return to homepage The Bohr Magneton Why is the Bohr Magneton not equal to the measured magnetic moment of the electron? In experiment, we find that the values are off by .1%. QED has no simple answer for this. I do. QED proposes to explain the error by once again pouring Diracís virtual sea on the problem and once again waving the magic wand. The electron is said to be interacting with virtual photons, giving it a precession and thereby a g-factor. All this is just one more fudge, however. Anytime you see the word ďvirtualĒ in modern physics, it means you have left the path of reason. I will show the simple mathematical reason for the error. The Bohr magneton was first proposed by Procopiu in 1913. It is not a particle, but rather an expression of the magnetic field created by the individual electron. We have a simple equation for it:
The Bohr Magneton
Why is the Bohr Magneton not equal to the measured magnetic moment of the electron? In experiment, we find that the values are off by .1%. QED has no simple answer for this. I do.
QED proposes to explain the error by once again pouring Diracís virtual sea on the problem and once again waving the magic wand. The electron is said to be interacting with virtual photons, giving it a precession and thereby a g-factor. All this is just one more fudge, however. Anytime you see the word ďvirtualĒ in modern physics, it means you have left the path of reason. I will show the simple mathematical reason for the error.
The Bohr magneton was first proposed by Procopiu in 1913. It is not a particle, but rather an expression of the magnetic field created by the individual electron. We have a simple equation for it:μB = eh/2π2me
Unfortunately, as I said, this gives us a number that fails by about .1%. To see why, we must study this equation more closely. We can do this by looking at angular momentum, and the easiest way to do that is by returning to Bohrís simple math, which I first critiqued in another paper.
L = nh/2π = rmv
We let n = 1, since we are studying the first electron in the hydrogen atom. So,μB = eL/2me = erv/2
We find that this equation yields the wrong number. Why? Because the math is wrong. As I showed in my paper called Bohrís First Big Mistake, the equation L = rmv is wrong. Current theory tries to cover this by never including momentum or velocity variables, but they were there in the beginning. Look at what tangential velocity we get, for starters:μB = erv/2
v = 2μB/er = 2(9.274 ◊ 10-24 JēT-1/(1.602 x 10-19C)(5.29 x 10-11m)
v = 2.19 x 106m/s
Mirroring current assumptions, I used the Bohr radius for r. That is well under c. Why? Why doesnít the electron maximize its orbital speed? I will be told that it is because v is not the orbital speed, ω is. Well, v = rω. So,ω = 4.14 x 1016/s
Is that over c? Nobody knows, because nobody understands angular speed.
You canít just multiply or divide by a radius to make a linear velocity into an angular velocity. That doesnít make any sense, mathematically or mechanically. Look at the equations for momentum and angular momentum closely:
p = mv
L = rmv
If the radius is greater than one, the effective angular velocity will be more than the linear velocity. If the radius is less than one, the effective angular velocity will be less than the linear velocity. That is a flagrant example of illogical scaling.
The history of physics fudges over this problem by creating a moment of inertia, but the moment of inertia is a ghost. It is the attempt to hide the fact that v = rω is wrong. You would not have a moment of inertia without v = rω.
Where does that equation come from? It comes from 2πr/t. If v = 2πr/t, and ω = 2π/t, then v must equal rω. But, as I have shown, v ≠ 2πr/t. In the historical derivations, v is defined as the tangential velocity. But 2πr/t is not the tangential velocity; it is the orbital velocity. The orbital velocity curves and the tangential velocity does not. The tangential velocity is a straight line vector with its tail on the curve, but it does not follow the curve.
If v = 2πr/t, then v is already an angular velocity. An orbital velocity and an angular velocity are the same thing. They both curve. Therefore, in going from 2πr/t to ω, you arenít really going from a linear expression to an angular expression. You are going from one angular expression, expressed in meters, to another angular expression, expressed in radians.
None of the angular momentum equations in books make any sense, so I developed my own equation to do this, going back to Newton to find the method. You can see my derivation in my paper on a = v2/r. In my equation, v really is the tangential velocity, and therefore it is not equal to 2πr/t. It is equal to x/t.
ω = √[2r√v2 + r2) - 2r2]
This equation is logical, because using it we find that the angular velocity is always less than the tangential velocity. We donít have any misdirection with moments of inertia, and we donít have the illogic of having the variables change in different ways for different values of r. We have a logical progression, since as we get larger, the angular velocity approaches the tangential velocity. Obviously, this is because it loses it curvature as it increases, becoming more like the straight-line vector. By the same token, at small scales, the angular velocity gets very small compared to the tangential velocity, and this is because the curvature is so great.
Using my new equation for ω,
L = mω = h/2πμB = eω/2
Now we just solveω = 2(9.283 x 10-24)/1.602 x 10-19
ωe = 1.16 x 10-4m/s
r = √[ω4/(4v2 - 4ω2)]
If we use c for v, we find,
re = 2.244 x 10-17m
Of course, I have redefined the variable ω here. It is no longer measured in radians. Strictly, it stands for the orbital velocity measured in meters, not the angular velocity measured in radians. My new equations simply separate it from the tangential velocity, since, as you have seen, we need both. We don't need to be able to measure circular motion in radians, but we do need to measure it both as a tangential velocity and as an orbital velocity. Therefore, I have jettisoned the old angular velocity in radians as a useless concept.
And the velocity must also belong to the electron. That is, it belongs to the spin, not to the orbit. The velocity in this equation is not a velocity of the electron in orbit, it is the velocity of the spin. It is the tangential velocity on the surface of the spin, or the linear velocity a point on the surface of the spin border would be going if it werenít going in a circle. The magnetic moment, like the charge, belongs to the electron, not to the orbit!
I will be told that the orbit must have a momentum of its own, angular or otherwise. Yes, there must be orbital energy, but we need not calculate an angular momentum. The only thing in the orbit is the electron, so there is no mass inside the sphere. The only thing inside the orbit is the nucleus, and it is not moving relative to the orbit, so it creates no angular momentum. This being true, we only have to look at mass and momentum at the tangent. If we do that, we donít have to be concerned with angular anything. The electron transmits energy from the tangent, by emission, and this emission is emitted during a very small interval of the orbit. The emission leaves the electron in a straight-line vector, so we donít have angular momentum involved. We can use the tangential velocity directly, and compute the momentum linearly, with p = mc. But this momentum does not contribute to the magnetic field. This momentum contributes to the linear energy of the emission, not its spin energy. Magnetism is not caused by linear energy, it is caused by spin energy. So we donít have to be concerned with the orbital energy.
Now, the electron does interact with the field outside the orbit, but this is not a virtual field. It is the emission field of other quanta. The vacuum is awash with emission, and this emission acts as a friction on the orbit. But this doesnít cause a precession or a g-factor. It causes the electron to have an outer spin that is opposed to its orbit, like a set of spinning cogs. The electron has a linear momentum, or tangential momentum of
p = mc = 2.7 x 10-22m2/s
The outer spin has an opposite momentum of about 10-34. This acts as a slight drag on the linear momentum, but only in the 12th decimal point. So it could not cause a .1% change in the magnetic moment. In fact, it changes nothing in the magnetic field, since the magnetic field is under the electric field. Only the orbital momentum and the electric field could be affected.
But why is the experimental number for the Bohr magneton .1% wrong? Is it just that Bohr's numbers were different than current numbers? Is it a problem of the virtual field, explained by the g-factor? No. It is caused by the unifed field. In another paper I derive a solid number for the summed charge field of the Earth. This is not the electrical field of the Earth or the magnetic field. It is what I call the foundational E/M field, caused by the emission of photons by all matter in the field. It causes the electric and magnetic fields, but is not equivalent to either one. It's direction is straight out from the Earth, radially; and it is always repulsive. All bodies create this field, and it is always in vector opposition to gravity proper. This field, with gravity, makes up the unified field. The average field strength at the surface of the Earth for this field is .009545 m/s2. This number was arrived at by rather simple math, by comparing the fields of the Earth and Moon. I will not repeat the math here: you will have to take the link to see it. The important thing here is that the number just quoted gives us almost precisely a .1% correction to the unified field, and therefore to the Bohr magneton. We just divide that number by 9.8 to find a .1% correction, you see. We get .0974%, which is close enough for me in this problem. The reason this solves the problem of the Bohr magneton is that the experiments have all been run on the surface of the Earth, in a field not known to exist until now. This charge field has been hidden in Newton's equation, as part of the gravitational field. Newton's equation gives us the total field, but not the constituent fields. Likewise for Einstein's field equations. The charge field is ignored at the macro-level. But since I have proved that there is a charge field in vector opposition to gravity existing at all points on the Earth, this gives us a simple explanation of the error in the Bohr magneton. This charge field must obviously affect the magnetic field of the electron directly, by straight bombardment of charge photons. This gives us the simple mechanical cause of the .1% error, with all the necessary math. As a matter of fact, all the historical and current experiments on the electron that show this .1% error are now proof of my theory. I predicted a .1% variance several years ago in the linked paper, before knowing of or studying the Bohr magneton. I have now found the pre-existing proof of it here, and have explicitly shown the necessary connection of the two numbers.
Finally, letís check that value for the electron radius. Actually, what I found above is the radius of the outer spin. The electron in orbit has both an axial spin and an x-spin. Therefore the radius of the electron proper is:
re = 1.122 x 10-17m
But the x-spin radius, 2.244 x 10-17m, must be the effective border of the electron, since due to the end-over-end spin, the mass will inhabit this entire radius, during motion. In another paper, I found the radius of the proton to be about 10-13m, and the proton is known to have a mass of about 1836 times the electron. Using those numbers, we get
r = 5.45 x 10-17m
Which is very close. We can use my number to re-estimate the radius for the proton, assuming it has the same density as the electron.
rP = 4.11 x 10-14m
We can fine tune that as well. Since we are finding radius here, not mass, we can use my spin equations from my paper unifying the electron and proton. In a nutshell, we find we need the transform 1822, not 1836. The number 1836 is a sort of mass transform, which means it must be a unified field transform. Mass is always a unified field number. But radius is not a unified field number. For that reason, we can just use the Dalton, 1822, which transforms size but not mass. In that case we get the number 4.09 x 10-14m for the proton radius. That is very interesting because it is the square root of the proton mass. See my new paper on proton mass for more on this.
Of course this means the Bohr radius is wrong as well. Bohrís math is completely compromised by now, so everything has to be redone. The problem with angular velocity has infected all the math, and nothing will stand. Letís correct the Bohr equation:
mv2/r = ke2/r2
First of all, a ≠ v2/r . If we want to use the angular velocity, we must use this equation
a = ω2/2r
Which gives us,
mω2/2 = ke2/r
Using Bohrís method, we find
L = mω = h/2π
ω = h/2πm
hω/4π = ke2/rω = 4πke2/hr
h/2πm = 4πke2/hr
r = 8π2mke2/h2
This gives us the inverse of Bohrís radius. You can see that because of the correction to the equation L = rmv, we get the radius on the wrong side of the equation, skewing the math. This means that Bohrís math depends on using that false equation. If you use the right math for angular momentum, the rest of Bohrís math fails. It fails because L = h/2π applies to the electron, but Bohr is trying to apply r and mω to the orbit. So these substitutions we are making canít work.
Think of the orbit like a big spinning particle, of radius r. That big particle has an angular momentum. The electron also has an angular momentum. Bohr has conflated the two. His equations are a mixing of both values.
So he has made two big errors. One, he has used the wrong equation for angular momentum, based on a mistaking of tangential and orbital velocity. Two, he has a false equality. If our first equation (mω2/2 = ke2/r) is correct, then the angular velocity ω must apply to the orbit, not to the electron. If it applies to the orbit, then mω ≠ h/2π. This is because h/2π applies to the electron in orbit, not to the orbit.
We will also see in subsequent papers that Schrodingerís equations do not solve this problem. Schrodinger has Bohrís principal quantum number and also a separate angular momentum quantum number, but he does not assign these physically. Because we donít get the mechanics, the math is unclear. And Schrodingerís equations retain the errors of Bohr in going from linear to angular velocity. That is, Schrodinger still uses a false angular momentum equation. L = rmv was not corrected by Schrodinger, and it has never been corrected since.
Can we still find a Bohr radius? Let us assume that the first equation is right, after correcting the momentum equation.
mω2/2 = ke2/r
But we have two unknowns, r and ω, and only one equation. We canít solve without another equation, and Bohrís momentum equations are false. Let us first try using c for v. We will assume the tangential velocity of the electron is maximized.
So we simply return to the equation ω = √[2r√v2 + r2) - 2r2], using c for v. Since the electron is bigger than the photon, it must have a limit just under c, but since that limit is in the fifth decimal point of c, we will ignore it here.ω = √[2r√c2 + r2) - 2r2]
mω2/2 = ke2/r
r√c2 + r2) - r2 = ke2/mr
We can simplify that by noticing that the left side will be dominated by c, allowing us to omit values of r.
cr = ke2/mr
r = √ke2/cm] = 9.19 x 10-4m
Thatís way too large, so we know something is still wrong. Letís try using Bohrís radius to find the angular velocity.ω = √2ke2/mr] = 3.09 x 106m/s
Interesting that that is almost what we got for the tangential velocity v using Bohrís math above: remember that using the Bohr radius, we found v = 2.19 x 106m/s. But if we use the right velocity equations, we find
r2 = ω4/(4v2 - 4ω2)
v = ω2√[4r2 + ω2] = 3 x 1019m/s
So that canít be right either. From these calculations, it would appear that the Bohr radius is either greater than or equal to about a millimeter, or we are using the wrong values for the electron, or the equation is still wrong.
It turns out that the equation is still wrong. The problem is simple: the constant k doesnít apply at the quantum level. Coulombís equation is for use at the macro-level, and the constant is a scaling constant.
Just as I showed with G in another paper, k takes us from one level of size to another, so that we can compare fields that have different mediating particles or accelerations. Coulomb was working with little pith balls, not electrons, and his balls were nine orders of magnitude larger than the orbital radius of the electron, as you will see.
I showed that G is a scaling constant that takes us from the size of emitted photons to the size of the atom. Yes, the B-photon is G times smaller than the proton. We find it in Newtonís equation, of all places, because Newtonís equation is actually a unified field equation in disguise. It contains both the gravitational acceleration and the foundational E/M field (or charge field). Well, the same applies to Coulombís equation. It looks just like Newtonís equation because it is the same unified field equation in a different disguise. Newtonís equation is hiding the E/M field, and Coulombís equation is hiding the gravitational field. Because neither Newton nor Coulomb understood the fields under their equations, they only provided us a math that works. Their equations work because they compress the unified field into one field, and the transform between the two fields is the constant.
Since in Bohrís equation the field is the actual field the electron is moving in, we donít need a transform or a scaling constant. The electron is already moving at the proper scale. In the little illustration in the book, we see the electron circling the nucleus, and the electron and the orbital radius are in the same field. We have to do very little scaling (between the charge and the field it is in) in order to draw the picture, and this is not beside the point. The proton is actually repulsing the electron down that very radius.
And so we get this very simple equation:
r = √(e2/mc) = 9.69 x 10-9m
That is the corrected Bohr radius. The value of Coulombís constant is 9 x 109. That is a scaling transform that takes us directly from the Bohr radius to our own world. But Coulomb's balls weren't one meter in radius. His pith balls were about 6mm, which is about 170 times smaller than 1 meter. That number 170 is not a coincidence either, since I have just found that the Bohr Radius is 177 times larger than we thought. In fact, if we divide 1 meter by 177, we get 5.65 mm. Coulomb himself tells us that his balls were "2 to 3 lines in diameter",1 which is 4.5 mm to 6.8 mm.
What all this means is that Coulomb's constant is not a constant. It is taking us from one size to another, so it cannot be applied across a range of sizes. This was to be expected, since I have shown that Coulomb's equation, like Newton's equation, is a unified field equation that includes both the E/M field and the gravitational field. Furthermore, I have shown that the two fields are not the same relative size to eachother, as we scale the equations up and down.
So the connection between Coulombís constant and the Bohr radius is not a coincidence. Although the current numbers are wrong, it is no coincidence that the current Bohr diameter is thought to be about 1/k meters. It is the mistake in the Coulomb equation that led directly to the mistake in the Bohr radius, and they are connected both mathematically and historically. It is also not a coincidence in my new math, since if you multiply the old Bohr radius and the real diameter of Coulombís ball by 177, you get my new Bohr radius and 1 meter. For more on this, see my paper on Coulomb's equation.
I have already shown that a misreading of the scattering equations means we have the atomic size about 100 times too small, so my new equation also fits that prediction and correction very well. The Bohr radius is 177 times larger than we thought, and you can now see all the math and logic behind the correction.
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