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

How a force carrier particle gets a message to its owner particle?

  1. Sep 19, 2013 #1


    User Avatar

    Can somebody please give me a little more detailed explanation of a simple Feynman diagram? It's not clear to me why the owner particle experience any force?

    Let's check out 2 electrons interacting by a virtual photon.
    1) 1st e is moving along, throwing gazilions of virtual photons but nothing is happening yet to its momentum. Even when it throws the photon that eventually will reach 2nd e, nothing happening just yet, right?
    2) A virtual photon of 1st e reaches the 2nd e and 2nd e got its momentum changed.

    Q: At which point 1st e got its momentum changed? I see several scenarios - please tell me which one is the correct one.
    a) at the moment when the photon is thrown by 1st e (despite this photon - like gazilions of other virtual photons thrown by this e - didn't interact with 2nd e yet, the 1st e somehow knows that it will, so, no point to loose time - that's QM for you!);
    b) after the photon will interact with 2nd e, 1st e will somehow instantaneously know it happened and will turn away;
    c) after the photon will return back to the 1st e;
    d) this process doesn't affect 1st e; it's up to 2nd e to throw another photon and hit 1st e to change 1st e momentum. Then 2 photons are needed to push 2 e away from each other.
    e) your version.

  2. jcsd
  3. Sep 19, 2013 #2


    User Avatar
    2017 Award

    Staff: Mentor

    What do you mean with "owner particle"?
    That is not a useful model of particle interactions. The exchange of virtual particles is a symmetric process, and it is meaningless to ask about the physical reality of particles on timescales smaller than those processes.
  4. Sep 19, 2013 #3


    User Avatar

    "owner" is the one that throws the force carrier particle. In my example it's the 1st e. It's still matters which one of 2 e thrown this specific photon that appears on the diagram, right? The photon doesn't just exist, it's moving from one e to the other one, right?

    My scenario d) is somewhat symmetric but requires 2 photons.

    I was afraid of "it is meaningless to ask..." reply but I anticipated it... If "throwing gazilions of virtual photons" (as a quantum representation of EM field) is not useful approach then you have to say that 1st e knew upfront where the 2nd e was and thrown just 1 photon in its direction - that sounds less likely to me...
  5. Sep 19, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    No, that is not a working model. The electrons don't "throw" virtual photons. The whole Feynman diagram is a single process, it is pointless to consider individual components as events.

    The whole approach to describe virtual photons like that does not work. They are virtual and not real, they are just a tool to calculate a process in quantum field theory.
  6. Sep 20, 2013 #5
    you should be careful with feynman diagrams.They are not drawn simply in the way you like.One of the subtleties is that if you consider two electrons interacting by exchange of virtual photon,then you should not say which one is emitting the photon and which one is absorbing it.There is no consideration of this,when you will consider the S-matrix way for writing this process,you already will have a time ordering which will take care of both cases in which first is emitting or the second one is emitting.It is taken care in a single feynman diagram.
  7. Sep 21, 2013 #6
    Besides, virtual photons do carry momenta even if they turn out to be reabsorbed by the same electron. The question seems to imply the opposite when it says
    something is happening to its momentum. the virtual photon's momentum is (temporarily) being subtracted from it. Momentum conservation applies to all vertexes of the diagram.
  8. Sep 22, 2013 #7
    One can more or less think of individual components as events, so long as one remembers that this is quantum mechanics and we are summing up over all the different ways/orderings in which these events can happen. And that the standard on-shell masses are the most resonant/probable, but are not otherwise required by intermediate states.
  9. Sep 24, 2013 #8
    Look at it the opposite way: starting at fields.

    Classical electromagnetic field includes freely propagating electromagnetic waves, and their emission and absorption. These DO possess energy, momentum, angular momentum.

    And they are observably quantized into photons.

    Now, classic electromagnetic field ALSO includes evanescent waves. As well as electrostatic and magnetostatic fields.

    And these are devoid of energy or momentum.

    Yet exert appreciable forces and torques...

    Evanescent waves and static fields are supposed to be represented by virtual photons.

    What signs do they show of being quantized in the first place?
  10. Sep 26, 2013 #9
    Right! But not really.
    When an electron throws a virtual photon, it has a "virtual recoil". If the particle lives, so does its recoil. Don't get caught up in the notion that the electron has a definite location in space.
  11. Sep 26, 2013 #10


    User Avatar
    Science Advisor

    AFAIK, you can't apply Newtonian physics to particles in a Feynman diagram. Concepts like force are meaningless.
  12. Sep 26, 2013 #11
    The systems center of gravity will not shift - but it can be significantly indefinite.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook