

You move too quickly to (B5) Richard. If you look at the intervening lines you will see that I actually get equation (B4) from combining (B2) with (B1).

It is here that the disconnect between the SR version and the (Andre) version occurs, because if you look at how (B4) is arrived at, you will also understand how the ratio gamma/gamma is arrived at, and how tempting it is to simplify it to 1/1 before proceeding.

If this simplification is carried on, and if it was assumed (as you do) that p is relativistic by default, then you end up with the SR version E2 = p2 c2 + (mo c2)2, that cannot account for the relativistic/electromagnetic mass increment.

But from a strict mathematical viewpoint, if you simplify the gamma/gamma ratio to 1, you actually remove it entirely from the equation since only Newtonian occurrences of mo remain in equation (B4) --->>>> including in the inner definition of p, since from the second line on, the Newtonian mov group had been mathematically separated from the gamma factor.

This is why the gamma/gamma ratio must not be simplified to 1/1 but rather squared in the following line (to match the exponent level of the left side of the equation), so each gamma occurrence can be reunited with its Newtonian mo companion for the mass to become relativistic again: m=gamma mo.

I hope this clarifies the sequence.

Do not worry about me losing patience. This won't happen.

Best Regards

Andre



which is consistent with the conventionalrelativistic energy momentum equation E^2 = p^2 c^2 + (mo c^2)^2 as long as we agree to define m as m=gamma mo. But thenyou unaccountablydropped the originalγ from your earlier expression (5.65) for relativistic momentum p =γmo v , and now say (above in red) you are incorporating the newγ into "the mo v definition of the p term”, so that your new equation for p as p=γ mo vis the SAME as your original equation for p=γ mo v, even though you have incorporated anadditionalγ into it. So until this point (unaccountably dropping the originalγ from p=γ mo v (5.65) in your derivation, before incorporating the newγ)is cleared up, it’s difficult for me to continue to follow your line of reasoning.

Thanks for your patience.
Richard
 
On Dec 10, 2017, at 12:56 PM, André Michaud <srp2 at srpinc.org> wrote:
 
Hi Richard,

I do not "insist" that E2= p2c2+mo2c4 is Newtonian. I "observe" by calculation that it is Newtonian in the sense that it doesn't account for the relativistic mass increase.

Even if you are convinced that I am mistaken, my Casio hand held calc tells me I am not mistaken.

This has nothing to do with the 3-spaces hypothesis.

We could argue like this in circle for years Richard.

This is very simple. The momentum energy that propels an electron is p= gamma mo v2, and the relativistic mass being propelled is gamma mo. It is as simple as that, as demonstrated by Kaufmann, gamma mo having been measured transversally.

Please take a hand calculator and "observe" the difference.

Equation (5.70) from which you subtract mo will give you the same energy as the Coulomb equation for the electron on the rest orbital of the hydrogen atom, while the standard SR equation minus mo will give you only its momentum energy.

The proof to me that equation (5.70) is right, is that it gets the same energy as the Coulomb equation (4.359743805E-18 j) in excess of the energy of the electron rest mass for the hydrogen rest orbital, which is the actual total energy of the system, as confirmable with the Coulomb equation.

The fact that the SR version does not, tells me that there is a clear disconnect between SR and Maxwell, because the Coulomb equation emerges from Maxwell's first equation (Gauss equation for the electric field), and that the SR version does not give me the total energy of the system on my hand held calculator that the Coulomb equation gives.

I have no other argument in support.

Please do the calculation.

Best Regards

André
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André Michaud
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On Wed, 6 Dec 2017 16:44:30 -0800, Richard Gauthier wrote:

Hello André,

I don’t know why you insist thatE2= p2c2+mo2c4 is Newtonian when it is constructed from the relativistic relations E =γmoc2 and p = γmov and the rest massmo. I think that it is you who are mistaken here, and not Wikipedia as well as and many standard textbooks that now often setmo=min the first equation above as in present high-energy physics usage. Seehttps://en.wikipedia.org/wiki/Energy–momentum_relation, which states: "Inphysics, the energy-momentum relation, or relativistic dispersion relation, is therelativisticequationrelating any object'srest (intrinsic) mass, totalenergy, andmomentum."If in this equationγis set equal to 1 (as you mention), so that v=0, the equation becomesmo2c4= 0 +mo2c4, which is pretty useless and is NOT a Newtonian equation. Using the above relativistic energy-momentum equation for very small (non-zero) velocities compared to c, we get KE = E -moc 2 => 1/2mov2 which IS useful as non-relativistic kinetic energy and is the standard KE formula for small velocities compared with c.

But for the sake of argument let's use YOUR DEFINITION m= γ mo(I know that many others use this equation also when talking about relativistic mass or inertial mass.) With this definition, the relativistic energy E and momentum p terms in the above relativistic energy-momentum equation become E=mc2 and p=mv, whilemo= m/γ. The standard relativistic energy-momentum equation above becomes

E2= p2c2+(m/γ)2c4

= p2c2+ m2c4/γ2

Multiply both sides of the equation by γ2 and we get:

γ2E2=γ2p2c2+m2c4 (I)

This corresponds to the equation above (B5) in your Appendix B:

γ2E2-γ2p2c2=m2c4 where m = γ mo .

But when you bring the second term -γ2p2c2above over to the right side of the equation, you intentionally drop theγ2,saying (in your explanation to me) thatγis incorporated into the momentum term p=mov . But in the equation (I) above , p is already p=mv =γmov so you can’t incorporate thisγinto the momentum equation p=mv =γmov a SECOND TIME— you will get p=γm v =γ2mov, which has no obvious relativistic momentum meaning. So although your derivation at the top of Appendix B starts out with relativistic expressions E=γmoc2, and p= γ mov, these RELATIVISTIC starting expressions for your derivation were apparently forgotten during the derivation at equation (B5), where you incorporatedγinto the NON-RELATIVISTIC expressions E= moc2 and p= mov at this point in your derivation. Then you arrive at a NEW relativistic energy-momentum equation:

E2=p2c2+m2c4 (5.70) or E2=p2c2+γ2mo2c4 since you definem=γmo .

So clearly your equation (5.70), with m=γmo ,is not the standard relativistic energy-momentum equation that you claim to be deriving in Appendix B. No one will object if you DEFINE a new E by this new, non-standard equation (5.70) where you define m=γmo, but this new E equation is not correctly derived from the standard relativistic energy-momentum equation above. Your new E is no longer given by E=mc2 =γmoc2, whichever you like since you define m=γmo. Rather, your new E is bigger than the standard E=γmoc2, maybe in just the way you need it to be bigger than the standard E for your tri-space hypothesis, which carries extra kinetic energy in its electric and magnetic spaces.


I’ve tried to be as clear as I can in pointing out what I see as serious errors in your derivation of the relativistic energy-momentum equation. If I am mistaken here I will be pleased to hear how.

best regards,

Richard

 
 
 
 
 
On Dec 4, 2017, at 8:36 AM, André Michaud <srp2 at srpinc.org> wrote:
 
Hi Richard

My comments inline in red for consistency.

Best Regards
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André Michaud
GSJournal admin
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http://www.srpinc.org/

On Sun, 3 Dec 2017 22:00:45 -0800, Richard Gauthier wrote:

Hello André,
In yourarrival at equation (B5) during your derivation of E 2= p 2c 2 +m 2c4in your book’s Appendix B, you made a simple algebra mistake by apparently accidentally dropping the factor ofγ2from theγ2p2c2term in the line aboveequation (B5):
γ 2E 2 -γ2p 2c2 = (mc 2)2 where m=γmo

Not apparent nor accidental. I did not drop it at all. I incorporated it to p.

I thought this inclusion was obvious, so I did not number this line, since m=γmoby definition. In this line, I treated only the right side of the equation to replace ymo by its relativistic equivalent m.

If you look at second line before this line, you will see that the right side of the equation is (γ2mo 2c4) which is the same as (γmo c2) 2 and since ymo=m, then (γmo c2) 2 is the same as (mc2) 2 which gives you the form you find in the equation before equation (B5), note the comment to the right of the equation: ("where m=γmo").

In other words, it is where m was established as being the same as ymo.


when you moved that term from the left side to the right side of the equation to obtain your equation (B5):
 
γ 2E 2 =p 2c 2 + (mc 2) 2 where p=γmov (B5).
 
Without this algebra mistake your equation (B5) would read
 
γ 2E 2 =γ 2p 2c 2 + (mc 2)2 where p=γmov (B5)
No. Same explanation as above.

In this line, I transfer (γ 2p 2c 2) from the left side of the equation to the right side (changing its sign), and incorporate the gamma factor into the (mo v) definition of (p): since p=mov, then (γ 2(mov) 2c 2) is the same as ((γmov) 2c 2) which is the same as ((mv) 2c 2) and finally the same as (p2c 2) which gives the form you find in equation (B5)

Sincep=γmovis included in your original statement at the top of this appendix, stating the same equation in (B5) has no effect on your derivation and doesn’t correct the algebra mistake mentioned above.
The y factor was separated from all occurrences of mo from the get go at the start of the derivation. Line (B5) is the first time it is combined to mo to establish relativistic form m.

This is where m is mathematically established as being the same as ymo even in p.

Then you deletedγ2from the termγ2E2inyour equation (B5) (now you wouldhave to delete both of theγ2's in the corrected equation (B5) ), stating thatγis dimensionless (though it is certainly not equal to one unless v=0 which you imply that it is not because the equation derived from this equation (B5), you claim, is the relativistic energy-momentum equation that fits yourtri-space theory.
No. I do not delete it. I "incorporate it" into the value of E. In reality you have y2 (E(before))2 = (E(after))2. Which means that the value of E(after) is the same as the product of yE. Since y is dimensionless, this multiplication does not change the fact that the value remains in joules.

 
You also state that E 2 = p2c2 +mo2c4is the non-relativistic energy-momentum equation or the Newtonian equation (which it is not, it is the correct relativistic energy-momentum equation.)
It is non-relativistic and moreover it provides only the momentum energy, not the total energy of the system. It is equivalent to K= ½ mv2.

Note here that to obtain the energy in excess of the rest mass, this rest mass must be subtracted from the total energy obtained:

E=sqrt(p2c2+ (moc2)2) - moc2 with the SR equation

And

E=sqrt(p2c2+ (mc2)2) - moc2 with equation (5.70)


If you calculate the total energy of the system with this equation, you get only half the total energy that can be calculated with the Coulomb equation at the corresponding axial distance between the electron and the proton in the hydrogen atom for example: (2.179871902E-18 j)

But if you do the same calculation with equation (5.70) you obtain the same energy as with the Coulomb equation (4.359743805E-18 j) which is proof enough to me that equation (5.70) is correct and that the SR version where the y/y term of equation (B4) is simplified to 1 is wrong.

So when you deletedγ2 fromthe term γ2E2your equation (B5) you should also have deletedγ2 from the term γ2p2c2in the corrected equation (B5) , andyou should not forget to delete the impliedγ2 from the third term (mc2)2 in (B5) which equals (γmoc2) 2 = γ2(moc2)2 in your deriviation, since for you, m =γmo
 
Deletingγ2from each of the three terms in your corrected equation (B5) leaves the following equation:
 
E2 =p2c2 + (mo c2)2
which is what you call the non-relativistic equation or the Newtonian equation.

Actually the above equation, as mentioned above, IS the correct relativistic energy-momentum equation, here written withmoinstead of the usual m which equalsmo (and not gammamo).
 
What you call therelativistic energy-momentum equation:
 
E2 =p2c2 + (γmoc2)2 or E2 =p2c2 + (mc2)2 where you say m = gammamo is not a correct equation at all, since the right side of your “equation”is larger than the left side of the equation due to the presence ofγ in the right side of your“equation”. E and p are defined (by you) in the standardway as E=γmoc 2 and p=γmov .
 
You conclude your appendix:
 
"At face value, fusing the last occurrence of the squared γ factor with the energy (γ2E2) may seem to be problematic, but considering that this factor is a dimensionless quantity, it can be multiplied with the energy component without any adverse effect for the integrity of the equation."

 
Yes,yourcalculations above, and your statement above, are unfortunately quite problematic, and lead to an incorrect relativisticenergy-momentum equation which is not even an equation, due to the two sides being unequal.

I hope you will take my remarks above in the helpful spirit in whichthey are intended, and make any necessary corrections accordingly.

Dear Richard, I do take your remarks in the helpful spirit that you intend, but there is one thing that supersedes all arguments. It is mathematical proof by calculation:

If you can mathematically prove that SR equation (E2 =p2c2 + (mo c2)2) gives the correct energy value (4.359743805E-18 j) that the Coulomb force induces at mean rest orbital distance from the proton (5.291772083E-11 m) in a hydrogen atom, then I will concede.

In the mean time, I observe by actual calculation that relativistic equation (E2=p2c2+ (mc2)2) gives the right value (4.359743805E-18 j)

Best Regards

André

 
On Nov 30, 2017, at 11:10 PM, André Michaud <srp2 at srpinc.org> wrote:
 
Hi Richard,

Forgotten added embedded inline comment in blue below, regarding where where the extra 1/2 mv^2 of kinetic energy of the particle go when the particle slows down to v=0 and gives up its standard kinetic energy 1/2 mv^2 to its environment.

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On Wed, 29 Nov 2017 18:40:12 -0500, André Michaud wrote:


Hi Richard and all,

I understand that your composite photon model involves no extra energy associated with the internal superluminal motion of the charged half-photons.

But the thing is (not meant here as a negative point) that by structure, in your model, the half photons longitudinally progress on spiral paths at some distance from the axis of motion of the linearly moving photon, they have to move "by structure" at a velocity slightly higher than c otherwise, they could not keep up with the axial linear motion that is defined as being c. Just geometric logic irrespective of any math considerations.

I'll try to explain more clearly about what I understand of kinetic energy, but to remain with the photon structure, by contrast to your "longitudinally" spiralling half photons (moving charges), in the trispatial model, the two charges are oscillating "transversally" with respect to the direction of motion, which is in sync with the established understanding that electromagnetic energy oscillates transversally.

As the related "kinetic energy" of which the half photons are made cyclically transfers from twin-particle state (charges state) to single magnetic particle state in the trispatial model, it accelerates from zero transverse velocity at maximum transverse extent to transverse velocity c at mid transfer to zero transverse velocity again when completely transferred to magnetostatic space. Then in reverse motion, from zero transverse velocity at maximum magnetic spherical extent to transverse velocity c at mid transfer to twin electric charges state to zero transverse velocity at maximum half photon state. Ready for the next transverse reciprocating cycle again. All the while, the complete photon translationally progresses at c, propelled by the other half of the photon's total quantum of energy, that remains permanently unidirectional while the transversally oriented half oscillates.

 
This is why c is never exceeded in the trispatial model.

Now to the nature of kinetic energy.

You mention that when I call kinetic energy a “substance”, this confuses you.

I know. I observed that it confuses everybody.

I have a question that may help you seeing what I mean. In you own photon model, what do you think your half photons are made of really? Charges, would you say! What are charges then? We don't really know.

But whatever what we name "charges" may be, don't you think that a physically existing "substance" has to be involved, that they must be quantities of "something" that physically exists?

Now what could this "something" be?

Remember that we talked about the 13.6 eV electromagnetic photon that is evacuated when an electron is captured by an ionized hydrogen atom (a proton)?

You certainly observed already that this amount of energy corresponds exactly to the kinetic energy that can be calculated due to the acceleration of the electron until it reaches the velocity related to the Bohr radius, which is 2187691.253 m/s classical (2187647.561 m/s relativistic).

Since the electron is now "immobilized", we know that this energy is indeed the energy corresponding to the maximum momentum that the electron had before "hitting the wall" so to speak. Now instead of converting to "potential energy" as classical physics assume, "we observe" that it escapes as an electromagnetic photon.

This means that unidirectional momentum related kinetic energy that propelled the electron is of the same nature as electromagnetic energy, and that in fact the complete complement of any electromagnetic quantum is actual kinetic energy "that-did-not-convert-to-potential-energy", half of which transfers to transverse orientation to henceforth electromagnetically oscillate, in transverse reciprocating motion (in the trispatial geometry), and spiralling about the axis of motion in your model.

This is why I concluded that "kinetic" energy is a "physically existing substance".

When 1.022 MeV photon sare made to convert to massive electro-positron pairs, this means that even the mass of electrons and positrons can also only be the very same substance also. What else could it be since the mother photon was entirely made of only this kinetic energy substance. Whatever it may turn out to really be in reality. Whatever it really is, it has to be a really and physically existing "substance", that we need to study. That's my conclusion.

Now to the extra fact or 2 that you mention,

You mention "KE=moc2(γ-1) wheremois usually just written m"

To remain strictly consistent with mathematical representation, m is the relativistic mass of the electron that was measured by transverse interaction by Kaufmann 100 years ago, which is made of mo + [KE=moc2(γ-1)]/c2. Here [KE=moc2(γ-1)]/c2 is the relativistic mass "increment".

Now KE=moc2(γ-1) also happens to also be equal to the unidirectional momentum related kinetic energy amount that propels the electron at velocity v (embedded in the gamma factor).

Consequently, the complete complement of kinetic energy that an electron possesses at any velocity in excess of its own mo energy value is KE=moc2(γ-1) + [KE=moc2(γ-1)]/c2 both of which physically exist (my conclusion), and can be summarily calculated with:

KE=2moc2(γ-1)

This stems from correlating the Kaufmann electron deflection experiment with Marmet's discovery that the magnetic field of a moving electron increases in sync with its velocity. Explained at the beginning of the "From Classical to Relativistic Mechanics via Maxwell" paper, but as analyzed completely in this other paper:

http://www.gsjournal.net/Science-Journals/Essays/View/2257


You ask: "Are you claiming that a moving massive particle has twice as much kinetic energy as is predicted by standard relativity kinematics?"

My answer is YES.

Grounded on the Kaufmann experiment PLUS Marmet's converging derivation, that is one half remaining unidirectional (momentum kinetic energy) plus the other equal amount transferring to transverse orientation to henceforth oscillate electromagnetically in reciprocal swing between twin component electric and single component magnetic states, displaying omnidirectional inertia (mass) like the invariant rest mass mo of the electron.

Hope this helps understanding how I see "kinetic energy".

Best Regards

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André Michaud
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On Mon, 27 Nov 2017 22:15:30 -0800, Richard Gauthier wrote:

HelloAndré (and all),

Thank you.
 
I want to clarify that in my composite photon model, there is no extra energy associated with the internal superluminal motion of the spin-1/2 charged half-photons composing the composite photon model, whose superluminal energy quanta both move helically at c sqrt(2). The energy of each half-photon is given by E=hf = h/lambda where f is the frequency of the half-photon (here E is not the energy of the composite photon which is 2E) and lambda is the wavelength of the half photon. Remember that the half-photon makes two helical turns per half-photon wavelength, so its apparent frequency 2f (due to its double-looping, like the zitterbewegung frequency) is twice its energy-related frequency f, and its apparent wavelength lambda/2 (due to its double-looping per wavelength lambda) is half of its energy-related wavelength lambda.
I’ve been working on understanding your tri-space approach to the photon and electron. I am trying to understand your approach to kinetic energy of a massive particle. In one place you call kinetic energy a “substance”, which confused me. What does this mean? Is this substance different from energy?

Also, in standard special relativity kinematics, the formula for the kinetic energy of a massive particle is

KE=moc2(γ-1) wheremois usually just written m

while in your article "From “Classical to Relativistic Mechanics via Maxwell”athttp://www.gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/3197

you write: "So from equation (42) we can now directly calculate the associated kinetic energy even if we know only the relativistic velocity of a particle

K=2moc2(gamma-1) (43)

You write the same equation in a slightly different form in your article “On De Broglie’s Double-particle Photon Hypothesis”

at: https://www.omicsonline.org/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.php?aid=70373:

“ From equation (39) can be derived the following equation that allows calculating the kinetic energy that must be communicated to an electron for it to move at relativistic velocity v, when only this velocity is known:

x = 2a(γ-1) (40)

Where “x” is the added kinetic energy, “a” is the energy making up the rest mass of the electron and γ is the Lorentz gamma factor. Any relativistic velocity plugged into the gamma factor will allow obtaining the amount of kinetic energy required for the particle to move at this velocity.”

So my question is, where does this extra factor of 2 come from in thetheoretical derivation of your formula for the relativistic kinetic energy of a massive particle as a function of gamma, compared to the standard relativistic formula for kinetic energy, which is well-established experimentally?Are you claiming that a moving massive particle has twice as much kinetic energy as is predicted by standard relativity kinematics?

Richard

On Nov 27, 2017, at 4:51 PM, André Michaud <srp2 at srpinc.org> wrote:
 
Hi Richard,

I agree with you here that this is an incremental progress over the previous view, Moreover it is now in sync with de Broglie's original twin halph-photons hypothesis.

As I already mentioned, possibly the best that can be had in 4D geometry.

Note that In the trispatial geometry, all of this is accomplished without the energy involved ever exceeding the speed of light.

Best Regards
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On Mon, 27 Nov 2017 16:15:08 -0800, Richard Gauthier wrote:

HI Ray (and all),
You are right that an electron logically can’t be composed of a photon which is composed of an electron-positron pair (or other dipole particle structure) where each half of the dipole is composed of a photon which is composed of… ad infinitum. I think that an electron that is composed from its beginning as a spin-1/2 charged half-photon (not a photon) solves this logical dilemma. I think most of us started out thinking that an electron, if it is composed of a light-speed object, must be composed of a uncharged photon of spin 1 which somehow curls up to become a charged spin-1/2 electron with rest mass m. But I now think that this approach has become a dead-end. Rather, a pre-electron (a spin-1/2 charged half-photon) produced with pre-positron in e-p pair production from a sufficiently energetic spin-1 photon (having net charge zero) starts out already electrically charged and having spin 1/2, so it doesn’t have to change from a spin-1 uncharged particle into a a spin-1/2 charged particle by curling up and losing half its spin while gaining its charge and mass due to its being curled up in a double-loop. Rather the spin-1/2 charged photon retains the spin-1/2 and electric charge e and inertial mass (and its rest mass m) that it started with coming (with a positron) from a photon in e-p pair production. This light-speed (in its longitudinal direction) charged spin-1/2 particle composing an electron is not curled up anyway when the electron it composes is moving very highly relativistically at v < c. Then the light-speed particle composing the relativistic electron retains its charge e, its spin-1/2 and its mass m while moving at light-speed c (longitudinally) along its helical trajectory with forward helical angle theta given by cos (theta) = v/c and with an electron velocity (along its helical axis) of v < c. The spin-1/2 charged photon maintains its spin-1/2 at highly relativistic velocities because it retains the same internal superluminal speed c sqrt (2) (and transverse momentum component value p= h/lambda and its helical radius value R= lambda/4pi) that it had while composing the composite photon from which it emerged in e-p pair production. Only now lambda is much shorter when the electron is very highly relativistic. So at highly relativist velocities the charged photon’s spin remains Sz = R x p = lambda/4pi x h/lambda = h/4pi = hbar/2 = spin-1/2 . To me this understanding represents real, if incremental, progress in understanding photons and electrons (and particles in general.)
Richard
On Nov 27, 2017, at 12:54 PM, Ray Fleming <rayrfleming at gmail.com> wrote:
Richard,
 
As I mentioned. The De Broglie quote in your recent paper expresses that the half-wavelength photons are Dirac Fermions. So a photon can be treated a a series of pairs of Dirac Fermions.
 
So I see these models of an electron as a photon are saying that the electron is made of a pair of Dirac Fermions while not showing why the electron is matter with negative charge rather than the opposite. Then there is the circularity issue an electron is made of a photon which is an electron-positron pair, which are both photons, which are both electron-positron pairs,...
 
Of course Dirac Fermions will behave like Dirac Fermions, but unless we break the circular logic we cannot achieve anything definitive. For my part I see The Dirac Fermion, the electron as what is fundamental. That said, I do not have a model for it. I do appreciate all the fine work on the problem.
 
One thing a lot of papers have in common is talking about the point like nature of an electron, while several including my own derivation of electron mass show that it relates to a size around the Compton wavelength. If we are to understand the true structure of an electron we need to perform scattering experiments with electrons and photons rather than protons, and at lower energies that allow us to see what is going on an the Compton scale. As I say in a couple of my books, proton scattering to find the size of an electron is simply an attempt to find the diameter of the center of a hole.
 
Ray
On Mon, Nov 27, 2017 at 2:25 PM, Richard Gauthier <richgauthier at gmail.com> wrote:
Hello John, Martin, Vivian, Chip,André, Grahame, Albrecht, Rayand all,

Three of our members, that I know of, have derived the de Broglie wavelength in different ways from our double-looping-photon-like-object electron models having spin-1/2: John (and Martin), Vivian and myself. I don’t know if Grahame, André, Chip or Albrecht have derived the de Broglie wavelength from their electron models (and if so, where), but I would like to know.

The three de Broglie wavelength derivations from the above electron models are at:
 
 
1. Is the electron a photon with toroidal topology?”, J.G. Williamson and M.B.van der Mark, http://www.cybsoc.org/electron.pdf, section 6, pp15-16.
2 “A Proposal on the Structure and Properties of an Electron”, VNE Robinson,https://www.academia.edu/10819172/A_Proposal_on_the_Structure_and_Properties_of_an_Electron , section 9, p.13

3. “Electrons are spin 1/2 charged photons generating the de Broglie wavelength”, Richard Gauthier,https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength, section 11, pp 9-11. What I call “charged photons” in my article I am now calling “charged half-photons”, but this does not affect the derivation.

Since we are focusing on the validity of the de Broglie wavelength relation in this email thread, I would like to know if anyone, besides myself, sees any serious error in any of the three de Broglie wavelength derivations above. If there is an error in the derivation in my electron model, I would certainly like to know what it is, and I think that the others feel the same about theirs. Thanks!

 
Richard

On Nov 24, 2017, at 3:41 PM, André Michaud <srp2 at srpinc.org> wrote:
Hello Chip,

You touch an important point by highlighting that at α*cvelocity, would-be dilation or contraction tiny at these velocities, since it lies in the very low relativistic velocity range.

A note however regarding motion at such velocity on a "trajectory" about the nucleus of the hydrogen atom, Heisenberg concluded that the electron did not have to translate at any velocity to remain captive on the ground state orbital. From my analysis from the trispatial perspective, I tend to agree with him. Considering how both electrons have to remain by structure midway between the two protons in a hydrogen molecule for their covalent bounding to be logically explainable, this seems to be factual from my perspective.

Best Regards
---
André Michaud
GSJournal admin
http://www.gsjournal.net/
http://www.srpinc.org/

On Fri, 24 Nov 2017 17:13:54 -0600, "Chip Akins"wrote:
Hi Albrecht and Andre
 
First, Albrecht, I agree that the de Broglie wave, as envisioned by de Broglie, leaves much unexplained, and may well be simply wrong, even though it sort of fits partial mathematical descriptions which seem to be shedding some light on the atomic orbitals in a narrow set of circumstances. This failure of the de Broglie hypothesis to work in all circumstances is a part of the reason I started looking into this subject more. The de Broglie wave also seems to fit double slit experiments, but does not really offer a foundation of physical cause. It just seems to work that way without really disclosing the physical reasons for the de Broglie wave itself. So I think we should look for a better foundation, a causal and concrete explanation, instead of building elaborate theoretical structure on speculation which still remains unexplained. So, yes, there are occasions where the speculation of de Broglie can be applied, and we get the right numbers, but that does not mean the theory is correct and we should stop looking for the actual answers.
 
Andre is onto something when he looks for a relationship between the fields of the proton and the fields of the electron to sort out these issues of the quantization of orbitals. However I will have to do some more math to see if the magnetic field relationships can actually be the answer. At this point, prior to doing the requisite math, I think there is also the possibility that certain dynamics of the electric fields created by the proton and electron will explain the quantization of orbitals. But this is premature speculation. And is sort of a moot point because the dynamics of electric fields are the cause of magnetic fields.
 
There exists a beat frequency which is ¼ the de Broglie wavelength, and this beat frequency is a natural condition of the electron in the significantly sub-light speed orbital (example: a mean circular path at α*cvelocity), so it requires no speculation about dilation or contraction (which are tiny at these velocities).
 
So I think you are both quite correct to look into these issues. We have a lot to gain by reexamining our theoretical basis.
 
Chip
 
From:General [mailto:general-bounces+chipakins=gmail.com at lists.natureoflightandparticles.org]On Behalf OfAlbrecht Giese
Sent:Friday, November 24, 2017 4:25 PM
To:general at lists.natureoflightandparticles.org
Subject:Re: [General] Compton and de Broglie wavelengththe "error"
 
Hi André, Chip, and all,
if we discuss de Broglie's concept of a particle wave, we should in my view refer to his original work and not to others who have used the results (well understood or misunderstood) in other applications.
So, de Broglie in original:
It is of course correct that de Broglie did not just “assume” his wave but he has deduced it from considerations about relativity. But his deduction is based on a severe error as I have explained in detail earlier. So, let’s do it again.
De Broglie has seen a logical conflict between the Einstein- Planck relation (1) E=h*frequency and (2) relativistic dilation; because according to (1) the frequency has to increase at motion and according to (2) dilation will cause the frequency to decrease. But his concern is an error as this conflict does not exist. Because we have to look at an interaction of particles, which is the relevant situation. Any interaction sees frequencies which are increased by the Doppler effect. And the Doppler effect gives an over-compensation of the normal relativistic slow down so that both frequencies above will fit on their own. The same result is achieved if the temporal Lorentz transformation is properly applied. - For de Broglie's new wave no justification exists at all.
The comment of two of you that a single electron does not produce an interference pattern is of course correct. One electron only produces one dot on the screen. But if we assume that a bunch of electron flies to the multi-slit with same speed then the argument works. There will be an interference pattern behind the multi-slit. But if we transform the experiment into the frame of the electrons then the momentum of the electrons is zero, and so the wavelength is infinite, and seen from that frame no interference pattern can occur. But it does occur, also visible for a co-moving observer, and that shows that de Broglie's idea is erroneous. - I have shown in calculations (but not in this place) why under certain circumstances the impression occurs that de Broglie is correct. But in general it is wrong. De Broglie's approach violates Galileo's relativity as well as Lorentzian relativity.
You have mentioned the good results of the use of the de Broglie wave to determine the quantization of atomic orbits. It is true that it works, but it has a similar problem like for the scattering of electrons. Assume a hydrogen atom moving into axial direction with a similar speed as the speed of the electrons in the orbits. Then the resulting momentum of the orbiting electrons increases by about 40% seen from the frame at rest. So the de Broglie wavelength has to decrease by this factor and the energy of these states has to change accordingly. But in practice there will be a much smaller energy change. So also in this case de Broglie fails at a more thorough look.
In the mails there have been some considerations about what de Broglie did "have in mind". But what he had in mind he has written in his PhD thesis. Anything about the energy states of atoms came later and by others (like Schrödinger and Bohr).
Now I will be wondering about objecting arguments.
Albrecht
 

I thank you for your answers and arguments. I will now answer to it, of course. Which means to repeat my arguments of the last three weeks here where I have given argument which seem to have been overlooked.
 
Am 24.11.2017 um 01:20 schrieb Richard Gauthier:
Hi John,André, Chip and all,
Deriving the de Broglie wavelength of an electron model without superluminal motion is easy (in hindsight, since de Broglie did it using special relativity.) But try getting, without superluminal motion, the spin-1 of a non-pointlike photon model (for a photon-in-a-box or otherwise) AND the spin-1/2 of a highly relativistic non-pointlike electron model. In either case there will be some longitudinal momentum Plong, at light speed for a photon model and at very near light speed for a highly relativistic electron model, as well as some significant locally transverse linear momentum Ptrans (even if the net transverse linear momentum of the photon model is zero as in the double-helix photon model) that generates spin Sz = R x Ptrans = 1 hbar for a photon model or 1/2 hbar for a highly relativistic electron model . A longitudinal light-speed or near-light-speed linear momentum vector plus a significant local transverse linear momentum vector gives a diagonal local linear momentum vector with a corresponding diagonal velocity vector whose magnitude is greater than c. Putting a photon model’s or electron model's transverse oscillatory motion, that generates its spin, into two different transverse dimensional spaces is ingenious, but if the photon is to move along longitudinally as a whole and not leave the two transverse dimensional spaces behind, I think there will still be some diagonal superluminal motion. I would be happy to see a proved counterexample.
Richard
 

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