Particles and Antiparticles can annihilate, and they are completely destroyed in the process, which creates photons.

From wikipedia:

An unstable atomic nucleus with an excess of neutrons may undergo β− decay
n → p + e− + -νe
neutron decays into proton, electron, and electron antineutrino.    

Unstable atomic nuclei with an excess of protons may undergo β+ decay
p → n + e+ + νe
proton decays into neutron, positron, and electron neutrino.

So, combining these two reactions gives us

n → n + e- + e+ + -ve + ve
neutron decays into neutron, electron, positron, electron antineutrino, and electron neutrino.

When the electron and the positron or the electron antineutrino and the electron neutrino collide and are annihilated, photons are produced. So, taking it a step further

n -> n + ?γ
neutron decays into neutron and ? photons.

So my question is, sorry for the long lead-up, where are these photons are coming from? Also sorry if I broke physics :P

(Note: I know that in β− decay, the neutron actually emits a virtual W- particle which decays into an electron and an electron antineutrino, but thought that wouldn't be relevant. It's here if it is, though!)
(Another note: Sorry for the bad symbolization, SE doesn't seem to be accepting all of the characters)

  • $\begingroup$ One more thing , the proton -proton you cancelled is quite incorrect because both protons have different energies. $\endgroup$ – Anubhav Goel May 16 '16 at 15:17
  • $\begingroup$ Proton -proton? What? You meant the electron and positron? You do know what happens when particles meet their antiparticle equivalents, right? $\endgroup$ – rustphoenix May 17 '16 at 4:10
  • $\begingroup$ when you combined the two equations, (n → p + e− + -ν_e) and (p → n + e+ + ν_e) , you cancelled protons on both sides. The two protons in two equations have different energies, so they can't be cancelled simply. Like you didn't cancel neutrons on two sides. $\endgroup$ – Anubhav Goel May 17 '16 at 5:28

I had similar stupid doubt. It's coming from binding energy.

The equation you give:

Unstable atomic nuclei with an excess of protons may undergo β+ decay

$$p → n + e^{+} + \nu_{e}$$

proton decays into neutron, positron, and electron neutrino.

How do you think a proton can be converted to neutron which has greater mass? + you get positron and neutrinos.

The energy comes from binding energy of nucleons in an multi-proton atom.

When you sum up to

n→ n + $\gamma$

You are actually showing energy released from binding energy.

  • $\begingroup$ Okay, that makes sense. Thanks! I couldn't find anything on the internet $\endgroup$ – rustphoenix May 13 '16 at 18:19
  • $\begingroup$ @byteoflogic I also took my teacher help for same. It's so simple but you don't get it in first attempt. $\endgroup$ – Anubhav Goel May 14 '16 at 2:12
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    $\begingroup$ I don't get why Lubos got +6 , while I got 0. Is he having fan following which does not care for correct answer. $\endgroup$ – Anubhav Goel May 14 '16 at 2:14
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    $\begingroup$ @AnubhavGoel - your answer is a package of several confusions. First, in the beta-plus decay, there are no photons involved, so this decay has nothing to do with the question which was "where photons were coming from". Second, a free neutron in the vacuum cannot decay to a neutron plus something else (e.g. photon) due to energy+momentum conservation. Third, when the neutron is a part of a bound state, new processes become possible but one can't say that the new processes only depend on "binding energy". In the bound states, the particles have different binding as well as kinetic energies. $\endgroup$ – Luboš Motl May 14 '16 at 8:10
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    $\begingroup$ For 3rd comment, something silly is where science starts from. Never stop yourself from doing something silly. $\endgroup$ – Anubhav Goel May 14 '16 at 9:07

Your question is identical to a question posed to Feynman by his father. A two-minute monologue on that is here:


The photons are genuinely created at some moments – and they may be similarly destroyed, too. They're not coming from anywhere and they are going nowhere. The number of photons $N_\gamma$ is simply not conserved in Nature. (Just to be sure, the momentum, angular momentum, and energy are conserved, and this may reduce the number of photons or other particles that may be produced.) And quite generally, whenever a charged particle moves (especially with acceleration) or interacts with others, new photons are almost always created.

In modern physics, we describe this creation of photons by new photon lines attached to the rest of the Feynman diagrams via a Feynman vertex. Classically, we were talking about electromagnetic waves and those were simply created whenever electric charges were accelerating.

Feynman's strategy to explain why it's possible was in terms of an analogy with speaking. A photon is a quantum of light – similarly to the word's being a quantum of speech. If we speak, we can say the word "cat" as many times as we want. We don't have any "word bag" with the words "cat" that could get depleted. We just produce the word – sound waves with the pattern carrying the audio information about the word "cat" – as we speak. In the very same way, sources of electromagnetic radiation keep on producing photons out of nothing. There is no "photon bag" that could ever get depleted.

Similarly, in weak interactions, a W-boson may get created out of nothing – if the aforementioned conservation laws are obeyed (along with the electric charge conservation) and the W-boson may also be destroyed given the same conditions. There are Feynman vertices from which photons or very analogously W-bosons are coming.

Just like particles may be created and annihilated, the list of particle species present in the world may change in more general way. For example, when a neutron decays, $$ n\to p + e^- +\bar \nu_e,$$ it doesn't prove that the neutron contained the particles from the right hand side. If it did, it wouldn't be a proton but a (much larger) hydrogen atom (plus some strangely attached neutrino). Microscopically, the decay is due to the quark-level decay $$ d \to u + e^- + \bar\nu_e $$ which changes one fermion into three others. This single "four-fermion interaction" may be further reduced to a Feynman diagram with a virtual W-bosons. The latter diagram has two vertices. In one of them, the down-quark is destroyed and replaced by an up-quark and a negative W-boson. In the other vertex, the W-boson decays to the electron-neutrino pair. This general creation or transformation of particles doesn't violate any valid law of physics.

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    $\begingroup$ Thanks, though I personally did not need this level of detail. I know what photons are, and understand that beta decay is caused by (or, more accurately, is) quark decay. I was simply wondering where the energy was coming from. So, while my question may be similar to his, my motivations were different. :) $\endgroup$ – rustphoenix May 13 '16 at 7:31
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    $\begingroup$ Where the energy is coming from is an entirely different question than the question you asked in this thread, however, and the answer is different, too. Energy is conserved in all these processes. One has to carefully compare the energies of the initial and final state including the $E=mc^2$ latent energy, including the kinetic energy, plus all forms of potential energy between all particles etc. and the conservation law holds in all cases. It's only the total energy that is conserved, however, and one can't "trace individual joules" and say "which joule" goes "where" in the final state. $\endgroup$ – Luboš Motl May 13 '16 at 7:41
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    $\begingroup$ To be more accurate, there is no "word bag", but there is an "air bag". You could not continually speak without ever breathing in. So I guess I was asking when the atom is "breathing in" the energy used to create the particles and antiparticles. And, secondly, where this is stored, it's proverbial "lungs". (weak force?) $\endgroup$ – rustphoenix May 13 '16 at 7:45
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    $\begingroup$ But, if I had wondered how energy can be transformed into photons, and vice versa, I would have asked you why neon signs glow (electron energy state) and why dark surfaces absorb some photons, and never emit them back. $\endgroup$ – rustphoenix May 13 '16 at 7:56
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    $\begingroup$ Yup. So if you could get out the same thing plus extras, it would no problems. Wait... In case you didn't realize, that was sarcasm. What you answered is not the question that was written down, it is the straw man you came up with. Can't see the first through the trees, eh? $\endgroup$ – rustphoenix May 14 '16 at 9:12

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