Reformulation instead of Renormalizations

[Bob_for_short]

$$\frac{1}{c^2} \dddot{\vec{A}}^*(\vec{k}) + k^2 \dot{\vec{A}}^*(\vec{k}) = \frac{2 q^2}{\epsilon_0 k^2} \epsilon(k) \left[ \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} + \vec{g}(\vec{A_{ext} }(\vec{r})) - [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]$$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.

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]

$$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \ddot{\vec{A}}(k)} = \frac{d}{dt} [q^2 \epsilon(k) \vec{A_{ext} }(\vec{r}) - \frac{\epsilon_0}{2 k^2 c^2} \ddot{\vec{A}}^*(\vec{k})] = q^2 \epsilon(k) \dot{\vec{A}}_{ext}(\vec{r})\dot{\vec{r}} + q^2 \epsilon(k) \vec{g}(\vec{A_{ext} }(\vec{r})) - \frac{\epsilon_0}{2 k^2 c^2} \dddot{\vec{A}}^*(\vec{k})$$

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:
$$\frac{1}{c^2} \dddot{\vec{A}}^*(\vec{k}) + k^2 \dot{\vec{A}}^*(\vec{k}) = \frac{2 q^2}{\epsilon_0 k^2} \epsilon(k) \left[ \frac{\partial \vec{A}_{ext}}{\partial t}(\vec{r}) + \vec{g}(\vec{A_{ext} }(\vec{r})) - [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]$$
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:
$$\frac{1}{c^2} \dddot{\vec{A}}^*(\vec{k}) + k^2 \dot{\vec{A}}^*(\vec{k}) = \frac{2 q^2}{\epsilon_0 k^2} \epsilon(k) \left[ \frac{\partial \vec{A}_{ext}}{\partial t}(\vec{r}) + \vec{g}(\vec{A_{ext} }(\vec{r})) - [\ddot{\vec{A}} (\vec{k}) + \ddot{\vec{A}} ^*(\vec{k})] \cdot \vec{f}(\vec{A_{ext} }(\vec{r})) \right]$$
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
$$\frac{\partial r_x}{\partial R_x} = 1,\frac{\partial \dot{r}_x}{\partial \dot{R}_x} = 1$$That gives:
$$-q [\frac{\partial}{\partial r_x} \ \phi_{ext}( \vec{r})] +q\dot{\vec{r}} \cdot [\frac{\partial}{\partial r_x}\vec{A_{ext} }(\vec{r}) ] = \frac{d}{dt} \left[ m \dot{R}_x +q A_{ext,x}(\vec{r}) \right] = m \ddot{R}_x +q \frac{\partial}{\partial t} A_{ext,x}(\vec{r}) +q (\dot{\vec{r}} \cdot \vec{\nabla}) A_{ext,x}(\vec{r})$$
Thus:
$$m \ddot{R}_x = q \left[ -\frac{\partial}{\partial r_x} \ \phi_{ext}( \vec{r}) - \frac{\partial}{\partial t} A_{ext,x}(\vec{r}) +\frac{\partial}{\partial r_x}[\dot{\vec{r}} \cdot \vec{A_{ext} }(\vec{r}) ] - (\dot{\vec{r}} \cdot \vec{\nabla}) A_{ext,x}(\vec{r}) \right]$$
The other components of R follow similarly, so combining to make the cross products to come easier to see:
$$m \ddot{\vec{R}} = q \left[ -\vec{\nabla} \phi_{ext} (\vec{r}) - \frac{\partial}{\partial t} \vec{A}_{ext}(\vec{r}) +\vec{\nabla} (\dot{\vec{r}} \cdot \vec{A}_{ext}(\vec{r}) ) - (\dot{\vec{r}} \cdot \vec{\nabla}}) \vec{A}_{ext}(\vec{r}) \right]$$
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
$$m \ddot{\vec{R}} = q \left[ -\vec{\nabla} \phi_{ext} (\vec{r}) - \frac{\partial}{\partial t} \vec{A}_{ext}(\vec{r}) + \dot{\vec{r}} \times (\vec{\nabla} \times \vec{A}_{ext}(\vec{r})) \right]$$
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:
$$\frac{1}{2}\int d^3k [\vec{j}^*(\vec{k})\cdot\cdot\vec{A'}(\vec{k})+\vec{A'}^*(\vec{k})\cdot\cdot\vec{j}(\vec{k})]$$
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:
$$\frac{1}{2}\int d^3k [F^*(\vec{k}) G(\vec{k})+G^*(\vec{k}) F(\vec{k})] = \int d^3r f(\vec{r}) g(\vec{r})$$

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.