Talk:Electron–positron annihilation

Mass and energy

If a negatron and a positron both have mass, how is it that their annhilation produces two photons that DON'T have mass.

Is it that the photon does have mass OR that the mass is somehow lost during the annihilation OR that the charge of the electrons IS the mass?

I believe that $E=mc^{2}$ is responsible - the mass of the converging particles is converted to energy. The mass is fully converted to energy during the annihilation process.

Chris 1127 10:21, 11 January 2006 (UTC)[reply]

-Yeah, Chris is right, matter-antimatter reactions are 100% energy conversion in cooperation with Einstein's $E=mc^{2}$ . The positron and electron both have mass, that is converted to pure energy (read more about the conservation of matter/energy). --Mirag3 20:58, 23 February 2006 (UTC)[reply]

thanks for your replies.

I was wondering whether the mechanisms for this energy matter conversion are understood at all? ie how quarks and leptons fit into the expanation or even what carries mass. thanks in advance--Trevor M 12 may 2006

It depends what you mean by "understand". We have a very good understanding of how elementary particles behave. For instance, we know their masses, we know which ones interact with which other ones and with what probabilities. However, I think that it would be a stretch to say that we really know what mass is. This is too complicated a topic to discuss here. I suggest that you read up on it on Wikipedia or elsewhere. --Strait 20:18, 2 August 2006 (UTC)[reply]

Angular momentum conservation

Can anyone explain how angular momentum is actually conserved if it is conserved at all? Also, an explanation of why two photons have to be created is needed.--Hpesoj00 14:46, 14 June 2006 (UTC)[reply]

Angular momentum is conserved. I have added the explanation you asked for. --Strait 20:18, 2 August 2006 (UTC)[reply]

e(-) + e(+) that doesn't yield annihilation?

As is implied by the image, if either the electron or positron has much more energy than the other then they may interact without annihilation. Is this the case? If so, that should be included in the article. --HantaVirus 13:37, 24 July 2006 (UTC)[reply]

Done. --Strait 20:18, 2 August 2006 (UTC)[reply]

m=E/c2 does not mean destuction

Einstein and many others know and accept that the equal sign (=) in this equation does not mean destruction. It simply means interconversion. It would be best to represent it with two opposing arrows. Bvcrist 16:41, 2 August 2006 (UTC)[reply]

No, it is much better written as it is. It is meaningless to say that an equal sign means destruction. It is an algebraic equation and could not be used with a double arrow. It is, in fact, the special case of a more general equation E² = m²c⁴ + p²c² for p=0.

Your changing the word "destruction" to "conversion" is fine. However, "destruction" is correct. There used to be an electron and a positron, now there isn't. Most people would call that destruction. Frankly, the word "annihilation" means "destruction." If you don't think of it that way, that's fine, but it's semantics, not science.

--Strait 20:01-09, 2 August 2006 (UTC)

Nature can be defined by various laws but is never controlled by them

Conservation laws are not true laws because modern day physicists turn them on and off for various reference frames. They are better called postulates.

Even if we accept conservation laws as laws, they do not control nature or the physics of nature. They are observations of what nature does or does not do, therefore we can not state that the conservation laws define or control anything. We can say that various processes follow the conservation laws, but we can say that nature is defined by a set of rules or laws that we might envision as explaining what nature does or does not do. I hope that the members who are helping to write Wikipedia will excercise correct logic in their writing especially in the hard sciences. Bvcrist 16:51, 2 August 2006 (UTC)[reply]

You have also posted this in my user talk. I will reply there. --Strait 20:18, 2 August 2006 (UTC)[reply]

conservation of energy

Does anyone know how the conservation of energy happens in a electron-positron anhilation(creation), also how is the three momentum conserved? --129.108.211.121

Well, I'm not sure what you mean by "how", but I'll try to answer the question in the standard manner. You can always choose a frame in which the total three momentum of the electron and positron is zero. In this frame, the photons emerge back-to-back with equal energy. So they also have zero three momentum. Each photon's energy is equal to the energy of either the electron or the positron, so energy is conserved as well. Does that answer your question? --Strait 19:00, 23 August 2006 (UTC)[reply]

Electron-positron equation

Wouldn't it be helpful for the following equation to be a part of this article?

e⁺ + e⁻ → 2γ + 1.02 MeV

If this is the correct equation for this page, then I think that it would be benificial for it to be added. Please comment on whether this would be a good addition, or tell me why this doesn't apply. Thx! --Dimblethum 03:28, 19 October 2006 (UTC)[reply]

That is basically correct, but it is not standard in physics to write the energy as a separate term (the photons have the energy, so it looks funny to write the energy as though it is a separate object), so usually it would be written just as e⁺ + e⁻ → 2γ. --Strait 05:09, 19 October 2006 (UTC)[reply]

I just copied the equation from the Proton-proton chain reaction page. I don't know a whole lot about physics. I just was looking for how much energy was released by electron-positron annihilation. Also, do you have an opinion of whether this equation, in either form, should be added to the page? --Dimblethum 03:55, 20 October 2006 (UTC)[reply]

Probably not a bad idea. --Strait 04:11, 20 October 2006 (UTC

May I add that each particle symbol carries quantum numbers with it. A symbol like e means a fixed mass, charge, spin etc. Equality in equations involving these symbols implicitly means equality of all these attributes on the two sides of an equation. The energy likewise is implicit once the symbol is introduced.

Linear Momentum Conservation

According to the article for the low energy case, "the gamma rays are emitted in opposite directions". I assume this is relative to the reference frame of the system. However if the system is moving with respect to the observer (i.e. the two particles have a nonzero momentum component perpendicular to the line between their centres), what are the implications for the emitted gamma rays? I assume they won't be moving in exactly opposite directions as far as the observer is concerned, since they will have a component of motion in the direction the electron-positron system was originally moving. Is this true? --202.191.103.123 06:01, 5 September 2007 (UTC)[reply]

e^- + e⁺ annihilation, second image wrong

The second of images is probably wrong. It shows electron emitting photon on the upper half, and positron emitting photon at the lower half, both changing into something depicted by the straight line between them (what's that? :P)

Here's another image depicting annihilation, also on English Wikipedia: http://en.wikipedia.org/wiki/Image:Electron-positron-annihilation.svg and it appears to be right: it shows electron and positron approaching each other, and transformation into a photon in the place of their meeting. Actually there should be two photons, but I saw that image also in my books and in other sites (like this one: http://www.egglescliffe.org.uk/physics/particles/parts/Image14.gif ), so it's probably correct. --sasq777

Both images (the one in question and the one additionally presented here) are incorrect. That said image in question, which is the second on this article, seems to give an undefined, imaginary particle between the electron and positron, with additional electromagnetic interactions which fail to give any product. The other presented image - http://en.wikipedia.org/wiki/Image:Electron-positron-annihilation.svg - is also incorrect, as it presents both the electron and positron with the same charge. Neither give the correct end result, which is a charm and anti-charm. Pastaguy12 (talk) 22:03, 26 March 2013 (UTC)[reply]

I concur! Feynman diagrams are drawn with time increasing to the right and space increasing upward. That way it makes it a lot easier to identify particles and antiparticles:

* If the arrow points in the direction of increasing time, it's a matter particle 
* If the arrow points in the direction of decreasing time, it's an antimatter particle

The following link demonstrates why we should stick with convention. The author of the question saw a diagram herein and became confused enough to pose a question to "Physics Stack Exchange": Confusion resulting from the odd orientation of a Feynman Diagram on Wiki (The chosen answer should indicate that the image should be mirrored before rotating, which shows how far out of whack the image is.)

See http://www.vega.org.uk/video/subseries/8, Lecture 3: "Part 3: Electrons and their Interactions." for a definitive source.

Hpfeil (talk) 17:42, 29 January 2014 (UTC)[reply]

I find this Feynman diagram completely useless. There is NO standard model particle that can represent the line from the electron and positron. That is probably why it is not labelled. In other versions of the diagram I find on the web it is labelled as an "e" or "e*". Which is of course not a standard model particle, but rather represents some sort of mixed state. But what exactly is that mixed state?

An electron (or positron) can emit a photon at anytime. That is known as bremsstrahlung radiation. However, emitting that gamma ray only changes the spin of the electron (or positron). It is still a charged particle. So it seems almost immaterial to this diagram, as emitting a photon is not going to change an electron to an "e" or "e*" whatever those are.

We really have two direct possibilities for interaction between the two particles. They can interact directly, with no line in-between and produce a Z boson... Or they can each emit a neutrino and interact indirectly via a W+ or W-. That is really it for the interaction that will change an electron (or positron) to something else. Now it is quite possible there are loop diagrams where the neutrinos are virtual. Or a Z decay channels that results in a chain of virtual particles and ultimately results in two photons. The first seems possible, the second seems highly improbable, as I would expect most decays to follow the Z decay lineshape not resulting in dual photons. Feynman diagrams that showed the interactions that can lead to dual gammas would be useful. The included diagram is not. In fact it makes me question if it is even correct to say dual gamma rays with 100% of the energy and momentum would be a possible outcome. Bill C. Riemers (talk) 16:04, 25 October 2015 (UTC)[reply]

Oh I think I understand what "e" or "e*" means. It means it is a virtual electron (or positron). It has non-physical momentum/energy total so that it can react with a positron (or electron) to produce simply a gamma ray without violating conservation of energy and momentum. So essentially is a mixed state as I suspected, but it is not a mixed state of bosons and neutrinos like I thought. Flip the axis on the diagram, and the appropriate labeling and then it would be correct. There should be text in the article that describes what the e/e* is. Bill C. Riemers (talk) 16:47, 25 October 2015 (UTC)[reply]

Fundamental interaction involved

I wonder which fundamental interaction is involved in the electron-positron annihilation. Weak? Thanks, Warut (talk) 17:25, 17 January 2008 (UTC)[reply]

For annihilation to two photons, it is electromagnetic; that should be clear from its Feynman diagram. The weak interaction would be responsible for annihilation into a Z⁰ at high energy. 68.34.20.71 (talk) 22:23, 6 December 2009 (UTC)Nightvid[reply]

Merge with annihilation

I see no reason to have a special article on electron-positron annihilation any more than muon-antimuon, proton-antiproton, or any other. There is a large amount of overlap and both articles are rather short. 69.140.13.88 (talk) 17:47, 22 December 2009 (UTC)Nightvid[reply]

I agree. More than one overlapping articles, practical for the same theme have negative effect in easy use of Wikipedia. Only if an article grows too much (at least 30kB), we must start to think to saperate it in more than one and only with a very good connection between the particles.

--Vchorozopoulos (talk) 23:27, 18 January 2010 (UTC)[reply]

I agree also. This article's content should be merged into the Annihilation article. CosineKitty (talk) 02:11, 21 January 2010 (UTC)[reply]
I suggest that in that case Annihilation get cleaned up and improved before the merge happens. Otherwise a relatively well written article, Electron–positron annihilation, is just being dumped into a relatively poorly written one, Annihilation. The material about renormalization in Annihilation seems like it was just stuck in there because someone wanted it there. There is no logical development of the topic of renormalization in the context of the article. Since WP articles are also supposed to be written for a general audience, it seems goofy to me to have this highly technical field-theory stuff stuck in the article in random places. There are other, similar problems with the Annihilation article, e.g., sticking in "(coherent superposition)" in one place. The whole thing reads like it was written by a first-year grad student who just came home from his first field theory lecture, but hasn't understood it well enough to explain it to a general audience.--75.83.69.196 (talk) 22:04, 23 January 2010 (UTC)[reply]
I disagree with the merge. While article length is a valid consideration, the electron-positron annihilation deserves its own article. This particular reaction is very important in high-energy astronomy. I am concerned that additional annihilation reactions and the reverse reactions haven't been included in the Annihilation article. Reversal of the electron-positron annihilation reaction should be included in it as well. I hope this helps. Marshallsumter (talk) 23:37, 24 April 2010 (UTC)[reply]
I disagree with the merge. Annihilation is a large topic, of great importance to physics. Electron-positron annihilation is only one part of a broader and more complex set of annihilation phenomena. To think that all annihilation is e+e- is very simplistic. Instead I suggest that the article on Annihilation should reference other articles on some of the more prominent other annihilation effects. Strangely, I see that such articles did once exist e.g. Proton antiproton annihilation, because the links are still there. But the articles themselves have been deleted. Strange. John Pons (talk) 07:28, 13 September 2011 (UTC)[reply]

I miss a more technical section...

It would be nice to have some formulae of the process: the expression of the first order feynmann amplitude, the cross section and its non- relativisic limit... I'm not confident enough to write it myself, though. — Preceding unsigned comment added by Mcasariego (talk • contribs) 20:11, 7 September 2014 (UTC)[reply]

"Pair Annihilation" listed at Redirects for discussion

The redirect Pair Annihilation has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2023 December 27 § Pair Annihilation until a consensus is reached. Jason Quinn (talk) 19:46, 27 December 2023 (UTC)[reply]