return to homepage Abstract: I develop a firm number for the local wavelength of the photon, show how it is created by the spin of the photon, and how this local spin is stretched by the linear motion into the wavelength we see and measure. I prove that the transform from the local wavelength to the measured wavelength is c^{2}, showing why this term is found in the famous equation E = mc^{2}. Before we get into that, I would like to give credit to Descartes, who knew that color was caused by photon spins in the early 1600's. I believe he made this proposal in Dioptrique, around 1638. This was before Newton's corpuscles, so the fact that Descartes was able to intuit an answer so close to correct at that time is extremely laudable. If I remember correctly, Descartes got the energies reversed from red to blue, but connecting color to spin was a masterstroke in itself. We will see that color is due to the size of the photon, which is more a matter of radius than spin velocity, but the two are inextricably linked. It is spin collisions that cause the spin radius, so we would obviously have no spin radius without spin. Since I recently published strong evidence against Descartes' theory of reflection by the back surface of raindrops to create rainbows, I felt compelled to add this paragraph to this paper, showing he was also often right against very long odds. From a previous paper, we know that the radius of the Bphoton is G times less than the radius of the proton. This gives us a photon radius of 2.74 x 10^{24}m. The zspin is 8 times the radius, so we should find a basic wavelength of 2.2 x 10^{23}m. Obviously, we don’t find photons with a wavelength that small. Why? Simply because the wavelength we measure has been stretched out by the velocity of the photon. The photon would be measured to have a wavelength of 2.2 x 10^{23}m only if it were at rest. You will say, “Even if we accept that the photon is spinning, how can the zspin be stretched? The spin would give us a spin radius, which is just a length. A length cannot be stretched by motion, unless you are proposing some kind of relativity here.” I am not proposing relativity as the solution here. The answer to your question is that a spin is not just a radius, and is not just a length. A spin is a motion: a motion that takes time. Even if the photon were spinning at velocity c, one rotation must take some real time. We know that the linear velocity of light is not infinite, so we must assume the speed of spin is also not infinite. If it is not infinite, it must take time. If it takes time, then it will be stretched by the linear motion. While the surface of the photon is spinning, the photon as a whole is moving some linear distance x. So how much does the velocity stretch out the wavelength? We can discover that most easily by using this simple equation: E = mc^{2} = hc/λ λ = h/mc Let us take an infrared photon, as our first candidate. The mass equivalence of the infrared photon is 2.77 x 10^{37 }kg, so we just solve: λ = h/mc = 8 x 10^{6}mIf we compare that to the wavelength at rest, we find the wavelength has been stretched out by a factor of about 3.63 x 10^{17}. Since that is very nearly c^{2}, we assume that the transform is in fact c^{2}, and that the difference is a difference between the size of the Bphoton and of the infrared photon. Remember that we developed the atrest wavelength from the Bphoton and the moving wavelength from the infrared photon. Our assumption is borne out by the numbers, since if we divide 8x 10^{6} m by c^{2}, we get 8.9 x 10^{23} m, which is almost exactly 4 times our Bphoton wavelength. We may assume that the infrared photon is about 4 times larger than our Bphoton. But why c^{2}? Let’s look at the mechanics, to find out. As usual, nothing esoteric is going on here, so we can analyze just as if we were analyzing pool balls. If the photon is spinning while it is going c, and the radius is being stretched by a factor of c^{2}, then that must mean that the photon is spinning at a velocity of 1/c. It takes one full rotation to create a single wave, since the rotation is physically creating the wave. And we want to find one full rotation while the photon is moving 8 x 10^{6}m. That will give us the wave we measure. At speed c, the photon goes that far in 2.67 x 10^{14} seconds. So the photon has spun once in that time. By my new kinematic circular motion equations, the circumference is 4 times the diameter, so a point on the surface of the photon travels 8.8 x 10^{23} m. If we divide the circumference by the time, we get 1/c. Which begs another question: Why would the spin be the inverse of the linear velocity? Because both are dependent upon the same fundamental factor: size. The smaller the quantum is, the faster it goes. The photon goes c precisely because it is so small. It can maximize its speed because it can dodge most other quantum traffic. But this size also determines its spin rate. Notice that we have found it to be spinning extremely fast: 1 cycle every 2.67 x 10^{14} seconds, which is equivalent to 3.7 x 10^{13} cycles each second. That is extremely fast, from our point of view. But, as I have just shown, from the photon’s point of view the surface is moving incredibly slowly: 3 x 10^{9} m/s. That is because one cycle is such a tiny distance. With such a tiny circumference, the photon can move with a tangential velocity of 1/c, and still achieve an astonishing local frequency. That being true, it still doesn’t provide a mechanical link between c and 1/c. I have shown that the photon can move very slowly, in its own realm, and still create the necessary wavelengths and frequencies we see, but “very slowly” and 1/c are two different things. Why 1/c, precisely? Because, as we know, velocity is a relative measurement, and we are measuring the spin velocity relative to the linear velocity. I will show this with a diagram:
You may say that c is not really a “very large number”, but relative to the radius of the photon, it is. We have 31 orders of magnitude between the two, which is a very large number by anyone’s reckoning. So, as the linear velocity gets larger, the tangential velocity gets smaller relative to it. And the difference between the two is so large that the relationship becomes nearly a perfect inverse relationship. In other words, c^{2} is so large that the tangential velocity can be estimated very successfully with just the transform 1/c. This also happens to be the reason we can ditch the ½ in the equation E=mc^{2}. The two equations E=mc^{2} and E=½mv^{2} are supposed to be analogous, at least when the first one is applied to a photon and the second one is applied to an electron, say. Nearly all the energy of a photon is kinetic energy, so the equation E=mc^{2} applied to a photon is an expression of kinetic energy, with m as the mass equivalence of the photon. But the equation has no ½. The reason it doesn’t is that because c^{2} is so large, ½ becomes negligible. One half of almost nothing is still almost nothing, so we can ignore it as a factor. I explain this in more detail in another paper.
So, I have explained the motion of the photon in a simple manner, providing us with not only a spin but a rate of spin. I have developed actual numbers, first for the radius and mass of the photon, and now for the rate of spin and the tangential velocity and orbital velocity. I have shown how this spin rate creates the visible or measurable wavelength of light. The radius of the photon’s spin creates the actual wavelength, and then this length is stretched out by the linear motion, giving us the measured wavelength of the light. But now we must move on to ask why and how photons express different wavelengths. Electromagnetic radiation, in the form of photons, comes in a wide range of wavelengths, as we know. How is this achieved? It is achieved by a wide variation of stacked spins. As I began to show in a previous paper, it turns out that photons can maintain a linear velocity very near c over a wide range of sizes. The photon does not reach a size limit that causes slowing until it approaches the spin radius just beneath the electron. At that limit, the largest photons begin absorbing the smallest photons, and the mass increase snowballs. This turns the nearly massless photon into the smallmass electron. The most common photons appear at the size range of 1821^{3} less than the proton mass and size. This is where we find the infrared photons, as I showed previously. But the small mass of the photon allows it to stack spins over a wide range of radii. In this, it is unlike the electron or proton. The proton cannot add extra spins above the zspin without creating instability. This is why “mesons” over the baryon size are not stable. The extra spins begin interfering with the energy of the inner spins. But with the photon this appears not to be the case. Extra spin levels do not cause appreciable slowing, nor do they cause appreciable instability. We may theorize that smaller photons would be more stable, but the difference in small photons and large ones is not easily measured from our level. What this means, specifically, is that if we give the infrared photon a zspin as its outer spin, we can find a smaller photon whose outer spin is the yspin. We can also find a larger photon with another axial or xspin on top of the infrared’s zspin. In this way, we find not only stacked spins, we find stacked levels. In other words, we find spins of a_{1}, x_{1}, y_{1}, z_{1} and a_{2}, x_{2}, y_{2}, z_{2} and a_{3}, x_{3}, y_{3}, z_{3} and so on. By this analysis, a_{2} has twice the spin radius of z_{1}. In fact, each spin has twice the radius of the spin under it. This means that photons do not come in a continuous spectrum. No, they come in stepped levels, each level double the one under it. For example, we found a wavelength of 8 x 10^{6} m for our infrared photon. You will remember that number comes from (c^{2}) 8.8 x 10^{23} m. If we want the next photon larger than that, we double the spin radius to 1.76 x 10^{22} m and multiply by c^{2}, which gives us 1.6 x 10^{5} m. There can’t be any wavelength between 8 x 10^{6} m and 1.6 x 10^{5} m. If we measure light with an average wavelength in between those numbers, we must have a mixture of photons. A single photon cannot be found with a wavelength that is in between those numbers. It must have one wavelength or the other. This explains many optical effects, including rainbows. Rainbows do not have a continuous spectrum, and neither do prisms. We see bands of certain colors. Photons only come in certain colors, and they do not come in the colors in between. In between colors have to be created by mixing photons. It is known that light is quantized, but no one before me has proposed an explanation of this sort for it. Photons have always been considered point particles, with no radius and no mass. For this reason, it was thought that spin could not be a possible cause of quantization. Therefore we have gotten many complex mathematical explanations of quantization, with no mechanical explanation at all. I have just shown you the simple mechanical cause of quantization.
