Angular Velocity
and Angular Momentum by Miles Mathis

One of the greatest mistakes in the history of physics is the continuing use of the current angular velocity and momentum equations. These equations come directly from Newton and have never been corrected. They underlie all basic mechanics, of course, but they also underlie quantum physics. This error in the angular equations is one of the foundational errors of QM and QED, and it is one of the major causes of the need for renormalization. Meaning, the equations of QED are abnormal due in large part to basic mathematical errors like this. Because they have not been corrected, they must be pushed later with more bad math.

Any high school physics book will have a section on angular motion, and it will contain the equations I will correct here. So there is nothing esoteric or mysterious about this problem. It has been sitting right out in the open for centuries.

To begin with, we are given an angular velocity ω, which is a velocity expressed in radians by the equation

ω = 2π/t

Then, we want an equation to go from linear velocity v to angular velocity ω. Since v = 2πr/t, the equation must be

v = rω

Seems very simple, but it is wrong. In the equation v = 2πr/t, the velocity is not a linear velocity. Linear velocity is linear, by the equation x/t. It is a straight-line vector. But 2πr/t curves; it is not linear. The value 2πr is the circumference of the circle, which is a curve. You cannot have a curve over a time, and then claim that the velocity is linear. The value 2πr/t is an orbital velocity, not a linear velocity.

I show elsewhere that you cannot express any kind of velocity with a curve over a time. A curve is an acceleration, by definition. An orbital velocity is not a velocity at all. It cannot be created by a single vector. It is an acceleration.

But we don't even need to get that far into the problem here. All we have to do is notice that when we go from 2π/t to 2πr/t, we are not going from an angular velocity to a linear velocity. No, we are going from an angular velocity expressed in radians to an angular velocity expressed in meters. There is no linear element in that transform.

What does this mean for mechanics? It means you cannot assign 2πr/t to the tangential velocity. This is what all textbooks try to do. They draw the tangential velocity, and then tell us that

vt = rω

But that equation is quite simply false. The value rω is the orbital velocity—even according to current definitions—and the orbital velocity is not equal to the tangential velocity. The velocity may be labeled "tangential," but what is derived in the historical proofs is the orbital velocity.

I will be sent to the Principia, where Newton derives the equation a = v2/r. There we find the velocity assigned to the arc.1 True, but a page earlier, he assigned the straight line AB to the tangential velocity: "let the body by its innate force describe the right line AB".2 A right line is a straight line, and if Newton's motion is circular, it is at a tangent to the circle. So Newton has assigned two different velocities: a tangential velocity and an orbital velocity. According to Newton's own equations, we are given a tangential velocity, and then we seek an orbital velocity. So the two cannot be the same. We are GIVEN the tangential velocity. If the tangential velocity is already the orbital velocity, then we don�t need a derivation: we have nothing to seek! If you study Newton's derivation, you will see that the orbital velocity is always smaller than the tangential velocity. One number is smaller than the other. So they can't be the same.

The problem is that those who came after Newton notated them the same. He himself understood the difference between tangential velocity and orbital velocity, but he did not express this clearly with his variables. The Principia is notorious for its lack of numbers and variables. He did not create subscripts to differentiate the two, so history has conflated them. Physicists now think that v in the equation v = 2πr/t is the tangential velocity. And they think that they are going from a linear expression to an angular expression when they go from v to ω. But they aren't.

This problem has nothing to do with calculus or going to a limit. Yes, we now use calculus to derive the orbital velocity and the centripetal acceleration equation from the tangential velocity. But Newton used a versine solution in the Principia. And going to a limit does not make the orbital velocity equal to the tangential velocity. They have different values in Newton's own equations, and different values in the modern calculus derivation. They must have different values, or the derivation would be circular. As I said before, if the tangential velocity is the orbital velocity, there is no need for a derivation. You already have the number you seek. They aren't the same over any interval, including an infinitesimal interval or the ultimate interval.

This false equation vt = rω then infects angular momentum, and this is where it has done the most damage in QED. We use it to derive a moment of inertia and an angular momentum, but both are compromised.

p = mv

L = rmv

Where L is the angular momentum. This equation tells us we can multiply a linear momentum by a radius and achieve an angular momentum. Is that sensible? No. It implies a big problem of scaling, for example. If r is greater than 1, the effective angular velocity is greater than the effective linear velocity. If r is less than 1, the effective angular velocity is less than the effective linear velocity. How is that logical?

To gloss over this mathematical error, the history of physics has created a moment of inertia. It develops it this way. We compare linear and angular energy, with these equations:

K = (1/2) mv2 = (1/2) m(rω)2 = (1/2) (mr22 = (1/2) Iω2

The variable "I" is the moment of inertia, and is called "rotational mass." It "plays the role of mass in the equation."

All of this is false, because vt = rω is false. That first substitution is not allowed. Everything after that substitution is compromised. Once again, the substitution is compromised because the v in K = (1/2)mv2 is linear. But if we allow the substitution, it is because we think v = 2πr/t. The v in K = (1/2)mv2 CANNOT be 2πr/t, because K is linear and 2πr/t is curved. You cannot put an orbital velocity into a linear kinetic energy equation. If you have an orbit and want to use the linear kinetic energy equation, you must use a tangential velocity.

The derivation of angular momentum does the same thing

L = Iω = (mr2)(v/r) = rmv

Same substitution of v for rω. Because v = rω is false, L = rmv is false.

But this angular momentum equation is used all over the place. I have shown that Bohr uses it very famously in the derivation of the Bohr radius. This compromises all his equations.

Because Bohr's math is compromised, Schrodinger's is too. This simple error infects all of QED. It also infects general relativity. It is one of the causes of the failure of unification. It is one of the root causes of the need for renormalization. It is a universal virus.

The correction for all this is fairly simple, although it required me to study the Principia very closely. We need a new equation to go from tangential or linear velocity to orbital velocity, which I am calling ω. Newton does not give us that equation, and no one else has supplied it since then. We can find it by following Newton to his ultimate interval, which is the same as going to the limit. We use the Pythagorean Theorem. As t→0,

ω2 → v2 - Δv2
and, v2 + r2 = (Δv + r)2
So, by substitution, ω2 + Δv2 + r2 = Δv2 + 2Δvr + r2
Δv = √ v2 + r2) - r = ω2/2r

v = √[(ω4/4r2) + ω2]
ω = √[2r√v2 + r2) - 2r2]
r = √[ω4/(4v2 - 4ω2)]

Not as simple as the current equation, but much more logical. Instead of strange scaling, we get a logical progression. As r gets larger, the angular velocity approaches the tangential velocity. This is because with larger objects, the curve loses curvature, becoming more like the straight line. With smaller objects, the curvature increases, and the angular velocity may become a small fraction of the tangential velocity. And if v and ω diverge greatly, as with very small particles, this equation can be simplified to

v = ω/r

Yes, it is just the inverse of the current equation.

This means that the whole moment of inertia idea was just a fudge, used to make v = rω. Historically, mathematicians started with Newton's equations, mainly v = 2πr/t, which they wanted to keep. To keep it, they had to fudge these angular equations. In order to maintain the equation v = rω, the moment of inertia was created. But using my simple corrections, we see that the angular momentum is not L = mvr = Iω. The angular momentum equation is just L = mω. We didn't need a moment of inertia, we just needed to correct the earlier equations of Newton, which were wrong.

Of course I have changed the definition of ω above. It is now the angular velocity measured in meters and v is the tangential velocity measured in meters. We can ditch angular velocity measured in radians, since it is basically useless. All it does is confuse the math, since we don't need to measure orbital motion in radians. But we do need to be able to write it as both a tangential and an angular velocity.

In fact, we have conspicuous and longstanding data proving I am right. According to the current equation v = rω, a greater orbital radius should give us a greater velocity. According to my new corrected equations, a greater radius should give us a lesser velocity. What do we find?

velocity of Mercury 48
velocity of Venus 35km/s
velocity of Earth 30km/s
velocity of Mars 24km/s
velocity of Jupiter 13km/s

Need I go on?

Maybe I do, because some readers will say, "That is assuming ω is a constant. The current math doesn't assume that." No, it doesn't. In fact, it finds different values for ω for all the planets, like this:

ω of Mercury 8.3 x 10-7rad/s
ω of Venus 3.2 x 10-7rad/s
ω of Earth 2 x 10-7rad/s
ω of Mars 1 x 10-7rad/s
ω of Jupiter 1.6 x 10-8rad/s

But with my equation, those given velocities are already orbital numbers, so they are already angular. And we find the tangential velocity is only different from the angular velocity by about 3 parts in a million, at the distance of the Earth (using this equation, v = √[(ω4/4r2) + ω2]). That is why this problem only comes up with quanta. At large scales like planetary orbits, the divergence in tangential and angular velocity is minimal. The larger the orbit, the more nearly equal the two will be, since the curvature decreases at large scales. So we think we have visual confirmation of Newton's old conflation of the two. In celestial mechanics, the two velocities have been taken as the same, and we see why. But if we apply my new equations to quanta, we get a large divergence of angular velocity and tangential velocity. This is one of the things that is skewing quantum math, as I show in my paper on the Bohr Magneton. That is why I link this problem to the Bohr and Schrodinger equations above.

*Addendum: August 2010. Many have not understood my variables here, even after all this. They have replied that my new equation v = ω/r cannot be correct simply due to units. They say that my v has the dimensions of an angle over a time over a length, which is not a velocity. But my ω is no longer an angular velocity measured in radians, you see. My angular velocity is the same as the orbital velocity, since they are basically equivalent. An angular velocity was always a curve, so it was always an acceleration. Therefore it was always a cheat or mistake to express angular velocity in radians/s. Since an angular velocity must be a curve, it must be expressed as an acceleration. Well, if ω is an acceleration, then v is a velocity, and the units do resolve. You will say, "What do you mean, the units resolve? They don't, since if we let ω be an acceleration, we get v = 1/s2. That still isn't a velocity!" Ah, but it is, as long as we remember that time and length are inverse parameters in all such equations. If you visit my paper on time, you will be reminded that L=1/T. In other words, as a matter of operation, length and time are really equivalent entities; we just put time in the denominator to create a ratio. For this reason, we can either multiply an acceleration by a time or divide it by a length: either way we will get a velocity.

In fact, it is the current equation that is senseless regarding units. In the equation v = rω, v is a velocity only if we give an angle no units or dimensions. But of course an angle is not really dimensionless. An angular velocity is a displacement just as much as a linear velocity, so we have motion. If we have motion, we must have units. An angular velocity is not "nothing per second," is it? The current equation acts as if the radians can just be dropped, but what we really have, as you now see, is radians x meters/seconds. Is that a velocity? I don't think so. Once we do the full analysis, it is my equation that makes sense and theirs that does not.

Others have complained that if L now equals mω, and if ω = vr, then the equation L = mvr is saved. My argument appears circular. But it isn't, because my v is different from their v. My v is the tangential velocity, and their v is the orbital velocity. They will say that their v is labelled vt, which is tangential velocity, but that labelling is false. Yes, they label it that way, but they do not use it that way. They substitute v = 2πr/t into that equation, and 2πr/t is not a tangential velocity. Their v is vo, not vt. The working equation is either L = mrvt or L = mω = mvo, but L ≠ mrvo. The last equation is the current one, so it is false.

You will say that if my ω is not measured in radians/s, I shouldn't label it ω; but since measuring angular velocity as radians/s is a fudge, there will be no use for that in the future. I am free to capture that variable and use it as I will.

For more on this, see my long paper on Newton.
You may also see my new paper on the electron radius for an interesting use of these new equations.

1Newton, Principia, Section II, Prop. IV, Theorem IV, Cor. 1.
2Newton, Principia, Section II, Prop. I, Theorem I.

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