This Â
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
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 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)
-Yeah, Chris is right, matter-antimatter reactions are 100% energy conversion in cooperation with Einstein's . 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)
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
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)
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)
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)
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)
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
Wouldn't it be helpful for the following equation to be a part of this article?
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)
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.
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)
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)
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)
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)
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)
I wonder which fundamental interaction is involved in the electron-positron annihilation. Weak? Thanks, Warut ( talk) 17:25, 17 January 2008 (UTC)
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
-- Vchorozopoulos ( talk) 23:27, 18 January 2010 (UTC)
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. â Preceding unsigned comment added by Mcasariego ( talk ⢠contribs) 20:11, 7 September 2014 (UTC)
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)
This Â
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
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 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)
-Yeah, Chris is right, matter-antimatter reactions are 100% energy conversion in cooperation with Einstein's . 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)
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
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)
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)
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)
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)
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
Wouldn't it be helpful for the following equation to be a part of this article?
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)
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.
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)
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)
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)
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)
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)
I wonder which fundamental interaction is involved in the electron-positron annihilation. Weak? Thanks, Warut ( talk) 17:25, 17 January 2008 (UTC)
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
-- Vchorozopoulos ( talk) 23:27, 18 January 2010 (UTC)
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. â Preceding unsigned comment added by Mcasariego ( talk ⢠contribs) 20:11, 7 September 2014 (UTC)
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)