# Mass and momentum of a photon

## Contents

## General

The photon has a mass from the Conservation of mass, even if it has no rest mass according to Einstein. Accordingly, according to the equivalence of space and time, mass and momentum are equivalent. So a photon also has an impulse. This corresponds to the elemental mass multiplied by the speed of light. This applies to the elementary frequency. Otherwise one can calculate the mass of a photon with m = h * f / c². Learning without frustration and the University of Ulm have calculated this based on this equation, which I have always fought for, and arrive at a value for an average photon of:

m (photon) = 10 ^ -36 kg

The momentum is correspondingly p = h * f / c = m * c:

p (photon) = 10 ^ -28 kg * m / sec

Since virtual masses, etc ... or not accepting any impulse would be wrong. This is also shown by quantum theory. The Heisenberg uncertainty principle assumes that even if a test object is bombarded with a photon, this makes the measurement blurred. This is due to the fact that the photon with its mass and its momentum shifts the test object, so that the location can no longer be precisely determined.

## Mass and rest mass of the photon

Some scientists say that the photon has no mass because it has no rest mass. The rest mass is just a theoretical term introduced by Einstein, which is defined as m (0) = m * relativistic root (= root of (1 - v² / c²)). The rest mass is the art term. According to the theory, it describes the value that a particle assumes if it were absolutely at rest. Since, according to the theory of relativity, nothing occurs in absolute calm, it is only of a theoretical nature. But as I said: The rest mass is the art term, the actual mass depends on the formula mentioned above. According to Einstein, the rest mass of the photon is 0. Nevertheless, the photon has a true mass that is greater than 0. Others assume this too, but call it relativistic or kinetic mass. This reminds a bit of Lorentz, one of the founders of the theory of relativity, who speaks of the longitudinal and transverse mass of the electron. Today we are safe and only talk about the mass of the electron.

The best counter-evidence against equating mass and rest mass is again a thought experiment. According to the previous view, if you wanted to accelerate an object, for example a person weighing 80 kg, with mass to the speed of light, you would need an infinite amount of energy. If we now assume in a thought experiment that one has infinite energy and accelerates a person from 80kg to the speed of light, then he weighs an infinite amount of kg and not just the 80kg rest mass. Therefore the concept of rest mass can hardly correspond to the concept of gravity or mass.

## Mass of the photon not defined?

It used to be said that the photon has an undefined mass, since it is divided by zero according to the connection between mass and rest mass. m = m (0) / relativistic root. Since both tends towards zero for the photon or becomes zero, it was assumed that the mass of the photon was not defined. But that you can define the Division by Zero, you can read here under Division by Zero. One could also say, if one disregards this, that the limit value tends towards zero, but the term becomes 1, which corresponds to the elemental mass. This is due to the fact that both the rest mass and the relativistic root tend towards zero and this evenly. It is a mystery to me why this question is ignored in recent literature and why it is claimed per se that the photon has no mass. The answer to this question is not that simple.

## Photon and Gravitation

According to Einstein, the photon, i.e. sun rays, is subject to gravity. He demonstrated the deflection of the sun's rays during a solar eclipse, in which the moon moves in front of the sun. The rays that came by anyway, the sun then forms a wreath in the sky, were deflected exactly as he had calculated. I also take that as an indication that the photon has a mass. Massless particles are not subject to gravity, even if Einstein defines gravity rather as the curvature of space-time and considers it detached from mass. Nevertheless, as is the case with gravity, every force, according to F = m * a, has a mass that is accelerated or, in this case, deflected. Without this mass, the force would also be 0 and therefore not exist. Or, assuming the force, with a mass of zero the acceleration would be infinite, which of course is not the case. With Einstein's exciting new equations, one should keep in mind that there are well-formed definitions that cannot be simply negated.

## The proof

Yes, for the derivation, not everyone believes my new mathematics like under Division by Zero and Complex numbers, I am forced to prove it with conventional mathematical means. As shown in the article relativistic root, this can easily be replaced by the velocity v (red) / c, which sweeps over several places. This is then again equal to the time flow T (0) / T. So we get: m (photon) = m (0) / v (red) / c or m (photon) = m (0) / ( T (0) / T * c). According to the unit system, I now leave out c as one, mathematically nothing changes. Then we look at the term m (photon) = m (0) / time flow. Depending on v. For limes (v -> c) the rest mass m (0) decreases until it becomes zero at c according to previous mathematics. Likewise, the time flow decreases, things move slower and slower until it becomes zero. Both functions are strictly monotonically decreasing for limes (v-> c). For faster than light speed, i.e. limes (v -> infinite), the function goes into the imaginary range, as Einstein and Minkowski prophesy. Here it would have to either continue to fall or rise, but not contain a turning point in order to be able to apply L'Hospital's theorem. According to this, the limit value 0/0 is defined when the limit value for a derivative, be it the first or the second, is defined. For faster than light the rest mass m (0) becomes negative, the time flow imaginary. But none of this really helps, today I had the brilliant idea. And in retrospect, it turns out that the question was never whether a photon has no mass or does. Einstein's question was, why does the photon have a mass according to E = mc² if 0 is divided by 0? I thank the anonymous whistleblower here who referred to L'Hospital's sentence, and my old school friend Schaper, you know him from the inflated crowd and the Division by Zero who put me on this trail.

Ultimately, we square the Einstein mass-rest mass relationship. Then we have m² = m (0) ² / (1 - v² / c²), because the root falls away below. Now we derive twice for v, because according to L'Hospital the limit value of 0/0 is given when the derivatives have a limit value. Below we can see that we get - 2 / c². So definitely a real number. For the expression above we then need some physics. The Einstein energy-momentum equation has to be simplified first. Then it is called m (0) ² * c ^ 4 + c² * p² = m² * c ^ 4. That means the rest mass is given by the term m (0) = sqrt (m² - p² / c²).

Since we have squared in our proof, the numerator is m² - p² / c². We had already proven that the denominator tends towards a negative number in the second derivative. We now multiply this expression for the rest mass by c ^ 4. (Don't worry, the c ^ 4 do not change the basic result in the denominator. Then the numerator is m² * c ^ 4- p² * c². According to E = mc² this is then E² - c²p². So now we turn to the Expression 2 * L'Hospital an. In this sense, E is equal to mc² and therefore does not depend on v, so it is omitted as a constant. -C² * p², however, depends on v. If you derive twice from v: -2 * Here I would actually have broken off because I have m again and according to the previous logic that could also be zero, so that the expression according to L'Hospital would also tend towards 0. Ultimately, however, nobody prevents me, including the derived ones To use formulas physics. In the first proof, I then applied the energy-momentum equation in a complicated way and thus proved it. I could also simplify it with E = mc² and I do that here now. Ultimately, this expression corresponds to - 2 E² / c² But since a photon definitely has an energy according to E = h * f, a negat also comes here iver real value out. Thus the expression 0/0 for the squared mass - rest mass relationship tends towards a positive real value when the speed goes against the speed of light. But if m² is a positive real value, then m is a positive real value or also a negative real value, but in both cases different from 0 in any case. Since negative masses made no sense in previous physics, it is valid that the mass of a photon is a positive real value, the elemental mass.

q.e.d.

## Addendum

Even if you take the equations from Wikipedia, which I expressly believe to be wrong and which are based on the fact that in a book by Einstein he forgot a 0 at the m later in his life (or whoever included the equations at the time Einstein certainly no longer wrote himself), I would only have one more negative E (kin) term in the numerator, which becomes a negative number with the second derivative. So that doesn't change the evidence. However, it should be noted that a total energy greater than mc² makes no sense. Anyone who claims this has not understood the theory of relativity, especially the equivalence of space and time. If you are also interested in Einstein's problem with the charge of photons, just click.

## Rest mass of the photon

Since people on the Internet are now claiming that it was always clear that the photon has mass and that Yang-Mill's theory was not about that, which, I make it clear, must definitely not be true I now dig deep into my bag of tricks. It's about nothing less than proving that the photon also has rest mass. Ultimately, one does not get any further with the root 0 consideration, since it is equal to 0 according to normal mathematical understanding. With the Division by Zero and with the Complex numbers I have shown that calculating with zero still offers surprises. But we can avoid this problem.

Ultimately it remains that m_0 = m * root (1 - v² / c²). But now it is the case that the relativistic root also corresponds to the compared time flow, so T '/ T: T' = T * root (1 - v² / c²) applies. Now one always claims that time stands still at the speed of light, because here the term becomes the root of 0. T 'here means the time that has passed in the moving system, in this case of the photon at the speed of light.

Now let's think back to how time is measured in the first place: It is measured as the tremor of atoms, i.e. as a frequency. However, due to its frequency according to E = hf, the photon also trembles in this sense. This means that it would be possible to measure time with the frequency of the photon. So here we also have time flow in Einstein's sense. For a concrete calculation you have to determine the frequency of the atoms with which we measure the time and here form a quotient with the frequency of the light.

So we have a number for the gamma factor, the relativistic root. Since from these considerations the relativistic root is not 0 for the photon, and since we have already proven that the photon also has mass, this means: The photon also has rest mass. The far-reaching mathematical consequences of this cannot yet be overlooked.

## Theory of the mass of the photon

As you can see, Einstein's Theory of Relativity also prescribes a photon mass when calculating the term 0/0 in the mass - rest mass relationship on the various paths. A photon mass also necessarily follows from E = mc², since the equation means that total energy and mass are equal and not just equivalent, as Einstein suspected. And that photons have energy, nobody denies that. And according to the Conservation of mass of the absolute theory, the mass of the photon then necessarily follows and not just the momentum and everything becomes clear, uniform and free of contradictions.

== Links ==

@quantenwelt has convinced me that I will not get the paper in a scientific journal, so I am now publishing it here for a small donation: photon mass