Forgive me for loving the history of physics. I guess it attracts me because it must meet two premises. One is not to lie scientifically and another is to present a kind of verbal theorems that, read carefully, raise concerns.
What we have discussed before has joined in my head with some simple concepts of classical electromagnetism. The following emerged.
1. When in the classroom one asks how it is done so as not to confuse a thing with another, they tell us to look at the size of the analyzed system. When there is polarization, you can always find a Gaussian surface, smaller that the system, within which there is a non-zero net charge. The other condition is that there is always a Gaussian surface, which does not exceed the size of the system, within which there are pairs of symmetric collinear vectors, that give a contribution equal to zero in terms of net charge. I guess the the same conditions must be met when a wave propagates in a vacuum.
2. In the classrooms one also learns that if in ## \vec {D} = \vec {P} + \varepsilon_o \
\vec {E} ## we suppress ## \vec {P} ## we ran out of ## \vec {D} ## and the only thing it exists, physically and mathematically, it is ## \vec {E} ##. So, if ## \vec {D} ## is essential for the existence of the electromagnetic wave, ## \vec {P} ## is also essential. This detail does not raise any doubts in the case of a wave that propagates in a material medium, because they have explained to me how it works ## \vec {P} ## in that medium. In the case of the vacuum it causes me concern, because the obligation to have ## \vec {P} ## in order to have ## \vec {D} ## and, in consequence, to have a displacement wave in a vacuum, has never been analyzed in the classrooms where I have been.
3. When I try to analyze it, I understand that my analysis must take into account that a wave in a vacuum can be constituted by a single cycle. In this case, the measurement of the entire system in the spread direction is equal to wavelength. And within a Gaussian surface that in the spread direction does not exceed one wavelength, I must have pairs of vectors ## \vec {P} ## collinear and symmetric. If I take that seriously, vectors ## \vec {P} ## what I'm looking for can not be transverse, because vectorially the only thing on cross direction is ## \vec {E} ##, which corresponds to the part ## \varepsilon_o \ \vec {E} ## and not to ## \vec {P} ##. That forces me to look for a longitudinal ## \vec {P} ## that fulfills a wave function. All of that brings me to a rotary ## \vec {D} ## vector which, in the simplest case, has the following form.
\vec{D} = \hat{D} \left[\ \vec{x} \ \cos{\left(\omega t-kx \right)} +\vec{y} \ \sin{\left(\omega t-kx \right)} \ \right] that by the identity of De Moivre corresponds to the following.
\displaystyle \vec{D} = \hat{D} \ e^{i \left(\omega t -kx \right)} I have symbolized ## \vec {x} ## and ## \vec {y} ## to the unit vectors of the respective axes, in the orthogonal Cartesian coordinate system. And I have symbolized ## \hat {D} ## to the peak value of the displacement.
4. This rotative displacement has a non-zero divergence, that is, implies a ##finite \ and \ continuous ## charge density where the wave is present. Suppose that the implication of a finite charge density in vacuum did not surprise me, because to be surprised one must to know more than what I learned in the classroom. But all the people who have gone through a classroom knows what continuity means. There must be charge in the environment of each point, that is, in each infinitesimal volume. That directly means that I can't think of charged particles, neither real nor virtual, because they do not work to give a continuous charge density. I have to think directly about a state of the vacuum that occurs when a wave propagates and that, acquiring such state, the vacuum physically participates in the propagation.
5. I know I can cheat the story and add knowledge that the physics acquired much after Maxwell, saying that classical functions describe averages in probabilistic terms. But that does not convince me, for the following reason. Developing in classical terms the consequences of rotating displacement, I get a finite wavefront, which is the cap of a cylinder whose length is equal to one wavelength. The Maxwellian fields they exist only within that finite cylinder. Outside the cylinder there is no fields, nor is there any Poynting vector, nor energy, nor anything that we can specify. Honestly, all that worries me, because I do not see that it leads to an alley without departure. As far as I can see, development can continue. But if it continues, the only way to accept it is to be consistent with today's knowledge. To demand that coherence is not to cheat the story, for the following. If the consequences are really consistent with today's knowledge, then we will have something else to try to understand the mental landscape of scientists from that time, because we know that nobody publishes everything they think or all calculations made to analyze an issue. In case of inconsistency with today's knowledge we all could, including high school students, understand very simply the need to replace classical electrodynamics with a quantum field theory. When something is didactically useful, it interests me a lot, provided it is lawful. In the case of displacementt wave, I do not know what one should think.
