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Do all exchange bosons violate energy conservation?

  1. May 3, 2015 #1
    Hi, so my question is along the lines of the following:

    Since the strong and EM forces are mediated by massless exchange particles, due to Heisenbergs uncertainty principle these forces are long range. Well, ok. But the weak force is mediated by W and Z bosons which are massive hence they can only get so far before their time is 'ran out', thus short range force.

    If this is how things work, then is this to say that all exchange particles violate energy conservation, hence the reason when the uncertainty principle has any relevance to the situation?

    Also, if this is the case, when we have say Higgs=>ZZ=>4 muons, what does it mean to say that one Z boson is real and the other is virtual. Are they not both virtual given that a Z boson is short lived? To my understanding, virtual simply means that the particle is temporary and doesn't have to be created in the final state.

    Thank you kindly for any responses. I've been really confused about this all day!
     
  2. jcsd
  3. May 3, 2015 #2

    mfb

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    Staff: Mentor

    Nothing violates energy conservation. Energy is conserved exactly and everywhere.

    All short-living particles, and all particles that appear in internal lines in Feynman diagrams (virtual particles), can be off-shell, which means their energy and momentum do not fit to a particle with the "correct" mass.
    For Higgs -> ZZ* -> muons, one Z has an invariant mass of about 90 GeV, and one has significantly less.
     
  4. May 3, 2015 #3

    ChrisVer

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    virtual particle is the one for which [itex]p_\mu p^\mu =E^2 -p^2 \ne m^2[/itex] where ##m## is the particle's real mass... The invariant quantity [itex]p^\mu p_\mu = M[/itex] is still there, where [itex]M[/itex] is the invariant mass (rather than particle's rest mass).
    Obviously the Higgs to 2 Z's cannot give two real Z's, because that would violate the energy conservation (the Higgs mass is lower than needed to create two Z bosons with masses ~90GeV).
    The real Z will decay probably leptonically, showing a peak around its mass [itex]M_Z[/itex] with some width [itex]\Gamma_Z[/itex] (it's like a resonance).
    The virtual Z will also decay leptonically, however it will give no consistent peak. The virtual Z's are those whose mass (after reconstructing the lepton-products' invariant mass) has been outside the region of [itex][M_Z -\Gamma_Z , M_Z + \Gamma_Z] [/itex].

    I am not so sure that HUP can help you understand what a virtual particle is. It can only help you understand that the [itex]E^2 -p^2 = m_{particle}^2[/itex] (on-shell condition) can not hold for some short times.
     
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