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What is Chromoelectric dipole moment?

  1. Jan 28, 2015 #1
    Basically, I wonder what Chromoelectric dipole moment is and how it is formed...
  2. jcsd
  3. Jan 28, 2015 #2
    It is exactly the same thing as an electric dipole moment, but it's generated from color field. Therefore you must be careful about non-commutativity of the charges
  4. Jan 28, 2015 #3


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    The Electric Dipole Moment [itex]\vec{d}_f[/itex] of some particle [itex]f[/itex], is defined by its interaction with an electric field [itex]\vec{E}[/itex]:

    [itex] H_{EDM} = - \vec{d}_f \cdot \vec{E}[/itex]

    For a spin-1/2 particle, this corresponds to the effective lagrangian density:

    [itex] \mathcal{L}_{EDM} = \frac{-i}{2} d_f F_{\mu \nu} \bar{\psi}_f \sigma^{\mu \nu} \gamma^5 \psi_f [/itex]

    you can have a similar term if you look at quarks [itex]\mathcal{q}_r[/itex], by replacing the electromagnetism with strong ints:

    [itex] \mathcal{L}_{CEDM} = \frac{-i}{2} d_r^{CEDM} G_{\mu \nu}^a \bar{\mathcal{q}}_r \sigma^{\mu \nu} \gamma^5 \frac{\lambda^a}{2} \mathcal{q}_r [/itex]

    and that's how it arises... In general both EDMs and CEDMs generate P and T (so CP) violations.
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