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The effect of the formation of electron-positron pairs on lightspeed

  1. Jun 16, 2013 #1
    Hello physicsforum,

    As the topic name says, I would like to know if a photon forming electron-positron pairs spontaneously could or does affect it's speed. Correct me if I'm wrong, but photons do form these pairs (and heavier?) even without the correct amount of energy, right? And if they do, will they lose/gain speed because they turn in to particles mass for a virtually small amount of time?

    Or is the answer just something like: the energy uncertainty 'used' here is only possible in an amount of time that does not influence the universe at all?

    Only definite answers please, or most accepted theories, else I won't have a good view of what the world is thinking.

    Thanks in advance,
    Evert
     
  2. jcsd
  3. Jun 16, 2013 #2

    ZapperZ

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  4. Jun 16, 2013 #3
    I mean Heisenberg uncertainty relation, can't it violate the conservation of energy for a short amount of time, to form virtual electron positron pairs? I guess i forgot to mention virtual.
     
  5. Jun 16, 2013 #4

    Bill_K

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    Nope. Energy is exactly conserved, even for virtual processes. Even if you hurry up about it!
     
  6. Jun 17, 2013 #5
    What is the meaning of the uncertainty relationship then? I thought the uncertainty about the energy could make you say you have 'enough' energy to create an electronpositron pair, without actually having that when time is more uncertain?
     
  7. Jun 17, 2013 #6

    Bill_K

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    In quantum mechanics, a system may exist briefly with the "wrong" energy. The shorter its lifetime, the greater the range ("width") of possible energies it can have. Thus an electron positron pair may be formed with energy 1.00 MeV, which is not exactly the 1.02 MeV you expected. We say the particles are "off the mass shell".

    But that does not mean that energy has not been conserved, or that it has been "borrowed" from some cosmic account! Whatever formed the electron-positron pair lost the same amount of energy, 1.00 MeV. Energy conservation is an absolute principle of physics, and the Heisenberg uncertainty principle does not violate it, even momentarily.

    As a more extreme example, the W boson has a rest mass of 80 GeV. But it is so short-lived, its width is a whopping 2 GeV. Which means that any one particular W can be expected to briefly have a mass anywhere within the 79 to 81 GeV range.
     
  8. Jun 17, 2013 #7
    Thanks for clearing that up for me, the information available to me at school is, as I expected, wrong. So when a photon forms these particles off the mass shell, does it do this constantly? I mean does it happen (significantly enough) with visible frequency light, or do you need gamma radiation photons? =And if it does happen frequently, will that influence the speed of the photon , that wasn't a 'photon' the whole time but formed a pair that annihilated again?

    Thanks already for the clarification, Bill_K
     
  9. Jun 17, 2013 #8

    ZapperZ

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    Take note that you need a minimum of 0.511 MeV x 2 = 1.022 MeV photo energy to create an e-p pair. This is nowhere near the visible spectrum. Also note that if you have a photon of that much energy, the e-p pair created will have practically zero kinetic energy, i.e. they aren't moving much.

    In reality, the probability of e-p pair formation using such energy scale is very unlikely. That is why we often require GeV photons to create a workable quantity of e-p pair.

    Zz.
     
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