G-factor Relation to Energy Perturbations

160
0
The g-factor is related to the magnetic moment by our universal equation [tex]\mu= g \mu_B S/ \hbar[/tex] and the magnetic moment experiences torque which is related to energy as [tex]\Delta E = -\mu \cdot B[/tex]. In fact there are equations which describe the magnetic interaction energy [tex]\Delta E = g \mu_B M B[/tex].

The g-factor has an appearance of [tex]\frac{e}{2M}[/tex]. It differs only very small to what we expect from the notation of the Bohr Magneton. So the question arises whether the g-factor is intrinsically related to energy.

It is possible to satisfy for instance that:

[tex]\sum^{N}_{i=1} \vec{\mu}_i \propto \frac{eh}{2Mc}[/tex] 1.

Where [tex]\mu_B= \frac{eh}{2M}[/tex] is the Bohr magneton. Equation 1. is simply the sum of magnetic moments where it measures the gyromagnetic ratio of a particle induced in a magnetic field.

The energy is given by the Lande' g-factor as a magnetic interaction on the system. This can be given as

[tex]\Delta E = \frac{eh}{2Mc}(L+2S) \cdot B[/tex]

Since

[tex]\sum^{N}_{i=1} \vec{\mu}_i \propto \frac{eh}{2Mc}[/tex]

Then

[tex]\sum_{i=1}^{N} E_i = \sum_{i=1}^{N} (\frac{eh}{2Mc})_i (L+2S) \cdot B \propto \sum_{i=1}^{N} \mu_i (L+2S) \cdot B[/tex]

If the calculation is right, then one can easily assume that energy is related to the sum (of) some periodic functions of the appearence of a magnetic moment and the g-factor is proprtional the magnetic moment.
 
969
3
I'm not sure what you're asking.

Shouldn't
[tex]
\Delta E = -\mu \cdot B
[/tex] be [tex]
\Delta E = 2 \mu \cdot B
[/tex]?

The g-factor is related to energy in that the energy is proportional to the g-factor. If the g-factor is zero, and assuming there is no orbital angular momentum, then the energy in a magnetic field would be zero.

I think you can sum up the magnetic moments of each particle in a composite particle to get the total magnetic moment of the composite particle.

This is done for example to calculate the magnetic moment of a proton, which is made up of 2 u quarks of charge 2/3, and 1 d quark of charge 1/3. A spin up proton has spin 1/2, so one of the quarks must be spin down. It's fairly easy to see that it's twice as likely for the d quark to be spin down than for it to be spin up. Also the mass of either quark is taken to be a 3rd of the proton mass, so that the magnetic moment is:

[tex] \frac{2}{3}\left(2*\frac{2*(2/3) e }{2*1/3 m_p}*\frac{\hbar}{2}+1*\frac{2*(-1/3) e }{2*1/3 m_p}*\frac{-\hbar}{2} \right)


+\frac{1}{3}\left(\frac{2(2/3) e }{2*1/3 m_p}*\frac{\hbar}{2}+\frac{2(2/3) e }{2*1/3 m_p}*\frac{-\hbar}{2}+1*\frac{2(-1/3) e }{2*1/3 m_p}*\frac{\hbar}{2} \right) [/tex]

which equals

[tex]\frac{3}{2}\frac{e}{m_p}=3*2*\frac{e}{2 m_p}*\frac{\hbar}{2}=3g \mu_B S_z [/tex]

so that the g-factor of the proton is 3 times that for the electron (I might have messed up an hbar since I'm so used to setting hbar equal to 1).

A similar calculation shows the surprising result that the neutron also has a magnetic moment, and the g-factor is -2 times the electron's.
 
160
0
I'm not sure what you're asking.

Shouldn't
[tex]
\Delta E = -\mu \cdot B
[/tex] be [tex]
\Delta E = 2 \mu \cdot B
[/tex]?

The g-factor is related to energy in that the energy is proportional to the g-factor. If the g-factor is zero, and assuming there is no orbital angular momentum, then the energy in a magnetic field would be zero.

I think you can sum up the magnetic moments of each particle in a composite particle to get the total magnetic moment of the composite particle.

This is done for example to calculate the magnetic moment of a proton, which is made up of 2 u quarks of charge 2/3, and 1 d quark of charge 1/3. A spin up proton has spin 1/2, so one of the quarks must be spin down. It's fairly easy to see that it's twice as likely for the d quark to be spin down than for it to be spin up. Also the mass of either quark is taken to be a 3rd of the proton mass, so that the magnetic moment is:

[tex] \frac{2}{3}\left(2*\frac{2*(2/3) e }{2*1/3 m_p}*\frac{\hbar}{2}+1*\frac{2*(-1/3) e }{2*1/3 m_p}*\frac{-\hbar}{2} \right)


+\frac{1}{3}\left(\frac{2(2/3) e }{2*1/3 m_p}*\frac{\hbar}{2}+\frac{2(2/3) e }{2*1/3 m_p}*\frac{-\hbar}{2}+1*\frac{2(-1/3) e }{2*1/3 m_p}*\frac{\hbar}{2} \right) [/tex]

which equals

[tex]\frac{3}{2}\frac{e}{m_p}=3*2*\frac{e}{2 m_p}*\frac{\hbar}{2}=3g \mu_B S_z [/tex]

so that the g-factor of the proton is 3 times that for the electron (I might have messed up an hbar since I'm so used to setting hbar equal to 1).

A similar calculation shows the surprising result that the neutron also has a magnetic moment, and the g-factor is -2 times the electron's.
The magnetic moment is a vector quantity which is perpendicular to the current loop using the right hand rule direction.

Since we considered torque, the symbol [tex]\mu[/tex] is actually equivalent to [tex]IA[/tex] where [tex]A[/tex] denotes the area.

This soon comes to calculation which show the potential energy for the magnetic moment can be given as [tex]U(\theta)= -\mu \cdot B[/tex] which you will realize is identical tothe quantity in the first theory and as you will also see that yes, one can find that the change in the potential energy is equal to [tex]2 \mu B[/tex] which is might what you would thinking of?
 
969
3
one can find that the change in the potential energy is equal to [tex]2 \mu B[/tex] which is might what you would thinking of?
Yes, I got confused because you used the notation delta E, so I thought it meant change in energy or difference in energy levels. It could be the case that your reference for the energy is 0, but for half integral spins that reference is not achievable.

Anyways, to get the neutron magnetic moment, just swap the 2/3 and -1/3 in the calculation for the proton above. That should get you a g-factor of -2 times the electron's.

So basically you can say all Dirac particles have a g-factor of 2 (ignoring corrections of the order of the fine structure), and treat composite particles as taking the sum of the magnetic moments of the Dirac particles (each quark with g-factor 2) that comprise it (for hadrons).
 

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