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What are space-like and time-like virtual photons?

  1. Jan 23, 2010 #1
    I'm trying to get my story straight about virtual photons. So far, my understanding of them is that they are the particles that mediate the electromagnetic interaction, and they only exist a short time due to the Heisenberg uncertainty principle meaning that they can't be directly observed.
    I've heard of time-like and space-like virtual photons though, and google as I might, I just can't find out what these terms mean. Can anyone explain them to me?
  2. jcsd
  3. Jan 23, 2010 #2
    A spacetime interval is called timelike if there is a rest frame in which the interval separates events by only time, that is in this rest frame there is no space in the interval. In this case, the 4-length squared (which does not depend on the frame) [itex]\Delta s^2 = \Delta t^2 - \Delta x^2[/itex] is positive (with a choice of units such as the speed of light c=1 and the choice I made for the signature, time comes with a plus sign and space with a minus sign). A timelike interval can belong to the trajectory of massive particle (for the above rest frame I was referring to). The interval is called spacelike if its squared is negative, or equivalently if it can not be in the trajectory of a massive particle. The limiting case where the 4-length squared vanishes is called lightlike because it belongs to a lightcone, that is it could be the trajectory of a real photon.

    If you annihilate an electron-positron pair into a virtual photon, the virtual photon will be timelike (there is a referential in which the electron and positron have opposite momenta, and this is the rest frame of the timelike photon). If you scatter two electrons off each other and they exchange a virtual photon, the virtual photon will be spacelike.
  4. Jan 23, 2010 #3
    Can you find a simple argument to show that ?
  5. Jan 23, 2010 #4
    As far as I know the only instance in which a "virtual" pair can produce "real" photon would be in the form of Hawking Radiation. and the other photon that is lost in the BH could be described in terms of the past worldline of the "real" photon that escapes. It's past the event horizon after all. Obviously Humanino deleted his post, but in terms of what he said, I don't believe that is the most accurate intrepation of that interaction.
  6. Jan 24, 2010 #5
    I did not delete anything.

    One can find a simple argument by thinking of the mass of the exchanged photon in Moller scattering in the frame where one of the electrons is at rest.
    Last edited: Jan 24, 2010
  7. Jan 24, 2010 #6
    My bad, I was thrown by you quoting your previous post. Sorry!

    I can see where your argument leads, but does it actually happen that way in nature? I don't know, so this isn't a snarky or leading question. I didn't think that virtual pairs were terribly conducive to producing a "real" (space-like) particle under anything approaching normal circumstances. That said, what you said seems to make sense. I'm confused! (obviously)
  8. Jan 24, 2010 #7


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    Hi jeebs! :smile:
    No … virtual particles aren't real (the clue's in the name! :wink:).

    They're just a mathematical trick to help calculations.

    They exist only in the mind. :smile:

    (btw, virtual electrons and positrons are also needed, and since there's twice as many of them as of virtual photons in most Feynman diagrams, it's a little biased to say that it's only the virtual photons that "mediate" the electromagnetic interaction)
    Real photons are (obviously!) light-like, meaning that the magnitudes of their energy and momentum are the same. In other words, they move at the speed of light.

    Virtual photons (just a mathematical trick, remember) in the "position representation" Feynman diagrams are light-like.

    Virtual photons in the "momentum representation" Feynman diagrams are "off-shell", and so can be space-like or time-like.
  9. Jan 24, 2010 #8


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    I never get all this stuff on virtual photons, I think it is north-american folklore, something in the secondary school textbooks, isn't it?

    In the rest of the world:
    - electromagnetic interaction is mediated by the electromagnetic field.
    - Photons are the quanta of the electromagnetic field.
    - so photons appear in the diagrams of quantum electrodinamics.

    Now,in quantum field theory, the calculation is done for all possible values of energy and momentum. It means that only the external particles being calculated are on-shell, [itex]E^2-p^2=m^2[/tex]. Particles in the diagrams of the calculation do not need to meet this equation, they are called off-shell.

    This virtual particle thing... is the same that the off-shell particles?

    If so, are you telling that in the tree-level diagram exchange of a photon, the integration limits are such that always [itex]E^2 < p^2[/itex]? I am not claiming it to be false; it is only that I had not noticed the phenomena, can you point it more clearly by writing the equation? All this stuff of s,t,u channels always confuse me.

    Here i think you mean "annihilate an electron-positron pair into a virtual photon what disintegrates again into a electron-positron pair". The other thing is obviously timelike because it is not virtual at all.
  10. Jan 24, 2010 #9


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    Why virtual? When you annihilate an electron positron pair you get real photons... at least two of them. You can't get just one photon, virtual or otherwise, from this annihilation.

    Cheers -- sylas
  11. Jan 24, 2010 #10
    As I said earlier, the only instance I know of where a "virtual pair" becomes a single particle is at the event horizon of a black hole in the process of Hawking Radiation. Even then, there is a requirement that one of the pair be lost inside the event horizon, formulated as the past worldline of the "real" photon that is emitted. Otherwise I thought virtual particle creation/annihilation was the description of the quantum vacuum, and by definition a bit like renormalization in the "trick" sense that Tiny Tim mentioned.
  12. Jan 24, 2010 #11


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    Hmm now you mention it, I hope the discussion on timelike vs spacelike will not depend on the mass nor the spin of the carrier, will it? Photons are tricky.
  13. Jan 24, 2010 #12
    As opposed to the REST of particle physics...?! :tongue2:
  14. Jan 24, 2010 #13
    Please keep in mind that I am only trying to help the original poster, and as in kindergarden I used only Moller and Bhabha scattering. At this level, all particles could be scalars, it would not change their invariant mass ! All I was trying to say is that this virtual photon (in the diagram, time flows in the horizontal direction, space is the vertical direction)
    is timelike whereas this virtual photon
    is spacelike. In both cases, there are simple, almost geometrical, arguments to show what the invariant mass of the one-photon-exchanged is.

    If you want to annihilate an electron-positron pair into a photon pair, the tree-level diagram will have an electron exchanged. This is obviously irrelevant for the original poster.
    Last edited by a moderator: Apr 24, 2017
  15. Jan 24, 2010 #14
    That's a very lucid explanation; works for me. And no, I'm not being sarcastic.
    Last edited by a moderator: Apr 24, 2017
  16. Jan 24, 2010 #15
    I'm still not clear on this.

    You were on about an electron-positron pair annihilating, and I am happy with there being a rest frame where their momenta are equal and opposite. What I don't get is what events (plural) are being separated here?
    There's only one event as far as I'm aware, the particle annihilation...

    Or do you mean that the particles collide and annihilate, and then some time later a photon leaves that point where they collided? Is it not considered to be happening instantly?
  17. Jan 25, 2010 #16
    The creation and annihilation of the virtual photon. This is not supposed to be meaningful : as the virtual photon is in a well-defined eigenstate of momentum, it should be "everywhere".

    I'm taking a classical analogy "as if" this virtual photon would correspond to a classical particle. Please keep in mind that I'm suggesting this only to explain the etymology of "spacelike" and "timelike" virtual photons.

    If we really wanted to be realistic, we would have to take wavepackets for the incoming and outgoing particles. That is not what I suggest to do, as it is quite painful and not very illuminating. Yes, I do suggest to take the Feynman diagram as a "little drawing" of what is really happening, although it's only the second term (first order, the first term being at zeroth order and corresponding the identity) in a perturbative expansion of the amplitude, and by itself is meaningless.
  18. Jan 25, 2010 #17


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    virtual = off-shell ?

    Hi arivero! :smile:
    (I added the "{Internal}")

    No, the virtual particles in the position-space representation Feynman diagrams are on-shell. Only in the momentum-space representation are they off-shell.

    I've looked at how a couple of books deal with this …​

    Weinberg, in Quantum Theory of Fields Volume I, doesn't seem to use the word "virtual" at all, just "on the mass shell" and "off the mass shell" (or "off-shell").

    V. B. Berestetskii, E. M. Lifgarbagez, and L. P. Pitaevskii, in the free online http://books.google.com/books?id=oA...d=16#v=onepage&q="virtual particles"&f=false" at p.312, §79, say "virtual states" in the position-space representation and "virtual particles" in the momentum-space representation.

    Does anyone have any examples from other books? :smile:
    Last edited by a moderator: Apr 24, 2017
  19. Jan 25, 2010 #18
    @Tiny-Tim: I haven't, but that doesn't mean you're not right. I hope you are... it would make for better terminology than "virtual".
  20. Jan 25, 2010 #19


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    (EDIT: I did this post before reading #17 above (Hi!) but fortunately I kept myself in the momentum space representation)

    Ok, I sort of convinced myself. I was first looking a bit at http://en.wikipedia.org/w/index.php?title=Bhabha_scattering&oldid=332559874 , a calculation for which I have sad rememberances when undergradute; but in any case it is not even needed. It is enough to understand than in a channel the 4-momentum is exchanged, in the other is accumulated. So in the former case the carries takes a 4-momentum which is a difference, in the later it carries a sum. Now it is only plain relativity: take [itex]||E,p||=E^2-\sum p_i^2=m^2[/itex], wash, rinse, and repeat.

    To be specific, consider a pair of four-momentum vectors [itex](E_1,p_1), (E_2,p_2)[/itex], with respective masses [itex]m_1, m_2[/itex]. Both of them will fullfill the relativistic equation above. For simplicity you can choose a reference frame where [itex]p_1=(0,0,0)[/itex], so [itex]E_1=m_1[/itex]. But it is not compulsory.

    Now, questions are: what can we tell of the "square mass" of the sum [itex](E_1+E_2,p_1+p_2)[/itex] and difference [itex](E_1-E_2,p_1-p_2)[/itex]? We are looking for the relativistic equivalent of "triangle inequalities". What we found just by applying the energy-momentum formula and knowing that [itex]E_i >= m_i[/itex], is that

    [tex]M_+^2=|| (E_1+E_2,p_1+p_2)||^2 > (m_1+m_2)^2[/tex]

    [tex]M_ -^2=|| (E_1-E_2,p_1-p_2)||^2 < (m_1-m_2)^2 [/tex]

    - the s-channel (which is a sum) particle is always time-like.
    - the t-channel (which is a difference) is mainly space-like, but it could have a small contribution from time-like exchanges. If the rest mass of the input and output particle is the same, then the channel is completely space-like.

    This is for a vertex. As one interaction tree has at least two vertex, I guess that there are really two bounds in each channel, but on the other hand there is more kinematic in game.

    Finally: why do we call "virtual" to the particles here? Because [itex]M_\pm^2[/itex] is fixed from the values of external 4-momentum (remember we are calculating the probability of having such and such outputs with such and such inputs, then all the external are given as premises of the calculation). So they do not coincide with the mass "m0" of the interaction carrier (zero mass for the photon, 91 GeV for a Z, etc). In some cases we can still think that each value of M can "live" during a time h/(M-m0)c^2, but it is more precise to use the propagator formula. Furthermore, one always want to take into account the interference between all the possible diagrams.
  21. Jan 25, 2010 #20
    Hmmmm... ok, I see where you're going. That does make the "virtual" title more sensible.
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