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I understand that the photon needs to have enough energy to produce a lepton and it's antimatter partner, and that all of the properties are conserved, but why does the photon do this in the first place? What's going on "behind the scenes" to transform a single, neutral particle (the photon) into two charged particles (i.e. electrons)? Why is the nucleus necessary?

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    $\begingroup$ Three question marks is not a good sign. Stack Exchange is built on the assumption that you are asking one question; that way each answer can be evaluated on a single metric (how well it answer the question). $\endgroup$ Commented Mar 8, 2013 at 23:43
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    $\begingroup$ @dmckee Right, I'll keep that in mind for the next question. I tend to do that in my questions I guess. I think I'm trying to create a picture of the gap I'm trying to fill. Asking multiple questions, to me, helps define the gap. But it also makes it harder to answer as you said. I'll try to be more to-the-point next time. $\endgroup$
    – John
    Commented Mar 11, 2013 at 18:39

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This process is the result of the cooperation of two theories of nature:

(i) Special relativity: This is a huge topic to study but we shall only need a small part of it, and perhaps the most famous one, which tells us this

$E=mc^2$.

This equation shows us that matter and energy are equivalent and interchangeable. For example, if an amount of energy $E = 2m_ec^2$ becomes available in a very small region of space, where $m_e$ is equal to the mass of the electron (or positron,) then it is possible to convert it into two particles, the electron and the positron.

(ii) Quantum mechanics: This tells us that electromagnetic waves are represented by “particles called photons,” which carry the energy of the electromagnetic field. The amount of energy carried by a photon is given by the famous equation

$E=hf$ where $h=6.63\times 10^{-34}$Js and $f$ is the frequency of the photon. So if the photon caries the amount of energy $E=2m_ec^2$ as it is emitted by a source (usually $gamma$)-emitter then the following can happen:

As the photon travels in space, QM allows the creation of an electron positron pair which lives only for a very short time, because they annihilate again into the original photon – this process is called “vacuum polarisation.” These two particles exist in virtual states and cannot be separated just like that, as that would violate the principle of conservation of momentum.

However, if an atomic nucleus is near by, then it is possible that a second photon coming from the nucleus can separate the two particles, before they annihilate again to give the original photon that generated them. I.e. the Coulomb field of the nucleus “pushes” the positron away, while it “pulls” the electron towards it. Hence the two particles became real, and can be guided into magnetic fields for storage and further use.

These processes are “behind the scenes” of a pair creation. The part of physics dealing with these fascinating quantum phenomena is called Quantum Electrodynamics.

Expansion: By the paragraph that begins “However, if an atomic…” I mean the following:

Imagine a $\gamma$-photon with sufficient energy approaching an atomic nucleus at a very close range. As the photon creates the $e^--e^+$ pair, the positron is scattered away from the nucleus while the electron flies towards the nucleus, in a virtual state, where it absorbs a virtual photon, effectively interacting with the Coulomb field of the nucleus and gets scattered to a new momentum state. During this process the nucleus carries some of the momentum of the virtual electron away. The presence of the atomic nucleus facilitates the splitting of the pair while the principle of momentum conservation is obeyed.

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  • $\begingroup$ Can you expand on this part? "However, if an atomic nucleus is near by, then it is possible that a second photon coming from the nucleus can separate the two particles, before they annihilate again to give the original photon that generated them. I.e. the Coulomb field of the nucleus “pushes” the positron away, while it “pulls” the electron towards it. Hence the two particles became real, and can be guided into magnetic fields for storage and further use." Thanks! $\endgroup$
    – John
    Commented Mar 8, 2013 at 20:32
  • $\begingroup$ @John Please see the expansion in my edited answer. $\endgroup$
    – JKL
    Commented Mar 8, 2013 at 22:56
  • $\begingroup$ +1 for the part of your answer, which deals with quantum mechanics is very nice. $\endgroup$
    – aignas
    Commented Mar 9, 2013 at 11:46
  • $\begingroup$ Accepted because of the nice explanation of the electron-positron virtual particle situation with the photon. I think I'm going to ask a more pointed question about this soon. Thanks! $\endgroup$
    – John
    Commented Mar 11, 2013 at 18:46
  • $\begingroup$ @John You are very welcome. It gives me pleasure to know I have made some contribution to your understanding of this fascinating phenomenon. I shall be looking forward to your extended question. $\endgroup$
    – JKL
    Commented Mar 11, 2013 at 22:57
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In the beginning, let's investigate the pair production. We know from relativity that mass can be equivalent to energy and if we set the most important physical constants to one $\hbar = c = \varepsilon = 1$, then we have the following relation (assuming that we produce electron-positron pair):

$$ E = \omega = \gamma_1 m_{e^-} + \gamma_2 m_{e^+} $$

Where $\omega$ is the energy (or the frequency) of the photon and $\gamma$ is the relativistic factor.

In quantum mechanics, the things is, that anything can happen what is possible to happen. But the question is how often it will happen. However, the pair creation in vacuum is not allowed due to momentum conservation. The proof is as follows: we can transfer into a zero momentum energy of the particle-antiparticle pair, but the photon will not have zero momentum, as it can not be at rest.

Now the momentum conservation can be 'fixed' if we have any nuclei around, then they carry the required ammount of momentum, so that the process becomes kinematically allowed.

If that is not enough, could you clarify, what do you mean "behind the scenes"?

UPDATE: Annav has a very good explanation to the 'why' part. I just wanted to emphasize, that it happens, because the process is physically Like annav and twistor59 said, the problem is that the process is physically feasible and that is enough for the quantum mechanical nature of our world to make it happen. So rather than asking why, we should ask how often and and annav's answer shows, that there is a theory called Quantum Electrodynamics, which helps us predict that.

Also, I wanted to note, that although we can have a process where we indeed produce a pair of quarks, in such a process one needs to remmember, that lone quarks have not been observed in our world and this is because of quarks being entirely different 'beasts' than leptons (ie. $e^-$,$e^+$, $\mu^-$,$\mu^+$, $\tau^-$ and $\tau^+$ elementary particles). The quark behaviour is rather well explained by Quantum Chromodynamics, however, the exact mechanism for me is still unknown. All I wanted to say, that the quark-antiquark production case is much more complicated than a lepton-antilepton production case.

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  • $\begingroup$ Nice explanation, but all of that explains the "how" but not the "why". Why does a photon "split" into an electron and positron, and not just bounce off the nucleus? There's something more fundamental here... Does science have an explanation for this? $\endgroup$
    – John
    Commented Mar 8, 2013 at 19:34
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    $\begingroup$ +1 for the answer "because it can". I think that captures the essential quantum mechanical view of events quite nicely $\endgroup$
    – twistor59
    Commented Mar 8, 2013 at 19:35
  • $\begingroup$ @twistor59 So do we not have an answer for the "why"? Just a mathematical model that can predict the events we observe? $\endgroup$
    – John
    Commented Mar 8, 2013 at 19:36
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    $\begingroup$ There are certain conservation laws that must be respected, and provided this is done, then quantum mechanics predicts the probability that a system starting off as a photon and a nucleus will end up as an electron+positron+nucleus. Statistical statements about how likely this is are pretty much all you can say. $\endgroup$
    – twistor59
    Commented Mar 8, 2013 at 19:40
  • $\begingroup$ @twistor59 I guess if we could understand the "why", we'd have a much deeper understanding of Physics... $\endgroup$
    – John
    Commented Mar 8, 2013 at 19:43
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The answer by @gns-ank covers the kinematics of why. Below I tackle the

Why does a photon "split" into an electron and positron, and not just bounce off the nucleus

in your comment to his answer.

In general physics can answer "why" in a nested way, like russian dolls. In the end, the kernel answer is "because it does". In this case though we are in the middle of the nesting, and we can answer "why".

Photons interact electromagnetically with the field of the nucleus or also sometimes the electrons of the atoms according to well established solutions of the problem using quantum electrodynamics. This theory has been validated very meticulously, and it gives the probability of the photon bouncing off, as well as the probability to make an electron positron pair or a quark antiquark pair etc.

In this buuble chamber picture from the large number of photons passing through at the same time without interacting with the atoms or nuclei, one photon managed to interact with an electron, seen, and create a pair and go on to create a pair with higher energy on a nucleus (which is not seen not having enough momentum in the lab frame). Low probability, but it happened ( could also be a coincidence, two photons, but it does seem to point to the vertex)!

electron positron pair

We can then go into "why electromagnetic" which will take us to the standard model that has been validated , and then "why the standard model" and some think it is strings, some have other opinions, but we have reached the current kernel of nesting, and the answer is "because that is the way it is".

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  • $\begingroup$ Nice demonstration/illustration. $\endgroup$
    – JKL
    Commented Mar 8, 2013 at 20:53
  • $\begingroup$ Do you happen to know where this image is originally from? I realize it's been a couple years, and even back then it might've been a mystery, but I figure it can't hurt to ask. $\endgroup$
    – Pops
    Commented Mar 18, 2015 at 20:25
  • $\begingroup$ @Pops astro1.panet.utoledo.edu/~ljc/EINab02.jpg . Might be a manipulated version of this, sciencephoto.com/media/1298/view , which has original credits. Found it again and similar ones by googling "bubble chamber photon pictures" $\endgroup$
    – anna v
    Commented Mar 19, 2015 at 4:37
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Because its electromagnetic field is attracting with a proton or an electron, because it has a charge itself as a particle, so if it attracts with an electron, it will knock it out while transferring momentum and take the electron's position with a negative charge with attracting a proton at first,but then dodging it, with the proton flying out of the nucleus and turning by the atom's magnetic field while exiting it, and will cause some static shock in the atom with a static shock causing an electron(probably the same photon). And probably also a neutron, casing another atom to form nearby, with the leftover energy boosting the electron.

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AFAIK there is no explanation under canonical theories.
I found a plausible explanation in the book (monograph) of Douglas Pinnow Our Resonant Universe

.. model of particles, based only on Electromagnetism (EM), that has only one parameter (electron mass) and derives the particle properties to within 1% of their values (barion masses bellow 0.1%) and does not suffer of the barionic spin crisis. The model uses three building blocks: Electron, Pion and Muon. The concepts of Mass, Charge and Gravitational Force became clear.

In short: a positron+electron equate to EM : when they meet we have gamma rays and conversely two gamma rays collide head on and we have a pair. How can both light and massive particles be 'the same' ? Pinnow book answers 'the why'.

The aux nucleus is needed to trap a photon with itself ( an head on collision ). This way the photon is trapped in a resonant state (the particles), and looses his natural freedom :).

One can find photos of selected pages of the book related to this question.
edit add:
Pinnow career:
Douglas Pinnow received a bachelors in engineering physics from Cornell in 1961, and a PhD in physics from the Catholic University of America, as a NASA fellow, in 1967. He went on to become Supervisor of the Quantum Electronics Group at Bell Labs, Assistant Manager of the Chemical-Physics Department at Hughes Research Laboratory, and Director of R&D at Times Fiber. In 1985 he founded Universal Photonix, and later its subsidiary, Electronic Monitoring Systems.

Dr Pinnow is a Fellow of the Optical Society of America and has chaired major conferences of the OSA and IEEE, such as the 1983 Conference on Lasers and Electro-Optics He currently teaches graduate electro-optics at UC Irvine, and has over fifty technical papers and numerous patents to his name

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    $\begingroup$ This looks like a book about non-mainstream physics... Not the best way to answer a question about mainstream physics. $\endgroup$
    – John
    Commented Mar 11, 2013 at 18:44

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