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

Can you answer this virtual photon paradox?

  1. Apr 18, 2015 #1
    In order to get my head around virtual particles I've created the following question. Maybe some bright person can answer this. If I can answer this question I feel I would have the required understanding to explain virtual particles to other people.

    In the diagram below, A is a radioactive source emitting positrons and B is a radioactive source emitting protons. Both sources emit particles within a small energy range. The covers of A and B are lifted simultaneously at an agreed time for 1 minute and then recovered. During this minute some virtual photons travel from the positrons to the protons faster than the speed of light. The particles are detected at A' and B'.

    Question: Given that even though the Feynman propagator is non-zero for space separations (virtual photons travel faster than the speed of light):

    (a) If the distance d is greater than 1 light minute the distributions of particles detected at A' and B' are unaffected by each other. In other words no signal passes from the protons to the positrons faster than the speed of light.

    (b) If the distance d is less than 1 light minute the protons and positrons feel a repulsive electromagnetic force and the distributions are pushed away from the centre.

    (c) At a later time if the distributions of particles detected at A' and B' are compared, even if d is greater than 1 light minute will there be any mathematical correlation between the distributions?

    Try to answer this question using the Feynman propagator ΔF(x-y). And try to express the distributions of the particles detected at A' and B' as functions of the separation distance d.

    virtual_photon.png
     
    Last edited: Apr 18, 2015
  2. jcsd
  3. Apr 18, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Virtual particles are not real, they do not "travel".

    I don't see where such a correlation would come from.
    It is not necessary to use quantum field theory here - classical fields (which are a limit of QFT) are sufficient.
     
  4. Apr 18, 2015 #3

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    There's your problem. Virtual particles do not travel.
     
  5. Apr 18, 2015 #4
    Hi.
    Neither of these two answers are helpful. Specifically I want an equation derived from QFT which gives the distribution of particles at A' and B' making use of the Feynman propagator to shed light on the meaning of the nonzero space-like value of the virtual propagator. i.e. to prove that information cannot travel faster than light. Unless you can do this please do not answer.

    Whether virtual particles "travel" or not is irrelevant. Virtual particles appear in the Feynman diagrams for the scattering of particles in QFT and are used to calculate the electromagnetic repulsion of charged particles. It doesn't matter if you call this "travelling" or "zoobling" or "gigglebobbing". Just saying "virtual particles don't travel" adds nothing to the discussion.
     
    Last edited: Apr 18, 2015
  6. Apr 18, 2015 #5
    Don't all your queries rely on the premise 'virtual particles travel faster than light'?
     
  7. Apr 18, 2015 #6
    The queries relate to the fact that the Feynman propagator for a virtual particle DF(x-y) is non-zero for space-like separations. I want to understand how using this in the calculations results in solutions where causality is preserved. If you can answer this question then please go ahead.

    DF(x-y) is often explained as the amplitude for a particle to "travel" from x to y. Hence non-zero values at space-like separations represents a virtual particle travelling faster than the speed of light. I don't much care about the semantics of it (I leave that to philosophers). Just the mathematics.
     
  8. Apr 18, 2015 #7

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    The reason you think these answers are not helpful is that you want to hold on to a premise that is false. Until you give that up, you will not be able to increase your level of understanding.
     
  9. Apr 18, 2015 #8
    What exactly is the false premise that nuclearhead is holding onto? As he said in the follow up post, the positrons and protons interact with each other via QED which, to first order in Feynman diagrams, is depicted as the exchange of a virtual photon. The ontology of that diagram is irrelevant for what is a purely computational question. That internal line in the diagram represents the Feynman propagator, and so, should, in principle, be used in the calculation of...whatever it is he's trying to show (even if it's just indirectly by first showing how Coulomb's law, which I agree should be sufficient here, follows from the first order diagrams). I don't entirely follow what nuclearhead is asking in part (c)—and it seems the whole scenario is a bit overcomplicated—but it's pretty obvious to me that getting hung up on his description of the first order Feynman diagram as a virtual photon "traveling" from A to B is missing the point entirely.
     
  10. Apr 18, 2015 #9

    bhobba

    User Avatar
    Science Advisor
    Gold Member

  11. Apr 19, 2015 #10
    I suspect that the problem is with the last sentence of the o/p. The Feynman propagator is appropriate for calculating S matrix elements i.e. time symmetric processes. The problem, as formulated, isn't time symmetric. Perhaps a treatment involving the retarded propagator could be given.
     
  12. Apr 19, 2015 #11
    I think you could be right. I have a feeling that the Feynman propagator is a mathematical simplification that is only applicable if each end of the propagator can take on any value of time (and space?). Hence why in scattering processes we assume the particles have "come in from infinity".

    For example to fourier transform the momentum propagator 1/(k.k-m^2) to position space one needs to integrate over ALL space and time.

    (Perhaps instead a Feynman propagator we get a propagator that does not have nonzero spacelike values unless d << 1 light minute. And the longer we run the experiment for the more the propagator we use approaches the Feynman propagator? Perhaps the Feynman propagator is appropriate only if the experiment were to run for an infinite amount of time. This is just a guess).

    In my example I deliberately ran the experiment for a length of time such that no particle travelling less than the speed of light could get from the positron rays to the proton rays.

    Since this complicates the mathematics as we can no longer assume the particles "came in from infinity".

    I just wondered how one would translate my example into Feynman diagrams (if this is at all possible) and show that if d >> 1 light minute then no coulomb force is felt (since no signal can travel faster than light). I'm surprised I can't find a worked example of this anywhere on the internet it seems such a basic example.

    You can ignore part (c).
     
    Last edited: Apr 19, 2015
  13. Apr 19, 2015 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I think your experiment runs longer than a minute in some way - those boxes do not shield net charges. All you get is a redistribution of the charges for a minute. There are no charges that appear suddenly. Switching on an electron gun would have the same effect. And I have no idea how you would include this process in QFT.
     
  14. Apr 19, 2015 #13
    Thanks for being honest.

    Well you can assume that the sources contain neutrons (or anti-neutrons) which decay into protons and electrons for example.

    I think maybe the paradox is similar to the statement that "if you removed the sun, the Earth would continue on it's orbit for a further 8 minutes." Whereas in reality you can't just "remove the sun".
     
  15. Apr 19, 2015 #14
    Charge conservation isn't the answer; consider the case where instead of a photon we have a massless scalar. A scalar need not couple to a conserved charge.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can you answer this virtual photon paradox?
  1. Virtual photons (Replies: 1)

  2. Virtual photons (Replies: 1)

Loading...