Talk:Annihilation

Latest comment: 4 years ago by Tpth in topic Feynman diagram axes reversed?

swarm of secondary particle

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"When a proton annihilates an antiproton they produce gamma rays and a swarm of secondary particles, like pairs of top-anti-top quarks."

This seems to be incorrect, or at least misleading, since protons are much too light to create top quarks (the mass of a top quark is similar to that of a Gold nucleus). If the reference is intended to be to high-energy collisions in which the protons and antiprotons carry enough kinetic energy to create a t-tbar pair, this should be explained; in this case, it should also be mentioned that colliding e+-e- pairs can also yield arbitrary kinds of particles (via a highly virtual photon or Z boson).

Possibly Incorrect

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Can someone pleas check to see if and antiparticle and a particle need to be of the same type of in order to annihilate with each other? Thanks! scienceman 23:27, 23 March 2006 (UTC)Reply

They do. Otherwise the sum of their quantum numbers could not be zero. Strait 14:54, 2 June 2006 (UTC)Reply


The feynman diagram where the photon decays into a electron positron pair is somewhat misleading. This can not happen for a free photon (as shown). The reason should be that unlike the W boson or the Z+- it does not have a mass and therefore infinite lifetime.

There is no such diagram on the page. The closest one is the one captioned "A Feynman diagram of a positron and an electron annihilating into a photon which then decays back into a positron and an electron.", which is totally valid. --Strait 17:06, 5 January 2007 (UTC)Reply
There is one problem with that very same Feynmann diagram though - the arrows on the positrons are pointing the wrong way.

Question

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And how close together do they have to be for annihilation to happen?68.6.85.167 10:21, 9 May 2006 (UTC)Reply

Due to quantum uncertainty, such a question may not be meaningful. The probability per unit time will depend on the overlap of the wavefunctions of the two particles. Something on the order of the Compton wavelength, maybe? -- Xerxes 16:22, 9 May 2006 (UTC)Reply

Energy nonconservation?

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What's the deal with this line:

If by intermediate processes in quantum field theory we mean Feynman diagram vertices, then no, every vertex conserves energy. All that's temporarily violated is the energy-momentum relations of free particles. And if we're not talking about Feynman diagrams, then what does this even mean? Melchoir 07:23, 3 July 2006 (UTC)Reply

The Proton-Proton Chain Reaction

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Is it true that the exact reverse of annihilation is involved in the proton-proton chain reaction?--67.10.200.101 00:20, 3 November 2006 (UTC)Reply

I can't see the picture

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I have firefox and I can't see the picture of a positron and an electron annihilating into a photon. I thought I should tell all of you, Jeffrey.Kleykamp 23:22, 12 May 2007 (UTC)Reply

scope

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This article's great, but I feel there should be another article created to discuss NONE physics related annihilation, such as apocalypse. 82.18.44.72 (talk) 23:59, 29 March 2008 (UTC)Reply

Why?

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After much looking, I can't seem to discover the theory behind why antimatter and matter (on contact) annihilate one another. Is it something that is simply unknown, or is it just an omission? 79.71.190.207 (talk) 22:13, 17 February 2009 (UTC)Reply

The particles are simply opposite. They cancel each other out, and… yeah. See the rest of the article for a more detailed background… 24.130.10.143 (talk) 02:48, 21 May 2010 (UTC)Reply

Amount of energy?

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Forgive a non-physicist for lack of knowledge here, but could someone tell me roughly how much energy you get out of particle Annihilation? Hypothetically speaking (particularly in Science Fiction) if a fellow had a heap of Matter and a heap of Anti-Matter and he put the two together, what would he get? Enough power to make his car run, or blow up the earth? If you had an equal mass of, say, Anti-Hydrogen as you did Plutonium, which could you get more energy out of? I know in practice this is pretty much useless to scientists because Anti Matter doesn't just happen to be sitting around in piles, but I've seen it used in Science Fiction and I'm quite curious as to how much energy you get out of it. For that matter, the article could go a section on examples in fiction. 43.244.33.36 (talk) 13:29, 7 March 2009 (UTC)Reply

--The amount of energy released by the anihiliation of matter and antimater is equal to Etotal=mc2+Ekenetic. You take the total mass of the anti-matter and matter which anihilates(this will be equal to twice the mass of the anti-matter, assuming you have less anti-matter than matter), multiply that value by the speed of light squared (the speed of light is 299792458 m/s, so c2=8.98755179 × 1016 m2/s2) and then add the kenetic energy of the anti-matter and matter (this amount is normally negligable, but I add it for completeness.)24.162.159.61 (talk) 15:04, 20 June 2009 (UTC)Reply

Lepton Number not conserved??

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Sorry, but I'm not really knowledgeable in this field. But I came across this when studying for exams :) It states that when an electron and an anti-electron undergo mutual annihilation, the Lepton number does NOT have to be conserved. Thanks for reading! Krazywrath~ (talk) 02:51, 12 May 2009 (UTC)Reply

Article stops short

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The article includes the summary of Electron-positron annihilation taken from the relevant main article, but the summary of Proton antiproton annihilation from its main article is missing. — Preceding unsigned comment added by Rhsimard (talkcontribs) 22:41, 19 March 2011 (UTC)Reply

References?

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I just corrected a factual error (that was introduced and went unnoticed quite some time), but in any case the paragraph on electron-positron annihilation would be a worthwhile target for a rewrite. I find the complete lack of any serious reference even more grave (I myself am not too comfortable in quantum electrodynamics to be able to do that). Seattle Jörg (talk) 13:17, 1 March 2013 (UTC)Reply

Composite particle u anti-u d annihilation decay missing

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There is this annihilation of a up anti-up down baryon that's missing. (The baryon is created by a high-energy electron doing interaction with a quark, switching charges, but I can't find the name of the process.) But it would make an interesting addition to the other annihilation examples. I can't find where it is. Sorry.

Meson decay needs it's own section

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Meson decay is also an example of matter-antimatter annihilation. Plz give meson decays that direct convert into photons because of quark antiquark annihilation it's own section.

Meson

Be sure to include pions: pi mesons. The neutral pions annihilate directly into gamma rays as their most prominent decay mode: Pion#Neutral pion decays While the charged ones decay into muons. Pion#Charged_pion_decays

Photon/gamma ray distinction

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Most of the pages on annihilation refer to the products of annihilation as gamma rays. While this is a common statement, it is not strictly true. By most definitions, a gamma ray originates from the nucleus, which is not the case of annihilation photons. It is unquestionably correct to refer to the product as a photon; gamma is a term of convenience and in this context is controversial at best. There is a small section on the X-ray#Gamma_rays article describing the difficulty of distinguishing between types of photon radiation; perhaps it is worthwhile to add an explanation of this distinction in the article? Cjweber (talk) 00:51, 11 January 2020 (UTC)Reply

Feynman diagram axes reversed?

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I'm no expert, but afaik it's generally the convention for Feynman diagrams to use the X axis for time and the Y axis for space, rather than vice versa. This allows the interaction to be "read" from left to right, rather than from bottom to top. The diagram here is technically correct, in that the interaction it shows is correct, but the fact its axes are reversed makes it a little confusing/counterintuitive for beginners. A possible replacement would be something like this:

 

Like I said, I'm def no expert here, so do lmk if I'm missing something obvious! Tpth (talk) 04:33, 22 May 2020 (UTC)Reply