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This feynman diagram represents the elastic proton scattering of electrons.

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The $e^-$ has an initial momentum $k$ and a final $k^\prime$. The circle represents that the target(the proton that is) has a structure and $q$ is the momentum transfer between the electron and the proton, which is mediated by a photon.

What does the $p$ and $p^\prime$ lines mean?

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p roton and p' roton? –  David H Jun 30 '13 at 8:27
    
@DavidH: Thank you very much for your comment! So what does the circle mean? The proton is suppossed to be at rest before the scattering, but if $p$ is the proton before the impact, this diagram shows that it is a moving target. That's the point where I am confused! –  Thanos Jun 30 '13 at 8:37
    
There's a reason I ended my comment with a question mark. I know very little about QFT. –  David H Jun 30 '13 at 9:02
    
@DavidH: I see. I thought, it was too obvious for you! Sorry! –  Thanos Jun 30 '13 at 9:14
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I think you missed my point. The four-velocity of a particle can never be zero - it always has the value $c$. No Feynmann diagram would ever show a stationary particle because no such thing exists. –  John Rennie Jun 30 '13 at 9:30

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The proton has no single line, because it is no elementary particle. I would assume that the three quarks (up,up,down or short uud) that are part of the proton stand for each line. The p and p' stand for the in and outgoing proton.

The circle is some interaction that may not yet be important for your calculations and will be added later. That it was done in my course on QFT.

An example for this interaction would be deep inelastic scattering

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Thank you very much for your answer! The problem was(thank you @JohnRennie) that I didn't take into account the fact that feynman diagrams are dealing with 4-vectors. So even if a particle is at rest, it has mass, so it can't be "stationary". –  Thanos Jun 30 '13 at 9:43
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I should probably also note, that in Feynman diagrams you see the time and space axis. This means, even though that a particle has a line, it does not mean that he is moving. He may move along the time axis and stay at the same space position which implies that he is standing still. Also you can keep in mind that QFT is based on Lorentz invariance, so that you always can Lorentz transform your system and change a particle in rest to a moving particle. –  physicsGuy Jun 30 '13 at 9:49
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this is a heuristic Feynman diagram, i.e. you cannot calculate the cross section using it like a standard electroweak diagram, because the proton it has a content and size. The circle represents that the proton "bag" is not a simple case and the calculation will have to take into account form factors. see phys.spbu.ru/content/File/Library/studentlectures/schlippe/… –  anna v Jun 30 '13 at 11:02

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