Mar 17 2010

## Magnetic Moment of the Electron

Using Tipler’s notation starting on page 917, we begin with a particle of charge q and mass M, in a circular Bohr orbit of radius r.  Never mind that there is no such thing as a circular electron atomic orbit; the results are valid.

Particle velocity, v, and orbital period, T, are related by:

vT = 2πr

The charge traveling in a circle produces a current of:

I = q/T = qv/2πr

Multiplying this current by the enclosed area, the magnetic moment is obtained:

m = (qv/2πr) πr2 = (qvr)/2                    [1], 39-16, pg 918

Angular momentum for the mass is:

L = Mvr = M (2m/q)

Rearranging and writing magnetic moment and angular momentum vectors in bold type:

m = (q/2M) L                                        [1], 39-17, pg 918

Having used a circular orbit of a particle with charge and mass to obtain this formula, it can also be used when spin angular momentum is inserted for L.  For the electron, the spin angular momentum is known to be ћ/2.  Also, still following Tipler, we can now insert e for the charge of the electron in place of q, and me for the mass of the electron in place of the general mass M.  There is just one problem when doing this however.  “For electron spin, the magnetic moment is twice that predicted by this equation.  This extra factor of 2 is a quantum-mechanical result which has no analog in classical mechanics.” ([1], pg 918)

The extra factor of 2 is the electron spin g factor, and is more precisely found to be:

ge = -2.002 319 304 3622        [2]

The magnetic moment of an electron can then be written as:

m = ge (e/2me) (ћ/2),

or equivalently:

μS = geμB(S/ћ)         [3]

This extra factor of 2.002 319 304 3622 shows that the gravitons which pass through electrons, and are not absorbed, serve as conjugate wave functions that double the expected magnetic moment of the electron.

[1] Tipler, Paul A., Physics, Worth Publishers, Inc., 1976

### 6 Responses to “Magnetic Moment of the Electron”

1. Carlon 17 Mar 2010 at 1:05 pm

Is there a way to explain in lay terms how the gravitons affect the electrons’ spin? I can’t follow the math and I don’t know how the spin is affected by the graviton acting as a conjugate wave.

2. Kevinon 18 Mar 2010 at 7:06 pm

In the complex number system (a – bi) is the conjugate of (a + bi). For a wave function, e^-ikx is the conjugate of e^ikx, where:
e^ikx = cos kx + i sin kx, and
e^-ikx = cos kx – i sin kx
Basically, gravitons passing through an electron, proton, or neutron create potential wells for the standing waves inside these particles, which internal waves are in bound states. To borrow from Kronig-Penny Hamiltonian math, in the “well domain of the potential array” ([4], pg 295) we may have:
φI = Ae^ik(sub1)x + Be^-ik(sub1)x,
and in the “barrier domain”:
φII = Ce^ik(sub2)x + De^-ik(sub2)x, ([4], (8.65), pg 307)
The speed of a conjugate wave graviton may slow down to less than the speed of light in a vacuum as it passes through an electron or a nucleus, though its frequency would stay the same. It may also have a conjugate rotational component to that earlier written [5], in the well domain.
This is only a very general description. I do not pretend to be anywhere remotely close to writing down the internal wave functions of electrons and nuclei. Gravitons are coming in from all directions and it would be a very difficult endeavor.

[4] Liboff, Richard L., Introductory Quantum Mechanics, Fourth Edition, Addison Wesley, 2003
[5] http://www.fruechtetheory.com/blog/2008/04/01/wave-function-transfer-2-2/

ps: evidently, in the reply sections, subscripts just won’t take

3. Carlon 19 Mar 2010 at 11:10 am

Even the dumbed-down version seems to be above my experience in physics’ ability to elucidate. I find this stuff very fascinating though. Thanks for the reply.

4. Kevinon 19 Mar 2010 at 5:33 pm

I took some liberty to also put out a new idea. Sorry.
The letter k is used for wave number, which is in units of radians per meter. Since k1 has a larger numerical value than k2, the full wavelength is shorter in distance for k1 compared to k2. The electromagnetic wave would be compacted within a subatomic particle, though once again, keeping the same frequency. How fast this would all have to happen with a particle as small as an electron or a proton is mind boggling, – a graviton passing through a sub atomic particle.

5. Carlon 20 Mar 2010 at 12:31 pm

Ah, I understand that :)
So, it’s not just a “quantum mechanical result which has no analog in classical mechanics”, but rather can be explained by this interaction with gravitons?

6. Kevinon 20 Mar 2010 at 8:16 pm

Yes, it unifies the classical with the quantum. The force we have all been aware of our entire lives may be associated with the single source of all energy and mass.