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Compton Scattering: a mechanical explanation
by Miles Mathis
First posted August 25, 2010 Abstract: I show the holes in the current interpretation and math of the Compton wavelength, and then fill them with proper math and mechanics. I show that the electron radius is determined by the photon radius, with simple math. I then show how this confirms my spin model. Finally, I show how inverse Compton scattering and pair production both prove the spin model.
I have already published papers on the Compton Effect and on the Electron Radius. Now, if we use this new information we can come to a new understanding of the Compton wavelength and Compton scattering. Quantum mechanics currently has no mechanical understanding of the cause of the Compton wavelength. This wavelength is 2.43 x 10^{12}m, and is calculated from this equation
λ' – λ = h(1 – cosθ)/m_{e}c
where h/m_{e}c is the Compton wavelength.
There are many very big problems here. The first is that this equation is derived assuming the electron is at rest (“initial momentum is zero”) to begin with, which cannot be true. The reason this assumption has never been overturned, despite its obvious falsity, is that physicists have no way of measuring the initial speed of the electron in orbit. They have too many unknowns here, which prevents them from solving in this way. They can only solve using a false assumption, and then prevent anyone from pointing out the falsity of it. Just look again at the form of the equation. It can be written as
λ' – λ = λ_{e}(1 – cosθ)
Since the term (1 – cosθ) has no dimensions, this equation is assuming that the wavelength of the electron and the wavelength of the photon can be compared directly, without any transform or other constant or variable. It is assuming the change in wavelength of a tiny particle is equivalent to the change in the wavelength of a much larger particle. This could be true only if the particles had the same energy or the same speed. Since they don't, we can't possibly assume this.
To say it another way, remember that the wavelengths on the left side of the equation belong to the Xrays before and after scattering. This being so, we should label the wavelength on the right side as belonging to the photons as well. If you subtract two wavelengths belonging to one particle, your difference cannot be applied to a different particle. But that is what we have here! No, the only way that equation could work is if the electron were a photon of the same energy.
Current theory admits this, in a way, by defining the Compton wavelength of a particle as “the wavelength of a photon whose energy is the same as the resting energy of the particle.” But that is just a dodge, since it is another way of saying the particle would have that wavelength if it were going c. But if the particle is not going c, it is not clear where the number comes from, or how it applies to the particle. We are told that the Compton wavelength is a quantum mechanical property of a particle, which explicitly assigns the wavelength to the particle (electron). But if the particle is not going c, then how is that assignment possible? We need some mechanics here, not just the words “quantum mechanical.” But this is typical of the standard model: whenever it conspicuously lacks mechanics, it substitutes the word for the physics. Quantum mechanics is not mechanical, it is mathematical. Quantum dynamics is not dynamic, it is mathematical. We get the word but not the substance of the word.
What the equation needs is a way to relate the wavelength of the electron to the wavelength of the photon. The only way to do that is with motions. We need to know how fast the electron is going and how fast it is spinning. Only then can we compare the energy of these spins. We need to compare energies, not measured wavelengths.
But how can we do this? It appears very difficult or impossible, which would explain why the standard model doesn't bother with it. However, using my new numbers, we can solve this one. The current Compton wavelength belongs to the photon, not the electron, as I said. The form of the equation alone tells us that. So to find the wavelength of the electron, we must compare the velocities of the two particles.
λ_{e} = λcv^{2}/c^{2}
λ_{e}/v^{2} = 2.7 x 10^{29}
In this way we see the two unknowns, and their relation. Only if v = c will the electron wavelength be the same as the Compton wavelength.
The current derivation begins by assuming that the electron is at rest and then writes equations that show the electron going c. The problem is put into high focus when the derivation lets the energy of the electron equal E=mc^{2} (see Wiki for example, where the full math is posted). That cannot be right, because in this problem we need a kinetic energy or a momentum, not a mass/energy equivalence. The electron is not being burned in this equation, turning it to pure energy, as in a matter/antimatter collision. The electron is simply colliding with a photon. So we can't and shouldn't use E=mc^{2}. That equation gives us the rest energy due to mass, not the kinetic energy. Again, the electron here isn't at rest, so it is not the rest energy that concerns us; it is the energy due to motion.
Also notice that due to the form of the equation, the current value for the Compton wavelength is telling us a change in the photon wavelength. So the electron wavelength in the last equation must be telling us the same thing. Not a wavelength of the electron, but a change in the wavelength of the electron in this scattering problem. So I cannot simply insert my value for the electron radius of 1.11 x 10^{17}m. Since that is the initial radius, it cannot be the change. That would imply no change. So we still have two unknowns, just like the standard model. If we are going to solve without fudging, we have to keep looking for a logical method.
We can return to the original wavelength of the Xray involved in the scattering, 7.09 x 10^{11}m. If the change in wavelength is 2.43 x 10^{12}, that's a 3.43% change on the photon. Since the electron is larger, we would expect it to change less. How much less? Well, since the photon is about 7.1 x 10^{11} smaller, [1.11 x 10^{17}/7.09 x 10^{11}/c^{2}] the electron will change that much less (which, notice, is about G). But that is assuming the particles have the same speed, and we know they don't. So we also have to multiply by c^{2}/v^{2}. So the change on the electron would be about .0343Gc^{2}/v^{2} .
If the electron is travelling c, then the change would be 2.435 x 10^{12}, which means the local change in wavelength would be 2.7 x 10^{29}m. But we must remember that is the local wavelength change on the electron, and the electron's velocity stretches out its wavelength due to spin just like the photon's does. So if we multiply by c^{2}, we are back up to 2.435 x 10^{12}. We would measure a wavelength change of that size. This math confirms the current number, but does not confirm the current interpretation. As I showed, this number depends on the electron travelling at c, and it cannot do so. This number is also given to the electron wavelength, but it is actually a change in the electron wavelength. It does not tell us the wavelength of the electron or the electron's effective radius or anything like that. It is only a change in wavelength.
Two things are extremely odd here. One, I have used a variant math to get the same number as the standard model, but my method uses my own values for the electron and photon radius, and it solves by scaling between them. The standard model got this number without knowing either value. It doesn't even believe in a photon radius. You will say that the standard model got this answer by pure brilliance, but I have shown they got it by simply conflating the electron and photon wavelength. They made multiple mistakes in theory and postulate, and were forced to fudge their equations each time to push them back toward the answer they wanted. Two, it is very odd that the change in wavelength for the electron and photon is the same, despite the fact that the electron is so much larger (and going the same speed, according to the current equations). As I said, at the same speed, we would expect the electron to change very much less. Well, it is changing much less, in fact, although the numbers are the same. We can see this by looking at percentages. I have shown that the photon is changing by about 3%. How much is the electron changing? I have shown that the local wavelength of the electron is just its spin radius, and that this number is 1.11 x 10^{17}m. But if it is moving at c, then its measured wavelength is c^{2} times that, or 1m. Yes, the electron moving c would have a measured wavelength of 1m. Which means its change in wavelength in this problem is .000000004%. The electron changes very little and the photon changes a lot, but they both happen to change 2.43 x 10^{12}.
Because the change in wavelength is the same for both photon and electron, the current equation appears to work. The standard model appears to be vindicated. But the standard model is not vindicated, it has simply covered over one more problem with dishonest math. Because all the numbers have been misinterpreted, no one bothers to ask the questions I am asking.
The first question to ask is, “Why is the number the same for both? As stated, it seems like a very large coincidence. Is it?” No, it isn't. This “coincidence” is determined by the fact that 1/c^{2} is both the radius of the electron and the scale between Xray wavelength and Xray radius. In other words, we can multiply the Xray wavelength by the electron radius to get the Xray radius.
r_{x} = r_{e}λ_{x}
Which means we can write the electron radius in terms of the photon entirely.
r_{e} = r_{x}/λ_{x}
And this applies to all photons, not just the Xray.
r_{e} = r_{γ}/λ_{γ}
This is the “coincidence” that makes their math work. This is the coincidence that makes the change in electron wavelength the same as the change in photon wavelength. Their pushed math works only because of that invisible equality beneath it.
But all that is moot, since the electron cannot go c. IF the electron were going c, its measured wavelength would be about 1m. IF the electron were going c, the change in wavelength would be the Compton wavelength. But since the electron is not going c, the Compton wavelength can only apply to the change in the Xray wavelength. It cannot apply to the electron.
So let us insert a different number than c, to get a feel for this equation. In another recent paper, I showed that the electron was travelling .0057c, which is just the fourth root of 2G. That scales particle speed to size. At this speed, the change in the electron would be 7.5 x 10^{8}. The change in the electron is larger because it is going slower. It has less energy relative to the incoming photon.
This shows us more clearly that the Compton wavelength can apply to the electron only if the electron is travelling c. And even then, the Compton wavelength would only be a change in the electron wavelength of 1m, a very small change.
In closing, let us return to the equation r_{e} = r_{γ}/λ_{γ}. That is very interesting, because it may show not just an equality, but a determination. The radius of the electron appears to be determined by the radius of the photon. And if we study the equation even more closely, we find that the radius of the electron is determined by the ratio of the local wavelength of the photon to its moving wavelength. To say it another way, the electron radius is the photon radius measured by itself over the photon radius measured by us. In this way, the electron radius can be defined as a relativity transform! I will have more to say about this later.
We can also use these new findings to confirm the spin model once more. Without spin, none of my equations here would have worked. The transform 1/c^{2} works only with spin, since it is the difference between spin speed and linear speed.
Another clear confirmation of this is inverse Compton scattering, in which the photon gains energy and loses wavelength. With current math, this is very difficult to explain, but with spin it is very easy to explain. Without real spin, any real collision must be a transfer of energy due to linear motion alone. The current model accepts that Compton scattering is caused by a real collision, since the scattering is used as proof of the particle model of light. But if this is the case, then a transfer of energy cannot be negative. If we have both Compton scattering and inverse Compton scattering, then one or the other must be a negative energy transfer. Look at it this way: if you collide a baseball with a bowling ball, the baseball must either lose energy or gain it. It cannot do both. Depending on the type of energy transfer you propose, it either must always gain or always lose. The velocities will not matter. If it gains, greater velocity will make it gain more. If it loses, greater velocity will make it lose more. But nothing will make it switch. This is the problem quantum mechanics faces with inverse Compton scattering. But it is easily and immediately solved with real spin, since spin can be either CW or CCW. We then propose that with Compton scattering, the spins are opposite, and cancel some amount. With inverse Compton scattering, the spins stack, increasing the outer spin energy. To change Compton scattering into inverse Compton scattering, we just turn the photons upside down.
Pair production also confirms the spin model. The standard model has no mechanical explanation of pair production, but the spin model once again provides a simple explanation. The current model assumes that one photon is producing the pair, but that is just a bald assumption, with no data to back it up. A better assumption is that two photons produce the pair by colliding with a neutron and proton, or a proton and antiproton. Since I have shown that a reversed outer spin can be the only difference between neutron and proton, two photons colliding with a neutron and proton would produce a pair of opposites.
As simple proof of this, we only have to look to the known requirement that the incoming photons have twice the energy of an electron at rest. This is of utmost importance, though it is always overlooked. It is important because the ejected pair are not at rest. Each of the two ejected particles has a kinetic energy, which means that the total energy out is very much more than twice the rest energy of two electrons. How much more? Twice more, as it turns out, requiring that we have four electron rest masses ejected. We have the pair, which is two rest masses, and then the kinetic energies, which is two more. So we must have two photons creating the pair, not one. One could only create a pair at rest.
Take a neutron and proton next to one another, and send two similar highenergy photons into the gap. If they both gain energy (enough) from the collision, and one is given a left outer spin and the other a right, they will become electron and positron. I have already shown that adding spins is enough to turn a highenergy photon into an electron (and even given you the equation to calculate it), so this is no problem. The only problem is that the two photons cannot both gain energy in this way. One of them would have to be upsidedown to start with. This is difficult to explain, because the ambient charge field should make the photons coherent, even as regards up/down. But it may be explained by the charge field in the immediate vicinity of the nucleus. The proton is emitting a charge field and the neutron is not, so it may be that this field divergence flips one of the photons at the last moment, providing the proper alignment.
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