# The stability of matter: is Maxwell really wrong?

Anonimo
First of all: excuse me for my not good english written language.
I come to the question with an example.
Let us suppose two electrons are in space with initial velocity equal to zero.
Given the Coulomb force such electrons will be submitted to an equal and opposite accelerations (for the action reaction Newton law).
I expect that the complessive radiation emission at great distances will be null.
So the system will not loose energy.
But it is said that according to Maxwell theory this is not possible.
Where is the truth ?
I could give many other similar examples where, even if there are electric charges in accelerated motion, the system will not transfere electromagnetic energy to space.
Is really Maxwell's theory wrong at an atomic scale ?

Mentor
2021 Award
I expect that the complessive radiation emission at great distances will be null.
So the system will not loose energy.
But it is said that according to Maxwell theory this is not possible.
Where is the truth ?
I think you should do some actual calculations. I am not convinced that either of your assertions here are correct

Mentor
Let us suppose two electrons are in space with initial velocity equal to zero.
Given the Coulomb force such electrons will be submitted to an equal and opposite accelerations (for the action reaction Newton law).
I expect that the complessive radiation emission at great distances will be null.
So the system will not loose energy.
Why do you expect that there will be no electromagnetic radiation emitted in this case?
I could give many other similar examples where, even if there are electric charges in accelerated motion, the system will not transfere electromagnetic energy to space.
Is really Maxwell's theory wrong at an atomic scale ?
It is not, but you do have to be careful about analyzing atomic-scale interactions using classical mechanics - and if you're thinking about electrons as small charged objects moving under the influence of the Lorentz force and Newton's laws you're using classical mechanics. There's no problem with applying Maxwell's theory in that situation; the problem is that at the atomic scale electrons do not behave like small charged objects moving under the influence of the Lorentz force. Instead, you have to use quantum mechanics, which gives an entirely different picture of the electron and its interaction with the electromagnetic field and in which the classical notion of "accelerated charged particle" is absent.

Anonimo
Let us suppose I do not know quantum mechanics and I only know Maxwell theory. So from a phylosophical point of view I do not see problems in considering electrons as particles.
I ask myself: is it possibile to conceive motion configurations such that the electromagnetic emissions at great distances is zero ?
This is my phylosophical position.
I made calculations and I can assert that the cancellation of the total emission is nearly perfect.The reason is quite easy to understand: it is well known that the emitted electromagnetic field depends on the acceleration vectors; in our system the two acceleration vectors are opposite and equal in modulus.
So when I add the two electric vectors in every point at great distance from the system the result is with great and good approximation zero.
In an another way: the two emissions interfere each other in a destructive way.
In another way: the electric dipolus moment is Zero.
Now it is possibile that in a secondary approximation I have neglected something but what is certa in is that the electromagnetic emission will be enormously attenuated.
From an esperimental point of view (by analogy): take Two speakers and drive them in phase opposition.
The speakers are near each other.
The re will be no sound.

Mentor
2021 Award
it is well known that the emitted electromagnetic field depends on the acceleration vectors; in our system the two acceleration vectors are opposite and equal in modulus.
So when I add the two electric vectors in every point at great distance from the system the result is with great and good approximation zero.
That is just hand waving, not a calculation.

If you were to actually calculate it then by Poynting’s theorem the result would be that energy is conserved. The energy in the field would decrease as work is done on the charges. Your hand waving “phylosophy” doesn’t negate Poynting’s theorem.

Gold Member
Now it is possibile that in a secondary approximation I have neglected something but what is certa in is that the electromagnetic emission will be enormously attenuated.
While there is some kind of "attenuation" you qualitatively talk about, it turns out that the small radiation that remains is precisely what is measured. This small radiation is proportional to the square of the distance between the charged particles. The distance is indeed small, but it is not zero. If it was exactly zero (which it is not), then your qualitative argument that there should be no radiation would be correct.

• Dale
Anonimo
Dear Demistyfier,
Now, answer me please the following problem (which is only a little different respect to the previous).
Take two electric charges on a straight line.
The charges have opposite sign.
The distance between the charges is fixed and the charges move with arbitrary acceleration along the straight line.
Are you sure this system radiates at great distances ?
If it is so I can deduce that an arbitrary body with no electric charge, if accelerates, radiates.
Quite surprising and...impossible!
So if you consider little distances from the system surely an electromagnetic field exists but at great distances there will be no appreciable radiation.

Anonimo
I mean equal and opposite charges.

Mentor
2021 Award
Now, answer me please the following problem (which is only a little different respect to the previous).
You are still guaranteed, by Poynting's theorem, that energy is conserved. Any energy that goes into the field increases the amount of work required for accelerating the charges. Again, I recommend actual computations over hand-waving assertions.

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Gold Member
The distance between the charges is fixed and the charges move with arbitrary acceleration along the straight line. Are you sure this system radiates at great distances ?
It doesn't radiate, because in this case the charge is zero and all electric multipole moments are time-independent.

Anonimo
Thanks Dale.
I agree with you: energy theorem is always true.
And I do not see problems as regards this aspect: if I make work on the system what I expect is an increase of the near field energy.
As regards the calculations: I only made just estimations.
I must be clear: what I claim is that the emission is much smaller if compared to the emission of one single charge submitted to the same acceleration.
So what emerges is precisely this: Maxwell's theory allows to think possible an attenuation (although as you say not a perfect annullation) of the emitted radiation.
This is for me the crucial point.

Anonimo
Dear Demystifier,
The reason is that between the two examples I have shown there is no substantial difference: if I change the sign of the charge and the sign of the acceleration vector the situation is the same.
Please see Landau, Theory of fields (vol.2), Lenard-Wiechert E, B fields.
Thanks again.

Mentor
2021 Award
Maxwell's theory allows to think possible an attenuation (although as you say not a perfect annullation) of the emitted radiation.
Of course, yes. Many different configurations achieve this, e.g. a Faraday cage.

Your original post was very different than this new claim, I have no objection to this new one.

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Mentor
Your original post was very different than this new claim, I have no objection to this new one.

But the new one has nothing to do with the title question of this thread. Discussion of things like Faraday cages belongs in a separate thread on that different topic. Here the topic is the OP's mistaken understanding of Maxwell's equations and the stability of matter.

@Anonimo:

You opened a second thread on this same topic, here:

https://www.physicsforums.com/threa...lly-maxwell-wrong-part-2.954493/#post-6049931

You should not open a new thread on a topic that already has an open thread. Further, as you will see from my final post in that thread, you need to do some work on your background knowledge on this topic before discussing it further. The references I posted in the other thread would be a good start. This thread is therefore closed, as well as the second one (as I have posted there).

• Dale