QED meets GR

[kcdodd]

What is your distinction between physical and construct? As an example, there is something which we have chosen to model as a "coffee cup", which we treat according to a "coffee transport theory", and which we have demonstrated experientially to move coffee from the coffee maker to our mouth. But are these coffee cups physical, or just a construct that fits with experience?

[johanfprins]

I suppose this is the addendum you have talked about:

If one could construct a rigid dipole in which two charges are a fixed length apart, and if one were to only translate that dipole along it's axis then the two charges cannot move relative to one-another. Now, according to you there can not be any magnetic field generated because it does not satisfy your relative movement condition.

Correct since there is zero current flowing there cannot be a static magnetic field or else Ampere's law must be wrong.

However, this meets the requirement you listed for electric fields: two charges. Which means you should admit there is an electric field. Now, transforming the fields to a frame where the dipole is moving along it's axis, one can find (in a non-relativistic approximation) that B' = -vxE/c,

It is here where you make the mistake: A static electric field viewed from another inertial refrence frame does not generate a magnetic field. A static magnetic field is only generated when charges move relative to each other so that a current is flowing and this is also required in order to be commensusrte with Special Relativity. To ignore relativity, as you have just done, ist a violation of Maxwell's equations.

But let us return to the thought experiment we have discussed above: i.e. considering the electric-field energy around a solitary, stationary electron surrounded by a sphere with an equal distributed charg around the electron. You claimed it cannot be capacitor, while I claim that it must be one.

Consider a double plate capacitor with "infintely large" areas. One can use Gauss's law to derive the "electric field" just outside the positively charged plat as being (1/2) (sigma)/epsilon. But in order to calculate the energy of the actual electric-field between the plates you have to ADD the "electric-field" derived in the same manner for the other plate: So that the electric-energy field is jointly generated by the charges on both plates.

Now back to the electron which I maintain is NOT a point charge but a region in space having a distributed charge. Let us assume that the electron is a sphere within which the distributed charge-density is the same at every point. By using Gauss' law in differential format, one can integrate to calculate the "electric field" within and around this sphere. There is now no infinity involved whatsoever. This "electric-field" is zero at the centre of the electron and reaches a maximum value at its surface from where it drops off inversely with distsnce along any radial direction.

To calculate the actual field energy between the electron and the positive shell we have to add the "electric field" generated by the positive shell to that of the "field" generated by the electron. As you quite correctly pointed out, the "electric-field" from the positively-charged spherical shell cancels everywhere within the spherical shell: So we end up, in this case, that the actual electric field, generated by both the electron and the spherical shell, is equal to that which is generated by the electron alone.

Now calculate the electric field-energy outside of the electron up to thne positively charged spherical shell. The amount of energy that can be present within this spatial region is determined by the capacitance jointly generated by the electron-sphere and the spherical shell. When now allowing the surrounding spherical shell to expand, the capacitance drops and the amount of electric-field energy between the electron and the spherical shell must also drop. Thus by assuming that there is no spherical shell, there can be no electric-field energy around the electron: ergo: There cannoit be a electric field energy around a solitary electron. The only electric field-energy which remains, is the field within thne electron; and the energy of this field is obviously equal to the mass-energy of the electron.

This model dovetails smoothly with quantum mechanics since it is possible to have a localised matter-wave with a distributed charge within it. In fact the Chemists use this fact on a daily basis to calculate chemical bonds. Neither do you require Poincare stresses to keep the distributed charge from not exploding since the "stresses" keeping the ditributed charge together is the electron's wave intensity, which is time-independent as long as the boundary conditions do not change.

[kcdodd]

It is here where you make the mistake: A static electric field viewed from another inertial refrence frame does not generate a magnetic field. A static magnetic field is only generated when charges move relative to each other so that a current is flowing and this is also required in order to be commensusrte with Special Relativity. To ignore relativity, as you have just done, ist a violation of Maxwell's equations.

It is exactly the opposite of what you have stated here. The EM field is a covariant field object. A pure electric field in one reference frame will appear as a mixed electric and magnetic field in another frame. Likewise, a pure magnetic field will appear as a mixed electric and magnetic field in another frame. It is precisely because what you have said violates special relativity that I have provided this example. I provided a first order approximation to the transformation just for simplicity. First order is not the same as "ignored". Ignored is exactly what you have done. If you had not ignored special relativity you would know that your model violates it.

To calculate the actual field energy between the electron and the positive shell we have to add the "electric field" generated by the positive shell to that of the "field" generated by the electron. As you quite correctly pointed out, the "electric-field" from the positively-charged spherical shell cancels everywhere within the spherical shell: So we end up, in this case, that the actual electric field, generated by both the electron and the spherical shell, is equal to that which is generated by the electron alone.

Now calculate the electric field-energy outside of the electron up to thne positively charged spherical shell. The amount of energy that can be present within this spatial region is determined by the capacitance jointly generated by the electron-sphere and the spherical shell. When now allowing the surrounding spherical shell to expand, the capacitance drops and the amount of electric-field energy between the electron and the spherical shell must also drop. Thus by assuming that there is no spherical shell, there can be no electric-field energy around the electron: ergo: There cannoit be a electric field energy around a solitary electron. The only electric field-energy which remains, is the field within thne electron; and the energy of this field is obviously equal to the mass-energy of the electron.

This was not the original setup, but once again reality is exactly opposite what you just stated. Using the model of capacitance you provide, the total energy would be W = Q^2/(2C). So obviously as C decreases the energy increases; exactly opposite what you stated.

Doing a more exact calculation given two concentric spheres of q0, R0 and q1, R1, where R0 is the inner sphere of uniform charge density and R1 the spherical shell. And if one assumes that q0 + q1 = 0, the total charge is zero, then the total energy of the system is given by:

W = (k*q0^2/2)*(6/(5*R0) - 1/R1)

A calculation I did just for you. You see that as R1 is increased the total energy of the system also increases, not decreases as you have stated. The physical reason for this is that the field is zero outside of R1 (again, when the spheres are concentric). So as R1 increases there is more field, and so more field energy, not less.

Now, what this really has to do with anything I am not sure. You brought it up. But whatever you intended, once again you only showed you don't know what you are talking about.

This model dovetails smoothly with quantum mechanics...

I should hope it does not dovetail smoothly with QM, or else physics really is in trouble. Your statements appear to come from bizzaro land where everything is opposite from reality. This is seriously going to be my last post in response to you. You seem incapable of doing the most basic of calculations to understand your ideas are inconsistent with special relativity and EM theory. How it is you have a masters degree in physics and math, and even taught physics for a number of years (according to your website), is honestly beyond me.

[johanfprins]

It is exactly the opposite of what you have stated here. The EM field is a covariant field object. A pure electric field in one reference frame will appear as a mixed electric and magnetic field in another frame.

A very seemingly logical argument but unfortunately still wrong. When you transform this dipole field using the Lorentz transformation, you will find that there is no magnetic component that forms circles around the dipole path as would be the case when there is a current flowing caused by opposite charges that move relative to each other.

This was not the original setup, but once again reality is exactly opposite what you just stated. Using the model of capacitance you provide, the total energy would be W = Q^2/(2C). So obviously as C decreases the energy increases; exactly opposite what you stated.

How did you obtain this formula? There is not a single capacitor in the world which does not to go to zero capacitance when the distance between the oppositely charged surfaces goes to infinity. Work yourself up from a parallel plate capacitor to a cylidrical capacitor to a spherical capacitor. If you need some help go to Serway's elementary handbook; pages 742-744 in my copy

W = (k*q0^2/2)*(6/(5*R0) - 1/R1)

A calculation I did just for you.

Unfortunately you again ignored the fact that the electric-field energy must be an integral overt (1/2)*(epsilonzero)* E*E. etc.

Now, what this really has to do with anything I am not sure. You brought it up. But whatever you intended, once again you only showed you don't know what you are talking about.

Thank you! Time will determine whether it is I or you who "knows nothing what he is talking about." I will not put my bets on you.

I should hope it does not dovetail smoothly with QM, or else physics really is in trouble.

Your hope is in vain; and you are correct: Physics is in serious trouble and we should straighten it out as soon as possible.

How it is you supposedly have a masters degree in physics and math, and even taught physics for a number of years (according to your website), is honestly beyond me.

May I know what are your qualifications and experience?

[kcdodd]

It is exactly the opposite of what you have stated here. The EM field is a covariant field object. A pure electric field in one reference frame will appear as a mixed electric and magnetic field in another frame.

A very seemingly logical argument but unfortunately still wrong. When you transform this dipole field using the Lorentz transformation, you will find that there is no magnetic component that forms circles around the dipole path as would be the case when there is a current flowing caused by opposite charges that move relative to each other.

This was not the original setup, but once again reality is exactly opposite what you just stated. Using the model of capacitance you provide, the total energy would be W = Q^2/(2C). So obviously as C decreases the energy increases; exactly opposite what you stated.

How did you obtain this formula? There is not a single capacitor in the world which does not to go to zero capacitance when the distance between the oppositely charged surfaces goes to infinity. Work yourself up from a parallel plate capacitor to a cylidrical capacitor to a spherical capacitor. If you need some help go to Serway's elementary handbook; pages 742-744 in my copy

W = (k*q0^2/2)*(6/(5*R0) - 1/R1)

A calculation I did just for you.

Unfortunately you again ignored the fact that the electric-field energy must be an integral overt (1/2)*(epsilonzero)* E*E. etc.

Now, what this really has to do with anything I am not sure. You brought it up. But whatever you intended, once again you only showed you don't know what you are talking about.

Thank you! Time will determine whether it is I or you who "knows nothing what he is talking about." I will not put my bets on you.

I should hope it does not dovetail smoothly with QM, or else physics really is in trouble.

Your hope is in vain; and you are correct: Physics is in serious trouble and we should straighten it out as soon as possible.

How it is you supposedly have a masters degree in physics and math, and even taught physics for a number of years (according to your website), is honestly beyond me.

May I know what are your qualifications and experience?

You have got to be shitting me.

[johanfprins]

You have got to be shitting me.

Why would I need to do so. A Person like you who cannot even understand how a capacitor stores electric energy; and that this electric-energy is generated by both negative charges and positive charges, and not just by the charges on one of the electrodes, just does not know elementary physics.

I thus repeat that a solitary charge cannot store electric field-energy around it since its capacitamce is zero. By assuming that an electric-field can be present when there is zero capacitance, you are the one who is shitting yourself.

In mitigation, I must note that you are just repeating the mistakes which all the leading physicists have made for the past 110 years. This is why modern physics has serious problems; as we all know that it has.

[johanfprins]

What happened to you kcdodd? Why do I always find vicious attacks on the logical physics I am presenting; but as soon as the other person comes to a point that he/she has to admit there must be something wrong with the mainstream physics of the past 110 years, he/she just suddenly evaporates? An then I am blamed for "dicking people around"

[Betruger]

It looks like he finds some fundamental component of the conjecture in your latest posts so incredible that he doesn't deem it worthwhile to demonstrate how it's wrong.

Which is pretty anti-climactic - to "put up with" a conversation obviously headed for a very decisive conclusion (with reaching that conclusion to the benefit of his side of the debate as motivator) all the way thru the penultimate stage, and then stop.

[johanfprins]

It looks like he finds some fundamental component of the conjecture in your latest posts so incredible that he doesn't deem it worthwhile to demonstrate how it's wrong.

This is unfortunately how ALL people react when they have to reconsider their most cherished beliefs. Physicists are supposed not to react in this manner; but I can assure you that they are even worse than any other person. I think it will be easier for the Pope to accept that Jesus is not the Son of God, than for modern theoretical physicists to accept that there is no electric field-energy around a solitary electric charge.

Yes it is incredible to conclude that our physics heroes like Lorentz, Einstein, Feynman etc. have missed this simple mistake for more than 110 years. But this does not mean that one must react as if somebody has accused them of being rapists and murderers. We are all human beings and can make mistakes. In fact Einstein also made a mistake when he derived the Lorentz-FitzGerald contraction as being valid within the framework of his Special Theory of Relativity. The latter is simple to prove but impossible to get published.

For this reason physics, even well-accepted physics, must be constantly scrutinized, since it is just such a small mistake which can cause modern physics to fall flat on its face. It shows that all quantum physics, except just using Schroedinger's equation being solved cookbook-style, is fundamentally flawed: This should have been obvious from the time that the mathematics had to be fudged by processes like renormalization.

The probability, that concepts like the vector bosons, Higgs boson, graviton, etc. are hallucinations, is thus very high. This possibility must be approached and studied with open minds and without throwing insults around.

[Betruger]

Either of you could defuse such bad chemistry by simply focusing on the math. You've put forward an explicit math argument, so the ball's in his court. Given this chance to concisely get to the bottom of this disagreement, opting not to is inconsistent with any scientific intent.

[johanfprins]

Either of you could defuse such bad chemistry by simply focusing on the math. You've put forward an explicit math argument, so the ball's in his court. Given this chance to concisely get to the bottom of this disagreement, opting not to is inconsistent with any scientific intent.

Thanks for your logical argument. I have, however, found during the past 10 years, after reaching many such an impasse with physicists, that they rather go into total denial and then spread the word behind my back that "Johan just does not understand physics". I sometimes imagine that they then start to sleep in the foetus position.

I have found that nobody will stand up and agree when somebody points out that the Emperor is not wearing the finest clothes ever made but is in fact naked. I always wonder what happened to that little boy who dared to say this. He was probably verbally remanded if not assaulted. Maybe even killed! No wonder that Physics is Rotting.

[kcdodd]

Energy of the sphere system.

http://www.youtube.com/watch?v=W_6mGW7TbpI

Magnetic field around the moving electric dipole.

http://www.youtube.com/watch?v=WwqLUwTLKTs

[johanfprins]

Energy of the sphere system.

http://www.youtube.com/watch?v=W_6mGW7TbpI

Magnetic field around the moving electric dipole.

http://www.youtube.com/watch?v=WwqLUwTLKTs

Thanks for the two YouTube postings. I have news which you might find flattering, and I have bad news. First the flattering news. I am impressed that you took the time to argue physics and back it up with mathematics. One rarely finds this amongst the crackpots who are in charge of main stream physics at present. If I were still a professor at a university and you applied to do a PhD, I would have welcomed you with open arms.

Now the bad news: You are still wrong. I had a bit of problem to follow your handwriting, and am not sure whether you used the correct expressions to calculate the electric-fields within the central sphere and within the spherical shell. When you have a distributed charge it is better to start with divE=(rho)/epsi0. In the case of the central sphere you will then obtain that E(r)=(constant)r: i.e. the electric field is zero at the centre and reaches a maximum at R(0).

But let us assume that your derivation is correct, then next to compare two cases where you first let R(1) go to R(0) and then letting R(1) go to infinity is just not rational physics.

When you let R(1) go to R(0) you end up with the total charge being within a single sphere. Since the two charges are equal and opposite the total charge within the sphere is ZERO. If the charges mix evenly there will be no charge and thus zero energy.

But let us assume that for some reason the two charge distributions stay separate with a spherical boundary between them. You will then have a spherical dipole within the sphere: i.e. equal opposite charge distributions across a spherical boundary. Thus in addition to the original field you now also have a polarization field. And since the electric field must be zero at the center of the sphere and on the surface of the sphere, this demands that the polarization field completely cancels the original field.

This also means that the energy within the sphere is now much higher than the energy without the polarization field. Think about a dielectric between two capacitor plates. So you cannot compare this energy with the energy you obtain when you let R(1) go to infinity in order to conclude that there is an electric field-energy around a solitary, spherical charge distribution.

A very important point which many theoretical physicists (and nearly all of the most revered ones) miss is that what you can calculate mathematically is not always physically possible. By assuming this you end up with crackpot ideas like magnetic monopoles. Circularly moving coherent waves etc. I just noted that in these two examples even the mathematics is wrong. But a calculation which gives an electric-field energy around a solitary charge, even when mathematically correct, is physically impossible.

One must ask when can there be electric-field energy within a region of space. The answer: It is only possible when that region of space has electric capacitance. Obviously, if you could generate a material with zero heat capacitance it will not be able to store heat energy and will be at absolute zero temperature.

Now instead of doing your calculation shown on YouTube, it is easier to calculate the electric capacity C of such a central sphere and a surrounding shell. You will then obtain that: C=[R(1)*R(0)]/[k(R(1)-R(0))]. Now let R(1) go to infinity. Clearly the capacitance goes to zero. This is incontestable proof that the solitary charged-sphere that remains cannot store electric field-energy in the space around it.

Your second YouTube demonstration lost me right from the start. I just could not figure out how you can draw a dipole as two circular rings. I think what you should do is to rather consider a parallel-plate capacitor moving past you and then transform the moving electric field lines. And please use the full Lorentz equations without assuming a low speed.

[kcdodd]

I have checked my derivation several times and am fairly confident it is correct. When R1 goes to R0, what happens is like a Twinkie where you have a sphere with a uniform charge density on the inside of q0 and a shell of surface charge density of q1. There is no funny business there.

Now instead of doing your calculation shown on YouTube, it is easier to calculate the electric capacity C of such a central sphere and a surrounding shell. You will then obtain that: C=[R(1)*R(0)]/[k(R(1)-R(0))]. Now let R(1) go to infinity. Clearly the capacitance goes to zero. This is incontestable proof that the solitary charged-sphere that remains cannot store electric field-energy in the space around it.

Firstly, I would caution you about continuing to use the capacitor analogy in this example because there is not a single voltage difference between the outer sphere and the inner one. The charge is distributed about the inner sphere and so there is charge at different potentials. If the inner sphere was also a shell, then it would be ok, but it is not which is why I don't calculate what the capacitance would be. However, I will now respond to you as if it where.

Secondly you must have your definition of capacitance backwards. C = Q/V. So, if the capacitance of a configuration was truly zero, then you would have either zero charge or infinite voltage. The energy of such a capacitor is defined then as W = (1/2)C*V^2 = (1/2)Q^2/C. Now assuming Q is not zero. The first form would be zero*infinite. Doing a proper limit you would find the W = infinite in the first form since V = Q/C. The second form is clearly infinite at first glance when C = 0. So such a lone sphere would have infinite energy, not zero!

However, luckily the capacitance is not zero. Please just check the equation you posted: C=[R(1)*R(0)]/[k(R(1)-R(0))]. I can rewrite this equation as C=[k(R(1)-R(0))/R(1)*R(0)]^(-1) = [k(1/R(0)-1/R(1))]^(-1). Now, setting R1 = infinite then 1/R1 = 0. and So: C = [k(1/R(0))]^(-1) = R(0)/k. Which clearly is not zero as long as R0 is not zero. You should really check to make sure the equations you post at least support your own argument.

Your second YouTube demonstration lost me right from the start. I just could not figure out how you can draw a dipole as two circular rings. I think what you should do is to rather consider a parallel-plate capacitor moving past you and then transform the moving electric field lines. And please use the full Lorentz equations without assuming a low speed.

The circular rings represent the electric field of a dipole field. I am sorry you could not follow it I don't know how to remedy that.

I somewhat figured you would not accept it until you see the full lorentz transformation of coordinates. I plan on doing that video soon. However it gets worse for you because the fields (magnetic field included) becomes arbitrarily strong as v goes to c.

[johanfprins]

Hi kcdodd,

You have had me severely worried for a while: When I had trouble reading some of your mathematical symbols on YouTube I became frustrated and acted in haste by even writing down a capacitance formula which obviously does not support what I was claiming. My time has been limited by the fact that my company shows healthy signs of recovery and my book sales are flourishing, thus requiring a lot of my time and attention.

I thus made the mistake to rely on my memory instead of delving into my archives of 5 years ago. This is a very dangerous thing for somebody of my age. I am fast approaching the situation that when I stop on the stairs to look at something I will not be able to remember whether I was going upstairs or downstairs. Thus I apologize for the booboo with the capacitor equation; but am still not in agreement that your calculation proves any physics; even though you claim that:

I have checked my derivation several times and am fairly confident it is correct. When R1 goes to R0, what happens is like a Twinkie where you have a sphere with a uniform charge density on the inside of q0 and a shell of surface charge density of q1. There is no funny business there.

The fact is that one can make an assumption and then derive a mathematical formula which seems quite valid but is not valid. There are two assumptions inherent in your derivation which I want to analyze:

1. You assume that the outer shell can have zero thickness and has thus zero volume. This requires that the charge-density must be infinite. What you should do is to also assume a charge density for this opposite charge. There is thus also electric-field energy within this shell. What now happens when you let R(1) go to R(0) is that the electric-field energy between the shell and the central sphere becomes zero (One can even see that from your equation). What remains is the electric-field energy within the central sphere plus the electric field within the shell having a thickness of R(2)-R(0)>0. Not a Twinkie as you call it above. Nonetheless, the fact remains that it is not physically possible to generate such a shell of distributed charge within "free space". So the whole derivation is only a mathematical game with no physics-relevance.

2. The other assumption is that the central sphere contains a distributed charge with a constant charge density: This is also physically irrelevant. But, let us return to the reason why you have assumed this in this discussion: You allowed for the possibility that this might be relevant for an electron. After all what we are really arguing about here is whether there is an electric-field in space surrounding a solitary electron. I have now gone into my archives and hope that I will be more clear on this aspect. We both seem to agree that if I can prove that there is not such a field then, as you have yourself has stated, physics is in trouble. So let us first talk physics by looking at what we know about a single, solitary electron.

We have all been taught, and it is in the physics textbooks that an electron is a "particle'. The latter term has become so accepted in physics that we all believe that we understand what it means. But after I realized that physics is actually in trouble, I started asking physicists to define what a "particle" is. Not a single one of them could do so to date.

So can one state that an electron is "particle" when we do not even know what a "particle" is? A very perceptive person in the audience asked after J J Thomson claimed that "an electron is a particle": " Tell me, how can you discover a particle that is so small that nobody has ever seen one?" A pity he was laughed at as if his question was crackpot.

So what do we know about a solitary electron: We know that it has mass m, a charge -e, and that its center-off mass follows a classical path through space. Now what does mass means. Newton created this concept to include Galileo's inertia, and the latter is thus included in all equations of physics (even Schroedinger's equation). And what has Galileo taught us under the threat of being put to death? That a body with mass and no forces acting on it, must be stationary within one of all the possible inertial reference frames.

This means that the position of its center-of-mass manifests at a single point in space. It also means that its momentum must be exactly zero within that reference frame and, this in turn means that for any body with mass, including an electron, both its position and momentum manifest with 100% accuracy. But the "Copenhagenists" have been telling us for more than 80 years that this is not possible for an electron, since the intensity of an electron's wave function is a probability distribution! And we have lapped it up without seeing how illogical it is. It is clearly a crackpot idea.

But what does it mean for a body with mass to be stationary within an inertial reference frame? It means that it is within a lowest energy ground-state. If it tries to move from its stationary lowest energy ground-state there will a restoring force keeping it there. So what is the most probable wave function modelling an electron within an inertial reference frame? Obviously a wave with a Gaussian intensity distribution. Where the restoring force might come from etc. is discussed in my book, and is not directly relevant here.

Suffice to say that the charge-density within such a wave will not be the same at every point but change as the intensity of the wave changes. This concept is used every day when chemical bonding is calculated by using Schroedinger's equation.

Around and near the center of such a wave, the charge density can thus be approximated by a parabolic expression. One can thus write that (rho)=C*(1-(alpha)*r^2). Now let us return to the electric field around a solitary electron and not trust our belief in Coulomb's law, but rather use the differential expression divE=(rho)epsi0. Writing this in terms of spherical coordinates it is elementary to calculate the electric field within a sphere using the parabolic expression for the charge-density. You will then find that the electric field is zero at the origin, increases to go through a maximum and then decreases again to become zero at a critical distance R(0). You can now normalize your calculation by choosing C and alpha so that the total charge within the volume with radius R(0) is equal to -e.

When you now proceed to calculate the electric field around this sphere using the differential equation divE=(rho)epsi0, you will find that there is NO electric field around the sphere, since it starts off at a boundary conditions on which the electric-field is zero. The field around a suitable distributed charge can thus be zero even though the blind application of Coulomb's law suggests that there must be an electric-field. The total field-energy is thus within the electron's wave-intensity and determines the electron's mass-energy.

Therefore classical mechanics and wave mechanics dovetail smoothly so that we do not need crackpot ideas like wave-particle duality, complementarity, uncertainties, probability wave amplitudes etc., etc., etc.