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What is the feynman diagram for electromagnetic force between an electron and proton?

by question dude
Tags: diagram, electromagnetic, electron, feynman, force, proton
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question dude
#1
Jan6-13, 06:52 PM
P: 77
Would it just be the same as with two electrons? (or any other pair of particles with the same charge)




I'm kinda in two minds, I suspect that is wrong because wouldn't the fact that they attract each other (instead of repelling) means that the diagram would be drawn differently?
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alemsalem
#2
Jan6-13, 07:06 PM
P: 159
no that wouldn't change the diagram it just changes the sign in the feynman rules for the vertex (ie for electron and -ie for the proton, or the other way around)..
but im not sure if you can use this diagram since the proton has constituents,, I think it works in an "effective" description.
question dude
#3
Jan6-13, 07:12 PM
P: 77
Quote Quote by alemsalem View Post
no that wouldn't change the diagram it just changes the sign in the feynman rules for the vertex (ie for electron and -ie for the proton, or the other way around)..
but im not sure if you can use this diagram since the proton has constituents,, I think it works in an "effective" description.
so the diagram is correct?

DMESONS
#4
Jan6-13, 07:27 PM
P: 27
What is the feynman diagram for electromagnetic force between an electron and proton?

I think proton (field) should be considered as external source and denoted by (X) in feynman diagram. The above diagram is valid for scattering of fundamental particles (leptons& quarks)
question dude
#5
Jan6-13, 07:33 PM
P: 77
Quote Quote by DMESONS View Post
I think proton (field) should be considered as external source and denoted by (X) in feynman diagram. The above diagram is valid for scattering of fundamental particles (leptons& quarks)
I don't really understand that.

btw this was a question in my A-level textbook (school)
Polyrhythmic
#6
Jan6-13, 07:41 PM
P: 342
The diagram you show may be useful (but maybe even more so misleading) for a heuristic approach to the scattering problem. Apart from the remark that you would have to include arrows on the lines, one has to notice two important facts:
1.) In quantum field theory, scattering amplitudes are calculated in terms of elementary particles. In the case of the interaction of an electron with a proton, one would have to represent the latter in terms of quarks.
2.) Even if you replaced the p (proton) with q (quark), you should keep in mind that you would only have one of a number of different Feynman diagrams relevant to your scattering process. What we have here is just one of the tree-level diagrams, furthermore one also has to account for loop corrections.
Polyrhythmic
#7
Jan6-13, 07:42 PM
P: 342
Quote Quote by DMESONS View Post
I think proton (field) should be considered as external source and denoted by (X) in feynman diagram.
This makes no sense, could you please clarify?
question dude
#8
Jan6-13, 07:45 PM
P: 77
Quote Quote by Polyrhythmic View Post
The diagram you show may be useful (but maybe even more so misleading) for a heuristic approach to the scattering problem. Apart from the remark that you would have to include arrows on the lines, one has to notice two important facts:
1.) In quantum field theory, scattering amplitudes are calculated in terms of elementary particles. In the case of the interaction of an electron with a proton, one would have to represent the latter in terms of quarks.
2.) Even if you replaced the p (proton) with q (quark), you should keep in mind that you would only have one of a number of different Feynman diagrams relevant to your scattering process. What we have here is just one of the tree-level diagrams, furthermore one also has to account for loop corrections.
This isn't about scattering or any high level physics-sounding stuff. This is just a question in my school textbook, asking me to draw a feynman diagram showing the electromagnetic force between a proton and an electron.
DMESONS
#9
Jan6-13, 07:49 PM
P: 27
Quote Quote by question dude View Post
I don't really understand that.

btw this was a question in my A-level textbook (school)
I've no idea about your course structure..but this scattering is known as deep inelastic scattering in which electrons scatter with proton constituents (quarks)..Any way, if your A-level textbook doesn't treat proton as a composite particle then you can consider it as a fundamental particle (electron with positive charge-positron)!
question dude
#10
Jan6-13, 07:53 PM
P: 77
Quote Quote by DMESONS View Post
I've no idea about your course structure..but this scattering is known as deep inelastic scattering in which electrons scatter with proton constituents (quarks)..Any way, if your A-level textbook doesn't treat proton as a composite particle then you can consider it as a fundamental particle (electron with positive charge-positron)!
yeah it does go in depth into explaining quarks, but thats several pages later on after this current question I'm doing

so if I assume the proton as being a fundamental particle with the opposite charge to an electron, is the diagram I posted correct?
DMESONS
#11
Jan6-13, 08:00 PM
P: 27
Quote Quote by question dude View Post
yeah it does go in depth into explaining quarks, but thats several pages later on after this current question I'm doing

so if I assume the proton as being a fundamental particle with the opposite charge to an electron, is the diagram I posted correct?
Yes..it is correct...don't forget to show the time follow (for example from bottom to up). Also, you should say if we consider the proton as a fundemental particle then the EM interaction between electron and proton is .. one you posted)
Polyrhythmic
#12
Jan7-13, 05:38 AM
P: 342
In this case, just don't forget to draw the arrows! However, this seems to me like a careless introduction of Feynman diagrams within a pedagogical text.
andrien
#13
Jan7-13, 07:46 AM
P: 1,020
Would it just be the same as with two electrons?
No,proton is an extended particle.An easy way to avoid structure is to use form factors i.e. modify the vertex factor -ieγμ to another one by using argument of lorentz covariance and gauge invariance in terms of two form factors F1(q2) and F2(q2) which depend only on four momentum of photon.It was first used by rosenbluth.The vertex factor at proton site should be something like on the basis of above invariance proportional to
γμF1(q2) +(SOME FACTOR)F2(q2) σμVqv,which must be used at proton vertex.
LastOneStanding
#14
Jan7-13, 12:07 PM
P: 718
andrien, the asker is in high school. If you're just going to give a highly technical answer that is obviously beyond the level of someone asking a question, you might as well not respond at all. In fact, it would probably be preferable to do so since receiving explanations that have no awareness of the audience just gets people frustrated.
LastOneStanding
#15
Jan7-13, 12:51 PM
P: 718
Quote Quote by question dude View Post
I'm kinda in two minds, I suspect that is wrong because wouldn't the fact that they attract each other (instead of repelling) means that the diagram would be drawn differently?
question_dude, there have been a few good answers here, but I think your above question indicates a bit of confusion that hasn't been addressed. It's important not take Feynman diagrams too literally. The only thing that matters physically is which bits are connected to what. So, when we draw the lowest order diagram for two electrons interacting, the fact that the legs for the two outgoing electrons are drawn pointing away from each isn't physically significant. So, for two electrons, this diagram (read left to right) is the same however you choose to rotate and twist the different pieces of it. What tells us two electrons repel each other is when we actually use the diagram to calculate the probability that two electrons with particular initial momenta will scatter off each other into two electrons with two particular final momenta. If we do this, we learn that it is more likely that by exchanging a photon the electrons will be scattered away from each other than scattered towards each other—a repulsive force.

We can do the same thing with electron-positron scattering. A positron is a type of anti-matter, which means it has the same properties as an electron but opposite charge. Now, when you hear about anti-matter, you usually just here about it annihilating matter—so you might think the electron-positron interaction just destroys the two particles (and releases energy). That's one possibility, but they can also just scatter of each other by Coulomb attraction (or, rather, the quantum interaction that creates Coulomb interaction on large scales). The lowest level diagram for that process looks like this (read left to right). Notice how similar this looks to the other diagram! In fact, this diagram isn't quite right as the arrows on the two top legs should point the other way. Those arrows are supposed to tell us whether its an electron or a positron, (not which way its travelling. Unfortunately, this is the best I could find. In any case, the electron and the positron attract each other, even though it doesn't look like it here! To actually see that they attract, you again would need to use this diagram to calculate the scattering probabilities for different starting and ending momenta. You would see the final momenta corresponding to the two particles being attracted to each other are the most likely. We can also, by making some approximations and comparing with non-relativistic quantum mechanics, show that these two diagrams give rise to exactly the same "Coulomb's law" you learn in school for the force between like or different charges.

However, as someone else mentioned, these diagrams aren't the whole story. They're only what are called "lowest order" diagrams (and even then, not all of them). To calculate the total probability something will happen, we have add up the contribution of every possible way it can happen. These diagrams are just some of the simpler ones. For electron scattering, we also have this diagram (read it from bottom to top—I'm sorry, there's no consistent convention for this): two electrons exchange a photon which along the way spontaneously turns into an electron and positron which very quickly annihilate back into a photon. There are an infinite number of diagrams with more and more vertices all contributing to the same process—and, in theory, we have to add them all up. This isn't possible, so instead we just approximate by adding up a few of the diagrams with small numbers of vertices. The higher order diagrams don't contribute much anyways because they have very low probabilities of happening.

Finally, as other people have said, the diagrams for an electron scattering off a proton are not as straightforward because protons are made of quarks, and those are what the electrons actually interact with. Electron-positron scattering is much simpler and makes the same point. However, if you add up the contributions of the various low order (two vertex) diagrams for a proton and an electron, you can represent them in a sort of short hand with something looking like the electron-electron and electron-positron diagrams (though instead of arrows on the lower two legs, which mean electron or positron depending on the direction, you would just have to write 'proton'). Just keep in mind that, unlike the two examples above, this is not an elementary process but a combination of multiple processes.
samalkhaiat
#16
Jan7-13, 02:30 PM
Sci Advisor
P: 908
Quote Quote by question dude View Post
Would it just be the same as with two electrons? (or any other pair of particles with the same charge)




I'm kinda in two minds, I suspect that is wrong because wouldn't the fact that they attract each other (instead of repelling) means that the diagram would be drawn differently?
No, is the correct answer. However, the problem is A-level textbooks contain a lot of garbage and written by non-expert. So, if you answer by No, which is correct, you might not get any mark.

Sam
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Vanadium 50
#17
Jan7-13, 07:03 PM
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There's a lot of brush in this thread that's worth sweeping away.

1. These are not Feynman diagrams without arrows. Full stop.

2. While a proton is a composite particle, there's absolutely nothing that prevents one from calculating this using a (properly arrowed) diagram. This is an elastic process, and the fact that there are also inelastic (deeply so, in some cases) processes is irrelevant if one is calculating an elastic process.

3. At leading order (one photon exchange) there is no difference between same-sign and opposite-sign. You have the same motion scattering off a hump as off a well. It's only with multiple photon exchange that it makes any difference.
LastOneStanding
#18
Jan7-13, 08:41 PM
P: 718
Quote Quote by Vanadium 50 View Post
3. At leading order (one photon exchange) there is no difference between same-sign and opposite-sign. You have the same motion scattering off a hump as off a well. It's only with multiple photon exchange that it makes any difference.
This is absolutely false. You can take the non-relativistic limit of the lowest order electron-electron and electron-positron scattering processes—which are single photon exchanges—and show that, in the first case, you get a repulsive Coulomb potential and, in the second case, you get an attractive Coulomb potential. You certainly do not need higher order corrections for the sign difference to be manifest. Consult p.125 of Peskin & Schroeder if you need to refresh yourself on how the calculation goes.


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