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B Brownian motion from virtual photons?

  1. Apr 30, 2017 #1
    This a really simple question: If I have, say, 2 ions close to one another, and measure their repulsion very precisely, is the force constant, or is it a series of little pushes caused by individual virtual photons?

    I know there are many misunderstandings about virtual particles, but I'm not sure if this is one of them.
     
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  3. Apr 30, 2017 #2

    Drakkith

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    No, the force will be a smooth and continuous function of the distance between the ions. You will not see individual "pushes" by virtual photons.
     
  4. May 1, 2017 #3

    jfizzix

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    Virtual photons are an artifact of popular formulations of quantum electrodynamics. Their existence has not been established experimentally because they cannot be measured directly. They may be regarded as a mnemonic device for doing calculations of interactions of charged particles with the quantum electromagnetic field.

    In the popular formulation:
    The interaction of two ions repelling each other is described not just by their exchanging virtual photons, but it's much more.

    One can calculate the probability of a given change in the ions momenta (due to repulsion), and find that their mean momenta point further away from one another as time progresses.
    The actual calculation involves a sum over amplitudes of all conceivable processes that begin with a given initial state, and end with the given final state.
    These processes include:
    - the exchange of one virtual photon
    -the exchange of a virtual photon, which decomposes into a particle-anti-particle pair, only to annihilate again before reaching the second ion
    - the previous interaction, where the particle-anti-particle pair exchanges its own virtual photon
    -and so on
    and so on...

    The full calculation of these interactions involves an infinite series of such interactions, which get exponentially less probable as the interaction gets more complicated, but nonetheless contribute to the exact result.

    Altogether, these result in a smooth change in mean momentum over time instead of individual discrete kicks.
     
  5. May 1, 2017 #4
    It will be a smooth and consistent repulsive motion. All in precise accordance with the specific charges of the respective ions. Whether they be cations or anions.

    Besides, does not Brownian motion refer to particles suspended in a liquid medium?

    Which decidedly is not the case in your ion scenario.
     
  6. May 1, 2017 #5
    I was imagining a big, measurable ion being pushed around by a sea of invisible virtual photons.
    Anyway, since you all agree that the push/pull is continuous, I'll take your word for it, but honestly, it doesn't make ANY sense.
    I know that there are higher-order diagrams, but I'm not computing the actual value, so I just neglect those. Also I'm pretty sure that the first, simplest Feynman diagram contributes more than 99% of the force.
    So, Feynman diagrams don't compute probability of the exchange, but strength of the exchange? I've never heard of such a concept in QM.

    It would be awesome if someone could write an Insight showing the actual calculation :wink: but I guess I'll stop here :confused:
     
  7. May 1, 2017 #6

    Drakkith

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    Do you think you'd be able to understand the math used? It may be very advanced. I'm not sure I'd be able to get through such an article. :biggrin:
     
  8. May 1, 2017 #7

    PeterDonis

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    And what everyone is telling you is that that's not a good way to imagine it. Virtual particles are just a model; they work for some purposes, but not for others. This is one of the purposes for which they don't work well.

    If the right answer doesn't make any sense in the model you've chosen to use to imagine things, perhaps you ought to consider finding a better model.

    Do you see the contradiction between these two statements?

    Even putting that aside, there is a more fundamental issue here. Feynman diagrams, virtual particles, etc. are tools used in perturbation theory. Perturbation theory assumes that we are computing small changes around a "base" process, where the "base" process is that nothing happens. So the "base" Feynman diagram, the one with the largest amplitude, is the diagram in which nothing happens.

    What you are considering is a static force, and a static force is not accurately modeled as a small perturbation around nothing happening. So perturbation theory as it is usually used is not a good tool for this problem. (The fact that pop science treatments often describe static forces using virtual particles does not change that fact.) You can construct perturbative models in which the "base" process is not nothing happening; for example, I believe that in QED models used to predict the Lamb shift, the "base" process is the static field of the nucleus. A model something like this might work for looking at the static force between ions. (I don't know if such a model has been done.) But in that case, the "base" process is not an exchange of virtual particles; it's what happens when no virtual particles are exchanged at all. So virtual particles still aren't modeling the static force.
     
  9. May 1, 2017 #8

    ftr

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  10. May 1, 2017 #9
    I'd love to but I can't really remember seeing any other model explaining electro-magnetic interaction than the one with virtual photons.
    Is there some I should try to find, or should I create one?

    I was trying to say that the higher-order diagrams aren't really important for the issue I'm dealing with, since they affect the quantity, not quality of the interaction.

    Ok, I'll try to find how it is usually modelled in QM. I hope it is modelled.
    I went through 5 pages of Google results, no luck so far. John Baez has an article where he kind-of uses virtual photons to explain the repulsion between electrons.
     
  11. May 1, 2017 #10

    PeterDonis

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    Note that the article says "for bound states the method fails"? Bound states are an example of static forces ("static" because things don't change with time, the bound states just stay the same). It's not actually clear to me whether the OP intends the ions to be isolated and repelling each other, or in some kind of bound state like a metal.

    Also see the response I'm about to post to SlowThinker.
     
  12. May 1, 2017 #11

    PeterDonis

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    The Wikipedia article that ftr linked to in post #8 derives the Coulomb potential (not force) for the electrostatic case--basically, two charges sitting at rest relative to each other have an interaction potential energy between them that is positive for like charges and negative for unlike charges. The observed force between the charges is the gradient of this potential. This result is derived using the path integral (but not actually evaluating it--see below), so it can be interpreted as a model using virtual particles--but that interpretation has serious limitations (see below). Similar results are derived in many QFT textbooks (e.g., Zee's Quantum Field Theory in a Nutshell derives it in an early chapter).

    Note that I said "potential (not force)" above. The force between the charged particles is the gradient of the potential--but this is just like the ordinary classical case of a continuous potential energy leading to a force. In other words, QFT says that the force between charged particles is smooth, not "bumpy". That is one of the serious limitations of the "virtual particle" interpretation of the path integral--that it leads to a picture of what is happening (virtual particles "bumping" things) that does not match the actual prediction (or experiment). But nothing forces you to interpret the path integral using virtual particles; the only necessity in the model is the path integral itself.

    The issue I was referring to is not that higher order diagrams have to be included; it is that, for this particular path integral (as you will see if you look at the Wikipedia article referred to above), the concept of "higher order diagrams" doesn't really apply to begin with. That's because we aren't actually evaluating the path integral; we are only using it to derive the propagator ##D(k)##, and then integrating the propagator over all ##k## to obtain the potential.

    In other words, we aren't even calculating the amplitude for one of the ions to emit or absorb a virtual photon, which is what evaluating the path integral would give us, because that doesn't correspond to anything we can actually measure in this situation; instead, we are calculating the potential energy between the ions due to quantum fields, and then, as above, taking the gradient of that potential energy to obtain the force. This is the other serious limitation of the virtual particle interpretation in this case: that interpretation, to the extent it makes sense, only makes sense if we are evaluating the path integral to compute amplitudes that we are going to compare with experiment, and in this case we aren't even doing that.
     
  13. May 1, 2017 #12

    PeterDonis

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    As my previous post points out, this isn't actually what is done to compute the interaction potential energy between charged objects in QFT.
     
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