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Can we remove the self-interaction in classical electromagnetism with discrete particles?


Sergio_Prats

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In classical electromagnetism we have two space vector fields E and H for the electric and magnetic field and a four vector made with spatial charge density and current [ρ, J] with determines EM field together with external sources and whose dynamic is affected by those E and H fields according to the Lorentz force. Classical electromagnetism assumes that each point of the charge density interacts with fields E and H and moves independently on how the rest of the charge moves (except, of course because the dynamics of the rest of the charge will affect the future EM field).

However, in nature there are not infinitesimal charges that move freely, but discrete particles such as electrons and protons. In quantum mechanics at first order a particle EM field does not interact on itself. An example of this is the Hydrogen's electron Hamiltonian, the potential we see is the one created by the proton and there is no contribution from the electron itself. At higher levels, we may find interactions due to the vacuum but they are much weaker, for example, the Lamb Shift in the Hydrogen atom (4*10^-6 eV) is much smaller that the difference between radial levels 1 and 2 (10.2 ev), so I consider a good approach to say that in general, the EM field generated by discrete a particle does not affect itself. 

So, can we assume that a particle only interacts with the field created by other particles? In that case (still in the classic non-quantic theory), we would have N charged particles that require N Hilbert spaces to define the charge and current distribution for each of them. Each particle “sees” a different EM field because it has to exclude its own contribution to the EM field, so we would have N spaces with EM fields plus an overall field.

As far as I can see, we can have this way a theory with discrete particles that respects Maxwell equations, however, the field energy density should be reviewed because in absence of other charges, there is no work done to “bring together” the density of charge needed to build a particle since the own field does not affect particle. Explained from another point of view, we can have a charged sphere which generates an electric field, which should have energy, but we don’t have spent any energy to bring together its charge.

Let the formula for the EM field density of energy be:

u = (ε/2)E2 + (μ/2)H2

Therefore, some correction must be done to remove the term of the energy needed to build isolated particles and once this is done, I consider this approach would be consistent.

 

 

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1 hour ago, Sergio_Prats said:

So, can we assume that a particle only interacts with the field created by other particles? In that case (still in the classic non-quantic theory), we would have N charged particles that require N Hilbert spaces to define the charge and current distribution for each of them. Each particle “sees” a different EM field because it has to exclude its own contribution to the EM field, so we would have N spaces with EM fields plus an overall field.

 

Classically it doesn't work like that.

The underlying classical equations for this are Coulomb's law which whould be compared with Newton's law of gravity.

Both of these rely on at least two agents to generate an output.

If there is only one mass or charge, then although there is a notional field, the force at all points is zero.

So no work is done if we move that mass or charge about.

The force and work only arise when we introduce a second charge or mass.

In the case of gravity we normally consider the Earth so massive by comparison with the test mass that all the force and work is considered to be applied to the test mass alone, with the Earth considered immovable.
But really we should be considering the centre of mass of the system to get the full picture.

Because we disparity between the agents for charge is so much less, when we consider Coulombs law we (have to) introduce an external agent to hold one charge still.

Then we can attribute all the effects to the other charge.

 

Equally we could calculate the resultant field and the centre of charge and work that way.

No charge will 'see' a different field, they all 'see' the same one.

 

There is an equivalent magnetic theory which has stalled in the absence of magnetic monopoles.

 

 

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6 hours ago, studiot said:

The underlying classical equations for this are Coulomb's law which whould be compared with Newton's law of gravity.

The Coulomb law is the force caused between two static charged particles, it also represents the field that one particle creates at some point. I consider "classical" all the non quantum electrodynamics, so any electrodynamic system based on Maxwell equations is classical, I also find very useful the formulae of the generated by a punctual particle with speed v and acceleration a as they tell the field that this particle extert at any point.

 

9 hours ago, studiot said:

If there is only one mass or charge, then although there is a notional field, the force at all points is zero.

The fact is that under classical EM, if we have, for example, a uniformly charged sphere with radius R, any point at distance r from the radius will suffer a force ρ(r)*Q/rr where Q is the charge inside that r radius. That will make the sphere quite unstable unless you remove that force by ignoring the self generated field. 

 

9 hours ago, studiot said:

The force and work only arise when we introduce a second charge or mass.

This is exactly what is not derived from "pure" Maxwell equations. 

 

9 hours ago, studiot said:

In the case of gravity we normally consider the Earth so massive by comparison with the test mass that all the force and work is considered to be applied to the test mass alone, with the Earth considered immovable.
But really we should be considering the centre of mass of the system to get the full picture.

I agree, the small object also attracts the big one, but I don't see the relation with my concern, in gravity its very hard to know if a particle attracts itself because gravity constant is so small.

9 hours ago, studiot said:

There is an equivalent magnetic theory which has stalled in the absence of magnetic monopoles.

Which theory is it?

 

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12 minutes ago, Sergio_Prats said:

The Coulomb law is the force caused between two static charged particles, it also represents the field that one particle creates at some point. I consider "classical" all the non quantum electrodynamics, so any electrodynamic system based on Maxwell equations is classical, I also find very useful the formulae of the generated by a punctual particle with speed v and acceleration a as they tell the field that this particle extert at any point.

 

You seem to put great store by your personal definition of classical and appeared to me to want to discuss this only in classical terms.
But like most boundaries I think this one is blurred as well, where do you place retarded potentials etc?

That aside I agree with your comment in general unless you are thinking about Force.
I say this because you go on to say

18 minutes ago, Sergio_Prats said:

The fact is that under classical EM, if we have, for example, a uniformly charged sphere with radius R, any point at distance r from the radius will suffer a force ρ(r)*Q/rr where Q is the charge inside that r radius. That will make the sphere quite unstable unless you remove that force by ignoring the self generated field. 

Which is not the situation I mentioned.

(electric) forces are not exerted on points. The concept is meaningless.

(electric) forces are only exerted on other charges, and I specified "there is only one charge".

22 minutes ago, Sergio_Prats said:

This is exactly what is not derived from "pure" Maxwell equations. 

Please explain the connection.

24 minutes ago, Sergio_Prats said:

I agree, the small object also attracts the big one, but I don't see the relation with my concern, in gravity its very hard to know if a particle attracts itself because gravity constant is so small.

The use of the word particle implies zero dimensions.
In those circumstances the mass particle does not attract itself.
In the case of a distributed mass, we use the position and motion of the centre of gravity to determine mechanical output such as work, kinetic energy etc.

However this is missing my point as this was just a comparative example.
I also said that my analysis required an additional agent holding one charge in place, instead of relying on an overwhelmingly large mass difference as in the mechanical case.

It should be noted that if this agent does not exist then Earnshaw's theorem requires both charges to be in eternal motion.

 

Anyway I don't want to just quibble I would like to find out the point of your thread.

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23 hours ago, studiot said:

Which is not the situation I mentioned.

(electric) forces are not exerted on points. The concept is meaningless.

(electric) forces are only exerted on other charges, and I specified "there is only one charge".

In fact we are saying the same thing: that a charge distribution that belongs to the same particle does not exert force on itself, but as far as I know, if we follow strictly the Maxwell equations we cannot "trick" the Lorentz Force to ignore the contribution to the field caused by the own charge distribution: the field is the field no matter what particle has created it. However, it turns out that this is not a trick and Quantum Mechanics behaves this way: a particle's generated potential does not affect itself. So some adjustments must be done in the classical electromagnetism to take this fact into account.

I have seen several theories about the electromagnetic mass and the electro self-energy but as far as I have seen, they are not completely satisfactory, you can find a description of some of them here:

http://www.feynmanlectures.caltech.edu/II_28.html

But if we assume that a particle does not interact with itself, why just to remove the effect of that particle over itself?

As far as I know, the only thing that needs to change is the energy stored in the EM field. Basically, since a particle "alone in the universe" does not need any energy to bring together its charge, we have to remove from the field energy expression, the energy that each of these particles would have alone. I have a candidate formula that has been successful in comparing the electron’s potential energy with the field energy for the Hydrogen atom radial levels (n=1, l=0, m=0) and (n=2, l=0, m=0).

 

23 hours ago, studiot said:

The use of the word particle implies zero dimensions.
In those circumstances the mass particle does not attract itself.
In the case of a distributed mass, we use the position and motion of the centre of gravity to determine mechanical output such as work, kinetic energy etc.

When I say "particle", I do not mean zero dimensions. I would say "punctual particle" for that. I understand a particle such as an electron can have a wave function and therefore can be distributed through space. Since electros are fermions, they should not be mixed in the same space, so everyone of them would requiere a particular space even if they were puntual particles.

 

23 hours ago, studiot said:

However this is missing my point as this was just a comparative example.
I also said that my analysis required an additional agent holding one charge in place, instead of relying on an overwhelmingly large mass difference as in the mechanical case.

An additional charge would feel the overall EM field but, in fact that charge would also add another field that would add another field which would not be able to detect itself, so we are only able to detect the field created by the system but not the field created by the probe.

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A couple of quick points about (scientific) English.

A particle is always taken to be of negligable dimensions.
So in the calculation of planetary orbits, the radius of the Earth is negligable compared to radius of the orbit and we can consider the Earth as a particle.

The word 'punctual' in English means at the correct or appointed time, it is inappropriate to use with particles.

If you want to describe something of finite or non-negligable dimensions you would use the word 'body' or even 'rigid body' as appropriate to the situation.

1 hour ago, Sergio_Prats said:

An additional charge would feel the overall EM field but, in fact that charge would also add another field that would add another field which would not be able to detect itself, so we are only able to detect the field created by the system but not the field created by the probe.

 

I still think you are missing the point and as it is an important one I will try to say it in a different way in my next post.

Perhaps it is a language thing as you obviously have a better understanding of electromagnetism than your English allows.

 

Maxwell did not know about electrons, that was Thompson at least 30 years later.
Maxwell's equations are classical field theory equations in continuous functions.

 

As regards self interaction with its own field, relativity limits the horizon of this as shown by this animation from NASA.

field_a.gif.afacb435676c4932916e7171fd17639e.gif

 

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10 hours ago, studiot said:

The word 'punctual' in English means at the correct or appointed time, it is inappropriate to use with particles.

I used incorrectly the word "punctual", I would say "point particle" . When I say "particle" I refer to protons or electrons whose wave functions can be extended over the space.

 

10 hours ago, studiot said:

Maxwell did not know about electrons, that was Thompson at least 30 years later.
Maxwell's equations are classical field theory equations in continuous functions.

I agree with that, but we are still using formulae based on Maxwell equations in modern physics. For example, the QED Lagrangian contains a term that is -(1/4) FuvFuv, this term is the classical field energy. If the fact that electrons and protons does not exert electric field over themselves entails a correction in the field energy, then this QED term should be reviewed.

 

10 hours ago, studiot said:

As regards self interaction with its own field, relativity limits the horizon of this as shown by this animation from NASA.

The EM field (induced or radiated) propagates not instantly but with the speed of light. In a electrostatic scenario this fact can be ignored but it cannot be in the general electrodynamic scenario, however, this does not not affect my approach. We only need to remove the field create by the own charge distribution at d/t seconds when d is the distance of each point respect the point we are evaluating. Of course it makes the calculations harder but it is essentially the same.

 

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