Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Feynman Diagram of a Coulombic Attraction?

  1. Jun 9, 2010 #1
    Whenever I see examples of the diagram for the Coulomb interaction it always seems to be two electrons interacting via an electron and being repulsed. The diagram looks intuitive in terms of momentum conservation.

    I was wondering how (for example) a proton - electron interaction, and subsequent attraction, would be drawn? Does the electron still emit a photon in the direction of the proton?

  2. jcsd
  3. Jun 10, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi mk17! Welcome to PF! :smile:
    (you meant "via a photon" :wink:)

    "the diagram" ?

    There's no such thing as "the diagram" for an interaction … there are infinitely many diagrams for the same interaction, of which that one is simply the simplest.
    The Feynman diagrams would be the same.

    But the virtual photons and virtual electrons in the middle of the diagrams (twice as many electrons as photons in most of the diagrams) don't repel or attract, beacuse they don't exist

    virtual particles don't exist (the clue's in the name :wink:), they're just maths.
  4. Jun 10, 2010 #3
    There difference between virtual and real particles is frame-dependent. Those who deny the existence of virtual particles should also deny the existence of 'real' ones.
  5. Jun 10, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hi Dmitry67! :smile:
    uhh? :confused:

    Are you saying that there is a frame in which the virtual photons and virtual electrons in a Feynman diagram for electron-electron repulsion are real? :confused:
  6. Jun 10, 2010 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What do you mean? A virtual particle is one for which

    [tex] P_\mu P^\mu \neq m^2 [/tex]

    and this is a frame independent statement.
  7. Jun 10, 2010 #6
    In some accelerated frames they are real. Different accelerated frames don't agree on the number of real and virtual particles of the same macroscopic events. Check Unruh effect for example. Hawking radiation is another one (free falling body does not see that radiation).

    My personal opinion: Feynman Diagrams describe what is actually happening. Sum on infinite number of histories. However, as there is no collapse and measurement, there are no 'real' particles, no incoming particles and no particles which go away to be 'registered'. Our world is just one infinite Feynman diagram without borders.
  8. Jun 10, 2010 #7


    User Avatar
    Science Advisor
    Homework Helper

    Are you saying that there is an accelerated frame in which the virtual photons and virtual electrons in a Feynman diagram for electron-electron repulsion are real? :confused:
  9. Jun 10, 2010 #8
    No, because if they participate in the repulsion, then they go from one particle to another without any absorption -> they could not leave any tracks (there are cases however when virtual particles create some macroscopic effects)

    But, on the other hand, claim that 'they are just pure mathematics' is too strong.
  10. Jun 10, 2010 #9


    User Avatar
    Science Advisor
    Homework Helper

    Sorry, but that doesn't make any sense … what does "without any absorption" mean, and what does an accelerated frame have to do with it? :confused:
  11. Jun 10, 2010 #10
    Ok, I was trying to say that these virtual photons can not be registered directly, but their effects are registered indirectly. But it is not enough to say that they are 'not real' and 'pure mathematics'.

    For example, wavefunction itself can't be registered directly. However, in many interpretations wavefunction is real. So the claim 'virtual particles are pure mathematics' for me is equivalent to Copenhagen Interpretation (where it was valid, and as CI was dominating for a long time, people got used to that point of view).
  12. Jun 10, 2010 #11
    Feynman diagrams do not sketch a real world situation where particles collide. They are a pictorial way of representing complicated mathematical expressions in terms of a few primitive objects (propagator, vertex, external source).
  13. Jun 10, 2010 #12


    User Avatar
    Science Advisor
    Homework Helper

    hmm … I notice that you're now ignoring the virtual electrons in the same interaction …

    there are twice as many virtual electrons as virtual photons in most of the diagrams …

    I suspect that you've heard that the electromagnetic field is "mediated by" photons, and so the virtual electrons in the diagrams somehow don't "mediate" it.

    Well, as Dickfore :smile: says, Feynman diagrams are a pictorial representation of mathematical expressions. Virtual photons and virtual electrons participate in the same way, and since an infinite number of diagrams is needed, so an infinite number of virtual particles is needed … so how many virtual photons and electrons do you think there are when two real electrons repel, and when do they appear?
  14. Jun 10, 2010 #13
    Whats about the repulsion between, say, 2 electrons with not only 1 virtual photon, but including the loop of virtual e+/e-? and (on very short distances) even t+/t- (top quarks)? Or even ?+/?-, where ? is not-yet-discovered supersymmetric particle? Why that 'pure mathematics' mimics the 'real' world so literally?
  15. Jun 10, 2010 #14
    If we had a way of determining quantum amplitudes directly (scattering processes, the ground state of a system or just the propagation of a particle), we would never need the concept of a virtual particle. It's just a mathematical side of perturbation theory. For example, in some integrable quantum field theories (e.g. certain conformal field theories) you do not need perturbation theory so to whole concept of Feynman diagrams and 'intermediate' states is completely circumvented! The concept of a virtual particle (= 'intermediate' state) never pops up!

    Perturbation theory, in the end, is nothing but trying to solve an interacting theory using a non-interacting formulation -- completely similar to using a Taylor series to approximate an analytic function. The different terms in the Taylor expansion is similar to the different Feynman diagrams (which contain the virtual particles) in the perturbation series ...
  16. Jun 10, 2010 #15


    User Avatar
    Science Advisor

    The concept of Coulomb force is gauge dependent!

    In the Lorentz gauge you will not see a Coulomb potential at all. In Coulomb gauge (that's where it's name is coming from) there is a "photon term" plus a static Coulomb potential term" in the Hamiltonian. So the static Coulomb potential is not due to virtual photons but looks exactly as in electrostatics.

    The choice of the gauge is arbitrary, but it should fit to the problem you want to study. For (relativistic) scattering experiments the Lorentz gauge is nice, but e.g. for Lamb shift calculations it's awful.
  17. Jun 10, 2010 #16
    That gauge choice actually named after Ludwig Lorenz (no t!), and not Hendrik Lorentz ;-)
  18. Jun 10, 2010 #17


    User Avatar
    Science Advisor

    correct; sorry for that
  19. Jun 10, 2010 #18
    Re: Welcome to PF!

    I did indeed, oops.

    Anyway, thanks for the replies everybody, obviously I was thinking Feynman diagrams have a physical significance that they don't.

    Eventually, hopefully, I'll be competent enough to understand some of the mathematics behind them and actually be able to use them! :smile:
  20. Jun 10, 2010 #19


    User Avatar
    Science Advisor

    The idea of using diagrams for perturbation theory spread to different areas either, so quantum chemistry has e.g. Hugenholtz diagrams and Goldstone diagrams.
  21. Jun 11, 2010 #20
    In SM - yes.
    But it is just one interpretation.
  22. Jun 11, 2010 #21
    Kepler law can be used to calculate positions of planets. And it works. However, you can't claim that 'gravity is just pure math used to calculate the orbits' just because the results match the calculations.

    Another example: using the same logic you use, you can deny the existence of Gravitational waves, saying that 'they are just pure math used to calculate the effects of distant rotating masses on the macroscopic objects'
  23. Jun 11, 2010 #22


    User Avatar
    Science Advisor

    One could even argue that in reality only virtual particles exist, no real particles! Why? Because if we detect a particle (and this is the only way to see if it's here) the particle is absorbed by the detector and therefore it is an internal line, a propagator, ending at a vertex in the detector.

    => I agree with Dmitry67 that ontology based on mathematical concepts in physics is not always straightforward.

    I do not deny that virtual particles may exist - whatever this means - I only want to warn people not to take everything for granted which is written in popular books (we have some threads here discussing the violation of energy conservation due to vacuum fluctuations; a misleading concept).
  24. Jun 11, 2010 #23


    User Avatar
    Science Advisor
    Homework Helper

    Gravity is a physical effect, not a particle.

    This effect definitely exists, because it shows itself in the results (to use your word).

    The analogy in electromagnetism is that the electromagnetic field is a physical effect which shows itself in the results, and definitely exists, but the virtual particles are only in the maths.

    (and the continuation of the gravity example would be that, although gravitons may well exist, virtual gravitons don't exist and aren't "exchanged" in gravitational attraction any more than electrons or photons are in electromagnetic interaction :wink:)
    No, because gravitational wave theory says that the wave has a definite (ok, slightly fuzzy) position at any time … the theory itself (of gravitation) says that the wave exists: quantum field theory does not say that virtual electrons and photons exist!

    A more interesting example is, do vortices (in flowing water) exist? They have definite position (you can even literally pick them up and transfer them to a different stream!), but there's no "vortex particle", nor even a separate state of matter … so we could argue that they're just a mathematical effect (a singuarity) with no physical existence.

    However, virtual particles aren't even analogous to vortices …

    they don't have a definite position, not even a really fuzzy one.

    Nothing in quantum field theory even purports to describe where or when they appear, or in what numbers.
  25. Jun 11, 2010 #24


    User Avatar
    Science Advisor

    Back to the original question. I think that at the level of drawing a diagramm for an exchange of only one photon really makes no difference between attraction and repulsion. It would correspond to a scattering in the first Born approximation and the angular distribution of the scattered particles would be identical, whether they are of like or unlike charge.
    To see a difference between attractive and repulsive interactions, one has to consider also processes where more than one photon is exchanged.
  26. Jun 11, 2010 #25
    There is absolutely no difference between, say, G-wave affecting the detector, and 2 Casimir plates attracting. In both cases 2 macroscopic objects change their position because of something, not directly detectable.

    As well as real ones - they don't have definite position too. (You can argue that you can pinpoint their position exactly in the detector, but it is so Copenhagen and obsolete! That process is an interaction of a particle (wave) and the 'particles' of the detector, leading to some macroscopic events, which we interpret as 'measuring the position')

    Yes, you know that there are 3 quarks in a proton, but you can't say how many gluons are there. But not being able to count something cant deny the physical existence of that something. Do you deny the physical existence of Higgs condensate?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook