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Quadrupole moment

  1. Jul 12, 2009 #1
    In wikipedia http://en.wikipedia.org/wiki/Quadrupole

    Is this [tex]\delta_{i,j}[/tex] Kronecker delta?

    In my notebook I have relation:

    [tex]Q_{i,j}=\frac{3}{2}eQ_0(x_ix_j-\frac{1}{3}\delta_{i,j})[/tex]

    When direction of external field are the direction of symmetry axis [tex]Q=Q_0[/tex].

    In which book I can find more about this?
     
  2. jcsd
  3. Jul 12, 2009 #2

    Meir Achuz

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    [tex]\delta_{i,j}[/tex] is the Kronecker delta.
    The relation from your notebook is not quite right. It should be
    [tex]Q_{i,j}=\frac{3}{2}Q_0(\delta_{i,3}\delta_{j,3}-\frac{1}{3}\delta_{i,j})[/tex]
    for a symmetric quadrupole aligned along the z (or 3) axis, having quadrupole moment Q_0.
    There is a full discussion of quadrupoles in Section 2.4 of Franklin, "Classical Electromagnetism" (AW.com).
     
  4. Jul 13, 2009 #3
    Thanks! :smile:
     
  5. Jul 17, 2009 #4
    I found that in "Non relativistic quantum mechanics" of Landau. There is formulation:
    [tex]Q_{i,j}=\frac{3}{2}Q_0(n_in_j-\frac{1}{3}\delta_{i,j})[/tex]

    where [tex]n_i,n_j[/tex] are components of unit vector [tex]\vec{n}[/tex].
     
  6. Jul 17, 2009 #5

    clem

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    My formula is the reduction of Landau's when the unit vector n is in the z direction.
     
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