What is physical significance of g factor?

[xfshi2000]

In quantum mechanics, magnetic moment is equal to g factor times (gyromagnetic ratio x angular momentum). For orbital angular momentum, g=1, for electron spin angular momentum, g=2, for proton, it is 3.56. Is there any physical significance for this factor? why scientist introduce this factor? Why don't they just absorb this factor into the gyromagnetic ratio? thanks

 

[AEM]

In quantum mechanics, magnetic moment is equal to g factor times (gyromagnetic ratio x angular momentum). For orbital angular momentum, g=1, for electron spin angular momentum, g=2, for proton, it is 3.56. Is there any physical significance for this factor? why scientist introduce this factor? Why don't they just absorb this factor into the gyromagnetic ratio? thanks

There is a very nice discussion of a naive model of electron spin, and the gyromagnetic ratio in Peebles Quantum Mechanics , pages 198-201. The g-factor was introduced by Goudsmit and Uhlenbeck to account for the behavior of an electron in a magnetic field because naive models of the electron do not give the proper magnetic dipole moment. g is measured experimentally via the precession of the spin vector. As you probably know, Dirac's theory predicts g = 2. However, it actually is slightly larger than 2, these corrections are explained by quantum field theory of electromagnetism. Incidentally, my sources indicate that g for the proton is 5.59 rather than 3.56.

 

[AEM]

In quantum mechanics, magnetic moment is equal to g factor times (gyromagnetic ratio x angular momentum). For orbital angular momentum, g=1, for electron spin angular momentum, g=2, for proton, it is 3.56. Is there any physical significance for this factor? why scientist introduce this factor? Why don't they just absorb this factor into the gyromagnetic ratio? thanks

I must apologize. When I wrote my previous post it slipped my mind that there is a beautiful little book by Sin-Itiro Tomonga (as you recall ,winner of the Nobel Prize for QED along with Feynman and Schwinger) called The Story of Spin , published by U of Chicago Press. The first chapter is a discussion of the history of spin and tells how the g-factor was introduced by Lande (not Goudsmit and Uhlenbeck as Peebles said). As you can imagine since the whole book is about spin, if you have a chance to read it, you'll have you answer in more detail than you might have wished.

 

[Meir Achuz]

Why don't they just absorb this factor into the gyromagnetic ratio? thanks

The gyromagnetic ratio has complicated units and depends on the particle's mass.
The g factor is dimensionless and is useful in comparing the magnetric moments of different particles.

 

[xfshi2000]

thank you all.

 

[AEM]

thank you all.

You're welcome. However, I had a few free minutes on my hands and you question interested me so I did a little reading. This is a case where the history of physics can be helpful in sorting things out. Lande introduced his g-factor when attempting to account for the Zeeman effect and according to Tomonaga, Lande's g is just the ratio of the magnetic moment to the angular momentum. Now, when one wants to develop a Hamiltonian for the motion of an electron circulating about a nucleus (remember this work was done in the mid 1920's slightly before the invention quantum mechanics) it is helpful to work in analogy with classical physics. In classical case, when one considers a current loop in a magnetic field one finds the relationship between the magnetic moment of the loop and the angular momentum of the charge circulating in the loop to be

\bf \mu = ( \frac{q}{2mc}) \bf \l

One can show ( see Shankar Principles of Quantum Mechanics ) that the same relation holds when one constructs the Hamiltonian for the circulating electron. So what happened a few years later when the idea of spin operators and electron spin had been introduced and one wanted to model electron spin? Reading between the lines of the text I just mentioned, a first try probably was to take \mu to be

\bf \mu = g (\frac{-e}{2mc}) S

where g is a constant to be determined by experiment and S is the spin operator. I believe that this g is the g factor you were inquiring about. It would appear that it was originally introduced for convenience and only later was found to have a theoretical basis. So in the early days it was a fudge factor. The fact that it was measured to be equal to 2 and Dirac's theory predicted it to have the value 2 was a positive for the acceptance of Dirac's theory. In quantum electrodynamics corrections to the value of g can be calculated in a power series in \alpha, the fine structure constant. The close agreement between the calculated value of g and the value found by experiment gives one confidence that QED is correct.

A reasonable conclusion is that g gives one an indication of the difference between the classical approach and the quantum mechanical approach.

 

[granpa]

https://www.physicsforums.com/latex_images/20/2054609-0.png

this formula assumes that the spinning charge and mass have the same radius (and distribution within that radius). 'g' suggests that they do not.

 

[granpa]

or it (gyromagnetic ratio) suggests that the effective mass of the spinning electron is half its normal mass.

 

[Meir Achuz]

or it (gyromagnetic ratio) suggests that the effective mass of the spinning electron is half its normal mass.

No. That's the wrong interpretation. g=2 is a relativistic effect that comes directly out of the Dirac equation.

 

[xepma]

Hmm, I'm having to trouble to accept it as a relativistic effect. To me it seems that the fact that it comes out of the Dirac equation only proves a consistency of the g-factor with special relativity. The g-factor is, in the end, a reflection of the fact that the electron carries a non-trivial spin. Spin is also a concept consistent with relativity (i.e. irreps of the Poincare Group are labeled by their spin), but that doesn't mean that the effect (spin) is relativistic. Non-relativistic spin is perfectly fine.

In other words, by taking c\rightarrow\infty we eliminate the relativistic effects, but the concept of spin and g-factors still remain.

 

[granpa]

me too. in fact now that I think about it I don't really understand it at all.

g_l=1
and g_s=2
but j_z=m_l + m_s
not j_z=m_l + 2m_s

 

[TFT]

In other words, by taking c\rightarrow\infty we eliminate the relativistic effects, but the concept of spin and g-factors still remain.

Good point. Since SO(3) has spin representation, it is perfectly fine to talk about spin in the content of non-relativistic QM. While to get the specific value of the g factor, you do need QED, which has relativity intrinsically. I think this is a reflection of the fact that we do need relativity to describe the real world, since the idea of the g factor came from experiments.

 

[granpa]

me too. in fact now that I think about it I don't really understand it at all.

g_l=1
and g_s=2
but j_z=m_l + m_s
not j_z=m_l + 2m_s

I think the above formula gives the fine structure splitting (due to spin-orbit coupling) minus the lamb shift and with zero external magnetic field. (the energy levels are a function of j only). someone please correct me if I'm wrong.

m_j ranges from −j to +j in steps of one

http://en.wikipedia.org/wiki/Zeeman_effect#Weak_field_.28Zeeman_effect.29

apparently g_l and g_s are replaced by g_j when calculating the splitting of individual fine structure lines in the presence of a magnetic field . (the splitting is proportional to the total magnetic moment?)

the quantity in square brackets is the Lande g-factor g_J of the atom (g_l = 1 and g_s = 2) and mj is the z-component of the total angular momentum. For a single electron above filled shells s = 1 / 2.
(V_M) is the total shift in energy.

the above formulas don't work for hyperfine structure/splitting though. that apparently uses different variables.

 

[granpa]

this image gives the fine structure + lamb shift + hyperfine doubling.
the fine structure is a function of j and n.

http://www.pha.jhu.edu/~rt19/hydro/img194.gif

hyperfine structure uses:
https://www.physicsforums.com/latex_images/22/2211684-1.png

 

[Narayanan.S]

When an electron is bound in an atom it is subjected to spin - orbit coupling.The g-factor is the ratio of electron magnetic moment [eCre)xSqrt2/Bohr Magneton) + Bohr magneton] and this sum is again divided by Bohr magneton.C-spin velocity at "C" and re-classical radius of electron.
Stumbled upon this formula for electron anomaly from google search
May be it means something more.
Cheers

 

[ytuab]

Uhlenbeck tried to explain spin g-factor 2 using the spinning electron sphere model.
So, he thought that total charge e is distributed only over the surface of an electron.
(If the charge is uniformly distributed, the spin g-factor becomes 1, which is the same as the electron orbital g-factor).

But if the electron has the classical radius size, (this means that an electron is as big as an proton or an neutron.) the spinning sphere speed leads to more than 100 times the speed of light. Of course if an electron is smaller than an proton, the spinning speed becomes much faster than that.
So he gave up the idea that the electron is a real thing (as Pauli did.)
Then the spinor matrices were introduced.

It remains a mystery what the electron spin g-factor 2 means if the electron is a real thing.

 

[granpa]

g relates the angular momentum of an electron to its magnetic moment. it should be 1. instead its 2. I think this may have to do with 'effective mass'.

 

[Narayanan.S]

Re: what is physical significance of g factor?

--------------------------------------------------------------------------------

I find it hard to accept that electrons are close to point particle (<10^-18m) and thought of considering classical radius for the following reasons
1) It is well known that the magnetic moments of electron are greater than those of nucleons and therefore I thought even modern physics should fundamentally expect a larger radius for the electrons(re-classical).Though experiments may show smaller and smaller radius as more and more energy is imposed on it.Is it that the interpretation of experiment is not correct?
2) Does it not mean violation of "C" will not reflect on mass energy relation?
3) At the instant of a neutron beta decay into a proton and since electrons are close to a point like particles the distance between them is about (0.5x10^-15- the accepted emperical value for nucleon radius) is well within range of nuclear forces
of (1.7x10^-15m).The binding energy between them should be (>95Mev) and the beta decay is not possible?
4) The electron size is not precisely known yet, then how to accept a precise value for the magnetic moment of an electron?.
5) Could not find an answer for the very high electric field intensity (point like charged electron) seems not addressed?

Therefore did an exercise with classical radius of electron and found that it agrees extremely well with g-factor correction.And this value for electron magnetic moment is considerably less than Bohr magneton and could have turned out to be anomalous Zeeman effect.
Request can anyone help me in clarifying on the above "5" points
Thanks

 

[Narayanan.S]

g relates the angular momentum of an electron to its magnetic moment. it should be 1. instead its 2. I think this may have to do with 'effective mass'.

By the introduction of gyromagnetic ratio tha angular momentum gets cancelled.
What remains is magnetic moment which is the only measurable quantity and for me I had to work back from this magnetic moment by dividing this by a suitable gyromagnetic ratio.
This is the answer I got

When an electron is bound in an atom it is subjected to spin - orbit coupling.The g-factor is the ratio of electron magnetic moment [eCre)xSqrt2/Bohr Magneton) + Bohr magneton] and this sum is again divided by Bohr magneton.C-spin velocity at "C" and re-classical radius of electron.
The g then gets related to (1/sqrt2)=Cos45 degrees
Classical magnetic moment "Mu"=IxArea
g=2 for spin may have been arrived at from projected area at Bohr orbit (2pixrB^2) and for electron it is (2pixre^2) for hemisperical projected area,hence g=1 but the directional quantum number
Where (rB=Bohr radius and re=Classical radius of electron)
(ms=+-)1/2) is now (+-)sqrt2)
Stumbled upon this formula for electron anomaly from google search
May be it means something more.

 

[ytuab]

I find it hard to accept that electrons are close to point particle (<10^-18m) and thought of considering classical radius for the following reasons
1) It is well known that the magnetic moments of electron are greater than those of nucleons and therefore I thought even modern physics should fundamentally expect a larger radius for the electrons(re-classical).Though experiments may show smaller and smaller radius as more and more energy is imposed on it.Is it that the interpretation of experiment is not correct?

Why do you think of the classical electron radius (2.8 x 10^-15 m) ?

Foundations of Quantum Physics by Charles E.Burkhardt
In page 264

------------------------------
They imagined that the electron is a spherical shell having total charge e uniformly smeared over its surface, reminiscent of the model used to derive the classical radius of the electron in Section 1.2.5.
This spinning sphere creates a magnetic moment identical with that of a bar magnet.
Is this model consistent with the classical radius of the electron? No-- as can be seen by
equating the angular momentum of the spinning sphere to 1/2 hbar. Solving for the speed of a point on the sphere leads to a speed that is roughly 100 times the speed of light.
(For a spinning shell with radius equal to the classical radius of the electron(Re), equate the angular momentum Iw to 1/2 hbar to show the spinning speed would be 100 times the speed of light.)
--------------------------------

So If the spinning speed does not lead to the speed of light, the electron must be much bigger than the classical radius size, an proton (10^-15 m) or an neutron.

For example, such a big electron can be captured by the nucleus?

 

[granpa]

regarding 'g':
Does everyone understand that for any object if the mass distribution and the charge distribution are the same then g should equal 1?

the equations describing magnetic moment are essentually the same as the equations that describe angular momentum. (hence the similarity of the names). you just replace mass with charge. there is a factor of 1/2 but that just has to do with the fact that for historical reasons the units of magnetic moment and angular momentum are calculated differently (I think one of them involves an integral). g should therefore equal 1 but instead it equals 2. Thats why I suspect that it has to do with the spinning electron having a different 'effective mass' than one would expect.

 

[Narayanan.S]

Hi everyone
Regarding classical radius electric potential with a test point charge”q” equated to mass energy of electron. All others are verified constants except (Re) backed with Thompson experiment
Re= (kq^2/meC^2) =2.82x10^-15m.
Since angular momentum of electron a mechanical unit (even smaller than Planck’s constant by 100 times) is not measurable whereas “C’ has been verified. Hence violation of is “C’ may not be a correct approach.
Further (1/2) spin lead to equating an electron magnetic moment (100 times weaker) with Bohr magneton and later did not work for nucleons
Consider a smeared charged (q) hollow sphere of radius (Re) and spinning too about its axis at “C’. The mass energy if derived from electric field energy outside sphere the angular momentum for it is (2/3)mCRe and for a solid sphere (2/5)mCRe.
With the above the only way out was to find the spin factor in place of (1/2) with the help of a very accurately measured electron magnetic moment made by CERN and landed up at ((1/sqrt2) when coupled with orbital Bohr magneton. Thanks to web postings
The precession of the angular momentum vector for this is 45 degrees
May this gives some visibility into the concept. I had requested for help on other problems faced due to the point electron concept such as infinite field intensity from electric charge.

 

[Bob_for_short]

Have you ever heard of the quantum mechanical charge smearing? I hope so. Then why to apply classical models?

Electron is always coupled to the quantized electromagnetic field. Such a coupling smears the electric charge quantum mechanically. So what is observed is the magnetic moment of a compound, complicated system. Of course g is different from 2 then.

 

[granpa]

if a (free) charge is in a dielectric material then the polarization of the material (and the resulting bound change) will cause the field to be less than it would be without it. (in the extreme case of a conductor the field is zero). This makes it look as though there is less charge there than there really. (free - bound change) But even space itself has a permittivity. so maybe the actual spinning charge is slightly greater than what we observe. would that explain the difference from 2 or would it work the opposite direction and make he discrepancy worse?

 

[granpa]

if a (free) charge is in a dielectric material then the polarization of the material (and the resulting bound change) will cause the field to be less than it would be without it. (in the extreme case of a conductor the field is zero). This makes it look as though there is less charge there than there really. (free - bound change) But even space itself has a permittivity. so maybe the actual spinning charge is slightly greater than what we observe. would that explain the difference from 2 or would it work the opposite direction and make he discrepancy worse?

http://en.wikipedia.org/wiki/Vacuum_polarization

According to quantum field theory, the ground state of a theory with interacting particles is not simply empty space. Rather, it contains short-lived "virtual" particle-antiparticle pairs which are created out of the vacuum and then annihilate each other.

Some of these particle-antiparticle pairs are charged; e.g., virtual electron-positron pairs. Such charged pairs act as an electric dipole. In the presence of an electric field, e.g., the electromagnetic field around an electron, these particle-antiparticle pairs reposition themselves, thus partially counteracting the field (a partial screening effect, a dielectric effect). The field therefore will be weaker than would be expected if the vacuum were completely empty. This reorientation of the short-lived particle-antiparticle pairs is referred to as vacuum polarization.

apparently its a well known effect.

 

[Bob_for_short]

Some of these particle-antiparticle pairs are charged; e.g., virtual electron-positron pairs. Such charged pairs act as an electric dipole. In the presence of an electric field, e.g., the electromagnetic field around an electron, these particle-antiparticle pairs reposition themselves, thus partially counteracting the field (a partial screening effect, a dielectric effect). The field therefore will be weaker than would be expected if the vacuum were completely empty. This reorientation of the short-lived particle-antiparticle pairs is referred to as vacuum polarization.

apparently its a well known effect.

It is not so simple. As I said, interaction with the quantized EMF (photon oscillators) smears out the electron charge so it is not point-like and its field is not strong. Pair polarization is inessential because the resulting filed is rather weak and pairs of smeared electrons and positrons do not contribute much.

 

[granpa]

http://en.wikipedia.org/wiki/Gyromagnetic_ratio

Consider a charged body rotating about an axis of symmetry. According to the laws of classical physics, it has both a magnetic dipole moment and an angular momentum on account of its rotation. It can be shown that as long as its charge and mass are distributed identically (e.g., both distributed uniformly)...μ=L(q/2m)

μ is the magnetic moment
L is the angular momentum
(q/2m) is the classical gyromagnetic ratio

multiplying the angular momentum by (q/2m) is the same as replacing mass with change and multiplying by 1/2 to account for the fact that the units were historically determined by different means. the 1/2 has no real mathematical significance. its just a historical artifact.

for an electon L = ℏ and the measured gyromagnetic ratio is the classical gryromagnetic ratio times the g-factor (which is approximately 2)

so if the charge is slightly greater than what we macroscopically observe then its magnetic moment will be slightly larger than what we expect to see. therefore we would expect the g-factor to be slightly larger than expected.

the measured value of the g-factor is g = 2.0023193043617(15)
slightly more than 2.
(trictly speaking we expect the g-factor to be 1. we assume the existence of some unknown mechanism that makes the expected value 2. As I've said I think the unknown mechanism has to do with 'effective mass')

 

[granpa]

It is not so simple. As I said, interaction with the quantized EMF (photon oscillators) smears out the electron charge so it is not point-like and its field is not strong. Pair polarization is inessential because the resulting filed is rather weak and pairs of smeared electrons and positrons do not contribute much.

I know the charge is not point like. whether it is or is not is irrelevant to my point. my point is that the total amount of charge that we observe is slightly less than the true amount of charge that actually exists.

I know that the field resulting from the polarization of the virtual particles is weak. so is the difference from 2.
g = 2.0023193043617(15)

 

[Narayanan.S]

I find it hard to accept that electrons are close to point particle (<10^-18m) and thought of considering classical radius for the following reasons
1) It is well known that the magnetic moment of electron are greater than those of nucleons and therefore I thought even modern physics should fundamentally expect a larger radius for the electrons(re-classical).Though experiments may show smaller and smaller radius as more and more energy is imposed on it.Is it that the interpretation of experiment is not correct?
2) Does it not mean violation of "C" will not reflect on mass energy relation?
3) At the instant of a neutron beta decay into a proton(0.5x10^-15m)- the accepted emperical value for nucleon radius) and since electrons are close to a point like particles the distance between them is also about (0.5x10^-15m)- well within range of nuclear forces
of (1.7x10^-15m).The binding energy between them should be (>95Mev) and the beta decay is not possible?
4) The electron size is not precisely known yet, then how to qulitatively explain the magnetic moment of an electron?.
5) Could not find an answer for the very high field intensity (point like charged electron) seems not addressed?

Therefore did an exercise with classical radius of electron and found that it agrees extremely well with g-factor correction.And this value for electron magnetic moment is considerably less than Bohr magneton and could have turned out to be anomalous Zeeman effect.
Request can anyone help me in clarifying on the above "5" points
Thanks

 

[Narayanan.S]

When an electron is bound in an atom it is subjected to spin - orbit coupling.The g-factor is the ratio of electron magnetic moment [eCre)xSqrt2/Bohr Magneton) + Bohr magneton] and this sum is again divided by Bohr magneton.C-spin velocity at "C" and re-classical radius of electron.
Stumbled upon this formula for electron anomaly from google search
May be it means something more.
Cheers

 

[Bob_for_short]

Speaking of some size of electron means we can see it without perturbing it. In fact any interaction (scattering) from an electron is inelastic because it is coupled permanently to the quantized EMF. Electron and the quantized EMF form a compound system in which the electron charge is smeared quantum mechanically, mathematically similarly to atomic orbitals. The elastic picture is a cloud of rather big size (depending on the external field).We cannot speak of electron itself but only in some "bound" state determined with the proper quantized EMF and the external field.

 

[Narayanan.S]

When ever an electron is found it is a full electron and the g- factor applied when in spin-orbit coupled state.
Electron-Positron collisions are being conducted shows ther is some size for these particles.

We should find an answer with a broader perspective to Dirac’s (1/2) spin. A point particle approach is fine mathematically, but a qualitative approach is needed for clarity and visibility. Thought try a semi classical approach.
Is it not right then to take classical radius based on my prevous reply and that
The (electron mass – me), h and “C” brought in the famous TRIO unit lengths- Compton wavelength, Bohr Radius and the classical radius of electron (re) that has been left out but solves so many problems.
May be there is a need to find a right interpretation for the experimented results

 

[Bob S]

.

I know that the field resulting from the polarization of the virtual particles is weak. so is the difference from 2.
[For the electron,] g = 2.002 319 3043617(15)

What is more interesting is that the g factor for the muon (also a lepton), is larger:
g=2.002 331 8414

See http://en.wikipedia.org/wiki/G-factor

Bob S

 

[granpa]

What is more interesting is that the g factor for the muon (also a lepton), is larger:
g=2.002 331 8414

See http://en.wikipedia.org/wiki/G-factor

Bob S

fascinating. thank you.

 

[Narayanan.S]

fascinating. thank you.

You are right
Dirac’s spin appears to be limited two degree of freedom “spin up” and “spin down” and its correlation to spin angular moment magnitude and the component half spin along Z-axis seems incorrect since it fails for composite particles also.
For photons it should then be unidirectional “spin 1”. Am I correct?

 

[Bob S]

Both the muon and electron are (supposed to be) point structureless leptons. Therefore, the g factor due to internal QM should be the same. The g factor should then be
g = 2( 1 + α/2π) = 2.002322819454
including only the external vertex correction.

The electron g factor is 0.000 003 515 lower, and

the muon g factor is 0.000 009 021 higher.

These disparities are due to external radiative corrections. The muon is higher than the electron primarily because the electron-positron vacuum polarization (2 Feynman bubbles) correction is much higher in the muon case than in the electron case.
Bob S

signs corrected 11/19/09