Talk:Pair production

message: After reading this page, I could not possibly imagine this formula being useful to any reader, so I decided to go to the source ("Koehn, C., Ebert, U., Angular distribution of Bremsstrahlung photons and of positrons for calculations of terrestrial gamma-ray flashes and positron beams, Atmos. Res. (2014), vol. 135-136, pp. 432-465"). I assumed there would be no way someone could even correctly copy this formula into the page. I made it as far as I2 where there is a sign error in wikipedia compared to the paper. This clearly does not belong in the page, and should be removed, since anyone remotely interested in the equation would simply refer to the paper. Unfortunately, I believe this requires a rather substantial rewrite.  — Preceding unsigned comment added by 129.67.38.58 (talk) 19:44, 5 December 2014 (UTC)
thread_title:  Can we please remove this formula? 

message: This is a note explaning my revert of the edit that was just made.  The paragraph beginning with this text is not about pair production in the vacuum:
In nuclear physics, this occurs when a high-energy photon interacts with an atomic nucleus, allowing it to decay into an electron and a positron without violating conservation of momentum. It is the chief method by which energy from gamma rays is observed in condensed matter.
It refers to a real photon decaying to an electron and a positron.  Although there is no nuclear decay involved, a nucleus is required to conserve momentum.  If that were not true, photons with E > 1.022 MeV would decay extremely rapidly while moving through space; instead, it happens only in matter.  -- SCZenz 14:49, 13 September 2005 (UTC)
thread_title:  Revert regarding pair production in matter 

message: The photon need only have a total energy of twice the rest mass(me) of an electron (1.022 MeV) for this to occur as described above; if it is much more energetic, heavier particles may also be produced.
What are these "heavier particles"? - Quirk 09:03, 21 May 2007 (UTC)
thread_title:  Production of heavier particles 

message: Article says: photon-photon pair production may occur at 511 KeV, comparing it to photon-nucleus pair production. This leaves me wondering how the photon-photon version might work: it means photons could as well interact with other photons to produce particle-antiparticle pairs? -- Alessandro Ghignola (talk) 09:07, 25 April 2010 (UTC)
thread_title:  Curiosity on photon-photon pair production 

message: Why couldn't the newly created positron and electron absorb the momentum of the photon? How would the conservation law break without a nucleus? The article doesn't give a reason for this, just states this as a fact. Could someone shed some light on this? PAStheLoD (talk) 02:54, 8 June 2010 (UTC)
thread_title:  Conservation of Momentum 

message: There is a significant chunk of math in this article, would it be wise to break it down some and explain it? I don't have a problem with pretty math, but since wikipedia likes for the common man to be able to understand things, we should probably do something to make the math more presentable to a wider audience. 76.198.38.250 (talk) 20:00, 20 July 2013 (UTC)
thread_title:  Math 

message: The Feynman diagram does not conserve momentum, so the interaction shown is only possible for a virtual pair production (though the diagram shows free pair production). 76.181.69.229 (talk) 22:00, 2 December 2013 (UTC)
thread_title:  Feynman Diagram 

message: http://puu.sh/gMWBX/ac4241b70a.png
Am I the only one that sees a huge space here? Even if I am not, it's odd that there's something that large on the page. Is it a formula that didn't load? (I refreshed.)  — Preceding unsigned comment added by Mechanic1c (talk • contribs) 23:28, 23 March 2015 (UTC)
thread_title:  I hope this is just my computer going wrong 

message: I think we should relabel "Z" as "nucleus" instead in the diagram shown to avoid confusion with the Z boson of the weak interaction.  See "."  Nicole Sharp (talk) 17:08, 20 September 2016 (UTC)
thread_title:  Z 

message: There are some serious mistakes in the section on Basic Kinematics. The conclusion is that the electron and positron are produced back-to-back. This is clearly wrong. As anyone who has seen photons convert to e+e- in bubble chamber photographs, they are produced with an almost zero opening angle not 180 degrees.
The problem with the maths is easy to see. (1+cos(theta)) is really (1-cos(theta)). Tracing this back, the minus sign follows from "E"-"p" not "E"+"p" which is the Minkowski geometry of spacetime.
So this section needs to be re-done, preferably with some better explanations of the approximations being used. For example, the statement that the nucleus has negligible mass compared to the energy of the photon would only be true for photon energies above several hundred GeV.
Chris Bowdery  — Preceding unsigned comment added by 2A02:C7D:541A:1400:C595:135F:3FF6:64F3 (talk) 21:23, 23 October 2016 (UTC)
thread_title:  Basic Kinematics section has some serious mistakes 

message: Looking at the first sentences, I see an error - "Pair production is the creation of an elementary particle and its antiparticle. Examples include creating an electron and a positron, a muon and an antimuon, or a proton and an antiproton" - if it is the creation of an elementary particle and its antiparticle then a proton and an antiproton, which are composed of quarks/antiquarks, would not qualify. This needs to be clarified. --Nerd1a4i (talk) 13:02, 15 July 2017 (UTC)
thread_title:  Photon is not an elementary particle 

message: The section "Basic kinematics" is inaccurate. Specifically, according to the metric signature (−, +, +, +) and with the assumption that the modulus of the velocity of the electron and the positron is the same, and that \theta is the angle between the two velocity vectors (as in the article), the calculation gives
p_{e^-}p_{e^+} = -\left( \frac{E^2}{c^2} - \mathbf p_{e^-} \cdot \mathbf p_{e^+} \right) = 
-m^2 c^2 \gamma^2 \left( 1-\beta^2\cos\theta \right)
and hence
(p_\gamma)^2 = -4m_e^2\gamma^2(c^2-v^2\cos^2\theta/2) = 0
The equality can be true only if v=c and \theta=0, that is clearly impossible for electrons and positrons.
The underlying reason is that the equality \frac{E}{c}=p holds for massless particles like photons, but not for massive particles like electrons and positrons, unless they move at the velocity of light (that is impossible for massive particles). Therefore, it is not possible to assure energy and momentum conservation when a single photon decays in massive particles without the interaction with other particles. Szunino81 (talk) 13:19, 7 September 2017 (UTC)
thread_title:  Basic kinematics - Energy and momentum conservation 

message: The article mentions Hawking radiation (under the heading "Astronomy", which incidentally is ridiculous), but Hawking radiation does not involve a neutral boson.  More generally, there are lots of other forms of pair production that also do not involve a neutral boson - Schwinger pair production, pair production in plasma, pair production from the collision of two (high energy) photons, etc.  So, either the lede should be rewritten to include these other phenomena, or it should make clear that it is not discussing them and the reference to Hawking radiation should be deleted.  I think the first option is better.  Waleswatcher (talk) 21:26, 18 April 2018 (UTC)
thread_title:  Hawking radiation is not pair production by the definition given here 

message: This article could use a bit of history.  I'd imagine the first observation of pair production was quite notable and surprising, but it's not mentioned at all.  — Preceding unsigned comment added by 96.40.48.159 (talk) 14:05, 29 June 2018 (UTC)
thread_title:  History 

message: The photon must be near a nucleus in order to satisfy conservation of momentum, as an electron–positron pair produced in free space cannot satisfy conservation of both energy and momentum.
Actually it does not require a nucleus. It requires a second particle.
The problem is the free choice of the reference frame. The energy of the photon depends on the frame of reference. It can have any positive value, see Relativistic_Doppler_effect.
So we have an initial state, where we have only one photon with an arbitrary energy, depending on our choice of reference frame.
In the final state however we have two particles. These particles have rest mass, and so the total energy in the final state must be at least equal to the rest masses of the produced particles. This contradicts the free choice of the reference frame, and thus impossible.
I believe there is another explanation involving Feynman diagrams. There are rules for drawing Feynman diagrams, and there can not be a process where only one particle enters the diagram and more particles leave.
However these are both indirect explanations.
Cs.B. (talk) 19:36, 1 February 2023 (UTC)
thread_title:  About the conservation of momentum and energy 

message: As energy must be conserved, for pair production to occur, the incoming energy of the photon must be above a threshold of at least the total rest mass energy of the two particles created.  (As the electron is the lightest, hence, lowest mass/energy, elementary particle, it requires the least energetic photons of all possible pair-production processes.)
The explanation is not factually correct, since the electron is not the lightest particle. Neutrinos are lighter.
I can not provide explanation on why neutrino-antineutrino pair production does not occur (or does it?). It may be related to the interactions, since photons are only known to interact via electromagnetism and gravity, where neutrinos do not interact with electromagnetism. Maybe there is neutrino-antineutrino pair production involving other bosons.
Cs.B. (talk) 19:44, 1 February 2023 (UTC)
replies: text: I came here to ask the same question.  I'm guessing that Buday's explanation (neutrinos don't interact via the EM force) is right, but in any case the article should be corrected by someone who knows. Mcswell (talk) 15:06, 14 May 2024 (UTC)
thread_title:  Minimum energy required