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Magnetic moment

  1. Nov 13, 2008 #1
    Say I know the total angular momentum of my electron as J. If I write the total magnetic moment as [tex] \mu = \gamma J [/tex] then does [tex] \gamma = \gamma_{spin} + \gamma_{orbital} [/tex] ?
     
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  3. Nov 14, 2008 #2

    malawi_glenn

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    yes, you have one contribution from orbital motion around nucleus and one from its intrinisc spin. The g-factors and so on of course differs so one has to be careful.
     
  4. Nov 14, 2008 #3

    clem

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    No. Mu will not be in the direction of J, since the g factor for S and L are different.
    For a single electron, [tex]{\vec\mu}=(-e/2mc)[{\vec L}+2{\vec S}][/tex].
    This is the origin of the Lande g factor.
     
  5. Nov 14, 2008 #4
    My research suggests one can define a [tex]\mu[/tex] in the direction of J with a Lande factor

    [tex] g_J= g_L\frac{J(J+1)-S(S+1)+L(L+1)}{2J(J+1)}+g_S\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)} [/tex]

    if one is measuring the total angular momentum, say in a magnetic resonance experiment. But as clem said, [tex] \mu_J \neq \mu_S + \mu_L [/tex].
     
  6. Nov 14, 2008 #5

    malawi_glenn

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    Ok, maybe my answer was not careful enogh, what I meant with "yes" was not referring to your result [tex] \gamma = \gamma_{spin} + \gamma_{orbital} [/tex]

    I didn't know at what level you was asking. Sorry
     
  7. Nov 14, 2008 #6

    clem

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    Mu will not be in the direction of J for a single electron. The Lande g factor is for the
    component of mu in the direction of J. It follows by dotting my formula for mu with J and doing some algebra, leading to Andrew's (and Lande's) formula.
     
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