Reformulation instead of Renormalizations

[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 -<br /> \epsilon_0 k^2 \phi(\vec{k})^2 + \frac{1}{\mu_0} \dot{\phi}(\vec{k}^2 <br /> ] \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<br /> - \frac{1}{2}\int d^3k [\phi^*(\vec{k}) \rho(\vec{k})+\rho^*(\vec{k})\phi(\vec{k})] <br /> + \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\vec{A}(\vec{k})+\vec{A}^*(\vec{k})\cdot\vec{j}(\vec{k})]
\ \ \ <br /> + \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<br /> - \frac{1}{2}\int d^3k [\phi^*(\vec{k}) \rho(\vec{k})+\rho^*(\vec{k})\phi(\vec{k})] <br /> + \frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\vec{A}(\vec{k})+\vec{A}^*(\vec{k})\cdot\vec{j}(\vec{k})]
\ \ \ <br /> + \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<br /> - \sum_{\alpha&gt;\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})] ]<br /> ...(1)

- \sum_{\alpha&gt;\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&#039;}(\vec{k})+\vec{A&#039;}^*(\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}} + <br /> 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[<br /> \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} +<br /> \vec{g}(\vec{A_{ext} }(\vec{r})) - <br /> [\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.

 

[Bob_for_short]

\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[<br /> \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} +<br /> \vec{g}(\vec{A_{ext} }(\vec{r})) - <br /> [\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.

An external filed (a tension B_ext = rotA_ext = curlA_ext) can cause the electron radiation via variable current j(t). As you know, any acceleration makes the electron radiate. The source term is proportional to the particle charge q. No charge, no 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!

I have not verified your derivation yet. I will see it closer.

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

Yes, it is another concept. It does not contain the electron self-action part but "interaction" with the filed oscillators my means of being a part of them.

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

You know well, I never wrote the CED Lagrangian, so I admit it is still imperfect. But now you see that we obtain quite similar wave equations.

The CED equation is valid also for a constant current (a wire with a current). The latter creates a constant magnetic filed (the vector-potential harmonics do not depend on time then).

I tried to separate this case since the created constant magnetic filed can be written explicitly, no need to mix it with the radiation (propagating modes). If you take a time derivative of the CED equation, you will see that the time-dependent harmonics depend on the charge acceleration (which is determined with an external force thus the latter sources the radiation).

EDIT:

1) I see you left r as the filed argument. It should be replaced with its expression via R and the oscillator coordinates in order to derive the equations correctly. I have to verify your derivation.

2) The current j is proportional to the particle charge q and its velocity. The particle acceleration in an external magnetic (or electric) field is also proportional to the particle charge. As a result, the pumping term is proportional to the external field tension and the charge squared in both theories.

3) The external magnetic filed (curlA_ext) in my approach may appear from your \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}) (unclear expression, too many vectors).

4) My wave equation will anyway be slightly different from the CED one even if you put in the latter the current derivative expressed only via the constant external magnetic filed B_ext = curlA_ext (the exact formula for the current derivative in CED is expressed via the total EMF including the proper, radiated, unknown filed). It is so because the external filed in my approach depends on the filed variables too, but in a different from the exact CED way.

5) The term \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} does not contain the particle velocity. It should be just the external electric filed \dot{\vec{A}}_{ext}(\vec{r}), as in the usual CED.

 

[JustinLevy]

But now you see that we obtain quite similar wave equations.

THEY ARE NOT SIMILAR AT ALL.
The external field ITSELF can source more radiation in your theory. (even if the particle does not move)

1) I see you left r as the filed argument. It should be replaced with its expression via R and the oscillator coordinates in order to derive the equations correctly. I have to verify your derivation.

I took account that r is actually a function of R and A.
Please take time to do the derivation yourself.

2) The current j is proportional to the particle charge q and its velocity. The particle acceleration in an external magnetic (or electric) field is also proportional to the particle charge. As a result, the pumping term is proportional to the external field tension and the charge squared in both theories.

Reread what you wrote there. You aren't even making sense.

You are trying to claim that in general the velocity of a particle is proportional to its charge. That is nonsense.

You are really really stretching to try to explain away the q^2.
Your theory disagrees with experiment.

3) The external magnetic filed (curlA_ext) in my approach may appear from your \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}) (unclear expression, too many vectors).

I even put a subscript on the Del operator to make it clear what it was differentiating with respect to. Please do the derivation yourself, and if you have a different preferred notation, please let me know and we can use that.

5) The term \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} does not contain the particle velocity. It should be just the external electric filed \dot{\vec{A}}_{ext}(\vec{r}), as in the usual CED.

That term doesn't exist at all in CED. That term only exists in your theory because r depends on A(k). And yes, the particle velocity is in that term.

I have not verified your derivation yet. I will see it closer.

Please work it out yourself.
If you disagree, then please show your math here.

I will wait for you to do the derivation so we can be in agreement here.

 

[Bob_for_short]

THEY ARE NOT SIMILAR AT ALL.
The external field ITSELF can source more radiation in your theory. (even if the particle does not move)

If the particle does not move, the total force is equal to zero. The right-hand side of my wave equation is proportional to the total external force. No force, no radiation.

You are trying to claim that in general the velocity of a particle is proportional to its charge. That is nonsense.

I wrote "acceleration", not velocity. Acceleration in an external electric and magnetic filed is proportional to the particle charge. This follows from the Newton equations (Lorentz force).

You are really really stretching to try to explain away the q^2.

Take a time derivative of your CED equation, express the particle current derivative via the external forces explicitly and you will obtain q². Or show me the contrary in the standard CED, please.

...I even put a subscript on the Del operator to make it clear what it was differentiating with respect to.

Your expression ∂r/∂A is a vector, scalar or a tensor?

That term doesn't exist at all in CED. That term only exists in your theory because r depends on A(k). And yes, the particle velocity is in that term.

Then the dimension of this term is wrong.

 

[Bob_for_short]

\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}} + <br /> q^2 \epsilon(k) \vec{g}(\vec{A_{ext} }(\vec{r})) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k})

In fact, it should be simply:

\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}) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k})

where \dot{\vec{A}}_{ext}(\vec{r}) is the full time derivative. It can be expressed via the partial time derivative ∂A_ext/∂t and the g-term but no velocity appears at ∂A_ext/∂t. The partial time derivative is, as I said, the external electric field.

 

[JustinLevy]

In fact, it should be simply:

\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}) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k})

where \dot{\vec{A}}_{ext}(\vec{r}) is the full time derivative. It can be expressed via the partial time derivative ∂A_ext/∂t and the g-term but no velocity appears at ∂A_ext/∂t. The partial time derivative is, as I said, the external electric field.

I looked through the derivation, and yes you are correct about that term.

The field evolution equations for your theory is therefore:
\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[<br /> \frac{\partial \vec{A}_{ext}}{\partial t}(\vec{r}) +<br /> \vec{g}(\vec{A_{ext} }(\vec{r})) - <br /> [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]
where
\vec{g}(\vec{A_{ext} }(\vec{r}))=(\dot{\vec{r}}\cdot \vec{\nabla}_{\vec{r}}) \vec{A_{ext} }(\vec{r})
\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})

Where as for CED we have:
\frac{1}{c^2} \ddot{\vec{A}}^*(\vec{k}) + k^2 \vec{A}^*(\vec{k}) = \mu_0 \vec{j}^*(\vec{k})Notice that CED is LOCAL in reciprocal space. In your theory, mode A(k) depends on ALL the other modes A(k'), and can even be in source terms for the other modes. Doesn't this bother you?

---

Since there also seems to be confusion on notation, here's an explicit example.
Consider a function f of x,y,z,t, where x,y,z can be functions of time as well:
\frac{d}{dt} f(x,y,z,t) = \frac{\partial f}{\partial t}(x,y,z,t) + [\frac{\partial f}{\partial x}(x,y,z,t)] \frac{dx}{dt} + [\frac{\partial f}{\partial y}(x,y,z,t)] \frac{dy}{dt} + [\frac{\partial f}{\partial z}(x,y,z,t)] \frac{dz}{dt}
or more simply
\frac{d}{dt} f(\vec{r},t) = \frac{\partial f}{\partial t}(\vec{r},t) + [\frac{d\vec{r}}{dt} \cdot \vec{\nabla}_{\vec{r}}] f(x,y,z,t)

If the particle does not move, the total force is equal to zero. The right-hand side of my wave equation is proportional to the total external force. No force, no radiation.

Don't start resorting to talking points again. The evolution equation is what it is. Period. You gave the Lagrangian and specified the coordinates, there is no room for your talking points here. Let's stick to the math.

Since the f(A) term can be non-zero even if the particle is not moving or accelerating, then clearly you cannot claim your talking point "no force, no radiation". Are you claiming despite this, that by merely requiring acceleration=0, you can prove that the right side actually is zero? I'd like to see that, for that appears impossible for the reasons stated.

You are trying to claim that in general the velocity of a particle is proportional to its charge. That is nonsense.

I wrote "acceleration", not velocity. Acceleration in an external electric and magnetic filed is proportional to the particle charge. This follows from the Newton equations (Lorentz force).

You stated, and I quote "The current j is proportional to the particle charge q and its velocity." You then argued that written in terms of the external field, this becomes proportional to q^2. That is equivalent to claiming that in general the velocity is proportional to q.

All you need to do is consider the case where the external field is zero for a bit, but the current is not. THERE IS NO MAGNETIC FIELD produced by the current according to your theory!

Take a time derivative of your CED equation, express the particle current derivative via the external forces explicitly and you will obtain q². Or show me the contrary in the standard CED, please.

Sure, I'll give you an easy counter example. Consider a particle with no external field moving at a constant velocity. Despite this, the time derivative of j(r) is non-zero and still only proportional to q, not q^2. Also, as stated earlier, this produces a magnetic field in CED, but not in your theory.

Your theory doesn't match experiment. It is wrong.
The math makes this clear, so either change your Lagrangian yet again, or accept that your theory is falsified by experiment.EDIT: To the various people that have sent me PMs. Thank you for the encouragement, or for pointing out additional issues with this Lagrangian. But I already spend any 'forum time' I have with this discussion. If you have comments for Bob, please just post them here. Don't be afraid to join the discussion. Let him know your issues, and please don't expect me to go through everything and/or play proxy. Heck, if anyone actually believe's Bob's theory isn't refuted by experiment, I'd love to see that here too ... it would be nice to hear from the other side if it exists.

 

[Bob_for_short]

I looked through the derivation, and yes you are correct about that term.

Nice to hear that.

The field evolution equations for your theory is therefore:
\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[<br /> \frac{\partial \vec{A}_{ext}}{\partial t}(\vec{r}) +<br /> \vec{g}(\vec{A_{ext} }(\vec{r})) - <br /> [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]
where
\vec{g}(\vec{A_{ext} }(\vec{r}))=(\dot{\vec{r}}\cdot \vec{\nabla}_{\vec{r}}) \vec{A_{ext} }(\vec{r})
\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})

You still use a vague notation. Why not to express the space derivatives via curl(A_ext)? I have already asked you: "Your expression ∂r/∂A is a vector, scalar or a tensor?" Work it out better to avoid confusion.

Where as for CED we have:
\frac{1}{c^2} \ddot{\vec{A}}^*(\vec{k}) + k^2 \vec{A}^*(\vec{k}) = \mu_0 \vec{j}^*(\vec{k})

In order to cast this equation into a form directly comparable with my wave equation, I differentiate it once more, if you don't mind:

\frac{1}{c^2} \dddot{\vec{A}}^*(\vec{k}) + k^2 \dot{\vec{A}}^*(\vec{k}) = \mu_0 \dot{\vec{j}}^*(\vec{k})

Notice that CED is LOCAL in reciprocal space

Who said that? Have you expressed the current in CED via the filed variables in order to judge? You hurry and you make wrong statements. You will see it later.

In your theory, mode A(k) depends on ALL the other modes A(k'), and can even be in source terms for the other modes. Doesn't this bother you?

In your CED it is also the case. In CED it is called a self-action which is non-physical. My dependence is physical just because of one of "talking points": no external force, no radiation. My theory is constructed so on purpose. What you call "talking point" here is the right physics implemented in the Lagrangian. You should have noticed that the external filed acts on the charged particle. Without this action no coupling between the radiated modes and the center of inertia exists. Indeed, look at my Lagrangian:

\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]

If q = 0 or there is no external filed, the equations are decoupled, in particular, the filed subsystem is not sourced. That means there is no radiated field.

Don't start resorting to talking points again. The evolution equation is what it is. Period. You gave the Lagrangian and specified the coordinates, there is no room for your talking points here. Let's stick to the math.

So stick to the math in CED and make sure that there is a self-action in it. Don't be superficial or blind.

Since the f(A) term can be non-zero even if the particle is not moving or accelerating, then clearly you cannot claim your talking point "no force, no radiation". Are you claiming despite this, that by merely requiring acceleration=0, you can prove that the right side actually is zero? I'd like to see that, for that appears impossible for the reasons stated.

It follows from my Lagrangian. Work out the space derivatives to see the curlA_ext clearly. The right-hand side is proportional to the external force in my theory: the Lorentz force q[E_ext + vxB_ext/c]. You just have not derived it properly yet.

You stated, and I quote "The current j is proportional to the particle charge q and its velocity." You then argued that written in terms of the external field, this becomes proportional to q^2. That is equivalent to claiming that in general the velocity is proportional to q.

Yes, j ∝ qv. Its time derivative \dot{\vec{j}}^*(\vec{k}) needed for the CED equation (given above) is proportional to q² in virtue of the Lorentz force. So in CED there is q² too in the radiated field equations. Your "general" conclusion about the particle velocity is wrong.

All you need to do is consider the case where the external field is zero for a bit, but the current is not. THERE IS NO MAGNETIC FIELD produced by the current according to your theory!

Wrong. There is no the proper magnetic field involved in one-particle dynamics. The electric and magnetic fields created by a static or uniformly moving charge are present explicitly in the charge interaction term of my complete Lagrangian (post #64, formula (2)). I do not need to solve any field equation to obtain them. They are already here. If there is no other charges, the proper field drops out of the charge dynamics in my approach. This is a major result: the free charge motion is physical, unlike the CED result (runaway motion). In CED the charge electric and magnetic fields get into the charge equations, you did not know this? Did you forget about the charge equations? Do you imply the current as a know or unknown variable?

Sure, I'll give you an easy counter example. Consider a particle with no external field moving at a constant velocity. Despite this, the time derivative of j(r) is non-zero and still only proportional to q, not q^2. Also, as stated earlier, this produces a magnetic field in CED, but not in your theory.

But first, it is not a radiated field, Justin! I have this term in case of at least two interacting charges, no problem. And you? What are you proud of? What to do with your magnetic and electric fields in one-charge CED? To put them in the particle dynamics equation and get a self-action, as H. Lorentz and the others did. You get first infinity and next, after discarding it, a runaway solution for the particle and a huge source for the radiation! This is exactly what I avoid with my reformulation. This is an achievement , not a drawback. My article starts from the phrase: "The interaction term that causes the mathematical and conceptual problems is the so called self-action term."

Your theory doesn't match experiment. It is wrong.

I am fed up with your groundless statements, frankly. No, it is CED which is wrong if taken seriously. There are no runaway charges in experiment nor their radiation. I tried to advance a better physical theory.

The math makes this clear ...

You are good in math, Justin. Make sure there is a self-action in CED before blaming me for nothing.

EDIT: To the various people that have sent me PMs. Thank you for the encouragement, or for pointing out additional issues with this Lagrangian. But I already spend any 'forum time' I have with this discussion.

I did not know that there is a whole band of your supporters. Now I understand why the thread has so many visits.

Well, Justin left to celebrate his victory over the old Bob. Anybody else to discuss the issue?

 

[JustinLevy]

You continue to say things that are blatantly wrong. When I prove this with examples, you just repeat your talking points. Please, please, consider the possibility you are wrong. I can make mistakes (for example that one term you pointed out, or even look how many times I need to edit a post to remove typos). You can make mistakes as well. If you are unable or unwilling to consider the possibility that you are indeed wrong about your theory, then there is no point in pursuing a scientific discussion with you. So please, consider the possibility you are wrong.

Since so many things were said, let me focus on the easiest to show mathematically.

Yes, j ∝ qv. Its time derivative \dot{\vec{j}}^*(\vec{k}) needed for the CED equation (given above) is proportional to q² in virtue of the Lorentz force. So in CED there is q² too in the radiated field equations. Your "general" conclusion about the particle velocity is wrong.

I was not making a "general" conclusion. I was arguing against your general conclusion that dj/dt is always proportional to q^2. I proved this was wrong with a counter example.

Let me repeat it again in explicit math.

Yes, j ∝ qv. And yes, the Lorentz force law tells us (in CED and non-relativistically), that a ∝ q. This does NOT mean you can conclude dj/dt ∝ q^2.

Proportional signs are not equal signs. You can't just take the derivative on both sides.
More explicitly, here is what j is equal to (as you agreed yourself):
\vec{j}(\vec{r}) = \sum_\alpha q_\alpha \dot{\vec{r}}_\alpha \delta^3(\vec{r}_\alpha - \vec{r})
Even after considering that the acceleration is proportional to q, it should be clear to you that the time derivative of j is NOT proportional to q^2. It contains two terms, one with q and one with q^2.

You theory only allows a coupling of q^2.
This does not agree with experiment.

It follows from my Lagrangian. Work out the space derivatives to see the curlA_ext clearly. The right-hand side is proportional to the external force in my theory: the Lorentz force q[E_ext + vxB_ext/c]. You just have not derived it properly yet.

You continue to claim this, yet you refuse to show your own work. Have you even worked it out yourself? If so, then show it here. If not, then you have no foundation for claiming I am incorrect.Now, let's look at that force you keep talking about.
The lagrangian for R:
\mathcal{L} = \frac{1}{2}m (\dot{\vec{R}})^2 - q \phi_{ext}( \vec{r}) +q \dot{\vec{r}} \cdot \vec{A}_{ext}(\vec{r})

To keep the notation as obvious as possible, I will start with just looking at a single component of R:
\frac{\partial \mathcal{L}}{\partial R_x} = -q [\frac{\partial}{\partial r_x} \ \phi_{ext}( \vec{r})] \frac{\partial r_x}{\partial R_x} +q\dot{\vec{r}} \cdot [\frac{\partial}{\partial r_x}\vec{A_{ext} }(\vec{r}) ](\frac{\partial r_x}{\partial R_x})
\frac{\partial \mathcal{L}}{\partial \dot{R}_x} = m \dot{R}_x +q A_{ext,x}(\vec{r}) (\frac{\partial \dot{r}_x}{\partial \dot{R}_x})

Since
\vec{r} = \vec{R} + q \int d^3k \ \epsilon(k) [\dot{\vec{A}} (\vec{k}) + \dot{\vec{A}} ^*(\vec{k})]
then
\frac{\partial r_x}{\partial R_x} = 1,\frac{\partial \dot{r}_x}{\partial \dot{R}_x} = 1That gives:
-q [\frac{\partial}{\partial r_x} \ \phi_{ext}( \vec{r})] +q\dot{\vec{r}} \cdot [\frac{\partial}{\partial r_x}\vec{A_{ext} }(\vec{r}) ] = <br /> \frac{d}{dt}<br /> \left[ m \dot{R}_x +q A_{ext,x}(\vec{r}) \right] =<br /> m \ddot{R}_x +q \frac{\partial}{\partial t} A_{ext,x}(\vec{r}) <br /> +q (\dot{\vec{r}} \cdot \vec{\nabla}) A_{ext,x}(\vec{r}) <br />
Thus:
m \ddot{R}_x = q \left[ -\frac{\partial}{\partial r_x} \ \phi_{ext}( \vec{r}) <br /> - \frac{\partial}{\partial t} A_{ext,x}(\vec{r}) <br /> +\frac{\partial}{\partial r_x}[\dot{\vec{r}} \cdot \vec{A_{ext} }(\vec{r}) ]<br /> - (\dot{\vec{r}} \cdot \vec{\nabla}) A_{ext,x}(\vec{r}) \right]<br />
The other components of R follow similarly, so combining to make the cross products to come easier to see:
m \ddot{\vec{R}} = q \left[ -\vec{\nabla} \phi_{ext} (\vec{r}) <br /> - \frac{\partial}{\partial t} \vec{A}_{ext}(\vec{r}) <br /> +\vec{\nabla} (\dot{\vec{r}} \cdot \vec{A}_{ext}(\vec{r}) ) <br /> - (\dot{\vec{r}} \cdot \vec{\nabla}}) \vec{A}_{ext}(\vec{r})<br /> \right]
Using the vector calc identity,
\vec{\nabla}(\vec{A} \cdot \vec{B}) = (\vec{A} \cdot \vec{\nabla})\vec{B} + (\vec{B} \cdot \vec{\nabla})\vec{A} + \vec{A} \times (\vec{\nabla} \times \vec{B}) + \vec{B} \times (\vec{\nabla} \times \vec{A})
we can rewrite the particle evolution as
m \ddot{\vec{R}} = q \left[ -\vec{\nabla} \phi_{ext} (\vec{r}) <br /> - \frac{\partial}{\partial t} \vec{A}_{ext}(\vec{r}) <br /> + \dot{\vec{r}} \times (\vec{\nabla} \times \vec{A}_{ext}(\vec{r}))<br /> \right]
Using the definitions of the potentials we then have:
m \ddot{\vec{R}} = q \left[\vec{E}_{ext}(\vec{r}) + \dot{\vec{r}} \times \vec{B}_{ext}(\vec{r}) \right]

A few quick things to note:
1) We didn't get the ugly f(A) term that you claim is part of the force. So your argument that the right hand side of the field evolution is proportional to the force appears to be incorrect.
2) For calculating the R evolution, an external field is handled equivalently to the fields due to other charges. So if E and B are considered to be the fields due exclusively to all other charges, we can drop the external subscript
m \ddot{\vec{R}} = q \left[\vec{E}(\vec{r}) + \dot{\vec{r}} \times \vec{B}(\vec{r}) \right]
Calculating the field evolution though is complicated due to the dependence of r on A, but at least the R evolution has an easy form when calculated in the multi-particle case.

3) Because the evolution of R (the 'center of mass' coordinate) depends only on the fields at r, not R, the 'back reaction' due to radiation is easy to calculate in your theory.
m \ddot{\vec{r}} = q \left[\vec{E}(\vec{r}) + \dot{\vec{r}} \times \vec{B}(\vec{r}) \right] + \vec{F}_{back\ reaction}
where
\vec{F}_{back\ reaction} = m q \int d^3k \ \epsilon(k) [\dddot{\vec{A}} (\vec{k}) + \dddot{\vec{A}} ^*(\vec{k})]Since A(k) can be sourced by another particle, this means your theory is non-local in normal space (as well as reciprocal space as pointed out earlier).In your theory, a particle can be affected instantaneously by a particle light years away. This is due to the mixing of the particle coordinates with the field coordinates. It is unavoidable.

So your theory has some very pathological problems. It wildly disagrees with experiment.

I did not know that there is a whole band of your supporters. Now I understand why the thread has so many visits.

They are not "my" supporters. They are merely people interested in reading the discussion of YOUR theory. You should be flattered that people are interested in discussions of your theory. The fact that many agree with me is more a reflection on your theory apparently disagreeing with experiment, as I have shown mathematically. If that is not correct, then please show us mathematically instead of with talking points. I, and assumedly they as well, would be interested in seeing the details of that math.

 

[Bob_for_short]

So please, consider the possibility you are wrong.

I admit it all the time. That is why I verify my calculations many time from different points of view - physical and mathematical.

Concerning our conversation, I am interested in your comprehension of the matter. For that it is necessary that you yourself make sure that my equations predict something non-physical. It is not sufficient to have a glance and judge, as you did it before. I am glad that you arrived at q² already. I will answer the other questions later.

 

[JustinLevy]

It is not sufficient to have a glance and judge, as you did it before.

Glance? I have put much work and written much math here. You on the other hand continue to dismiss my comments without providing math of your own.

I am glad that you arrived at q² already.

What are you talking about!? I am arguing against your claim that your q^2 only coupling is correct. I have, and continue to, claim that there should be a q coupling as well.

The coupling in your theory is incorrect.

I will answer the other questions later.

There are no questions in my previous post. There are just statements about predictions of your theory in order to further discussion of its viability.

If you disagree with a statement that I have backed up with math, please back up your counter statements with math. If instead you do not disagree with anything I have said, then we have finally come to agreement about the predictions of your theory: it is incorrect when compared to experiment.

 

[Bob_for_short]

You continue to say things that are blatantly wrong.

I say things the you estimate to be wrong, nuance.

I was not making a "general" conclusion.

Yes, you stated that according to me the velocity is generally proportional to q. It was your groundless statement.

I was arguing against your general conclusion that dj/dt is always proportional to q^2. I proved this was wrong with a counter example.

I hear that for the first time. I spoke about acceleration.

Let me repeat it again in explicit math.

Yes, j ∝ qv. And yes, the Lorentz force law tells us (in CED and non-relativistically), that a ∝ q. This does NOT mean you can conclude dj/dt ∝ q^2.

Proportional signs are not equal signs. (?) You can't just take the derivative on both sides. (?) More explicitly, here is what j is equal to (as you agreed yourself):

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

Even after considering that the acceleration is proportional to q, it should be clear to you that the time derivative of j is NOT proportional to q^2. It contains two terms, one with q and one with q^2. You theory only allows a coupling of q^2. This does not agree with experiment.

Yes, I agree here with your conclusion about existence of a term proportional to q in the current derivative. this term is also proportional to the particle velocity. That means in case of a small velocity but a strong external force (consider a strong constant B_ext with no E_ext) the q²-term dominates and in any case it determines the radiation. The magnetic filed due to velocity is not radiation and is not present in my theory of on-charge dynamics indeed. It is not a degree of freedom that carry energy-momentum independently of charge. It is a charge "feature". It cannot "fly away".

Next, the q- and q²-terms come in sum. That means the resulting filed is a superposition of a magnetic and radiated fields in CED. The resulting filed solutions serve to affect other charges, don't they? Unfortunately in CED you are obliged to put these fields in the original charge equations - because the CED Lagrangian is made so. This causes problems of infinities and self-acceleration even in case of one charge.

In my theory the magnetic field of a moving charge is explicitly present in a two- or more charge system. It is important in many-particle case because it is involved in the particle interaction. In case of one-charge system my theory deals only with the radiated field. It propagates far away and gets in the equation of motions of other, very distant charges. It does not get into the charge dynamics itself as a self-action.

There is a major difference between the quasi-static and radiated fields. We can have a neutral and compact system of charges whose electric and magnetic fields decay rapidly with distance. The only thing we can observe at far distances is their radiation. That is why it has a separate physical meaning. We may observe some radiation without knowing how and where it is created.

Thus the q² in CED exists well and determines the radiated field. The radiated power in CED is proportional to q⁴ (see the Larmor formula). This property is experimentally verified and preserved in my theory. You blame me for nothing.

Have you even worked it out yourself?

That's the question to you in CED. Apparently you do not recognize a huge problem existence in CED and this is why you criticize my efforts.

Now, let's look at that force you keep talking about. The lagrangian for R:

\mathcal{L} = \frac{1}{2}m (\dot{\vec{R}})^2 - q \phi_{ext}( \vec{r}) +q \dot{\vec{r}} \cdot \vec{A}_{ext}(\vec{r})

Using the definitions of the potentials we then have:

m \ddot{\vec{R}} = q \left[\vec{E}_{ext}(\vec{r}) + \dot{\vec{r}} \times \vec{B}_{ext}(\vec{r}) \right]

That's right! Bravo, Justin!

A few quick things to note:

1) We didn't get the ugly f(A) term that you claim is part of the force. So your argument that the right hand side of the field evolution is proportional to the force appears to be incorrect.

It was your ugly term and in the field, not mechanical equations. Your g- and f-terms should give vxB_ext, just like in the mechanical equation.

2) For calculating the R evolution, an external field is handled equivalently to the fields due to other charges. So if E and B are considered to be the fields due exclusively to all other charges, we can drop the external subscript
m \ddot{\vec{R}} = q \left[\vec{E}(\vec{r}) + \dot{\vec{r}} \times \vec{B}(\vec{r}) \right]
Calculating the field evolution though is complicated due to the dependence of r on A, but at least the R evolution has an easy form when calculated in the multi-particle case.

That's correct. I would just like to underline here that the fields here do not contain the particle proper field (no self-action).
Concerning a complicated character of the mechanical equation, we may consider the simplest case of a uniform electric (or gravitational) external filed. Then there is no term with B_ext and there is no argument in the external filed because the field is constant. So no R- and A-dependence intervene into the mechanical equation. The field and the center of mass equations are completely decoupled and can be solved exactly. Having the exact solutions for R and A, you obtain the charge coordinate r(t). Yes, it oscillates due to variable A(t). You may average it over time and see how close it is to <R(t)>.

3) Because the evolution of R (the 'center of mass' coordinate) depends only on the fields at r, not R, the 'back reaction' due to radiation is easy to calculate in your theory.

m \ddot{\vec{r}} = q \left[\vec{E}(\vec{r}) + \dot{\vec{r}} \times \vec{B}(\vec{r}) \right] + \vec{F}_{back\ reaction}

where

\vec{F}_{back\ reaction} = m q \int d^3k \ \epsilon(k) [\dddot{\vec{A}} (\vec{k}) + \dddot{\vec{A}} ^*(\vec{k})]

Correct. The electron is a part of oscillators in my model of the charge-field coupling, and its motion r(t) contains a "smooth" <R(t)> and oscillating addenda. For that it is not necessary to write an equation for \ddot{\vec{r}}. It follows just from definition of \vec{r}. This dependence is not dangerous as you can make sure in the simplest case outlined above (unlike CED with its run-away solutions).

Since A(k) can be sourced by another particle, this means your theory is non-local in normal space (as well as reciprocal space as pointed out earlier).

In my theory each electron has its own oscillators (as well as "its own" electric and magnetic quasi-static fields). They are labeled with α in your notations (post #66). So they cannon be sourced by "another" particle. Any other particle field serves for a given charge as an "external" field (I mean, labeled with β, etc.).

In your theory, a particle can be affected instantaneously by a particle light years away. This is due to the mixing of the particle coordinates with the field coordinates. It is unavoidable.

In electrodynamics a charge is a source of EMF "seen" by another charge. The EMF contains the Coulomb, magnetic, and radiated parts, even in CED. They come in superposition. What are you blaming me for? And "light years away" distance makes all of them so weak that there is nothing to measure.

So your theory has some very pathological problems. It wildly disagrees with experiment.

I consider this as your signature.

Glance? I have put much work and written much math here. You on the other hand continue to dismiss my comments without providing math of your own.

You started to deny my theory in post #22 in https://www.physicsforums.com/showthread.php?t=351345&page=2 without any math. Apart from a "glance" approach, you have also a huge prejudice against my theory. The only way to overcome this is to make you check each your doubt yourself. Besides, how could I do math here if I have not done any derivative for about thirty years now? And if I did, seeing you prejudice, would you believe me? When I talk Physics, you call it “talking points”. You are elementarily impolite with me. Where is your respect to my gray hair, my age, my experience, my patience to you, and finally to my efforts to get the thing straight?

What are you talking about!? I am arguing against your claim that your q^2 only coupling is correct. I have, and continue to, claim that there should be a q coupling as well.

And I speak of radiation which is proportional to q² in case of external electromagnetic fields. Answer clearly now: what result does CED give for such a radiation? Not the Larmor formula?

The coupling in your theory is incorrect.

The coupling in my theory is different. So far you just failed to appreciate its advantage.

...then ... it is incorrect when compared to experiment.

You are too quick to judge and execute. In fact, you are led by a desire to execute. That is why your judgments are unjust. You are unjust, Justin. So far you have not compared any particular case of radiation calculation in CED and in my theory to judge. Take, for example, a charge motion in a uniform external field (electric or gravitational) and make a comparison. Show me the difference between CED and NCED radiation and charge trajectories. (Or in a uniform external magnetic field, whatever.)

 

[DrFaustus]

Just one quick remark here. bob_for_short, it's not JustinLevy's task to compare trajectories of a charged particle in CED and NCED, but yours. You are proposing a modification to a 150+ years well established theory and you cannot just say "Hey people, that theory sucks, mine is better. Check that out for yourself, I'm sure mine is better." Even Einstein bothered to compute the precession of the Mercury when he proposed GR as a modification to Newtonian's gravity. And he recovered Newtonian's gravity in the appropriate limit.

It is your task to
(a) recover CED in some limit (because no matter how much you dislike it, all the electrical devices around you and that you use are based on it),
(b) show us in what specific physical situation (experiment) one could measure a relevant difference with respect to CED.

And this has got nothing to do with your age, your gray hair or what not. But that you did not do a derivative in 30 years might be a problem here... And I do not know JustinLevy, but I'm firmly convinced that if you put forward the above two points by using rigorous maths, he will accept it. Just like everyone else. The task will then be left to experimentalists. You hopefully agree that a theory's existence is only tied to it's ability to make predictions, don't you? Unfortunately, or fortunately - you judge - the times when a Faraday could make huge progress in physics without ever writing a single formula are over. So if you cannot back your theory up by maths, there's no argument that will be able to turn things around. Especially when counter arguments are mathematically sound.

As a last remark, you should be actually more than happy that someone did make the effort to (try to) understand your theory and not keep telling him that he's got prejudices.

 

[JustinLevy]

Bob_for_short,
Please understand this from my point of view. This is very frustrating to me because I take a considerable amount of time (it often takes me over an hour to type up all the math and remove typos), and YOU dismiss my comments at a glance ... and without math. It is even more frustrating on top of this that you then turn around and claim I am not taking your claims seriously ... despite the fact that I have backed MINE with math, and YOU have not.

So I apologize that I get frustrated, and that I get impolite. But my conclusions are NOT cursory; they are backed by math ... where as yours are not. Can you really blame me and the others reading this for not believing your talking point rebutals given this situation?

The entire point of you mathematically specifying your theory was so that we could stop disagreeing on the precise predictions. But we still are because you refuse to accept the outcome of the math because of preconceived notions.As an example, let me pause for an aside here on one of your worst preconceived notions:

That's correct. I would just like to underline here that the fields here do not contain the particle proper field (no self-action).

You have NOT removed the self-action. I will prove this to you mathematically in a bit.

In my theory each electron has its own oscillators (as well as "its own" electric and magnetic quasi-static fields). They are labeled with α in your notations (post #66). So they cannon be sourced by "another" particle. Any other particle field serves for a given charge as an "external" field (I mean, labeled with β, etc.).

As I explained before, a j.A term BOTH sources A and causes the particle in j to feel a force caused by A. Therefore you still have self-action, despite your duplication of A for each particle.

In more explicit math, you have this term in the Lagrangian:
\frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\cdot\vec{A&#039;}(\vec{k})+\vec{A&#039;}^*(\vec{k})\cdot\cdot\vec{j}(\vec{k})]
which requires some explaining, so I will just copy your remarks

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

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.

So let me start by explicitly writing out the subscripts:
\frac{1}{2}\int d^3k \sum_{\alpha, \beta \neq \alpha} [\vec{j}_\alpha^*(\vec{k})\cdot\vec{A}_\beta(\vec{k})+\vec{A}_\beta^*(\vec{k})\cdot\vec{j}_\alpha(\vec{k})]

There is a mathematical identity such that for any functions f(r), g(r) and their reciprocal space versions F(k),G(k) we have:
\frac{1}{2}\int d^3k [F^*(\vec{k}) G(\vec{k})+G^*(\vec{k}) F(\vec{k})] = <br /> \int d^3r f(\vec{r}) g(\vec{r})

Therefore I can write your integral above in the form:
\int d^3r \sum_{\alpha, \beta \neq \alpha} [\vec{j}_\alpha(\vec{r})\cdot\vec{A}_\beta(\vec{r})] = \sum_{\alpha, \beta \neq \alpha} q_\alpha \dot{\vec{r}}_\alpha \cdot \vec{A}_\beta(\vec{r}_\alpha)

Sure, you "labelled" the field with a beta, but that doesn't change the outcome of the math. Vary j.A with respect to the particle coordinates and you get a force on the particle due to the fields. Vary j.A with respect to the field coordinates and you get a source for the fields due to the particle.

YOU STILL HAVE SELF-ACTION in your theory.

I also did this aside to make it abundantly clear that the point you dismissed off hand to counter the fact that your theory has instantaneous interactions over light-years, well that my point is still in fact correct. And no, the interactions are not weak just because the distance is large ... look at the math, not your preconceived notions of what you'd like the math to say. (The interactions are not weak because the force in normal space is given by a mode in reciprocal space, so it doesn't have the usual fall-off with distance and instead is independent of distance.)

Besides, how could I do math here if I have not done any derivative for about thirty years now? And if I did, seeing you prejudice, would you believe me? When I talk Physics, you call it “talking points”.

I'm not sure why you are saying this. Of course you can do a derivative and a derivation. Of course you know the math. If you are to seriously propose a radical change to the way field theory is done, then you must be able, at the minimum, to derive the consequences of this theory.

Math is precise. If when I worked out the math, it showed that your theory did all the wonderful things you claim, that would be wonderful! However it doesn't. If I am somehow making an error, you could have cleared this up ages ago by showing me the math. And yes, if your math was correct, then that would clear this up. Countering math instead with 'talking points' is not helpful.

And your 'talking points' are NOT physics. They are just what you hope your theory will do. It is hard to discuss your theory with you when I show you something explicitly with math, and then you get upset when I don't dismiss that result just because it doesn't match your talking points -- doesn't match what you hoped your theory would do.

You talking points are not even self-consistent. You claim your theory only couples A_alpha to the motion of charge q_alpha. You agree this coupling occurs via the equation with a coupling of q^2 (and it is missing the term proportional to q that is in CED). You agree this missing q term is related to the magnetic field. Yet you claim your theory has correct magnetic interactions once you consider more particles. But if only q_alpha sources the A_alpha, then adding more particles cannot add that missing magnetic term back in according to your very own talking point. So your talking points are not consistent. Please stop repeating them, and let's stick to the math.

I repeat, if you are going to disagree with my claims, then respond to them with math.

 

[Bob_for_short]

Dear DrFaustus,

Don't prevent Justin from learning and making sure that he is right. And do not teach me what to do. I am not going to report to you.

I am fed up with highly advertised ink-drinks, infinite bare particles and infinite counter-terms. I want changes towards a good sense:

http://www.youtube.com/watch?v=krcqHiiPGy0&feature=related

http://www.youtube.com/watch?v=PuQ4Y_MnaFc&feature=related
.
.

 

[Bob_for_short]

Justin, do not be unhappy with doing this math. Theoretical Physicists are not unhappy about it. I hope you are one of them.

I will consider your last post. Meanwhile do me a favor: calculate the radiation and the charge trajectory in CED and in NCED in the simplest case at your choice. This is a direct test of difference between theories and this difference exists indeed. It has an interesting physical explanation. Only you can do this calculation amongst all of us.

 

[Bob_for_short]

This is very frustrating to me because I take a considerable amount of time (it often takes me over an hour to type up all the math and remove typos), and YOU dismiss my comments at a glance ... and without math. It is even more frustrating on top of this that you then turn around and claim I am not taking your claims seriously ... despite the fact that I have backed MINE with math, and YOU have not.

Do not get frustrated. I am much more experienced than you, admit it. I, as the author of my approach, know better its properties and solutions. I took me much more time to figure out how to reformulate QED to bypass difficulties. Your attacks are incomplete and often wrong. I do need to do much math to tell apart the right and wrong derivations/conclusions. Be critical but have more confidence in my words. If I had wanted to get rid of this discussion, I would have done it long ago. I like and respect you and your efforts in finding out the truth. If my theory is proven to be wrong, it will be OK with me. I am myself finding out the truth.

In post 77 you managed to derive a correct mechanical equations for R (see my post 80). As soon as the Lagrangian depends on R in the same way as on A_rad, the right-hand side of the wave equation (driving force) is also the same as in mechanical equation – it is the external force. Unfortunately you have not listened to me and insisted on wrong statements although the corresponding derivation is quite the same. I added it as an Appendix to my “Reformulation instead of Renormalizations” article: http://arxiv.org/abs/0811.4416.

I see that I was wrong when decided to start this thread in PF. No discussion of new physics implemented in my construction has been carried out. In about 90 posts we are still deriving elementary equations. I ask PF mentors to lock this thread. Those who might be interested in the subject may find discussions in my research group http://groups.google.com/group/qed-reformulation or on my web log http://vladimirkalitvianski.wordpress.com .

 

[Vanadium 50]

We're done here.

In my view, Justin's pointing out specific problems by pointing out exactly where in the Lagrangian they occur trumps Bob's claim that he's right because he is so much older and wiser than Justin.