(and a Refutation of Newton's

Lemmae VI, VII & VIII)

("(Formal disproof of these Lemmae can be found here")

© Miles Mathis

Please note that this paper is a simplification by me of a paper or papers written and copyrighted by Miles Mathis on his site. I have replaced "I" and "my" with "MM" to show that he is talking. All links within the papers, not yet simplified, are linked directly to the Miles Mathis site and will appear in another tab. (It will be clear which of these are Miles Mathis originals because they will be still contain "I" and "my".) The original papers on his site are the ultimate and correct source. All contributions to his papers and ordering of his books should be made on his site. (This paper incorporates Miles Mathis' avr paper, avr2 paper, angle paper and lemma paper.) |

Newton

Newton published his *Principia* in 1687. Except for Einstein’s Relativity corrections, the bulk of the text has remained uncontested since then. It has been the backbone of trigonometry, calculus, and classical physics and, for the most part, still is. It is the fundamental text of kinematics, gravity, and many other subjects.

To rigorously analyze all the historical proofs of the
orbital equation a=v^{2}/r,
including the proofs of Newton and Feynman, showing they all
contain fundamental errors. This current equation is shown to be
false in that its denominator should be 2r rather than r thus a=v^{2}/2r. Also the equation for the orbital velocity v=2πr/t is also
shown to be false.

It is assumed by most that Einstein's corrections to Newton's gravitational equations all but completed the necessary analysis of the problem. Einstein fine-tuned an already highly successful mathematics, and almost nothing is left to be done. That is current wisdom.

Of course work continues on the mechanism of gravity, since it is still completely unknown. But the mathematics of gravity is considered to be finished. No one is working on the field equations of General Relativity because they are assumed to be correct.

This paper shows that this assumption cannot be maintained. MM has uncovered a basic error of math in one of Newton's fundamental equations. The equation, and Newton's derivation of it, has stood unquestioned for centuries. The equation is used today in many esoteric theories, including the derivation of the Schwarzchild radius, the predicted intensity of a gravity wave, and on and on. It is imported into these derivations as a known fact. Furthermore, the equation is used in General Relativity. It is one of the basic preconditions of several parts of various tensors. MM shows that all these derivations and computations are fatally compromised by this.

We all learned this equation in high school, in regard to uniform circular motion. It states the relationship between an orbiting velocity and centripetal acceleration. The reason the equation is used so often in contemporary physics is that it is also assumed to describe the relationship, in its simplest form, between an orbiting body and the force of gravity felt by that body. It is basic physics, and yet it appears as if no one has looked hard at the equation in a very long time. Certainly no one has had the perspicuity, or the gumption, to question it in a high school physics class. By the time a student of physics reaches college such equations are not interesting anymore—they are outgrown toys—ones to be used if needed, but never closely examined.

MM was led to examine this equation due to problems that have cropped up in several fields. MM will not get into theory in this paper: suffice it to say that the fundamental concepts of gravity seemed to me a bit attenuated in several areas. What was necessary, in MM's opinion, was not more esoteric math—as in pursuing superstring theories and the like—but rather a closer look at the theories and concepts that supported gravitational mathematics, and especially the simple algebra that lay under most of the higher math. In doing so, MM has discovered many errors that can only be called astonishing. This paper relates one of them.

Newton used the equation a = v^{2}/r to tie his famous equation of universal gravitation to Kepler's Third Law. That is,

F = Gm_{1}m_{2}/r^{2} where

(F is the force between the masses;
G is the gravitational constant (6.674×10−11 N · (m/kg)2);
m1 is the first mass;
m2 is the second mass;
r is the distance between the centers of the masses)

becomes

t^{2}/r^{3} = 4π^{2}/Gm_{2}

only by assuming that

a = v^{2}/r (where v is the velocity and r is the radius)

The full derivation that is shown in all college textbooks will be shown in Part 3 after Newton's derivation here.

Amazingly, a = v^{2}/r has been a bedrock equation from the beginning. Newton treated it almost as an axiom himself. He "proved" the equation in a very early part of the *Principia* (Section 2, Proposition 4). It is not actually "proved" but actually introduced as a corollary, with only the outline of a proof.

Corollary 1 is but one sentence embedded in a theorem. This is corollary 1, in full: "Therefore, since those arcs are as the velocities of the bodies, the centripetal forces are in a ratio compounded of the duplicate ratio of the velocities directly, and of the simple ratio of the radii inversely." In modern English, that is, "Since the arc describes the velocity, the acceleration is the square of the velocity over the radius." Newton might justly reply that his sentence contains no implication of exact equality. He simply said that the force was proportional to the inverse of the radius. Therefore, if the equality is not exact the mistake was never his.

In fact, in the previous paragraph, he had said, "These forces tend to the centers of the circles and are one to another as the versed sines of the least arcs described in equal times; that is, as the squares of the same arcs applied to the diameters of the circles." MM has this to mean that, according to the trigonometry applied to the problem, it is the diameters that take the proportionality, in the first instance. Newton then elides from diameter to radius simply by saying, "since the diameters are as the radii." To me this proves beyond a doubt that he is talking in this section of proportionalities, not equalities. Both the radius and the diameter are equally proportional, since proportionality does not take into account first-degree magnitudes. If you are proportional to 2x then you are proportional to x.

The current derivation of the equation never mentions Newton's method using the versed sine, presumably since knowledge of versed sines is no longer common. Or it may be that the method is not mentioned because it is very difficult to penetrate. MM will show that it comes much closer to solving the problem than the current derivation. This should not be too much of a surprise, considering the author. However, by standing on Newton's shoulders modern science should have seen farther eventually, one would have hoped, given the 300 years it has had to perfect his work. Instead it appears that it has only used its time to become myopic, replacing a slightly flawed derivation with one that is a mathematical embarrassment.

Proving below that the current derivation fails badly, MM also analyzes Newton's original derivation, to show that science does not, at least, have the ignominy of replacing a correct derivation with an incorrect one. Both Newton and the current derivation make fundamental errors in analyzing circular motion.

Newton proposes a body that moves from A to B, is there compelled by a force which turns it, and continues on to C. It had a constant velocity to start with, therefore AB = Bc = BC. Newton postulates that c is where the body would have gone without the force. He seeks the size and direction of the force to turn the body from c to C. He assumes that d is the acceleration vector caused by that force, since it is the difference in the two velocities.

In trigonometry, the versed sine is simply the outer section of the radius, when the radius has been cut by a line dropped from the far end of the arc. Newton never draws this line in his *Principia* diagrams, which is interesting. Newton liked to hide his math, for whatever reason. It is assumed that the reason was to keep the competition guessing, but in this case it appears to me to be a bit of obfuscation. Hiding good math may be cleverly cloak-and-dagger for some, but hiding bad math is always something less than that. What Newton is hiding may have been clearer in the 17th century, but it is very arcane now. The versed sine approaches zero very fast for very small angles, so that it may take on what is called the *sagitta* equation:

Newton proposes that, at the limit, h = the arc. And, since the versine is proportional to the centripetal force, the acceleration must be proportional to arc^{2}/2r. Furthermore, he says, the arc is equal to the velocity, so that *a* is proportional to v^{2}/2r. But, the versine is only half the force, he says [see Prop. I, Corollary IV], so that the full acceleration becomes *a = v ^{2}/r*. You can see that the

This is Newton's hidden math, such as it is. It is finessed in several ways, one of which is his use of Lemma VII. In Lemma VII, Newton states that at the limit (when the interval between two points goes to zero), the arc, the chord and the tangent are all equal. But if this is true, then both his diagonal and the versine must be zero. According to Lemma VII, everything goes to either equality or to zero at the limit, which is not helpful in calculating a solution. Neither the versine equation nor the Pythagorean theorem apply when we go to a limit by Newton's definition. MM shows below, with a very simple analysis, that the tangent must be allowed to remain greater than the chord at the limit; only then can the problem be solved without contradiction.

Before this, it is interesting to note that Newton nearly achieves the correct answer, despite some faulty lemmae. The versine will give us the correct answer, provided we analyze the correct interval. The versine becomes equal to *a* only if we are considering the arc length from A to b. Newton has been considering the arc length from A to C. We must drop the perpendicular from b instead of C, in order to achieve the correct versine. If we do this, we do indeed find that versine = *a* at the limit.

Once we have found *a* in this way, there is no need to double it though, since in finding the versine we used the angle θ and the arc length from A to b. That must therefore be our interval. You may say that the only difference in Newton's method and MM's correction is that he finds the force over the interval from A to C, whereas MM finds the force from A to b. His force is twice mine, and his arc is twice mine, therefore everything should stay the same. But it is not quite that simple.

What we find by Newton's method once we discover d, is the force required to move the body from c to C over the interval B to C. MM agrees that this force is:

d = 2a = v^{2}/r

Newton then spreads that force out over the interval from A to C, and we have our current equation. Obviously, the force to take the body from A to C is twice the force to take it from A to b. MM admits that *a* = v^{2}/2r and that d = v^{2}/r. But there remains one very big problem. Newton has gone to the limit to find d. MM has gone to the limit to find *a*.

They are both supposed to be at the ultimate ratio, but Newton has found the solution over not one but two intervals. He begins Proposition I with this: "For suppose the time to be divided into equal parts, and in the first part of that time let the body by its innate force describe the right line AB. In the second part of that time, the same would proceed directly to c, along the line Bc equal to AB." So he has postulated two time intervals. You cannot postulate two time intervals and then postulate that you are at the ultimate interval. The ultimate interval is the last interval in the series. It cannot be further subdivided, by a time variable or by anything else. Therefore d = v^{2}/r must apply to two time intervals. It is the force required to move the body twice the ultimate arc distance, by Newton's own reasoning.

Perhaps you can already see that it is much more logical simply to let Ab be the ultimate interval, so that the arc Ab is compounded of the vectors AB and Bb. Then we can solve for *a* using either a versine or the Pythagorean theorem—which is what MM does below. In either case we find that over the ultimate interval, *a = v ^{2}/2r*.

An important point is that Newton above is saying that the arc was the velocity, as derived by his method and by his equations (which still stand today). This means that the variable v in all final equations must be understood to be the orbital velocity. It is not the tangential velocity.

The tangential velocity is shown by a straight-line vector along the tangent. That means that it moves in that direction. That is what the vector stands for. The tangential velocity does not curve, and it does not follow the curve of the arc. In the diagram above, the tangential velocity over the first interval is AB and the orbital velocity is Ab. Newton gives us the tangential velocity to start with, when he gives us AB; then we seek the orbital velocity. The velocity that follows the curve of the arc is the orbital velocity, and it is the velocity variable in Newton's final equation *a = v ^{2}/r*.

Historically, physicists have not kept these two velocity variables separate, but you must learn to do so as you follow the arguments and diagrams in this paper. The two velocities have become conflated, and when we get to modern equations like v = rω, there is confusion about what v we are talking about. Contemporary textbooks tell us that the v in that equation is tangential velocity, but it isn't. It is orbital velocity.

In further analyzing this problem, MM will also prove that the arc does not describe the velocity—or any true velocity—and that we require a further equation to express *a* in terms of the tangential velocity. The tangential velocity and the orbital velocity are not the same thing—although by Newton's Lemma VII they have been taken for the same thing throughout history. The tangential velocity is the tangent and the orbital velocity is the arc. Lemma VII says that they are the same length at the limit. MM will prove that this is false. **Since an arc is a curve, it cannot describe a velocity, since by definition a velocity cannot curve. A curve describes an acceleration, as we all know.** The orbital velocity is a velocity only over the ultimate interval—where it becomes straight. But even there it is not equal to the tangential velocity, as will be proved.

It may also be worth pointing out that the basic linear equation for acceleration is v^{2} = v_{0}^{2} + 2ar. That is in chapter one of most physics books. It took me several years *after* writing this paper to remember that that reduces to

v^{2} = 2ar

v^{2}/2r = a

Amazing, really, that no one thought to connect those two equations.

Newton provided a mathematical proof that was both slender and dense, but current textbooks offer a slightly more explicit derivation. What is shown here is the standard mathematical derivation of

*a* = v^{2}/r. The steps below from a recent college textbook.

This then is the accepted derivation of a = v^{2}/r:

Let v_{o} be the initial tangential velocity, as shown in the first illustration.

Since v_{o} and v are both perpendicular to r, the two angles θ must be equal; therefore the triangles shown are similar;

therefore as t→0

Δv/v = Δl/r

Δv = vΔl/r

a = lim __ Δv __

t→0 Δt

= lim __ vΔl __

t→0 rΔt

And since the speed, v, of the object is

lim __ Δl __

t→0 Δt

a = v^{2}/r

The book says, "When Δt is very small, Δl and the angles are also very small, so v will be almost parallel to v_{o}, and Δv will be essentially perpendicular to them. Thus Δv points toward the center of the circle."

The mistakes here are many. Disregarding the conceptual setup and the calculus for a moment, let me hit the most important problem first. In its derivation the textbook assumes that the variable v in the last equation is the same as the one in the first. But it's not. In the first line of the derivation, the variable v stands for the tangential velocity. Four lines later we are told "the speed of the object, v, is

lim __ Δl __ ."

t→0 Δt**But that is the orbital velocity!** The variable v has switched. You can see that v in their first diagram is not limΔl/Δt, since limΔl/Δt is the curved velocity from A to B: v is only a component of that velocity; **v is a straight line!** But the book substitutes one for the other.
That is,

__vΔl __

rΔ

magically becomes

v^{2}/r

But, if

v≠
lim __Δl __

t→0 Δt

Then the substitution must fail. It does fail, and the derivation falls with it.

A closer analysis of the situation shows that v is the tangential velocity, Δl/Δt is the orbital velocity, and they will never be equal—not over any interval, including an infinitesimal interval. The book needs subscripts to differentiate the two, like v_{t} and v_{orb} (for v orbital).

v_{orb} = Δl/Δt but vt ≠ Δl/Δt

So the equation a = v^{2}/r should read a = v_{t}v_{orb}/r, if the book is following its own method very closely.

v_{t} v_{orb}/r ≠ v^{2}/r.

It is finally unclear whether v in the current equation applies to orbital or tangential velocity, since the derivation makes both assumptions.

For those who are already confused, let me state that in a slightly different manner. This modern derivation is a piece of prestidigitation, or sleight of hand. Like a magician who has crossed over, MM will uncover the magic for you. Return to the illustration and notice that they have labelled the two tangential velocities as v_{0} and v. Why? The two vectors are both tangential velocities, they are just in different positions. But it is the length or numerical values of the vectors we are interested in, not their positions. The numerical values are the same, so the vectors should both be labelled the same. In *value*, v_{0} = v, so labelling them differently is just a trick. It is this trick that allows the magicians here to push you from v_{0} to v, and to complete this dishonest proof. Look again at the equation

Δv/v = Δl/r

Ask yourself, shouldn't that be

Δv/v_{0} = Δl/r

That is where the switch was made. That is where the hand is quicker than the eye. As you see, v_{0} has shifted to v, so that when the arc is *also* defined as v, the two will look the same on the page. Then the magicians can substitute one for the other, and achieve the desired result.

Shocking, really, to find such ham handed cheating in foundational math and physics.

But there are even more problems. Notice that the magicians allow themselves to make substitutions in a limit equation such as the equation

a = lim __ Δv__

t→0 Δt

They substitute vΔl/r into that. But you can't do that, because those variables are captured by the limit sign. That equation reads, "The limit as t goes to zero of change in v, etc." That is not the same as simply "change in v." The substitution is disallowed.

After the substitution, you see, you have Δl going to the limit, whereas before you had Δv. To make the substitution, you have to assume that the two delta variables go to the limit in the same way, but you cannot assume that. The specific reason you cannot assume it is here is because the two deltas are not equivalent. Δl, like Δt, is a simple interval. But Δv is a change in velocity, which is not a simple interval. A change in velocity is already an acceleration, by definition, which means it is not the same sort of variable that Δl is. In the calculus, you must differentiate lengths and velocities and accelerations, usually by primed variables, but here we have none of that. An acceleration looks just like a length here, with no difference in notation.

And the problems continue.

The entire (b) part of the illustration is false.

Δv ≠ v - v_{0}, because this math is concerned with values. The numerical value of v is the same as the numerical value of v_{0}, so Δv could only be zero here. A body in orbit does not change the numerical value of its velocity. It has a constant velocity. The difference between v and v_{0} is only an angle here. So solving in this way can only be called perverse. Even Newton didn't try to subtract one tangent from another. Look back above at his derivation. He analyzes lengths and velocities and acceleration in the *same* interval, not in subsequent intervals. The vector v should be no part of this analysis, and using it to manufacture a proof here is just flamboyantly bad math.

Some will say that this is bad of example of the proof from a poor textbook. But this is not so as the critique of a proof of the equation by Richard Feynman will show. Feynman is famous for explaining difficult problems lucidly and concisely, we are told. In the proof from *Six Not-so-Easy Pieces*, Feynman varies the problem a bit, for good measure. This will get us out of our rut, perhaps. He postulates any velocity along a curve, not necessarily a circle.

As you can see from his diagrams, the tangential velocity varies and this gives us two components to the acceleration. [In the first diagram, you can ignore the little r vectors and the origin o: they do not really come into this problem. And the vector for Δv is drawn in the wrong place on purpose by Feynman, but he "corrects" it in the second diagram.]

He says, "The acceleration tangent to the path is of course just the change in length of the vector." This is correct, but it is important to note that, in uniform circular motion, this component of the acceleration will be zero, since the length of the vector is not changing. He then calculates the other component, the acceleration at right angles to the curve. You can see the break down in his second illustration. This illustration clears up some of the points left vague in the textbook illustrations. You can see, for one thing, that the right angle made by Δv_{┴} is where it intersects v_{2}, not v_{1}. He then says that Δv_{┴} = vΔθ, where "v is the magnitude of the velocity." He does not specify which velocity, but it is clear that he means v_{1} , since that is the hypotenuse. That is the only way the equation even looks like working, at a glance.

But we already have major problems.

Δv_{┴} = vθ is a false equation. Hopefully, you can see that it should be

Δv_{┴} = vs inθ

There is more. Let us say that the tangential acceleration happens to be zero over this interval of the curve. We are back to circular motion in that case. We keep Feynman's second diagram but lose

Δv_{║}. But you can see that makes v_{2} shorter than v_{1}. Feynman's triangle must be a right triangle for his last equation to work. But the lengths don't add up. v_{2} - Δv_{║} ≠ v_{1} . This means that to calculate a perpendicular acceleration, he must have a negative parallel acceleration. Logically impossible.

Feynman gets into even more trouble. In the next step he lets a_{┴} = v_{┴}/Δt. That should be a _{┴} = vs inθ/Δt. But to get Δθ/Δt, he says, if "at any given moment the curve is approximated as a circle of radius r, then in a time Δt the distance s is of course vΔt where v is the speed. "This would give us

Δθ = vΔt/r

Δθ/Δt = v/r

a_{┴} = v^{2}/r

Thus as you can see, he made exactly the same "mistake" in that step as the textbook.

You must go back to his first illustration to find s. Once you find it you see that s curves: s is a distance along the arc.He says that

s = vΔt

But this is simply not true.He has already defined v as the magnitude of v_{1}. Therefore vΔt describes a straight line length along that first vector v_{1}—not a length along the curve. He has confused tangential velocity and orbital velocity. His variables have shifted in the same way the textbook's did. Nor does he tell us at the end what v stands for in the final equation. A reader has no idea, since he has defined it both ways.

Furthermore, he has hidden a step previous to Δθ = vΔt/r above. He needs an equation for the arc length s, which would be

Δθ/2π = s/2πr

Δθ = s/r

Notice that this is not the same as sinθ = s/r, therefore the substitution fails. Remember that we had Δv_{┴} = vsinθ from above. You can see why Feynman suppressed that step. He didn't want it to suffer any scrutiny. **Besides, the equation Δθ = s/r only works if θ is measured in radians.** But Feynman cannot measure the angle in radians and hope to achieve a solution this way. He wants a velocity measured in m/s, not in radians/s.

Also notice the strange notation for the angle. An angle is usually represented simply by a variable, not by a delta variable. That is, θ not Δθ. An angle is already understood to be a change in direction: the delta is superfluous. But in these derivations both Feynman and the textbook use the delta notation. Why? Probably, in order that the derivation would seem to work. That delta sign makes the variable seem more than it really is. It is a sort of mystification. You look at that the first time and say to yourself, "What does that mean? Δθ? Does it mean something more than θ. You do not know, but maybe Feynman knows. This proof must work, since we already know that a = v^{2}/r, so it is best just to pretend to understand what is going on here." The problem, you now see, is that no one knows what is going on here.The proof is a whitewash.

And finally, notice that Feynman has simply done a copy job on the textbook's ridiculous illustration. Specifically, he has drawn v_{2} sticking off the far end of the arc. He glibly warns us that vector additions work "only when the tails of the vectors are in the same place" and so he moves the tail of v_{2} to that of v_{1}. What he forgets to warn himself is that vector additions work only when the vectors are in the same interval. Importing a vector wildly-mildly from a subsequent interval just because you need it to finesse an equation is no better than drawing open-tailed vectors or three-headed vectors.

Feynman's proof fails. It fails in at least three places in a half-page derivation. How could this be? Is it possible that he and everyone else cannot follow short steps of high school geometry or whether there is a conspiracy to hide these errors.It is beyond belief that the greatest minds of the 20th century cannot see these errors. This sequence of steps in Feynman's book is so obviously finessed that it seems to imply that it was done purposefully.

You may think the arc length equation was left out because everyone knows it by heart, but it appears to be a way to hide the unequal substitution of sinθ for Δθ

If Feynman and others like him cannot see these mistakes, it is a very bad sign. If they can, it is a very bad sign, since it means that we are the victims of some sort of Jesuitical casuistry—being lied to for our own benefit. We must believe that science is not another house of cards, lest we run screaming into the night. Therefore we must take these absurd derivations on faith. *Credo qua absurdum*.

It is possible that this derivation is used simply because it was used by Newton, and we have never been able to improve on it. It yields the correct experimental numbers, so who cares that it is full of holes?

MM shows two fudged proofs of a=v^{2}
/r on his site: We Watch Maxwell
Finesse an Equation and an "engineering" proof of this equation at youtube.

We have seen three different failed derivations, from the likes of Newton and Feynman, no less (although one should probably not put Newton and Feynman in the same sentence). MM will not critique Huygens derivation here, since MM considers it equivalent to Newton's. He, like Newton, correctly showed the proportionality of the acceleration, the radius and the "velocity." But he did not incontestably show the equality. MM will do so now, by a very transparent method.

[MM has drawn v flying off like that only to mirror the textbook illustration, to show how absurd it really is. In MM's solution, that vector is superfluous.]

In MM's drawing, a right triangle is formed by the radius, the tangential velocity, and Δv added to another radius. No matter how short or long you make the tangential velocity vector the right triangle pertains. Also notice that in MM's illustration Δv is always pointing at the center of the circle. It does not point at the center only when t or Δv or arc d is zero or if v_{o} is very long or infinitesimal. Now all we need is the Pythagorean theorem.

v_{o}^{2} + r^{2} = (Δv + r)^{2} becomes Δv^{2} = v_{o}^{2} - 2Δvr

To solve, we can then treat that last equation as a quadratic equation, and use the standard method for finding the square root (see your math textbooks).

Δv = a = ((±√4r^{2} + 4v_{o}^{2}) - 2r)/2 . If we assume a positive motion around the circle, that reduces to

a = √ v_{o}^{2} + r^{2}) - r

First of all, notice that Δv = a. That vector is the centripetal acceleration. That vector is the number we are seeking. Having initially followed the book and Feynman in putting off a full description of what Δv applies to, but it is clear that Δv is not a velocity vector. It is an acceleration vector. Of course the form alone should tell us the difference. A delta v vector is not the same as a v vector. A delta v vector is an acceleration vector, clearly. Feynman and the textbook imply that it is the difference between one tangential velocity and the next: a difference between velocities is an acceleration. But MM has shown that the acceleration vector Δv can be calculated from a single tangential velocity, given the radius. It is the difference between the tangential velocity and the orbital velocity, measured over the same interval. In this way MM's analysis mirrors that of Newton, who said the same thing. See above where Newton defines the length d as *the difference between the tangential velocity and the orbital velocity, measured over the same interval.* My illustration also mirrors his, as you can see. Just flip his illustration over, and his d is MM's Δv. The only difference is that MM points his vector at the center of the circle.

Some will say, "That won't work. You need to differentiate. You need to find your values at a dt. As it is, you will get a different value for a depending on whether you solve at Δt = 1, Δt = 5, or Δt = dt. A change in the length of your v_{o} vector will change the length of your a vector."

No, it won't: v_{o} is a constant in the case you are offering. If you are just varying times to make v_{o} change in length, you are talking about a particular given circle. You are not talking about *any* circle. Therefore, if you increase the Δt from one to five, for example, you are also increasing the distance along the vector: therefore the velocity stays the same. A shorter velocity vector in that case is not a different velocity; it is the same velocity measured over a shorter time. If you make the move from Δt = 5 down towards dt, and the triangle gets smaller, the value for v_{o} does not get smaller. Velocity equals x/t, remember. The length of the vector expresses only the x, but the t is always implied.

We don't need to differentiate, because differentiation would yield a change in that vector. It would require us to consider the vector Δv a velocity, and we would be calculating a change in that velocity, a ΔΔv, during an infinitesimal interval dt. Not only is that unnecessary, it is absurd. If the vector were a velocity, it would not change over any interval, not a large interval or a tiny interval dt. Therefore *a* ≠ dv/dt nor dΔv/dt. Those equations would only yield *a* = 0. Δv is already a differential—it is the difference between two velocities—therefore it would be redundant to differentiate it. The Pythagorean theorem works at any t, even dt. But there is no limit here, since the value for *a* is the same whether you calculate it at any real interval (a large triangle) or near zero (a tiny triangle).

Think of it this way: the equation *a* = dv/dt describes a ratio of change between v and t. If v does not change as t changes, then v is a constant. The derivative of a constant is zero. Therefore it makes no sense to differentiate a constant velocity, even if it happens to be labeled Δv.

You may say, "OK, but is all that legal? Can you combine different vectors in a vector addition? Isn't there some rule about mixing acceleration vectors and velocity vectors?" Yes, there are rules. The length of the vector stands only for its numerical value: that's why you must keep careful track of angles. But no one has ever had any problem with the way that distance vectors and velocity vectors were combined in this problem, historically. The radius of the circle is obviously not a velocity; it is a distance.

But both the textbook and Feynman use the radius and the velocity vectors as values that can be put in the same equation. If you can do that, why not use acceleration vectors as well? The answer is, you can, and Newton, the textbook and Feynman all do that, too. They just don't call attention to it. They solve this problem without ever defining their variables. The trick is, apparently, to fail to define anything: then everyone will accept it without question. But Feynman's vector Δv must also be an acceleration vector, just like mine. Why do you think it is labeled with a delta? A delta v is an acceleration. The vector makes no sense as a velocity, not in his diagram or mine. If Feynman had defined it as a velocity vector, then notice that that vector does not change in length all the way around the circle—if the motion is circular. If there is no change in that velocity vector, then a_{┴} must be zero. Neither the orbital velocity nor the tangential velocity (nor the vector Δv) change in magnitude over any interval, so calculating any change in any v, or an a that was a change in v, would only give us the number 0. The number we have always achieved in the equation *a* = v^{2}/r for *a* can only signify the acceleration vector that MM has just found.

Feynman says that *a*_{┴} = Δv_{┴}/Δt. But Δv_{┴} is always the same in uniform circular motion, by definition. It is a constant in his diagram and mine. Therefore in his equation, *a* = 0. And we see yet another way that he finessed this proof.For that equation to work, v_{┴} would have to be a velocity vector. Otherwise the form of the equation makes no sense. In the vector diagram he labels the tangential velocity v_{1}.Then he labels the perpendicular velocity Δv_{┴}. One is a variable and one is a delta variable. For what reason? They are equivalent types of vectors according to this equation. If that is so, then Δv_{┴} should be labeled simply v_{┴}. He does it to confuse the issue. He needs a number for Δv_{┴} to put into this equation: *a*_{┴} = Δv_{┴}/Δt. And he does gets a number. That seems to imply that the equation will yield a non-zero number for a_{┴}. But by his notation, what we really need to make the equation a real equation is this a_{┴} = ΔΔv_{┴}/Δt. We need a change in his velocity variable—which he labeled Δv_{┴} for no reason. A change in his perpendicular velocity would then read ΔΔv_{┴}. But ΔΔv_{┴} = 0.

If Feynman admitted that Δv_{┴} is an acceleration vector to start with, then MM could answer that his equation does not work that way either. *a*_{┴} = Δv_{┴}/Δt is false, since you would then have a_{┴} = a_{┴}/Δt. The rest of his substitutions also get skewed if he defines Δv_{┴} as an acceleration vector. But it must be one or the other. It is either an acceleration vector or a velocity vector, but neither works in his proof.

You may say that MM's Δv is a constant, too. Yes, it is a constant **acceleration**. But it is not zero, since M never differentiates it.

As a final proof that MM's analysis of a = Δ v is correct, go back to the beginning of Feynman's proof. Remember that he said, "The acceleration tangent to the path is of course just the change in length of the vector." Aha! No differentiating or putting the change in length of the vector over Δt there. If you translate his quote into a mathematical equation, it reads, *a*_{║} = Δv. That is all. If the acceleration tangent to the path is figured in that way, why would the acceleration perpendicular be figured in some convoluted way? The answer: it is not. It is figured in exactly the same way.

Besides, with the tangent acceleration in his example, you can differentiate if you want: it doesn't matter as long as you use the correct velocity variable. If you differentiate v_{1} in his diagram (not Δv_{║}), you get a_{║}.You also get Δv_{║}, since *a*_{║} = Δv_{║}.In other words,

a_{║} =dv_{1}/dt = Δv_{║} ≠ dΔv_{║}/dt

likewise *a*_{┴}≠ dΔv_{┴}/dt

You cannot differentiate Δv_{┴} in order to calculate a_{┴}, since Δv_{┴} does not change over time. And you cannot use Feynman's other tricks, since they are all dirty tricks.

Someone may notice that MM's equation gives the wrong notation for an acceleration. Yes, that is true. In using the Pythagorean theorem on the lengths of the vectors, MM has lost their full notation. MM only found the length of the acceleration vector, which is to say its number value. However, this is not crucial. MM will also show that the notation in the current equation is incorrect.

As a final complaint, someone may notice that the vector Δv does not curve: how can it be an acceleration? A vector diagram is a conceptual simplification. The vector's length stands for the Δx and the direction stands for the direction, but nothing can show the change in time. It is understood that the same change in time underlies all the vectors. All the vectors in the diagram exist during the same time interval. But the t-variable is completely ignored. If you put an acceleration vector into a diagram, the same thing holds. The t-variable is ignored. But if you have an acceleration and the t-variable is ignored, then the vector does not curve. It looks just like a velocity vector. An acceleration curves on an x, t graph because you are plotting x against t.In these illustrations we are not plotting against t, we are ignoring t. Therefore, it is possible to have an acceleration vector in an illustration that does not curve. It is not possible to have a curve that is not an acceleration, but it is possible to have an acceleration that is not a curve.

Besides, we have accepted for centuries that the centripetal acceleration points to the center of the circle at every instant. Every time it is drawn in textbooks it is drawn as a straight line vector. If history has drawn it as a straight line vector, then MM should not be taken to task for it.

The next thing to notice is that MM's new equation yields very similar proportions to the current equation between a, r, and v_{o}. If you think that MM's equation looks completely different from the current equation, it can be verifies by putting some numbers into it. It will yield different values in almost all situations, but those values change in almost precisely the same way as the current equation. Meaning that as r and v_{o} change, the value for a increases or decreases at the same rate as the current equation. This seems logical.

Now let's get back to MM's proof. His last equation is the relationship between acceleration and tangential velocity.What if we want orbital velocity?

As t→0, d→b and the triangle formed by v_{o}, Δv, and b approaches becoming a right triangle, with the right angle at point B.You can see in MM's illustration that the angle at B is obtuse. But as the arc d gets shorter, the angle diminishes, reaching a limit at 90^{o}. In that case,

b^{2} + Δv^{2} = v_{o}^{2}

As t→0, b becomes the orbital velocity vector v_{orb}, which is what we seek.

v_{orb}^{2} + Δv^{2} = v_{o}^{2}

From above, v_{o}^{2} + r^{2} = (Δv + r)^{2}

So, by substitution, v_{orb}^{2} + Δv^{2} + r^{2} = Δv^{2} + 2ΔVR + r^{2}

2ΔVR = v_{orb}^{2}

Δv = v_{orb}^{2}/2r**a = v _{orb}^{2}/2r**

Δv = √ v_{o}^{2} + r^{2}) - r= v_{orb}^{2}/2r

v_{orb} = √[2r√ v_{o}^{2} + r^{2}) - 2r^{2}]

As should be clear, MM has just solved the problem using Newton's own idea of an ultimate ratio. He has applied the Pythagorean theorem over the last interval of the series—an interval which is not zero. MM treats it as a real interval, not as a mystical infinitesimal interval, nor as an "evanescent" (Newton's word) interval. It is a normal interval* and there is nothing to keep one from using the Pythagorean theorem over that interval. (See "A Redefinition of the Derivative and the Integral")

Note that at the limit, the angle at B (between b and Δv) is 90^{o}, but v_{o} ≠ v_{orb}. For v_{o} to equal b, the angle at B would have to go past 90^{o}. It would have to be slightly acute. But that implies a negative time interval. B therefore cannot go past 90^{o}. 90^{o} is the limit. And when the angle is at 90^{o}, v_{o} > v_{orb}.

You can now see that this solution stands as a refutation of Newton's Lemma VII. Newton states that at the limit, the arc, the tangent and the chord are all equal. MM has just shown that at the limit the arc and the chord approach equality, but the tangent remains greater than both. Newton applied his limit to the wrong angle. He applied it to the angle θ in MM's illustration above, taking that angle to zero. MM has shown that the limit must apply first to the angle at B. That angle hits the limit at 90^{o} before θ hits zero. Therefore, θ never goes to zero, and the tangent never equals the arc or the chord. This is why the acceleration never goes to zero (and neither does the versine, for those keeping score). If θ went to zero we could not calculate an acceleration.

Newton's states that BC is compounded of (is the vector addition of) Bc and cc, in the first illustration above. If this is true then the tangential velocity and the orbital velocity cannot be equivalent. That would make BC and Bc equivalent at the limit. That cannot be, since that would completely nullify cc at the limit. But cc is the acceleration vector. You cannot nullify that vector at the limit and then claim to derive it. The arc and the tangent cannot be equal. The orbital velocity and the tangential velocity are never equal.

If Lemma VII is false, Lemmae VI and VIII must also be false, since they both concern taking the angle θ to zero. MM has that θ is not zero at the limit.

When working in the ultimate interval: we cannot take quantities all the way to zero, since then our variables start to disappear. We do not take B all the way to A or let θ equal zero. We are in the last interval in the series; we are not at zero. Even the last interval must have dimensions, no matter how small they are. Some time must pass; some distance must be crossed. We seek the dimensions at the end of that first interval, not at the beginning of the first interval. The beginning of the first interval is zero. The end of the first interval is not. At the beginning of the first interval, sinθ = 0. At the end of the first interval, sinθ = Δv/v_{o} ≠ 0. This is now generally understood, in some form or another. What is not understood, apparently, is that this dooms not only Lemmae VI, VII and VIII but all current derivations of circular motion and a = v^{2}/r. The foundations of the calculus have been rebuilt since the time of Newton, but many of Newton's assumptions have remained standing within the old walls. They have never been thoroughly examined.

Circular motion is, at bottom, a rate of change problem. We have two changes happening at the same time. While the body is moving in a straight line it is also being accelerated toward a central point. But a rate of change problem implies change. Change happens only over definite intervals. If we take our variables to zero we cannot solve, since the changes have gone to zero and therefore the ratios have gone to zero. Forces also cannot act over zero time intervals. There is no such thing as an instantaneous force, by physical definition. A force must act over an interval. The definition of kinetic energy, as related to the force, makes this clear. A force must act through some time or distance interval in order to do work, which work is the transference and equality of the kinetic energy. The same thing applies to acceleration, although this is never made as clear in current definitions. A force must move a body through some time or distance interval in order to impart an acceleration. There is no force at a point or instant. Every force and every acceleration must act through some period of time. This is what Newton did not fully comprehend and what current physics and calculus cannot comprehend.

MM has solved this problem using Newton's idea of the ultimate ratio, since it mirrors the current conception of the calculus in most ways. MM has corrected Lemma VII, but this has not affected MM's ability to use the historical concept of the limit. In MM's opinion, this concept of the limit is still overly complicated. There is an even easier method to the solution of any curve, without using limits or "infinite" series. However, use of that method requires knowledge of MM's paper on the foundation of the calculus, knowledge of which MM cannot take for granted in this paper. Hopefully. MM's arguments here are clear enough in the present form, making it unnecessary to include that method in this paper.

We have seen that no matter which velocity you assign v to in the final equation—either orbital or tangential—

a ≠ v^{2}/r.

You may ask, "Well, which is it? Which of your new equations are you proposing as a replacement for a = v^{2}/r?" Notice that one factor in this decision might be supplied by Newton himself, for in the proof mentioned in the beginning for deriving Kepler's third law from his gravity equation, he uses this step:

v = 2πr/t

where t is the period of the orbit, and 2πr is the circumference of the orbit.It is clear that this v is the orbital velocity here. For his derivation to work, v in equation a = v^{2} must be the orbital velocity. And according to his math, MM's correction would not affect his derivation of Kepler's law. It only changes the constant:

t^{2}/r^{3} = 2π^{2}/GM instead of 4π^{2}/GM

However, that leads us into the final embarrassment in this whole ignominious history of blunders.

Not only is it now clear that Δv is not a velocity, v_{orb} is also not a velocity. Newton's orbital velocity is not a velocity. This should come as no shock, since a velocity cannot curve. Every calculus curve tells us this. The whole history of circular motion tells us this. A curve is an acceleration. The orbital "velocity" is a complex motion made up of the tangential velocity and the centripetal acceleration.

Feynman and the textbook should have already known this, since it is one of the conclusions of the whole problem, but for some reason they kept calling it a velocity and treating it as a velocity in the derivation of a. They differentiated it as a velocity; put it into acceleration equations as a velocity; notated it as a velocity. They acted for some reason as if the orbital velocity was known, and we were deriving the acceleration from it. They acted this way because they had a number for it from the equation above, v = 2πr/t. It was easy for them to calculate: orbital "velocities" are the easiest thing in the sky to calculate from visual data. Therefore they thought they understood it. But they didn't, as is clear from these failed derivations.

Newton didn't understand it either, for he substitutes v^{2}/r for a as if v is a velocity. In the Kepler proof he lets

ma = mv^{2}/r.

That looks very familiar, in a dangerous way, if you don't know that the v is really an acceleration. That could lead us into kinetic energy problem meltdowns that equal this one. There is no reason why we should still be labeling the orbital velocity as a velocity.

A reader may well ask how this finding can be commensurable with current engineering and applied physics. We have loads of empirical evidence that the historical equation is true. Newton was trying to derive an equation that he already knew was true from data he had on hand. Therefore the historical equation a = v^{2}/r is deriving for us a relationship between the centripetal acceleration and 2πr/t. We have need of this relationship and can make use of this relationship in physical calculations, despite the fact that the second value is not a velocity.

For the sake of convenience, we have labeled it a velocity and gone on with our business. These two equations work together, because they are correct relative to each other. That is to say, if a = v^{2}/r, then v = 2πr/t. But the fact is, neither one is true. The orbital acceleration is really a_{orb} = 2√2πr/t, and the relationship is a_{┴} = a_{orb}^{2}/2r.

What this means is that our current equations are just straightforward heuristics. We use them all simply due to the fact that it is easy for us to measure 2πr/t. That is our basic data, data that we like and always have liked, and whether 2πr/t is a velocity, an acceleration, or none of the above, has never really mattered to us.

Newton was trying to develop an equation that contained this data, and he did that. He developed a true equation that relates a and 2πr/t. Unfortunately, he labeled 2πr/t as the orbital velocity, and it is not the orbital velocity. Nor is it the orbital acceleration. It is just a relation of two variables and a constant. Because Newton derived a true equation, it has not mattered much (in most situations) that his variable assignments were sloppy. As long as we remember that the v variable in the equation a = v^{2}/r is equal to 2πr/t, we cannot go wrong. But we have not always remembered this. Bohr and quantum mechanics forgot this!

MM has shown that the circle describes not a velocity, but an orbital acceleration. This acceleration is the vector addition of the tangential velocity and the centripetal acceleration. To find it we use the equation **a _{┴} = a_{orb}^{2}/2r**. Using this equation, we find that

(2πr/t)

a + r = √(v

a

This is another very useful new equation for tangential velocity. It will allow us to calculate velocities and energies that have so far eluded us, such as the energy of a photon emitted by an electron in orbit.

The answer to the question, "which of MM's equations must replace the current one?" is therefore the first one:

a = √ v_{o}^{2} + r^{2} ) - r. If we want an equation that relates the velocity of an orbiting object to its centripetal acceleration, we must use this equation, since it is the only equation with a true velocity variable in it.

This works out in other ways, too, since the proportions of the variables in that equation remain the same as the historical equation, while they don't in the equation a = v_{orb}^{2}/2r

Which brings up another problem. You can see that physics has never had a way to measure tangential velocity. "Orbital velocity" can easily be measured by circumference per time of one revolution. But tangential velocity must be calculated. You can calculate using MM's new equations, but before this paper there was no equation from orbital to tangential velocity. Neither Feynman nor any of the textbooks were even clear on the difference. This implies that no one was clear on the difference.[Notice that Newton also could not calculate from one to the other, since according to lemma VII they were the same thing.]

Both theoretical scientists and engineers should understand that such mistakes as this one ultimately lead to ruin. In the short term they may lead to simple engineering failures, which is bad enough. But in the long term they always lead to theoretical dead-ends, since a sloppy equation is the surest of all possible ways to stop scientific progress. A correct equation is almost infinitely expandable, since its impedance is zero. Future scientists can develop it in all possible directions. But a false or imprecise equation can halt this development indefinitely, as we have ample proof. Mislabeling variables is not a semantic or metaphysical failure. Is it failure of science itself.

Not only does this come as a shock, but it is also the opening of the floodgate of errors. By putting physics under a microscope, MM will be expose the sloppy algebra and calculus that appears to be pandemic, infecting the entire field, including the highest levels. Modern scientists and mathematicians have proven themselves more interested in juggling complex matrices and other higher math than in mastering high-school algebra. Which is precisely why such an error in derivation was left standing for so long.

Much work is left to be done in basic physics, despite the hubristic claims of many that the field is nearly complete. In MM's opinion, the most important and immediate work to be done is in conceptual analysis—in combing the mass of theoretical and mathematical work already done, and making it consistent. This will be achieved not with so-called higher math, in which the original concepts get lost; nor with the esoteric theories of the scientific avant-garde, in which the production of paradoxes becomes a sign of distinction; but with simple algebra, in which the concepts are kept near the surface at all times.

Philosophical subtleties can always be dismissed as arbitrary or subjective or metaphysical, but it is time for the world to see that science fails from simple math mistakes and fudges.

Another mathematical proof proving that the current equation is wrong is in MM paper on the virial on his site. It shows that the 2 in the equation 2K = -V is an outcome of this bad equation. We are told that the potential energy in the virial is twice the kinetic energy, but that has always been illogical. By correcting the equation a = v^{2}/r, the equation 2K = -V can be corrected making it K = -V. This not only makes the virial logical, it acts to confirm MM's correction in this paper. The two corrected equations confirm one another.

*First written November 2005 Taken from lemma.html*

In this paper Miles Mathis presents a simple and straightforward disproof of one of Newton’s first and most fundamental lemmae, a lemma that remains to this day the groundwork for calculus and trigonometry. This correction is important—despite the age of the text being critiqued—due simply to the continuing importance of that text in modern mathematics and science. This correction clarifies the foundation of the calculus, a foundation that is, to this day, of great interest to pure mathematicians. In the past half-century prominent mathematicians like Abraham Robinson have continued to work on the foundation of the calculus (see Non-standard Analysis). Even at this late a date in history, important mathematical and analytical corrections must remain of interest, and a finding such as is contained in this paper is crucial to our understanding of the mathematics we have inherited. Nor has this correction ever been addressed in the historical modification of the calculus, by Cauchy or anyone else. Redefining the calculus based on limit considerations does nothing to affect the geometric or trigonometric analysis shown here.

The first lemma in question here is Lemma VI, from Book I, section I (“Of the Motion of Bodies”). In that lemma, Newton’s provides the diagram below, where AB is the chord, AD is the tangent and ACB is the arc. He tells us that if we let B approach A, the angle BAD must ultimately vanish. In modern language, he is telling us that the angle goes to zero at the limit.

This is false for this reason: If we let B approach A, we must monitor the angle ABD, not the angle BAD. As B approaches A, the angle ABD approaches becoming a right angle. When B actually reaches A, the angle ABD will be a right angle. Therefore, the angle ABD can never be acute. Only if we imagined that B passed A could we imagine that the angle ABD would be acute. And even then the angle would not really be acute, since we would be in a sort of negative time interval. Newton is using A as his zero-point, so that we cannot truly cross that point without arriving in some sort of negative interval, especially since we are talking about the motion of real bodies.

(To visualize this, you must slide the entire line RBD toward A, keeping it straight always. This was the visualization of Newton, and it is not changed it here. His physical postulates are not changed, but his his geometry is being analyzed with greater rigor than even he achieved.)

If we are taking B to A and may not go past A, then the angle ABD has a limit at 90^{o}. When ABD is at 90^{o}, the angle BAD may not be zero. This will be crystal clear in a moment when we look at the length of the tangent at the limit, but for now it is enough to say that if angle BAD were zero, then ADB would also have to be 90^{o}, which is impossible to propose. A triangle may not have two angles of 90^{o}.

In Lemma VII, Newton’s uses the previous lemma to show that at the limit the tangent, the arc and the chord are all equal. To was just disproved this by showing that the angle ABD is 90^{o} at the limit. If ABD is 90^{o} at the limit, then the tangent must be greater than the chord. Please notice that if AB and AD are equal, then ABD must be less than 90^{o}. But also as can be shown **ABD cannot be less than 90 ^{o}**. B would have to pass A, which would put us in a negative time interval. If B cannot pass A (A being the limit) then the tangent can never equal the chord, not when approaching the limit and not when at the limit.

This verifies MM's previous assertion that the angle BAD cannot go to zero. If the tangent is longer than the chord at the limit, then this is just one more reason that the angle BAD must be greater than zero, even at the limit. If AD is greater than AB, then DB must be greater than zero. If DB is greater than zero, then the angle BAD is greater than zero.

All this is caused by the fact that the angle ABD goes to 90^{o} *before* the angle BAD goes to zero. The angle ABD reaches the limit first, which keeps the angle BAD from reaching it. BAD never reaches zero.

Of course this means that B never reaches A. If B actually reached A, then we would no longer have a triangle. The tangent and the chord are equal only when they both equal zero, and they both equal zero when the interval between A and B is zero. But the 90^{o} angle at ABD prevents this from happening. When that angle is at 90^{o}, the tangent must be greater than the chord. Therefore the chord cannot be zero. If the chord is zero, then the tangent and the chord are equal: therefore the chord is not zero. To put it into a more proof-like form:

1) If the chord AB is zero, then the tangent AD is also zero. 2) zero = zero 3) If AB = AD, then the angle ABD must be less than 90 ^{o}.4) The angle ABD cannot be less than 90 ^{o}.QED: AB does not equal AD; AB does not equal 0. |

In fact, this is precisely the reason that we can do calculations in Newton’s “ultimate interval”, or at the limit. If all the variables were either at zero or at equality, then we could not hope to calculate anything. Newton, very soon after proving these lemma, used a versine equation at the ultimate interval, and he could not have done this if his variables had gone to zero or equality. Likewise, the calculus, no matter how derived or used, could not work at the limit if all the variables or functions were at zero or equality at the limit.

Some will say that the claim that B never reaches A is like the paradoxes of Zeno on the Miles Mathis site. Does this imply that Achilles never reaches the finish line? No, of course not. The diagram above is not equivalent to a simple diagram of motion. B is not moving toward A in the same way that Achilles approaches a finish line, and this has nothing to do with the curvature. It has to do with the implied time variable. If we diagram Achilles approaching a finish line, the time interval does not shrink as he nears the line. The time interval is constant. Plot Achilles’ motion on an x/t graph and you will see what MM means. All the little boxes on the t-axis are the same width. Or go out on the track field with Achilles and time him as he approaches the finish line. Your clock continues to go forward and tick at the same rate whether you see him 100 yards from the line or 1 inch from line.

But given the diagram above and the postulate “let B go to A”, it is understood that what we are doing is shrinking both the time interval and the arc distance. We are analyzing a shrinking interval, not calculating motion in space. “Let B go to A” does not mean “analyze the motion of point B as it travels along a curve to point A.” It means, “let the arc length diminish.” As the arc length diminishes, the variable t is also understood to diminish. Therefore, what MM is saying when he says that B cannot reach A is that Δt cannot equal zero. You cannot logically analyze the interval all the way to zero, since you are analyzing motion and motion is defined by a non-zero interval.

The circle and the curve are both studies of motion. In this particular analysis, we are studying sub-intervals of motion. That subinterval, whether it is applied to space or time, cannot go to zero. Real space is non-zero space, and real time is non-zero time. We cannot study motion, velocity, force, action, or any other variable that is defined by x and t except by studying non-zero intervals. The ultimate interval is a non-zero interval, the infinitesimal is not zero, and the limit is not at zero.

The limit for any calculable variable is always greater than zero. By calculable MM means a true variable. For instance, the angle ABD is not a true variable in the problem above. It is a given. We don’t calculate it, since it is axiomatically 90^{o}. It will be 90^{o} in all similar problems, with any circles we could be given seeking a velocity at the tangent. The vector AD, however, will vary with different sized circles, since the curvature of different circles is different. In this way, only the angle ABD can be understood to go all the way to a zero-like limit. The other variables do not. Since they yield different solutions for different similar problems (bigger or smaller circles) they cannot be assumed to be at a zero-like limit. If they had gone all the way to some limit, they could not vary. A function at a limit should be like a constant, since the limit should prevent any further variance. Therefore, if a variable or function continues to vary under a variety of similar circumstances, you can be sure that it is not at its own limit or at zero. It is only dependent on a variable that is.

If AB and AD have real values at the limit, then we should be able to calculate those values. If we can do this we will have put a number on the “infinitesimal.” In fact, we do this all the time. Every time we find a number for a derivative, we put a real value on the infinitesimal. When we find an “instantaneous” velocity at any point on the circle, we have given a value to the infinitesimal. Remember that the tangent at any point on the circle stands for the velocity at that point. According to the diagram above, and all diagrams like it, the tangent stands for the velocity. That line is understood to be a vector whose length is the numerical value of the tangential velocity. It is commonly drawn with some recognizable length to make the illustration readable, but if it is an instantaneous velocity, the real length of the vector must be very small. *Very small but not zero*, since we actually find a non-zero solution for the derivative. The derivative expresses the tangent, so if the derivative is non-zero, the tangent must also be non-zero.

Some have said that since we can find sizeable numbers for the tangential velocity, that vector cannot be very small. If we find that the velocity at that point is 5 m/s, for example, then shouldn’t the velocity vector have a length of 5? No, since by the way the diagram is drawn and defined, we are letting a length stand for a velocity. We are letting x stand for v. The t variable is not part of the diagram. It is implicit. It is ignored. If we are letting B approach A, then we are letting t get smaller. A velocity of 5 only means that the distance is 5 times larger than the time. If the time is tiny, the distance must be also.

There is another way to analyze Newton's problem, and it may be the most interesting of all (for some). In the *Principia*, Newton's actual language in describing this problem (Lemma VI) is this: "if the points A and B approach one another. . ." Two things bear closer attention here. One, A cannot approach B without messing up the geometry. If we start moving the point A, we destroy our right triangle. What he means is: Let B approach A. To be rigorous, we should let one point remain stationary and let the other point move. If we let both move, we create unnecessary problems.

The other thing to notice is the word "approach". Newton is postulating motion. As confirmation of this, we need only look at his title for this section: "Of Natural Philosophy". Natural philosophy is not pure math, it is physics. Newton is describing a philosophy or study of nature, which we now call physics. Nature is not pure, it is physical. Therefore this lemma must be a part of what we now call applied mathematics. If this is so, then time must be involved.

As asserted above, Newton is studying a diminishing interval in order to analyze curved motion. He uses this analysis immediately afterwards to apply to an orbit, for instance. So both motion and time are involved in Newton's analysis. For this reason alone, his angle BAD cannot vanish. That would be taking the problem to a zero time interval, and there is no such thing as a zero time interval in physics.

You cannot study motion and then postulate a zero time interval, since motion is defined by a non-zero time interval. If you have a zero time interval, you have no motion, by definition. Simply by using the word "approach", Newton has ruled out a zero time interval. His interval can get smaller and smaller, to any extent he likes, but it cannot vanish. By definition, "approach" and "vanish" are mutually exclusive.

But it gets even more interesting. Using the limit concept alone, this problem cannot be solved at all. Meaning, if we let our angle at R equal θ, then BAD = θ/2 and ABD = π/2 + θ/2.

If we let θ go to zero, then BAD and ABD approach the limit in the same way. The limit concept does not support MM's analysis. No, it supports Newton's analysis, since historically it grew out of his analysis. The limit concept fails to explain why we find non-zero solutions at the limit for both the chord and the tangent, and it fails because its analysis is faulty. The limit analysis treats the entire problem as an abstract or pure-math problem, whereas it is a physical problem.

Motion and time are both involved here. What that means is that we must have a necessary time separation between A and B. Since we have motion, we cannot have a zero interval. If we do not have a zero interval, then we must have a time separation. Stated that way, we arrive at. . . yes, Relativity.

If this is a physical problem, then A and B cannot exist the same time, operationally. An event at B cannot be fully equal to that same event as seen from A. If we think of the measurement of an angle as a physical event instead of an abstract geometric quantity, then angles in a diagram like this must be analyzed from a physical point of view.

Some that is overcomplicating the problem, or inventing esoteric solutions, but consider this fact: Newton's gravitational studies and proportionalities came out of this same book, the *Principia*, indeed this same section. Is it not strange that Einstein's Relativity corrections have been applied to gravity but not to the orbit? The diagram above is a preliminary study of the orbit, and underlies a=v^{2}/r, and yet it has never benefited from a Relativity analysis until now. We think that gravity causes the orbit, and yet we do a Relativity analysis of gravity but not of the orbit. Very strange.

The way that Relativity solves this problem once and for all is that it gives us a way of separating θ/2 at B and θ/2 at A. According to the limit analysis, both angles should diminish in the same way. But because they are spatially separated, they cannot act the same. According to Relativity, we must pick a point and measure everything from there. We must study the problem from A or B, but we cannot study the problem from both places simultaneously.

Since we have given the motion to point B, we must let that be our point of measurement. In other words, in this problem, we exist at B. The event is at B. Let that event be π/2 + θ/2 going to the limit. θ goes to zero, so ABD goes to 90^{o}. Of course BAD is also going to zero, but there is a time lag. As seen or measured from B, information from A must be late, and vice versa. Therefore, as measured from B, the limit at B must be reached before the limit at A. Or, since MM has shown that limits are never really reached anyway, especially when those limits are at zero, it would be more rigorous to say that θ/2 is smaller at B, as measured from B, than θ/2 at A. Given time separation, equal angles are not quite equal.

Some may find this analysis fascinating and others may find it to be gibberish. Still, the simpler explanation is preferred: a zero time interval cannot be proposed, therefore the angles cannot vanish, therefore the lines cannot be equal. No matter how small we go, in order to talk of motion we must have a real time interval. As long as we have a real time interval, we have a triangle. As long as we have a triangle, we have a tangent that is longer than the chord. We "approach" the limit, we do not "reach" the limit. The Relativity analysis is also correct.

Either analysis gets the right answer, using ideas that are physically correct and physically real. To be consistent, if we apply time separations to the gravitational field, we must also apply them to the orbit. Gravity cannot physically cause the orbit, Relativity applies to gravity but not to the orbit. Since Newton's whole section in question here is physical, we must either apply Relativity to all of it, or to none of it. Einstein updated Newton's analysis of gravity, and MM has now done the same for the orbit.

The finding in this paper of Miles Mathis affects many things, both in pure mathematics and applied
mathematics. It has been proved in a very direct fashion, that when applying the calculus to a
curve, the variables or functions do not go to zero or to equality at the limit. This must
have consequences both for General Relativity, which is tensor calculus applied to very
small areas of curved space, and quantum electrodynamics, which applies the calculus in
many ways, including quantum orbits and quantum coupling. QED has met with
problems precisely when it tries to take the variables down to zero, requiring
renormalization. My analysis implies that the variables do not physically go to zero, so
that the assumption of infinite regression is no more than a conceptual error. The mathematical limit for calculable variables—whether in quantum physics or classical
physics—is never zero. Only one in a set of variables goes to zero or to a zero-like limit
(such as the angle 90^{o}). The other variables are non-zero at the limit.

For QED, this
means that when the Planck limit is reached, length and time limits are also reached.
Neither time nor length variables may go to zero when used in momentum or energy
equations of QED. In fact, beyond the logic used here, it is a contradiction to
assume that values for energy would *not* have an infinite and continuous regression
toward zero, but that values for length and time would.

This is not to say that length and time must be quantized; it is only to say that in
situations where energy is found empirically to be quantized, the other variables should
also be expected to hit a limit above zero. Quantized equations must yield quantized
variables. Space and time may well be continuous, but our *findings*–our measurements
or calculations—cannot be. Meaning, we can imagine shrinking ourselves down and
using tiny measuring rods to mark off sub-areas of quanta. But we cannot *calculate* subareas
of quanta when one of our main variables—Energy—hits a limit above these subareas,
and when all our data hits this same limit. The only way we could access these
subareas with the variables we have is if we found a smaller quantum.

There has also been confusion on this point in the tensor calculus. In section 8
of Einstein’s paper on General Relativity, he gives volume to a set of coordinates that
pick out a point or an event. He calls the volume of this point the “natural” volume,
although he does not tell us what is “natural” about a point having volume. General
Relativity starts [section 4] by postulating a point and time in space given by the
coordinates dX_{1}, dX_{2}, dX_{3}, dX_{4}. This set of coordinates picks out an event, but it is still
understood to be a point at an instant. This is clear since directly afterwards another set
of functions is given of the form dx_{1}, dx_{2}, dx_{3}, dx_{4}. These, we are told, are the “definite
differentials” between “two infinitely proximate point-events.” The volume of these
differentials is given in equation 18 as

dτ = ∫dx1dx2dx3dx4

But we are also given the “natural” volume dτ_{0}, which is the "volume dX_{1}, dX_{2}, dX_{3}, dX_{4}". This natural volume gives us the equation 18a:

dτ_{0} = √-gdτ;

Then Einstein says, “If √-g were to vanish at a point of the four-dimensional continuum, it would mean that at this point an infinitely small ‘natural’ volume would correspond to a finite volume in the co-ordinates. Let us assume this is never the case. Then g cannot change sign. . . . It always has a finite value.”

According to MM's disproof above, all of this must be a misuse of the calculus, a misuse
that is in no way made useful by importing tensors into the problem. In no kind of
calculus can a set of functions that pick out an point-event be given a volume—natural,
unnatural, or otherwise. If dX_{1}, dX_{2}, dX_{3}, dX_{4} is a point-event in space, then it can have
no volume, and equation 18a and everything that surrounds it is a ghost.

In the final analysis this is simply due to the definition of “event”. An event must be
defined by some motion. If there is no motion, there is no event. All motion requires an
interval. Even a non-event like a quantum sitting perfectly still implies motion in the
four-vector field, since time will be passing. The non-event will have a time interval.
Every possible event and non-event, in motion and at rest, requires an interval. Being at
rest requires a time interval and motion requires both time and distance intervals.
Therefore the event is completely determined by intervals. **Not coordinates, intervals.**
The point and instant are not events. They are only event boundaries, boundaries that are
impossible to draw with absolute precision. The instant and point are the beginning and
end of an interval, but they are abstractions and estimates, not physical entities or precise
spatial coordinates.

This may appear to be an apology for Einstein, saving him from MM's critique.
After all, he gives a theoretical interval to the point. The function dX is in
the form of a differential itself, which would give it a possible extension. He may call it a
point, but he dresses it as a differential. True, but he does not allow it to *act* like a
differential. He disallows it from corresponding to (part of) a finite
volume, since this would ruin his math. He does not allow √-g to vanish, which keeps the
“natural” volume from invading curved space.

Newer versions of this same Riemann space have not solved this confusion, which is
one of the main reasons why General Relativity still resists being incorporated into QED.
Contemporary physics still believes in the point-event, the point as a physical entity (see
the singularity) and the reality of the instant. All of these false notions go back to a
misunderstanding of the calculus. Cauchy’s "more rigorous" foundation of the calculus,
using the limit, the function, and the derivative, should have cleared up this confusion,
but it only buried it. The problem was assumed solved since it was put more thoroughly
out of sight. But it was not solved. The calculus is routinely misused in fundamental
ways to this day, even (to say *especially*) in the highest fields and by the biggest
names.

*Abstract: MM will show how the arc and tangent are equal only in one specific place on the circle, and that place is not at zero or the limit. It is at 1/8th of the circle. This being so, we must rework all of Newton's orbital math. Once this is done, we can easily calculate a time for the centripetal acceleration. Yes, the acceleration is not instantaneous, and it can be proved, by showing you the number for the time.*

In the previous analysis of Newton's orbital math and the many problems with the historical proofs, it was concluded, among other things, that the orbital velocity cannot be equal to the tangential velocity. However, in the thesis "π (pi) is 4 not 3.14....", it was shown that the tangent is actually equal to the arc. In fact, they are equal not at the limit but at a finite real length. How can these two findings be commensurate?

When we take the orbital velocity as v = 2πr/t, then the tangential velocity cannot equal the orbital velocity. The tangential velocity cannot be expressed that way. Not only is it in the wrong form, it is the wrong number. The orbital velocity cannot be expressed that way either. So, NEITHER velocity equals 2v = 2πr/t. That expression is simply a heuristic ratio that we like, since it is easy to measure from visual data.

However, once we find new correct expressions for both the orbital velocity and the tangential velocity, they MAY equal each other under certain very specific conditions. The tangent DOES equal the arc, provided the tangent is the same length as the radius. In the thesis "π (pi) is 4 not 3.14...." it is shown that when the tangent equals the radius in length, the tangent equals the arc.

This does not confirm Newton and the standard analysis because the tangent and arc are not equal at the limit, they are equal *only* when the tangent equals the radius, in which case the radius, tangent and arc are all equal. Even in this situation, none equals the chord, so Newton's lemma was wrong. At the limit, the arc approaches the chord, but at the limit, the arc does not equal the tangent. So at the limit, the orbital velocity and tangential velocity are not equal. Newton's proof fails.

It seems strange that the tangent equal the radius, when the tangent is a velocity and the radius is a length, but this is actually true when the numbers are equal. If the velocity is 30, the radius must be 30. Stated differently, if a circle has a radius of 5m, the velocity would be 5m/s. They CAN be related kinematically, and in the case of an orbit, they are related by the the equation. We cannot have an arbitrary velocity at a given radius, we must have a specific velocity. A velocity that is too great will cause escape, and a velocity that is too small will cause a crash. The velocity is necessarily related to the radius, and if we get our equations and dimensions right, the velocity should equal the radius. All we have to do is get the time right.

Using the new equations, the velocity in orbit is 8r/t. Therefore, to make the velocity equal the radius, we just let the time period equal 1/8 of the orbit. In the diagrams π (pi) is 4 not 3.14...." the radius is 1/8 of the circumference; therefore, in 1/8 of the orbital period, the velocity must equal the radius, numerically.

Here is an example using the Moon. The orbital speed of the Moon is 1.022km/s, and the radius is 384,400km. This appears not to work because the orbital speed of the Moon is wrong, but that is because it is developed from a faulty equation. They need to be corrected. If the radius is 384,400km, then the distance travelled by the Moon in one orbit must be 8 times that, or about 3 million km. The orbital period is 2,360,534 seconds. That is a velocity of about 1.3km/s.

Still the velocity does not equal the radius, numerically or otherwise. The number value of the velocity is 1.3, and the number value of the radius is 384,400. However, the number equality between the radius and the distance travelled by the velocity in for only 1/8th of an orbit and thus the distance travelled at 1.3km/s in 1/8 of an orbit is 384,400km.

An onboard satellite speedometer, calibrated to work for straight-line distances, will not work in orbit, and engineers know this. That is precisely what caused the Explorer Anomalies on MM's site. The thrusts were set for velocities as calibrated here on Earth, in straight-line motions, then transformed into orbital motions by the common equations. Since the common equations were wrong, the orbits established from the thrusts were wrong. The satellites flew too high and were thought to be lost.

Not only is the velocity wrong, the acceleration of gravity is wrong. Using the current equations and the value of π, the acceleration of gravity is calculated from an equation that reduces to a=39.5r/t^{2}. But the correct equation is 32r/t^{2}. Therefore, the acceleration at the distance of the Moon is not .002725. It is .002208.

Some will look at those numbers and say, "That means that using the given numbers for the Moon, in these equations *g* would not be 9.8, but 7.947!" Well, if you assume that *g* balances the Moon's velocity by itself, yes. But that isn't the way it works. The Moon is moving in a three-body problem, at its simplest, not a two-body problem. If you calculate that number 7.947, you have left the Sun out of it. To see what MM means, you need to visit his paper "Charge replaces Lagrange points yielding 3-body stability"where MM runs the full unified field math on the Moon, showing how the distance and the velocity are caused by Earth *and* Sun.

As you see, the current equations are hiding many secrets. As another example of a secret being hidden, we can continue to study the arc and tangent. Let us return to the diagram from "π (pi) is 4 not 3.14....":

There it was proved that AD + DC = arcAC. This means that the tangent equals the arc. This works only when the tangent is equal to the radius, as stated above. The angle at O must be 45^{o}, so that DC = DB. If the angle is not 45, then the tangent cannot equal the arc, because AD + DC is not equal to AB.

This is important because if we assign the tangent to the tangential velocity and the arc to the orbital velocity, as Newton did, we find they are equal not at the limit, but only when the tangent equals the radius. In fact, in the "Formal Disproof of Newton’s Fundamental Lemmae, the tangent and the arc are NOT equal at the limit. At the limit, the tangent remains longer than the arc. And this means that the tangential velocity and orbital velocity are equal only when the length of the tangent is equal to the radius, or when the time is equal to 1/8th of the orbital period. An orbital velocity found by any other method will get the wrong answer. **This is why 2πr/t is wrong: 2πr/8 is not equal to r.**

With all this under our belts, we are now in a position to see that we may assign the acceleration a to the line segment BC. Furthermore, if a = v^{2}/2r, and AB = r, then r = v, and a = r/2. Consulting the diagram above again, that is also 2BC = CO. What this means is that we have a new way to find a centripetal acceleration, currently called gravity. The equation a = r/2 gives us a distance of acceleration over 1/8th of the orbit, so the total distance of acceleration over the entire orbit is 4r. Or, we can find the acceleration over any subinterval.

Say we want to find the acceleration of the Moon over 1 second. We are given that the period for the Moon is 2,360,534s. Since the Moon is orbiting at 384,400,000m, the total distance of acceleration over the orbit is 4 times that, or 1.538 x 10^{9}. Dividing gives us a = 1.538 x 10^{9}/2,360,534^{2} = .000276m/s^{2} over 1 second.

We can even show that the centripetal acceleration is not found at an instant. If the Moon's acceleration over 1 second is .000276, then an acceleration of .002208 cannot happen over an instant. We can even find the time, with simple math. With the same math, we find that a = .002208 when t = 8s. That is not an instant or an infinitesimal, since it is a calculable number. You cannot have a real acceleration over an instant: we should have known simply from logic that the centripetal acceleration we have always had could not be an acceleration over an instant or infinitesimal. Having just calculated the real time of orbit during the "instantaneous" acceleration, it was shown that Newton did not go to a limit or approach zero. In "A Redefinition of the Derivative and the Integral", it was shown that the calculus does not work by going to zero or a limit, it works by going to a subinterval, as was done here in a specific problem.

You might say, "But if we can find an acceleration at that small time, we should be able to find an acceleration closer to zero, over an even smaller time." Yes, we can, but that acceleration would not be the centripetal acceleration. In seeking a centripetal acceleration, we are not seeking an acceleration at or near zero time or length. We are seeking the derivative of the orbital velocity, and the derivative of any motion is found by going to a subinterval, as shown in "A Redefinition of the Derivative and the Integral". We have just gone to that subinterval mathematically, since MM can do calculus without calculus. What is found is the subinterval underneath the orbital motion where that motion becomes uncurved, which is defined as the derivative. And that gives us the acceleration we were seeking. Acceleration is not defined as the change in velocity at or near zero, it is defined as the change in the velocity, period. In the paper above you do not have to go toward zero to find any derivative, and, as shown here, the subchange is always happening over a real interval.

Now you may ask, "But that number, 8s, where is that coming from? You just calculated it, but it is not clear." Normally, the derivative is found at a subinterval of 1, since that is how it is (or should be) defined. The derivative is not found by going to zero or to a limit, it is found by going to a subchange or subinterval where we have a constant differential of 1. So normally, we would be looking for a derivative at 1 second. But in the case of the circle, this logic changes slightly. As you can see from the diagram above, the "calculus", or MM's calculations, over 1/8th of the circle. MM's solution is therefore over 1/8th of the circle. Therefore, to find a solution for the entire orbit,one has to multiply everything by 8. Even the time of the subchange, or what we call the derivative, must be multiplied by 8. That is why the time here is 8 seconds rather than 1 second.

You can see that it took a complete reworking of Newton's postulates to crack open the orbital math. Once thus is done, everything begins to make sense. We have always been taught that the centripetal acceleration is the acceleration at an instant, but that is illogical. Acceleration, like velocity, is motion, and you cannot have motion at a instant, by the definition of motion. Acceleration is also defined as a change in velocity, but you cannot have a change at an instant. Change requires an interval of change. Motion can only take place over a differential. This being so, we should have been able to find that differential. If the acceleration is not taking place at an instant, it must be taking place over some real time, and we should have been able to find that real time. Problem is, Newton couldn't solve this one, and no one else since then could either. They were looking in the wrong place. They were looking near zero, and the answer was hiding at 1/8th of the circle. The answer is found only when the tangent equals the radius. Because physicists could not solve this, they decided to hide it. Once again, they hid it in the instant. They buried it in the zero and covered it over with centuries of slippery math and slippery explanations. As you now see, the solution is simple.

*Background*

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.

Previously in this paper MM showed 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

v_{t} = 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 = v^{2}/r. There we find the velocity assigned to the arc. (*Principia*, Section II, Prop. IV, Theorem IV, Cor. 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". (*Principia,* Section II, Prop. I, Theorem I.) 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 v_{t} = 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.

To start with, look again at the basic equations

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) mv^{2} = (1/2) m(rω)^{2} = (1/2) (mr^{2})ω^{2 }= (1/2) Iω^{2}

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 v_{t} = 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)mv^{2} is linear. But if we allow the substitution, it is because we think v = 2πr/t. The v in K = (1/2)mv^{2} 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
= (mr^{2})(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. In the paper Bohr's Three Mistakes shows how 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,

ω^{2} → v^{2} -Δv^{2}

and, v^{2} + r^{2} = (Δv + r)^{2}

So, by substitution, ω^{2} +Δv^{2} + r^{2} =Δv^{2} + 2Δvr + r^{2}

Δv = √ v^{2} + r^{2}) - r = ω^{2}/2r

v = √[(ω^{4}/4r^{2}) + ω^{2}]

**ω = √[2r√v ^{2} + r^{2}) - 2r^{2}] **
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

r = √[ω^{4}/(4v^{2} - 4ω^{2})]

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 MM's 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 MM has 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 MM is right. According to the current equation v = rω, a greater orbital radius should give us a greater velocity. According to MM's 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

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^{-7}rad/s

ω of Venus 3.2 x 10^{-7}rad/s

ω of Earth 2 x 10^{-7}rad/s

ω of Mars 1 x 10^{-7}rad/s

ω of Jupiter 1.6 x 10^{-8}rad/s

But with MM's 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 = √[(ω^{4}/4r^{2}) + ω^{2}]). 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 MM's 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 MM shows in the paper Bohr's Three Mistakes. That is why MM links this problem to the Bohr and Schrodinger equations.

**Addendum: August 2010. *Many have not understood MM's variables here, even after all this. They have replied that MM's new equation v = ω/r cannot be correct simply due to units. They say that MM's 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/s^{2}. 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. From the paper A Revaluation of Time and Velocity and the Concept of Relativity, 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? No! Once we do the full analysis, it is MM's 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 MM's 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 v_{t}, 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 v_{o}, not v_{t}. The working equation is either L = mrv_{t} or L = mω = mv_{o}, but L ≠ mrv_{o}. The last equation is the current one, so it is false.

You will say that if MM's ω is not measured in radians/s,it should not be labeled ω; but since measuring angular velocity as radians/s is a fudge, there will be no use for that in the future. MM is free to capture that variable and use it at will.

You may also see MM's paper on the Finding the Electron Radius for an interesting use of these new equations.