
#1
Jan613, 06:52 PM

P: 74

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? 



#2
Jan613, 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. 



#3
Jan613, 07:12 PM

P: 74





#4
Jan613, 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)




#5
Jan613, 07:33 PM

P: 74

btw this was a question in my Alevel textbook (school) 



#6
Jan613, 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 treelevel diagrams, furthermore one also has to account for loop corrections. 



#7
Jan613, 07:42 PM

P: 342





#8
Jan613, 07:45 PM

P: 74





#9
Jan613, 07:49 PM

P: 27





#10
Jan613, 07:53 PM

P: 74

so if I assume the proton as being a fundamental particle with the opposite charge to an electron, is the diagram I posted correct? 



#11
Jan613, 08:00 PM

P: 27





#12
Jan713, 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.




#13
Jan713, 07:46 AM

P: 973

γ_{μ}F_{1}(q^{2}) +(SOME FACTOR)F_{2}(q^{2}) σ^{μV}q_{v},which must be used at proton vertex. 



#14
Jan713, 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.




#15
Jan713, 12:51 PM

P: 718

We can do the same thing with electronpositron scattering. A positron is a type of antimatter, which means it has the same properties as an electron but opposite charge. Now, when you hear about antimatter, you usually just here about it annihilating matter—so you might think the electronpositron 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 nonrelativistic 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. Electronpositron 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 electronelectron and electronpositron 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. 



#16
Jan713, 02:30 PM

Sci Advisor
P: 815

Sam 



#17
Jan713, 07:03 PM

Mentor
P: 15,570

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 samesign and oppositesign. 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. 



#18
Jan713, 08:41 PM

P: 718




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