Reformulation instead of Renormalizations

[Bob_for_short]

It seems to me you made a typos in the first equation: the field Lagrangian does not contain phi (the last term). The last term should be E².

The second Lagrangian contains the letter a at two places. So your oscillator equation is wrong: there is still no E_k in it. The oscillator equation should be a free oscillator with its proper frequency and a right-hand-side source, proportional to a. Such an oscillator does not radiate with power unbounded. You need initial conditions, for example E_k(0)=0. Then each oscillator will take only a finite part of energy, although each take will last different time (depending on proper frequency), so in total the EMF power grows.

 

[JustinLevy]

So the Lagrangian is this?
$$\mathcal{L} = \frac{1}{2}m (\dot{\vec{r}}_{CI})^2 - \phi_{ext}(\vec{r}}_{CI} + \sum_k \epsilon_k \vec{E}_k) + \sum_k [\frac{1}{2} \mu_k (\dot{\vec{E}}_k)^2 -(\vec{E_k})^2]$$

Now the equation of motion in a constant external electric field is
$$\mu_k \ddot{\vec{E}}_k = -a \epsilon_k - 2 \vec{E}_k$$

This cannot be correct.

I wish you'd just provide me the full Lagrangian to play with. If you don't want to include multiple particles, fine, but at least include arbitrary external field as well as sourcing of magnetic field. Then I can calculate your new 'version' of the five CED equations for comparison.

 

[Bob_for_short]

You forgot that a is a vector, it is E_ext. The external potential $$\mathcal\phi_{ext}$$ is a scalar product proportional to

$$\mathcal ( \vec{r}_{CI} - \sum_k \epsilon_k \vec{E}_k) \vec{E}_{ext}$$

There is no run-away solutions. An oscillator with a constant driving force has a physically reasonable solution.

Take your time. Do not try to catch me. Better think of physics.

 

[JustinLevy]

I think you may have seen a previous version of my post. The sign error is corrected. You are correct that 'a' should be a vector. Neither of these change the fact that the result is wrong compared to experiment.

I don't understand why you won't just provide the full Lagrangian for me to play with.
Please?

 

[Bob_for_short]

Now the equation of motion in a constant external electric field is
$$\mu_k \ddot{\vec{E}}_k = -a \epsilon_k - 2 \vec{E}_k$$

This cannot be correct...

I think you may have seen a previous version of my post. The sign error is corrected. You are correct that 'a' should be a vector. Neither of these change the fact that the result is wrong compared to experiment.

I don't understand why you won't just provide the full Lagrangian for me to play with.
Please?

Dear Justin,

In each your post you state that "this cannot be correct" or so. Meanwhile you hurry and make errors in elementary derivations the main part of which is contained in my paper. I do not understand why you are interested in the "full Lagrangian" if my approach is wrong in advance in your eyes.

Please explain why "the result is wrong compared to experiment." I feel uneasy to continue without clarifying this question.

 

[Bob_for_short]

Strange results of QFT

There are several threads discussing "rigorous results" of some QFT. For example, it is states that once the quantum fields are distributions, the renormalizations are "necessary" and inevitable. I completely agree with the following reservation: in the present QFTs with self-action, no physical results can be obtained without removing the self-action contributions. It is known since long-long ago, and many-many books have been written on this subject rewording this problem in different ways. The main conclusion of "happy with renormalizations" researchers is the following: the renormalization is not an issue.

At the same time I see a crying problem in such QFTs (QED for certainty): whatever momentum q is transferred to the electron, no soft radiation appears on the tree level in charge-from-charge scattering (Rutherford, Mott, Moeller elastic cross sections). Is it physical? Don't you view this as a problem?

So, in the first non-vanishing order the standard QED predicts events that never happen elastic processes). And it does not predict the phenomenon that happen always (soft radiation, inelastic processes). Don’t you consider this theory "feature" as a complete failure in the physics description? Isn’t it a too bad start for the perturbation theory?

QM predicts probabilities (or cross sections). If the probability of some event is equal to 1 and the theory predicts 0 in the first non-vanishing order, that means a complete missing the point in physics description by physicists in their trial theory. Proving in these conditions that there are no issues, everything is fine and well understood is fooling oneself and others.
No wonder such a practice fails on most cases including QG. When I attract attention to this in the appropriate threads, I obtain infractions. It seems I am the sole person who worries about it.

So I have a sole question to you: why you do not see such a mismatch in the probability prediction as a severe problem of theory formulation?

 

[JustinLevy]

In each your post you state that "this cannot be correct" or so. Meanwhile you hurry and make errors in elementary derivations the main part of which is contained in my paper.

I have been practically begging you to clearly state your full Lagrangian and what the coordinates are for many posts now. Instead you try to lead me around and complain when I interpret you wrong. Yes, I am 'hurrying' because I want to get to my original questions and you seem to be forcing me through an obstacle course before you won't just once and for all clearly lay out what the lagrangian and coordinates for your theory are.

In the calculations I made a sign error that I noticed immediately and fixed myself before you even finished writing your post. And yes, I accidentally didn't mark a constant as a vector. Is this really reason to continue to ignore my simple request?

Please provide the general Lagrangian (multiple particles, arbitrary external field, B_rad field, etc.) for your theory along with what you consider the coordinates to be.
If you don't want to include multiple particles, fine, but at least include arbitrary external field as well as sourcing of magnetic field. Then I can calculate your new 'version' of the five CED equations for comparison.

I do not understand why you are interested in the "full Lagrangian" if my approach is wrong in advance in your eyes.

I am trying to give you the benefit of the doubt, in that there is a possibility that I have a 'straw-man' view of your theory. I want to make absolutely sure I understand what your theory IS before I get into detailed discussion of it with you.

Please explain why "the result is wrong compared to experiment." I feel uneasy to continue without clarifying this question.

If the particle is accelerating in the x direction, are you really claiming there is no electric field in the y or z direction?

Infinite hyperbolic motion (well, parabolic here) can be confusing to imagine, so let me use a much more familiar problem involving radiation.

Imagine a charged particle at the origin, with the external magnetic field on the z axis being zero, and the external electric field $$E_x=E_y=0$$ and $$E_z = a \cos(\omega t).$$

So we can use $$\phi_{ext}(x,y,z) = -a z \cos(\omega t).$$
The charged particle should oscillate along the z axis.

For the x and y components of the electric field, we get the equations of motion:
$$\mu_k \ddot{E}_k^{(x)} = - 2 E_k^{(x)}$$
$$\mu_k \ddot{E}_k^{(y)} = - 2 E_k^{(y)}$$
while for the z components we get:
$$\mu_k \ddot{E}_k^{(z)} = a \epsilon_k \cos(\omega t) - 2 E_k^{(z)}$$

Nothing couples the x and y components of the electric field to the charged particle. This is not correct.
Furthermore, we don't see the correct distribution of radiation (this lagrangian gives only radiation with E_z non-zero). Experiment shows dipole radiation giving a field with radiation going like sin^2(theta) from the z-axis.Either that Lagrangian and set of coordinates is not your actual theory, or your theory is just plain wrong. I am giving you the benefit of the doubt, and assuming I just don't have the correct Lagrangian and set of coordinates. So please, provide the full lagrangian and state explicitly what your coordinates are so that everyone here may learn what exactly your theory IS.

 

[Bob_for_short]

Imagine a charged particle at the origin, with the external magnetic field on the z axis being zero, and the external electric field $$E_x=E_y=0$$ and $$E_z = a \cos(\omega t).$$

So we can use $$\phi_{ext}(x,y,z) = -a z \cos(\omega t).$$
The charged particle should oscillate along the z axis.

For the x and y components of the electric field, we get the equations of motion:
$$\mu_k \ddot{E}_k^{(x)} = - 2 E_k^{(x)}$$
$$\mu_k \ddot{E}_k^{(y)} = - 2 E_k^{(y)}$$
while for the z components we get:
$$\mu_k \ddot{E}_k^{(z)} = a \epsilon_k \cos(\omega t) - 2 E_k^{(z)}$$

Nothing couples the x and y components of the electric field to the charged particle. This is not correct. Furthermore, we don't see the correct distribution of radiation (this lagrangian gives only radiation with E_z non-zero). Experiment shows dipole radiation giving a field with radiation going like sin^2(theta) from the z-axis.

I want to point out that the pumped z-component of oscillator filed propagates mostly along X and Y axes. Do you agree?

Either that Lagrangian and set of coordinates is not your actual theory, or your theory is just plain wrong. I am giving you the benefit of the doubt, and assuming I just don't have the correct Lagrangian and set of coordinates. So please, provide the full lagrangian and state explicitly what your coordinates are so that everyone here may learn what exactly your theory IS.

OK, let me put it in this way:

If you take the classical CED equations and neglect the radiative friction effect (which is really small) you will obtain the same equations as mines, just instead of the known external force it will be the known charge acceleration - they are proportional so it is the same radiation field source. So my results do not differ in this sense from the CED ones if in the latter the radiative friction term is neglected. There is nothing to criticize in my theory. In the standard CED one has to neglect this term because with it the solution is at least difficult to find (and it is actually wrong - run-away solution). In my pet theory the radiative friction term is just absent in the "particle" equation so I obtain easily the right solution for CI and radiation. The electron coordinate r is highly fluctuating but on average it behaves smoothly, like R(t). Now you see, my theory gives nearly the same results as CED. (In fact, I advanced it for QED, not for CED, where the electron fluctuations are quantum rather than classical.)

You may safely use CED equations without self-action (with the radiative friction term neglected) and you will obtain my theory results for fields and averaged results for the electron coordinate. Estimate the relative contribution of the radiative friction in CED and you will see that it is an extremely small value. So the standard CED equations without radiative friction term is the answer to your question about my Lagrangian and charge equations in general, multi-particle case.

Now you see that my purpose was to exclude small but non-physical self-action from the theory and preserve the energy-momentum conservation laws in a more physical way.

Concerning the Noether's theorem. You know, in physics the equations came first and the least action (LA) principle came later. It is OK if the equations have physically meaningful solutions. Then the conserving quantities are well defined. But even in this case the artificial character of the least action (LA) principle is seen easily: after obtaining the equations of motion we never use the future coordinates x(t₂). We use the initial coordinates and velocities. The latter are absent in the least action principle. And x(t₂) are absent as "boundary" conditions since it is non-physical situation - to know future.

So the Noether's theorem derived from LA principle, may formally "work" even when the equation solutions do not exist in the physical sector. This is just the case with CED. An this is what I just fixed by advancing my own approach.

 

[JustinLevy]

I want to point out that the pumped z-component of oscillator filed propagates mostly along X and Y axes. Do you agree?

That Lagrangian only predicts radiation parallel to the x-y plane.
As mentioned, this disagrees with experiment. The fact that sin(theta) is maximum at theta=90 degrees does not mean you can ignore the rest of the distribution.

That Lagrangian disagrees with experiment.

So please, provide the full lagrangian and state explicitly what your coordinates are so that everyone here may learn what exactly your theory IS.

If you take the classical CED equations and neglect the radiative friction effect (which is really small) you will obtain the same equations as mine, just instead of the known external force it will be the known (!) charge acceleration (they are proportional so it is the same field source). So my results do not differ in this sense from the CED ones if in the latter the radiative friction term is neglected. There is nothing to criticize in my theory. In the standard CED one has to neglect this term because with it the solution is at least difficult to find (and it is actually wrong - run-away solution). In my pet theory the radiative friction term is just absent in the "particle" equation so I obtain easily the right solution for CI and radiation. The electron coordinate r is highly fluctuating but on average it behaves smoothly, like R(t). Now you see, my theory gives nearly the same results as CED. (In fact, I advanced it for QED, not for CED, where the electron fluctuations are quantum rather than classical.)

This is getting incredibly frustrating.
You keep repeating your talking points instead of ever telling me specifically what your theory is. If your theory is that lagrangian and coordinates then the simple example above already shows that your theory does not match CED or experiment closely at all.

So I request yet again:
Please, provide the full lagrangian and state explicitly what your coordinates are so that everyone here may learn what exactly your theory IS.

So the standard CED equations without radiative friction term is the answer to your question about my Lagrangian and charge equations in general, multi-particle case.

PLEASE! You know that is not an answer. We already agreed on what the five equations of CED are. There is no explicit "radiative friction" term. Furthermore, repeating your talking point here in no way is explicit enough for me to "guess" your Lagrangian and coordinates.

Please stop giving the run around.
Please just directly answer the question with as much specific math as possible.
Please provide the general Lagrangian (multiple particles, arbitrary external field, B_rad field, etc.) for your theory along with what you consider the coordinates to be.

This thread has shot up to > 3000 views since this conversation started. If you want people to learn your theory, please state the lagrangian and coordinates here so that everyone knows explicitly what your theory is. You keep claiming your theory is almost equivalent to CED, but I see absolutely no indication of that from what math I've seen.

 

[Bob_for_short]

That Lagrangian only predicts radiation parallel to the x-y plane. As mentioned, this disagrees with experiment. The fact that sin(theta) is maximum at theta=90 degrees does not mean you can ignore the rest of the distribution.

That Lagrangian disagrees with experiment.

Take the CED formulation in terms of E_k from post 28 and obtain the field equations as in post 42, please. I bet they are the same as mine.

Then we will move farther. Each thing in its time.

 

[JustinLevy]

Take the CED formulation in terms of E_k from post 28 and obtain the field equations as in post 42, please. I bet they are the same as mine.

Then we will move farther. Each thing in its time.

No, they will not be close at all.
First of all because there is a SOURCE TERM coupling the fields to the particle which the lagrangian above is missing and also because the Lagrangian of CED uses different coordinates than your theory.

There is no need to even do the calculations to see how much a difference these make.
For instance, the Lagrangian of CED leads to Maxwell's equations and the Lorentz force law. The Lagrangian from above does not.

Stop stalling me with more hoops to jump through "first" before you tell me what your theory is.
Please, provide the full lagrangian and state explicitly what your coordinates are so that everyone here may learn what exactly your theory IS.

This is a reasonable request and I've requested it in 9 different posts now. If you wish to discuss your theory in the independent research forum, you should be willing at a minimum to explain what your theory is ... in this case just providing a single equation along with explanation of what the coordinates are.

 

[Bob_for_short]

No, they will not be close at all.
First of all because there is a SOURCE TERM coupling the fields to the particle which the lagrangian above is missing and also because the Lagrangian of CED uses different coordinates than your theory. ... There is no need to even do the calculations to see how much a difference these make.

I still would like to see the CED equations for E_k, please.

This is a reasonable request and I've requested it in 9 different posts now.

Even more times you have declared my approach wrong whereas it was your misunderstanding.

If you wish to discuss your theory in the independent research forum, you should be willing at a minimum to explain what your theory is ... in this case just providing a single equation along with explanation of what the coordinates are.

I want to be understood, I want you to understand my approach. Each time I gave "a single equation", you hurried to declare it wrong physically. With difficulties we advance nevertheless. Now derive the equation for E_k in CED and let us see what is the difference. Without it we cannot advance. If you do not want to follow my advice, I will not be able to prove you anything. Then you may consider my attempt as failed, whatever, I will not care about your groundless opinion.

 

[JustinLevy]

I want to be understood, I want you to understand my approach.

Then give me the full lagrangian and state the coordinates for your theory already.

Even more times you have declared my approach wrong whereas it was your misunderstanding.

That is patently false.
I have been trying to ask you what your theory is for several pages now. That you won't tell me specifically what your theory IS doesn't mean you can blame me when I try to make guesses based on your talking points.

NONE of this would be a problem if you just gave me the Lagrangian and coordinates like I keep asking. Then it would be abundantly clear precisely what your theory is.

I still would like to see the CED equations for E_k, please.

Fine.
This is the last hoop though. I expect you to finally respond in kind to my simple request then.

I don't feel like showing every step, so a google search gave this which provides a similar path with more details
http://www.oberlin.edu/physics/dstyer/AppliedQM/photon.pdf

Let us look at the free field term in the Lagrangian:
$$F_{\mu\nu} F^{\mu\nu} = 2(B^2 - \frac{1}{c^2}E^2)$$
In the free field Coulomb gauge, Maxwell's equations in terms of potentials become:
$$\partial_\nu \partial^\nu A^\mu = 0$$
these four equations are each individually a coupled set of equations for the field 'coordinates' at each point in space. To make them a set of decoupled equations, we can go into the momentum space instead
$$\vec{A}(\vec{r}) = \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{r}} d^3r$$
$$-\frac{1}{c^2}\ddot{A}^\mu(\vec{k}) + k^2 A^\mu(\vec{k}) = 0$$
In the free field, these modes are completely decoupled and appear as harmonic oscillators.

$$\vec{E}(\vec{k}) = - \dot{\vec{A}}(\vec{k})$$
$$\vec{B}(\vec{k}) = i \vec{k} \times \vec{A}(\vec{k})$$

So the free field terms become:
$$\frac{1}{2}F_{\mu\nu} F^{\mu\nu} = -k^2 \vec{A}(\vec{k})^2 + \frac{1}{c^2}\dot{\vec{A}}(\vec{k})^2$$

Previously the CED Lagrangian was written as:
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\phi + q\dot{\vec{x}} \cdot \vec{A} - \frac{1}{4\mu_0} \int F_{\mu\nu} F^{\mu\nu} \ d^3r$$
where the 'coordinates' are the particle coordinates and the field coordinates $A^\mu(x,y,z)$
Now rewriting it in terms of field coordinates at each point in momentum space, we have:
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \epsilon_0 \dot{\vec{A}}(\vec{k})^2 - \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2] \frac{d^3k}{(2\pi)^3}$$


Unlike your Lagrangian, where the radiated field doesn't even couple with the particle if the external potential is linear in position, this clearly couples the radiated fields to the particle. Here, the particle can actually source some fields.

Also notice, unlike your lagrangian, this coupling in CED will allow the radiated electric field x and y components to be non-zero.

Each time I gave "a single equation", you hurried to declare it wrong physically.

The ONLY equation you have written here in this entire thread is when you copy-pasted an equation I wrote earlier (a Hamiltonian), then repeated many of your talking points as if that somehow clearly gave me the Lagrangian and coordinates.

I have spent a great deal of my time to satisfy you so that you'd just answer a single question of mine. Will you please, please, finally grant my request:
Please provide the general Lagrangian (multiple particles, arbitrary external field, B_rad field, etc.) for your theory along with what you consider the coordinates to be.

 

[Bob_for_short]

Now rewriting it in terms of field coordinates at each point in momentum space, we have:
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \epsilon_0 \dot{\vec{A}}(\vec{k})^2 - \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2] \frac{d^3k}{(2\pi)^3}$$

Unlike your Lagrangian, where the radiated field doesn't even couple with the particle if the external potential is linear in position, this clearly couples the radiated fields to the particle. Here, the particle can actually source some fields.

The radiated field should be caused with the particle acceleration determined in turn with an external force. And a particle in my approach is a part of oscillator. Push the particle and the oscillator gets excited. Very simple and physical mechanism of coupling. You see, you still do not understand or do not know what my approach tells about the relationship (coupling mechanism) of electron and the field. And you want "the total Lagrangian"! For whom I wrote my detailed articles?

Also notice, unlike your lagrangian, this coupling in CED will allow the radiated electric field x and y components to be non-zero.

Each radiated mode propagates along k. You hurry again to judge. I am afraid that you need "the full Lagrangian" solely in order to declare it "obviously wrong", in your understanding. And I want you to open your eyes. You are already close. Derive the equations for E_k. It should be an ordinary oscillator equation with a pumping source. Get it. You will achieve much more than you think.

I have spent a great deal of my time to satisfy you so that you'd just answer a single question of mine.

Me too. We should speak the same language first. What I am trying to get is your comprehension of the coupling mechanism in my theory. There is a conceptual difference to overcome. Then you will be able to judge and appreciate. There is no hoops, we are advancing strait ahead, thanks to my patient efforts.

 

[JustinLevy]

You continue to complain that I don't know what your theory is. Yet you refuse to state in clear math what your theory is. Complaining more to me and repeating talking points is not going to somehow magically impart any clear math about your theory.

If you stated the general lagrangian clearly and what you consider the coordinates, there would be absolutely no room for confusion. Anyone could work out the equations of motion.

I have asked for three pages for you to answer ONE REQUEST.
You are clearly not interested in discussing your theory if you are not willing to write a single equation that would have prevented three pages of "discussion".

Do you deny that if you wrote the general lagrangian for your theory and stated clearly what you consider the coordinates, that the theory would be precisely laid out in just that one equation?

As you have already been warned by one moderator, if you are going to have a thread in the Independent Research forum, you must respond to questions about your theory.

If you want someone to understand your theory, stop complaining and mathematically specify your theory. Please provide the general Lagrangian (multiple particles, arbitrary external field, B_rad field, etc.) for your theory along with what you consider the coordinates to be.

 

[Bob_for_short]

You continue to complain that I don't know what your theory is.

I do not "complain" but attract your attention. We were speaking of the simplest case of one charge. It is a correct methodological approach - explain the mechanism in a simple case and then generalize to many-particle case. We are still there.

Yet you refuse to state in clear math what your theory is.

It is not true. In my articles I clearly and repeatedly introduce this "math". It is a usual math for a compound system. I use the CI and relative coordinates with the corresponding conjugated momenta in the Hamiltonian formulation or velocities in the Lagrangian formulation.

I have asked for three pages for you to answer ONE REQUEST.

I have already answered it. It is a pity you missed it.

As a matter of fact, I wrote a general Hamiltonian (60) for QED; not for CED. It may describe as many particles as you like. CED is obtained as the inclusive result of QED.

You asked for a CED Lagrangian although "my CED" is obtained as the inclusive QED result. Yet I agreed to explain you what is what in principle in an elementary CED case. Even such an elementary CED case looks ridiculous from a classical point of view because the electron coordinate r(t) is highly fluctuating: it contains a smooth part R(t) and oscillating part because in my model the electron is a part of oscillators. On average one obtains R(t). In QED it corresponds to the inclusive picture which is more physically correct than just averaging the classical trajectory.

I do not have the most general CED Lagrangian. Lagrangian serves to obtain equations of motion. They are more important. We have them already, fortunately. Let us start from mechanical equations.

From practical point of view my approach corresponds to neglecting the radiative friction (jerk) in the charge equations of the usual CED and considering the charge positions as electronium's CI positions. The charge equations may contain only external fields - as the Lorentz force (i.e., in a usual way). This is a "mechanical part" of "my CED". So you have these equation already.

The radiated energy or power is entirely contained in the Maxwell equations since, according to my model, they are equations of the "internal degrees of freedom". The energy-conservation laws are already preserved perfectly in this model. We should not, unlike H. Lorentz, add a radiative friction term like jerk (2e²/3c³)da/dt in the charge equations because they are the CI equation in my model. So I removed the "uneasiness" in practising CED without radiative friction term. According to my model, the mechanical equations are more correct without it than with it.

So you have the Maxwell equations already. Together with mechanical equations they are "my CED", if you like. Of course, such a description is valid only in case of small quantum effects.

When you look for a charge trajectory in an external field, the Lagrangian contains the term L_int = jA_ext.

When you look for a field evolution with given sources, the Lagrangian contains the term L_int = j_extA. By the way, in this case the field equations can be formally solved and their solutions can be put in the mechanical equations of another charge, thus one excludes the field variables from consideration. This is clearly seen from the Hamiltonian (60) (four-fermion trem ∫∫jDj).

The self-induction is contained in this current-current term. It is a mutual effect of several charges, not a self-action.

For a self-consistent description in CED it is sufficient to use the ordinary equations without the radiative friction (jerk term) in the mechanical equations. You can use L_int = ∑{j_extA + jA_ext}, where the sum is done over all elementary charges and fields. Is it OK with you?

You see, there is a conceptual gap between your understanding of CED and my theory. It is not reduced just to different math. CED equations contain already the necessary math but in my model we have different physical meaning of variables given just above.

So take the CED Lagrangian and use the corresponding equations without radiative self-action (jerk contribution). That is my answer to your demand.

Now, derive the oscillator equations in case of CED, please. What is a source of radiation in "your CED"? I want to compare it with my theory. You said it is quite different. Show me that.

Regards,

Vladimir.

 

[JustinLevy]

It is not true. In my articles I clearly and repeatedly introduce this "math".

You do not state the lagrangian for this simple case, let alone for the general case in your paper. You forced me to guess based on your statements.

I have asked for three pages for you to answer ONE REQUEST.

I have already answered it. It is a pity you missed it.

NO YOU HAVE NOT! HOW DARE YOU MAKE ME WRITE PAGES WORTH AND THEN LIE STRAIGHT TO MY FACE.

I do not have the most general CED Lagrangian.

So you finally admit why you refuse to answer my question.

As a matter of fact, I wrote a general Hamiltonian (60) for QED; not for CED. It may describe as many particles as you like. CED is obtained as the inclusive result of QED.

I told you multiple times why presenting just the Hamiltonian is not enough. The Hamiltonian equations of motion are useless unless it is precisely clear mathematically what the conjugate momenta are ... if you feel you know these, then just take them and obtain your lagrangian.

It is starting to appear that you are either:
A) dragging me along without intent to EVER answer my question
or
B) don't have your theory mathematically well defined enough to even be ABLE to answer my question

I have spent much time and wrote many equations here.
The least you can do is try to answer my one simple request.
Once your theory is mathematically specified, then there can be no "confusion". Heck, at that point we don't even need to "interpret" anything. We just work the equations and if we do the correct math we will HAVE to agree on the answers.

So please finally answer my request.

From practical point of view my approach corresponds to neglecting the radiative friction (jerk) in the charge equations of the usual CED and considering the charge positions as electronium's CI positions.

I am getting very sick of your talking points.
Especially this one. First of all you already agreed to what the equations of CED were. There is no 'radiative jerk' in those equations. That is not a fundamental part of CED. It is misapplication of Abraham-Lorentz that cause many of these problems.

PLEASE, LET'S FOCUS ON WHAT THE HECK YOUR THEORY IS INSTEAD OF YOUR COMPLAINTS ON CED. We can return to that once we agree what your theory even is.

You see, there is a conceptual gap between your understanding of CED and my theory. It is not reduced just to different math.

NO! It does reduce to math.
If you mathematically state what your Lagrangian and what you consider the coordinates, THEN THERE IS NO ROOM FOR "CONFUSION". It is precisely defined. The answers follow by calculation and the "philosophy"/interpretation of the equations in this sense are meaningless metaphysics. It is good to have a mental picture, but the math must be first.

So take the CED Lagrangian and use the corresponding equations without radiative self-action (jerk contribution). That is my answer to your demand.

THAT IS NOT IN THE CED LAGRANGIAN!
DAMN IT. Please state MATHEMATICALLY what your lagrangian is.

Now, derive the oscillator equations in case of CED, please. What is a source of radiation in "your CED"? I want to compare it with my theory. You said it is quite different. Show me that.

You already agreed, for a point particle a (non-relativistic) Lagrangian that gives CED is:
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\phi + q\dot{\vec{x}} \cdot \vec{A} - \frac{1}{4\mu_0} \int F_{\mu\nu} F^{\mu\nu} \ d^3r$$
The coordinates are x and A^\mu, with the fields being a function of position.

I was sloppy in 'deriving' the momentum space version of the fields, as I was talking too much about the free field. This lead to me accidentally dropping one part. Here is the correct one
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2 - \epsilon_0 \dot{\vec{A}}(\vec{k})^2 - \epsilon_0 k^2 \phi(\vec{k})^2 + \frac{1}{\mu_0} \dot{\phi}(\vec{k}^2 ] \frac{d^3k}{(2\pi)^3}$$
Now the coordinates are x and A^u, with the fields being a function of momentum space.
You can derive:
$$(-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\phi(k) = -\rho/\epsilon_0$$
$$(-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\vec{A}(k) = -\mu_0 \vec{j}$$
Which is maxwell's equations in terms of the potentials in the Lorenz gauge.

And do you agree that your Lagrangian is:
$$\mathcal{L} = \frac{1}{2}m (\dot{\vec{r}}_{CI})^2 - \phi_{ext}(\vec{r}}_{CI} + \sum_k \epsilon_k \vec{E}_k) + \frac{1}{2}\sum_k [\mu_k (\dot{\vec{E}}_k)^2 -(\vec{E_k})^2]$$
With the coordinates being r_CI and E_k.

If we can agree on these things, I will agree to work out more math for you.

 

[Bob_for_short]

...It is starting to appear that you are either:
A) dragging me along without intent to EVER answer my question, or
B) don't have your theory mathematically well defined enough to even be ABLE to answer my question.

Calm down, take it easy, I have no bad intentions.

We see the CED differently, it is obvious. For example, you do not find there the jerk contribution. Let me tell you that here you are alone.

The whole point of my research and my model is to get rid of this jerk. It has severe consequences in QED. Look at my title. You have to understand that I was motivated by this problem. I don't hide my general Lagrangian from you. I work with QED, not with CED.
I am coming from QED reformulation, if you like this vision better. CED was not my concern because even for one particle it has ridiculous features. If you accept fluctuating electron coordinate - it is OK, we can advance in CED. But it is much better accepted in QED (quantum mechanical charge smearing), so I worked and work with QED actually.

Concerning classical things, I gave a quite detailed mechanical analogy in my paper where the oscillator is mechanical. Being a part of oscillator resolves the energy-momentum conservation problems in interactions. This is my fundamental result which I apply in QED just as in Classical Mechanics.

Concerning "my Lagrangian", the particle part is OK and the oscillator part is like yours but in the Coulomb gauge. It is nearly the same. The only difference is that the field is radiated one - the vector potential A(k) is transversal (orthogonal to k). It represents physical degrees of freedom that take and give away (exchange) the energy-momentum. (In other gauges there are non-physical degrees of freedom decoupled from matter.)

If you take a time derivative of your vector potential equation, you will obtain an equation for the electric field expressed via particle acceleration. The latter is proportional to the external force from the particle equation so they are interchangeable if there is nothing but an external force. My theory gives naturally the known external force as a field source (rather than an unknown particle acceleration in case of taking into account self-action). This excludes the self-action of the radiated field on "particle" motion in my model. This was my primary concern in my research.

 

[JustinLevy]

Calm down, take it easy, I have no bad intentions.

Okay. I was just very very upset that you still won't answer my question, but yet claimed you had answered it despite saying in the same post that you don't currently have an answer.

Concerning "my Lagrangian", the particle part is OK and the oscillator part is like yours but in the Coulomb gauge.

Is what I wrote for your Lagrangian and coordinates in the single particle case with no external magnetic field correct?
If not, then please write the correct equation and coordinates here.
Also, please specify here in this thread what r_CI is in terms of the radiated fields and particle position.
Then we will have everything that mathematically specifies your theory all here in one place for unambiguous discussion.

It is nearly the same. The only difference is that the field is radiated one - the vector potential A(k) is transversal (orthogonal to k).

I disagree strongly with your statement that it is nearly the same.

We need to be precise enough with the math that we can discuss this precisely. Once you commit to stating what your Lagrangian and coordinates are, then there can be absolutely no room for confusion of the consequences of your changes ... I want to finally move on to discussing these consequences and predictions of your theory.

As explained above, it appears to me that your theory does not give the correct radiation distribution for dipole radiation. If you disagree with me, fine. But let's agree on what the math of your theory is so that we can work on the calculations and come to an agreement.


As for my request, if you believe you know the Hamiltonian and conjugate momenta, please work this backwards to get the Lagrangian. I do not feel this is an unreasonable request considering how little work it should be for you.

 

[Bob_for_short]

OK, Justin, I will do it tomorrow. It's late now in Grenoble.

Regards,

Vladimir.

 

[Redbelly98]

As for my request, if you believe you know the Hamiltonian and conjugate momenta, please work this backwards to get the Lagrangian. I do not feel this is an unreasonable request considering how little work it should be for you.

Bob_for_short, please satisfactorily address JustinLevy's request as your next response. Otherwise, this thread will be "locked pending moderation". This has gone on too long.

OK, Justin, I will do it tomorrow. It's late now in Grenoble.

Regards,

Vladimir.

Okay.

 

[JustinLevy]

After your PM's, I decided I'd be willing to post ONE more post before your response. But I am not willing to be stringed along any further. Please finally respond to my request, stating here precisely and mathematically, the lagrangian and coordinates and any supporting mathematical definitions needed to define your theory. Then there can be no room for confusion about what your theory actually is, and we can move on to discussing predictions.

------------

Since you wish to work in the Couloumb gauge and in reciprocal space, for comparison, here is the Lagrangian for classical electrodynamics (CED) written that way.

Gauge condition:
$$\nabla \cdot \vec{A}(\vec{r}) = 0$$
in reciprocal space this is
$$\vec{k}\cdot \vec{A}(\vec{k})=0$$
so the vector field is purely transverse, and thus only has two free components.Non-relativistic, since the discussion has been non-relativistic so far, and for an arbitrary number of particles:
$$\mathcal{L} = \sum_\alpha \frac{1}{2}m (\dot{\vec{r}}_\alpha)^2 - \frac{1}{2}\int d^3k [\phi^*(\vec{k}) \rho(\vec{k})+\rho^*(\vec{k})\phi(\vec{k})] + \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\vec{A}(\vec{k})+\vec{A}^*(\vec{k})\cdot\vec{j}(\vec{k})]$$
$$\ \ \ + \frac{\epsilon_0}{2}\int d^3k [k^2 \phi^*(\vec{k})\phi(\vec{k}) + \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) -c^2k^2 \vec{A}^*(\vec{k})\cdot\vec{A}(\vec{k})]$$
where
$$\vec{\rho}(\vec{r}) = \sum_\alpha q_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$
$$\vec{j}(\vec{r}) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$

Since the vector $$\vec{A}(\vec{r})$$ is real, $$\vec{A}(\vec{k}) = \vec{A}^*(-\vec{k})$$, and similarly for the scalar potential. So the generalized coordinates are:
only half the reciprocal space k $$\phi(\vec{k}),\phi^*(\vec{k}),\vec{A}(\vec{k}),\vec{A}^*(\vec{k})$$ (only the transverse components for A), and $$\vec{r}_\alpha$$

1] Do you agree the above gives CED?

2] Please finally respond to my request, stating here precisely and mathematically, the lagrangian and coordinates and any supporting mathematical definitions needed to define your theory.

 

[Bob_for_short]

... in the Couloumb gauge and in reciprocal space, for comparison, here is the Lagrangian for classical electrodynamics (CED) written that way.

Non-relativistic, since the discussion has been non-relativistic so far, and for an arbitrary number of particles:
$$\mathcal{L} = \sum_\alpha \frac{1}{2}m (\dot{\vec{r}}_\alpha)^2 - \frac{1}{2}\int d^3k [\phi^*(\vec{k}) \rho(\vec{k})+\rho^*(\vec{k})\phi(\vec{k})] + \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\vec{A}(\vec{k})+\vec{A}^*(\vec{k})\cdot\vec{j}(\vec{k})]$$
$$\ \ \ + \frac{\epsilon_0}{2}\int d^3k [k^2 \phi^*(\vec{k})\phi(\vec{k}) + \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) -c^2k^2 \vec{A}^*(\vec{k})\cdot\vec{A}(\vec{k})]$$
where
$$\vec{\rho}(\vec{r}) = \sum_\alpha q_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$
$$\vec{j}(\vec{r}) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$

Since the vector $$\vec{A}(\vec{r})$$ is real, $$\vec{A}(\vec{k}) = \vec{A}^*(-\vec{k})$$, and similarly for the scalar potential. So the generalized coordinates are:
only half the reciprocal space k $$\phi(\vec{k}),\phi^*(\vec{k}),\vec{A}(\vec{k}),\vec{A}^*(\vec{k})$$ (only the transverse components for A), and $$\vec{r}_\alpha$$

1] Do you agree the above gives CED?

Not really. φ should not be involved in the filed Lagrangian, it's a mistake. φ has an explicit solution (∆φ ∝ ρ) in this gauge so it can and should be excluded. The inter-charge electrostatic interaction - the second term in your expression - is then described with the following sum: ∑_(α>β) q_αq_β/|r_α - r_β|. (There was no need to make a Fourier transform.) In case of one charge in an external filed Φ_ext(r_e), the latter is present in the Lagrangian as a potential energy. Similarly, there may be the term jA_ext describing an external magnetic field, for example. Both Φ_ext and A_ext are given function of space-time, not the dynamics variables.

2] Please finally respond to my request, stating here precisely and mathematically, the lagrangian and coordinates and any supporting mathematical definitions needed to define your theory.

Do you imagine me to be a boy to run your errands? Whose confusion we are trying to resolve? Exclude φ from the dynamics, please, and add external filed contributions. Then the dynamic variables are particle variables and the vector potential ones in the same sense as given in all textbooks. This is the standard CED. As soon as you get it ready yourself and understand what is what in it, we will be able to compare it with my theory.

 

[JustinLevy]

Do you imagine me to be a boy to run your errands? Whose confusion we are trying to resolve?

YOU decided to promote your theory here. And this is YOUR theory.
As explained to you multiple times now by forum moderators, you are expected to answer questions about your theory.

Yes there is confusion, because your theory is currently not mathematically explained well enough. When I or others state there is something wrong, you instead just claim we don't know the theory. When we try to understand better, you instead just give talking points. YOU NEED TO GIVE SPECIFIC MATH HERE, so that there can be no room for confusion.

You have promised me multiple times now that you would give me this SIMPLE response.

Not really. φ should not be involved in the filed Lagrangian, it's a mistake.

Then you are wrong.
Vary the coordinates and you will get the Maxwell's and Lorentz force law. Do you deny that? If so, vary them and prove it to me mathematically.

Yes, φ can be removed since there is no dependence on its time derivative in the Lagrangian. So you can define φ in terms of its 'equation of motion' in the Coulomb gauge and then plug that into the Lagrangian to remove dependence on φ explicitly. This is important when changing to the Hamiltonian, but we are not discussing this yet. The Lagrangian I gave IS in the Coulomb gauge, and it DOES give classical electrodynamics.

Since your Lagrangian looks like it involves placing a different coordinate in the evaluation of the scalar potential, I thought it better to leave it explicitly in for comparison. Either way, leaving it in does not violate the gauge condition nor the CED equations of motion.

please, and add external filed contributions. Then the dynamic variables are particle variables and the vector potential ones in the same sense as given in all textbooks.

To add external field contributions just replace $\phi$ and $A$ with $\phi + \phi_{ext}$ and $A + A_{ext}$ respectively in the Lagrangian. The generalized coordinates remain the same.1] If you still disagree that what I wrote in the previous post gives CED, please prove it mathematically.

2] Please finally respond to my request, stating here precisely and mathematically, the lagrangian and coordinates and any supporting mathematical definitions needed to define your theory.

 

[Bob_for_short]

OK, we are nearly here. I have got to go to work right now and you, please, just think of use of φ if it is not involved in radiation. It determines the instant Coulomb interaction and can be written explicitly. So the searched variables are in fact the particle ones and the radiated filed A. The rest is known. If you agree, I will write the Lagrangian without φ as a variable.

 

[JustinLevy]

Yes, I already stated that phi can be removed.
If you want to write your Lagrangian with it removed, fine. But please state explicitly what you define phi to be in your theory in order to remove it (so that one can fully understand what you mean by phi_ext as well as allowing calculations of the electric fields if one so chooses).

 

[Bob_for_short]

The Coulomb gauge is also called a radiation gauge. To a certain extent it is a gauge-invariant formulation (in terms of Dirac variables). The particle and quanta (radiated filed in CED) are the physical degrees of freedom that exchange with the energy-momentum.

Here is the Lagrangian for classical electrodynamics (CED):

$$\mathcal{L} = \sum_\alpha \frac{1}{2}m (\dot{\vec{r}}_\alpha)^2 - \sum_{\alpha>\beta} q_\alpha \cdot q_\beta | \vec{r}_\alpha - \vec{r}_\beta|^{-1} - \sum_\alpha q_\alpha \phi_{ext}( \vec{r}_\alpha) + \int d^3r \vec{j}(\vec{r})\cdot\vec{A_{ext} }(\vec{r})$$

$$+ \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\vec{A}(\vec{k})+\vec{A}^*(\vec{k})\cdot\vec{j}(\vec{k})] + \frac{\epsilon_0}{2}\int d^3k [ \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) -c^2k^2 \vec{A}^*(\vec{k})\cdot\vec{A}(\vec{k})]$$

where

$$\vec{j}(\vec{r}) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$

The generalized coordinates are:

$$\vec{A}(\vec{k}), \vec{A}^*(\vec{k}), \vec{r}_\alpha$$

The generalized velocities are their time derivatives. This is a classical Lagrangian including not only radiation but also quasi-static fields involved in the charge interaction. In particular, the self-induction is contained in the current-field term. If there is only one charge, there is no self-induction.

Two remarks:

1) The quasi-static magnetic interaction can also be separated from the total current-filed term explicitly. It will remind the quasi-static Coulomb interaction energy but will involve the charge velocities. In other words, we can separate the part of A which is the static equation solution of ∆A ∝ j.

2) The current-field term contains a self-action via the radiated field. The radiated field is time-dependent and propagating with c unlike a quasi-static magnetic filed due to non-zero charge velocity (I speak of one-charge case here).

I propose you to consider now the case when an external filed makes a charge oscillate and radiate in CED.

 

[JustinLevy]

All you did was just write down CED. That is not what I asked for, nor have been asking for the last few PAGES.

I propose you to consider now the case when an external filed makes a charge oscillate and radiate in CED.

Stop delaying. Stop demanding I do yet _more_ things before you answer a my question. I have been asking it for a long time now, and you promised multiple times and even to a moderator that you would answer.

Please finally respond to my request, stating here precisely and mathematically, the lagrangian and coordinates and any supporting mathematical definitions needed to define your theory.

 

[Bob_for_short]

All you did was just write down CED. That is not what I asked for, nor have been asking for the last few PAGES.

CED in the Coulomb gauge is important for comparison with my theory. As soon as you did not give objections to it, I consider it as understood by you.

Stop delaying. Stop demanding I do yet _more_ things before you answer my question. I have been asking it for a long time now, and you promised multiple times and even to a moderator that you would answer.

I am not stalling. I want us to speak the same language with the same notions. In particular, you claimed that the usual CED radiation was different from mine. The CED Lagrangian will serve you to prove it. It should be a Lagrangian accepted by you and me.

Now comes my Lagrangian:

$$\mathcal{L}_{NCED} = \sum_\alpha [\frac{1}{2}m (\dot{\vec{R}}_\alpha)^2 + \frac{\epsilon_0}{2}\int d^3k [ \dot{\vec{A}}_\alpha^*(\vec{k})\cdot\dot{\vec{A}}_\alpha(\vec{k}) -c^2k^2 \vec{A}_\alpha ^*(\vec{k})\cdot\vec{A}_\alpha (\vec{k})] ]$$ ...(1)

$$- \sum_{\alpha>\beta} q_\alpha \cdot q_\beta | \vec{r}_\alpha - \vec{r}_\beta|^{-1} + \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\cdot\vec{A'}(\vec{k})+\vec{A'}^*(\vec{k})\cdot\cdot\vec{j}(\vec{k})]$$ ...(2)

$$- \sum_\alpha q_\alpha [\phi_{ext}( \vec{r}_\alpha) - \dot{\vec{r}}_\alpha \cdot \vec{A_{ext} }(\vec{r}_\alpha)]$$ ...(3)

where

$$\vec{j}(\vec{r}) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha \delta^3(\vec{r} -\vec{r}_\alpha)$$

The generalized coordinates are:

$$\vec{A}_\alpha (\vec{k}), \vec{A}_\alpha ^*(\vec{k}), \vec{R}_\alpha$$ ...(4)

The electron coordinate is expressed via center of mass and "internal" variables of each "electronium":

$$\vec{r}_\alpha = \vec{R}_\alpha + \sum_k \epsilon_{\alpha}(k) (\dot{\vec{A}}_\alpha (k) + \dot{\vec{A}}_\alpha ^*(k))$$...(5)

The "internal", EM oscillator variables are time-dependent ones, not static ones in this expression.

Double product "⋅⋅" and the prime in the term j⋅⋅A' mean absence of the "proper" fields for each charge involved. In other words, for a given charge in j the filed A' is the field of all other charges (quasi-static and radiated). I did not represent this term as a doubled sum on charges and fields to be short. The whole second line disappears in case of one charge.

 

[JustinLevy]

In particular, you claimed that the usual CED radiation was different from mine.

Because it is.

First off, there are serious problems with your Lagrangian, since it depends on the second time derivative of your generalized coordinates.

Furthermore, simplifying for the case of one particle under the influence of an external field, we have:
$$\mathcal{L}_{NCED} = \frac{1}{2}m (\dot{\vec{R}})^2 - q [\phi_{ext}( \vec{r}) - \dot{\vec{r}} \cdot \vec{A_{ext} }(\vec{r})] + \frac{\epsilon_0}{2}\int d^3k [ \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) -c^2k^2 \vec{A} ^*(\vec{k})\cdot\vec{A} (\vec{k})]$$
note that
$$\dot{\vec{r}} = \dot{\vec{R}} + \sum_k \epsilon(k) (\ddot{\vec{A}}(k) + \ddot{\vec{A}}^*(k))$$


Because you have no j.A term, the field does not couple to the motion of q like is necessary for Maxwell's equations. Instead, the field is sourced by the $$q \epsilon(k)\ddot{\vec{A}}(k) \vec{A}_{ext}$$ term even if the particle does not move. This is wildly in disagreement with experiment.

----
The 'center of mass' equation seems to have a typo:
$$\vec{r}_\alpha = \vec{R}_\alpha + \sum_k \epsilon_{\alpha}(k) (\dot{\vec{A}}_\alpha (k) + \dot{\vec{A}}_\alpha ^*(k))$$...(5)
as written, that doesn't make sense. I see there is some coupling constant epsilon that depends on the magnitude of k, but not the direction. But what about A? And how do you sum over k? And I don't see how the coupling constant can depend on which electron we are talking about. Did you instead mean:
$$\vec{r}_\alpha = \vec{R}_\alpha + \int d^3k \ \epsilon(k) [\dot{\vec{A}}_\alpha (\vec{k}) + \dot{\vec{A}}_\alpha ^*(\vec{k})]$$
Regardless, if you could clarify here, that would be helpful.

 

[Bob_for_short]

...there are serious problems with your Lagrangian, since it depends on the second time derivative of your generalized coordinates.

note that
$$\dot{\vec{r}} = \dot{\vec{R}} + \sum_k \epsilon(k) (\ddot{\vec{A}}(k) + \ddot{\vec{A}}^*(k))$$

Yes, you are right. Instead of the electron coordinate we have to use its expression (5) in L_NCED. I wrote it as a separate line to be short. As soon as we find R(t) and A(t), we have r_e(t).

Because you have no j.A term, the field does not couple to the motion of q like is necessary for Maxwell's equations. Instead, the field is sourced by the $$q \epsilon(k)\ddot{\vec{A}}(k) \vec{A}_{ext}$$ term even if the particle does not move. This is wildly in disagreement with experiment.

You are right but if the charge does not move, the oscillator amplitude is equal to zero.
The oscillator equations have the external force as a pumping source in my theory. It corresponds well to the standard CED when the charge acceleration (proportional to the external force) is the oscillator pumping term. My "derivation" was wrong since it followed CED too much. My oscillator Lagrangian should contain the first and the second derivatives of A as coordinates and velocities (i.e., the electric fields E(k) and their derivatives). It should vanish for static fields A. I will rewrite this part of Lagrangian to be compliant with my Hamiltonian. My fault. It should not be difficult since

$$\vec{A}}(k) = -\ddot{\vec{A}}(k)/\omega^2$$

By the way, the external potential depends on both generalized coordinates too.

The 'center of mass' equation seems to have a typo:
$$\vec{r}_\alpha = \vec{R}_\alpha + \sum_k \epsilon_{\alpha}(k) (\dot{\vec{A}}_\alpha (k) + \dot{\vec{A}}_\alpha ^*(k))$$...(5)
as written, that doesn't make sense. I see there is some coupling constant epsilon that depends on the magnitude of k, but not the direction. But what about A? And how do you sum over k? And I don't see how the coupling constant can depend on which electron we are talking about. Did you instead mean:
$$\vec{r}_\alpha = \vec{R}_\alpha + \int d^3k \ \epsilon(k) [\dot{\vec{A}}_\alpha (\vec{k}) + \dot{\vec{A}}_\alpha ^*(\vec{k})]$$
Regardless, if you could clarify here, that would be helpful.

I admitted a sloppiness here: instead of sum over k I should have written an integral over d³k. Vectors A(k) are the same as in the previous lines of the Lagrangian. The charge q should have been written explicitly in front of epsilon:

$$\vec{r}_\alpha = \vec{R}_\alpha + q_\alpha \int d^3k \ \epsilon(k) [\dot{\vec{A}}_\alpha (\vec{k}) + \dot{\vec{A}}_\alpha ^*(\vec{k})]$$

 

[JustinLevy]

I will rewrite this part of Lagrangian to be compliant with my Hamiltonian. My fault.

I look forward to seeing the corrected version then.

 

[JustinLevy]

It's nearly midnight at Grenoble. I will do it tomorrow if you don't mind.

That is fine.

You can do it yourself: I gave the necessary elements for that.

That term I pointed out appears to source a field regardless of whether the particle moves. What you seem to be suggesting will not fix that. Therefore I will wait to see what your theory is explicitly in math before commenting further. That is the point of this, to have your theory explicitly stated mathematically so that there is no room for confusion.

 

[Bob_for_short]

I rewrote the one-particle Lagrangian in the following way:

$$\mathcal{L}_{NCED} = \frac{1}{2}m (\dot{\vec{R}})^2 - q [\phi_{ext}( \vec{r}) - \dot{\vec{r}} \cdot \vec{A_{ext} }(\vec{r})] + \frac{\epsilon_0}{2}\int d^3k [ \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) - \ddot{\vec{A}}(\vec{k})^*\cdot\ddot{\vec{A}}(\vec{k}) /\omega^2]$$

with

$$\vec{r} = \vec{R} + q \int d^3k \ \epsilon(k) [\dot{\vec{A}} (\vec{k}) + \dot{\vec{A}} ^*(\vec{k})]$$

The generalized field coordinates and velocities are $$\dot{\vec{A}}(\vec{k}), \ddot{\vec{A}}(\vec{k})$$ and their complex conjugates.

Let us try this. I think we have to derive the corresponding equations before making conclusions. The coordinates and velocities in the Lagrangian are unknown variables rather than solutions.

If the particle does not move, its velocity is equal to zero and the coordinate is a constant. It should make the source in the filed equations vanish but not before. Factually it is possible (to have a particle at rest) only in case of absence of any external filed and internal oscillations.

 

[JustinLevy]

You have inserted an $\omega$ in your equations. I will assume this is the usual $\omega = k c$.

Let's start by talking about the field equations of motion in a simplified case.
When
$$\phi_{ext} = 0$$
the lagrangian density (in phase space) determining the evolution of the field coordinates in your theory is:
$$\mathcal{L} = q^2 \epsilon(k) [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{A_{ext} }(\vec{r}) + \frac{\epsilon_0}{2} [ \dot{\vec{A}}^*(\vec{k})\cdot\dot{\vec{A}}(\vec{k}) - \frac{1}{k^2 c^2} \ddot{\vec{A}}^*(\vec{k})\cdot\ddot{\vec{A}}(\vec{k})]$$

So now finding the equation of motion for A(k):
$$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \ddot{\vec{A}}(k)} = \frac{d}{dt} [q^2 \epsilon(k) \vec{A_{ext} }(\vec{r}) - \frac{\epsilon_0}{2 k^2 c^2} \ddot{\vec{A}}^*(\vec{k})] = q^2 \epsilon(k) \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} + q^2 \epsilon(k) \vec{g}(\vec{A_{ext} }(\vec{r})) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k})$$
where
$$\vec{g}(\vec{A_{ext} }(\vec{r}))=(\dot{\vec{r}}\cdot \vec{\nabla}_{\vec{r}}) \vec{A_{ext} }(\vec{r})$$
and
$$\frac{\partial \mathcal{L}}{\partial \dot{\vec{A}}(k)} = \frac{\epsilon_0}{2} \dot{\vec{A}}^*(\vec{k}) + q^2 \epsilon(k) [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r}))$$
where
$$\vec{f}(\vec{A_{ext} }(\vec{r}))=(\frac{\partial \vec{r}}{\partial \dot{\vec{A}}(k)} \cdot \vec{\nabla}_{\vec{r}}) \vec{A_{ext} }(\vec{r})$$Thus the equation of motion for the fields in your theory is:
$$q^2 \epsilon(k) \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} + q^2 \epsilon(k) \vec{g}(\vec{A_{ext} }(\vec{r})) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k}) = \frac{\epsilon_0}{2} \dot{\vec{A}}^*(\vec{k}) + q^2 \epsilon(k) [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r}))$$
simplifying some
$$\frac{1}{c^2} \dddot{\vec{A}}^*(\vec{k}) + k^2 \dot{\vec{A}}^*(\vec{k}) = \frac{2 q^2}{\epsilon_0 k^2} \epsilon(k) \left[ \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} + \vec{g}(\vec{A_{ext} }(\vec{r})) - [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]$$Compare this to the CED result of:
$$\frac{1}{c^2} \ddot{\vec{A}}^*(\vec{k}) + k^2 \vec{A}^*(\vec{k}) = \mu_0 \vec{j}^*(\vec{k})$$Both look like a driven oscillator. However, in CED, the external field does not directly source any radiation.

In your theory, despite your repeated claims, the radiation looks nothing like CED. For example, the external field itself can directly source radiation, where as the current does NOT. Also notice the coupling is q^2 instead of q. Since one driving term is proportional to A(k) itself, it is likely you even have run away solutions even for a stationary particle!

Your resultant "modified" maxwell's equations are not little adjustments like you claim. They look nothing like the original, and wildly disagree with experiment.

So your choice is to either change your Lagrangian yet again, or admit that your theory is falsified by experiment.