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How can be explained the phenomenon of two related Gamma rays ending their courses by becoming an electron and a positron without the involvement of a photon and an anti-photon ? This question must be understood as "How can we explain the said phenomenon ?".

I am an autodidacte in Physics. I have developed a theory about the Universe that will be edited soon on YouScribe. It is based on the Ockham principle of simplicity and results in a full explanation without mysteries nor any unexplained subjects. And so, what is it worth ?

Roberto
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  • There isn't an anti-photon because the photon is its own antiparticle. Also you only need one gamma ray of sufficient energy to make a fermion-antifermion pair (it of course needs more energy than the combined rest mass of the particles) – Triatticus Apr 04 '18 at 21:29
  • There is no proof about a photon being it's own antiparticle. It is simply a theory based on another unproven theory which states that a photon is pure energy. So, you cannot explain the said phenomenon. – Roberto Apr 04 '18 at 21:46
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    It's a consequence of charge conjugation which arises from conservation of charge – Triatticus Apr 04 '18 at 22:00
  • What is the "mechanism" behind the concept of a charge ? It is unknown. Then, we are using a concept that should not be used for proofing. – Roberto Apr 04 '18 at 22:57
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    @roberto, you are wrong. the mechanism behind the conservation of electrical charge is well-known: it arises from the invariance of the wavefunction of the electron to global changes in phase. – niels nielsen Apr 04 '18 at 23:15
  • @Neils How do you explain the electron "dipole moment" ? – Roberto Apr 04 '18 at 23:41
  • @Neils Richard Feynman repeatedly stated that the "wave-matter" duality is a complete mystery. But, it is the founding of "quantum Physics". Hence, your statement proving me wrong is more than fragile. – Roberto Apr 04 '18 at 23:51

1 Answers1

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Any photon is also an antiphoton.

my2cts
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  • The properties of momentum and spin have nothing to do with being an anti-particle or not. – ACuriousMind Apr 04 '18 at 22:11
  • For momentum, agreed. For spin I am not so sure. – my2cts Apr 04 '18 at 22:29
  • The "absolute" spin - as in, whether the particle is a spin-0 or spin-1 or spin-1/2 particle - is the same for both a particle and its anti-particle, and any more specific value depends on the exact state the particle or anti-particle is in, but e.g. a positron is always the anti-particle of an electron no matter its spin state. Being an anti-particle is a property of the particle species, not of a specific state. – ACuriousMind Apr 04 '18 at 22:49
  • By definition, antiparticles have same mass, same spin but opposite charge. The spin should not be different for being antimatter. Could it then be that particles are of the Majorana type ? – Roberto Apr 04 '18 at 23:01
  • I changed my statement to be more acceptable to the commenters. I think the purest definition of an antiparticle is that which can annihilate a particle. In a particular case that involves requirements on the states of the particles. In general any photon is also an antiphoton. – my2cts Apr 05 '18 at 18:35
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    That is not the purest definition. A photon cannot annihilate another photon at tree level (i.e. there's no standard model vertex with two photons). If you allow loop diagrams, then a Higgs boson can annihilate with a photon the same way another photon would. Does that make the Higgs boson an antiphoton? – Chris Apr 05 '18 at 20:57
  • This argument is dependent on perturbation theory, which is a mathematical approach to solve an equation. Besides, in the perturbation theory of scalar electrodynamics there are two photon vertices. – my2cts Apr 16 '18 at 20:25
  • @my2cts Scalar electrodynamics is a toy model- it doesn't correspond with reality. The Standard Model has no two photon vertices. Does that make a photon not its own antiparticle? – Chris May 26 '18 at 19:58
  • In the squared Dirac equation there is a two photon vertex as well, describing annihilation. – my2cts May 26 '18 at 20:21