# The magnetic moment of a particle (and a neutrino)

Is the magnetic moment of a particle an intrinsic property or is there a formula to find it? What is its source?

Is there a formula or a general explanation that can account for the $\mu$ of known particles: electrons, protons, neutrons and neutrinos?

I searched the web and found different values for the neutrino, ranging from 10^-10 to 10^-19, how is such great discrepancy possible? Can we use the formula given for neutrinos (see here) for any other particle?

Can you briefly explain the genesis/rationale of that formula: $$3eG_Fm_\nu /8\pi^2 \sqrt{2}$$ and how it applies to other particles?They say in that article that its value is proportional to the mass of the neutrino, why so? In what way ismass related to it?

## 2 Answers

The magnetic moment was first defined for macroscopic classical magnets and is the first dipole term in an expansion of the magnetic field, whether it is a permanent magnet or moving charges creating a magnetic field.

More precisely, the term magnetic moment normally refers to a system's magnetic dipole moment, which produces the first term in the multipole expansion of a general magnetic field.

In classical electrodynamics it is defined as :

where one sees that a circulating current( or its equivalent) plays a role in the definition.

You ask:

Is there a formula or a general explanation that can account for the μ of known particles: electrons, protons, neutrons and neutrinos?

At the quantum mechanical level where electrons etc are defined, the magnetic moment will be an operator whose expectation value will give the size of the magnetic moment. In this reference this expectation value is given by : the expectation value of the z component of spin- magnetic moment of a free Dirac electron in positive energy state .

We see that for the electron and any dirac spinor if the charge is zero the spin dipole moment will be zero.

Classically for a particle of charge e moving with velocity v the magnetic moment We can see that the mass in the denominator does not allow reasonable magnetic moments for zero mass particles. The neutrino has a small mass, but it is excluded of having a magnetic moment to first order by the zero charge. Higher order corrections will introduce other terms of charged exchanged particles which may give rise to a magnetic moment but will be very small, due to the smallenes of the weak interaction constant. The discrepancy comes from how many terms are added in the perturbative expansion for calculating the magnetic moment.

Protons as composite particles will have a magnetic moment appropriate to the spins and angular momenta of its constituents.

From the nonrelativistic, quantum mechanical wavefunction for baryons composed of three quarks, a straightforward calculation gives fairly accurate estimates for the magnetic moments of protons, neutrons, and other baryons.

Higher order corrections will be present,but to first order the data agree with the calculations.

• this answer here explains the higher order terms physicsforums.com/threads/… . G_F must be the fermi coupling constant thphys.uni-heidelberg.de/~maniatis/Lecture06/stachel-mulife.pdf. You have to enter into the details of the calculations to see why it appears there. The handwaving explanation is that the neutrino interacts with the electroweak sector, and the loops in the feynman diagrams for the vacuum fluctuations of the neutrino into W e will involve it. – anna v Jan 11 '17 at 10:30
• The proportionality to the mass leads to a zero neutrino magnetic moment for zero mass, which is consistent. You need to find a theoretical physicist to enter further into the QFT calculations. See this arxiv.org/pdf/hep-ph/0601113v1.pdf – anna v Jan 11 '17 at 10:31

The spin of a particle is an intrinsic property. From experiments we can infer that a particle with the spin $\mathbf{s}$ has a magnetic spin moment $\mathbf{\mu_s} = g_s\cdot\mathbf{s}$. For electrons the so-called Landé factor $g_S$ is roughly 2.

If the particle also has an angular momentum $\mathbf{l}$, however it also carries a magnetic momentum $\mathbf{\mu_l}$ due to its angular momentum, much like in classical electrodynamics. So in total you have $\mathbf{\mu}=\mathbf{\mu_s}+\mathbf{\mu_l}$.

Hence, the source of the magnetic momentum of a particle are both its spin and its angular momentum. But the calculation of the Landé factor differs in classical quantum mechanics and in quantum electrodynamics, which is by far the more accurate theory.