Compton and Thompson scattering. Quantum theory
These two experiments actually describe the interaction of charged particles with photons, that is, with electromagnetic radiation.
Arthur Compton in 1923, in experiments with X-rays, discovered that when X-rays act on a free electron, the following phenomena occur. The experimental setup is quite simple. Figure 1.
X-ray radiation hits a crystal, in which electrons are bound rather weakly (practically they can be considered free) and is scattered by these electrons. The scattered photon hits the detector, where it is recorded in the form of altered radiation.
The energy of the incident photon is divided into parts. One part is emitted by the electron in the form of a photon of lower energy, and the second part of the energy is spent on the kinetic energy of the electron. The secondary photon is emitted not strictly in the direction of the incident photon, although this is possible, but at a certain angle to this direction. Approximately as shown in Figure 2, taken from Wikipedia.
Figures of the same nature are illustrated in other articles on Compton scattering. An incident photon and a scattered photon are plotted at the same frequency. And this is not in vain. After all, there is also the inverse Compton effect, in which the scattered photon has more energy than the energy of the incident photon. It depends on the quality of the electron on which the photon is scattered. In addition, scattering also depends on the frequency of the incident electromagnetic field. If the incident field is high-energy (gamma and X-ray radiation), then its scattering is accompanied by a change in its energy. This is Compton scatter .
Less energetic radiation (visible, radio and others) is scattered without changing the energy and scattered not only mainly forward, but generally scattered in a circular manner. This is Thompson scatter .
The total cross section for Compton scattering on a free electron is described by the Klein - Nishina formula:
The total Thompson scattering cross section is described by the formula:
On the other hand, it is clearly seen from the second formula that the cross section for scattering by a proton is negligible compared to the cross section for scattering by an electron (inversely proportional to the square of the mass). Why so, you can explain only by the fact that the electron repels the photon and its effective cross section is greater than that of the proton, which, if it does not attract the photon, is at least indifferent to it.
As we can see from a mathematical point of view, if we do not take into account the opposite effect, everything is fine with the Compton effect. But from a physical point of view, the explanation is somehow incomplete. Both general questions and questions concerning the specific experience of Compton arise.
What can be shown to an outside observer present during the experiment? Take a detector that captures two frequencies with wavelengths? λ and λ′. First place the detector in the path of the beam to the crystal. Enable installation. Switch on the detector at wavelength λ. The arrow of the device should crawl to the right. If you adjust the beam power, you can wiggle the beam power and show that the detector responds to this wobble. The greater the power of the photon flux, the greater the deflection of the arrow and vice versa. Next, the detector should be switched to the wavelength λ′ and make sure that the needle of the device is at zero, which will be if the frequency with λ is sufficiently monochromatic. This means that the beam contains no frequency corresponding to the wavelength λ′.
Let's move the detector to its rightful place. Let's turn on the installation again. We will see that the detector shows the presence of both radiation. In addition, it can be found by shifting the detector along the angle Θ that the radiation intensity is peaking. Placing various crystals into the setup, we see that with an increase in the atomic number of the scatterer, the intensity of the unbiased line increases, the intensity of the shifted line decreases. Probably, researchers will find many other regularities in these experiments and even describe them mathematically. But until the experimenter takes a quantum standpoint, he will never tell you why this is so.
“ The Compton effect is one of the proofs of the validity of the wave-particle duality of microparticles and confirms the existence of photons .”
Dear reader, do you find here the dualism of the electron? From the fact that the electron has changed its speed and, possibly, the direction of movement, are you ready to conclude that the electron is a particle and a wave at the same time? And what about the Compton effect? On the scattering of photons. If there are no photons, then there is nothing to scatter.
After turning on the installation, what flies through the collimators, filters and hits the crystal? Electromagnetic wave - there will be an answer. Will this wave be continuous or will it be pieces of an electromagnetic wave? There is no third. Let us assume that the wave in the ray is continuous. It hits an electron in the crystal and begins to interact with it, which consists in swinging the electron. How the swinging occurs, in the longitudinal, transverse or some other intermediate direction, we do not know for sure. The important thing is that as a result of this wobbling, the electron emits a new photon with a different wavelength. How did this new photon come about? Was the incident photon somehow deformed into a new one, or the incident photon disappeared and a new one appeared?
This experience, like many others, has everything but time. We have now assumed that the photon is infinite, naturally within the limits of experience. The question is, how long will it interact with the electron? After the contact of the first wave of a photon and an electron, the latter begins, according to the now accepted opinion, to oscillate. How many waves will participate in the oscillatory process? After all, someday, never, but the photon and the electron will have to scatter. What happens next? If the electron only deforms the photon, then we get some kind of combined photon. One part of it will be with one wavelength, and the other part will have a different wavelength. This does not seem to be observed in nature.
If the incident photon is absorbed, then only part of it is absorbed, for the same reason the divergence of the paths of photons and electrons. And where does the rest, which is also endless, go? We have to find it at the angle Θ = 0. Although it is not known how to distinguish a truncated photon from the original one.
It should be remembered that a photon, especially an elementary photon, is a snake that cannot be lengthened or shortened by deforming it in the longitudinal direction. (see articles: "Quantum, what does it consist of" , "Quantum of energy, how it works and how it moves" and "Photon" ).
In general, the assumption about the infinity of the photon is somehow unsatisfactory, so we will try to represent the photon in the form of a finite structure, for example, in the form of a soliton (a finite number of oscillations). But here the question immediately arises: how many oscillations should there be in this soliton? What amplitude are they and more? In general, if you operate with the generally accepted model of the photon, then certain difficulties arise.
Already at the very beginning of the description of the experiment, it is assumed that Compton is scattered by free electrons. More precisely, on almost free electrons, that is, electrons that are bound in an atom, but with a very weak bond. Let's look at the table borrowed from the independent researcher Grishaev. These are typical spectra when observing Compton scattering on various substances.
As you can see, scattering occurs both on metals and non-metals. P is primary radiation, and M is scattered (modified) radiation. From lithium to copper, the intensity of the scattered radiation decreases with increasing atomic number of the substance. And for silver there is no such radiation at all. The mirror should not distort the color gamut of the image; it simply reflects the light falling on it. But after all, silver is a good conductor of electric current, that is, it contains a lot of free electrons and, presumably, there are a lot of weakly bound electrons. That these electrons are poorly swayed by the primary radiation and therefore do not generate modified radiation? Why are these electrons worse than the electrons of copper or aluminum? Nothing. And in non-metals, free or weakly bound electrons, it is not at all easy to find. Apply voltage to silicon, the current will not be the same as in silver.
From this we can conclude that not only the electron, but also other elements of the atom is involved in the scattering process. Unfortunately, modern science does not have an acceptable model of the atom, and therefore some misunderstandings arise. Based on the model of the atom proposed in the article "Atom device" , then all misunderstandings will be removed.
In this case, the scattered radiation is represented as
The very first laser is a ruby ??cube. A green light emitting lamp is wrapped around the other four faces. That's the whole device. When the green lamp is lit, a red ray appears from the transparent mug of the face. How did green turn to red? One can, of course, say that the green photons rocked the electrons of the crystal and made them emit red photons. Exactly like Compton's. But somehow I can't believe it.
Why doesn't Thompson's wavelength change? That other frequencies do not swing electrons and, accordingly, do not emit other frequencies? Or swinging still goes on, but only those frequencies that are primary are emitted? So, why does the secondary radiation in this case have an almost circular scattering pattern?
And everything becomes clear if we admit that Compton, like a laser, has stimulated radiation, and these are already exchange photons of an atom. That is, without being in the composition of an atom, this secondary photon is simply nowhere to be taken by the electron.
In the case of Thompson scattering, an electron of an atom simply retransmits a photon, as in the case of reflection or propagation of light. Part of the incident radiation at Compton is also simply retransmitted, and even the photon that forces the exchange photon to be emitted is also retransmitted. And in order to understand why the Thompson scattering does not have a pronounced directionality, you should read the article "Dispersion in quantum representation" .