Magnetic moment

1. Nov 13, 2008

andrewm

Say I know the total angular momentum of my electron as J. If I write the total magnetic moment as $$\mu = \gamma J$$ then does $$\gamma = \gamma_{spin} + \gamma_{orbital}$$ ?

2. Nov 14, 2008

malawi_glenn

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.

3. Nov 14, 2008

clem

No. Mu will not be in the direction of J, since the g factor for S and L are different.
For a single electron, $${\vec\mu}=(-e/2mc)[{\vec L}+2{\vec S}]$$.
This is the origin of the Lande g factor.

4. Nov 14, 2008

andrewm

My research suggests one can define a $$\mu$$ in the direction of J with a Lande factor

$$g_J= g_L\frac{J(J+1)-S(S+1)+L(L+1)}{2J(J+1)}+g_S\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}$$

if one is measuring the total angular momentum, say in a magnetic resonance experiment. But as clem said, $$\mu_J \neq \mu_S + \mu_L$$.

5. Nov 14, 2008

malawi_glenn

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

I didn't know at what level you was asking. Sorry

6. Nov 14, 2008

clem

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.