Submissions | Add Your Comments | Physics Site Links | Home Page |
| Email: Miles Mathis |
In an earlier paper I refuted the current derivation of a = v2/r. In that paper I also thoroughly critiqued a proof of the equation by Richard Feynman in Six Not-so-easy Pieces. To further support my findings there, I will now refute Newton’s own derivation of the equation from The Principia.
In doing so I will also falsify two of Newton’s famous lemmae. These lemmae have stood as part of the groundwork of the calculus for 300 years. In this way my falsification will act as further support for my paper on the foundation of the calculus (also recently published here). I have claimed that Newton and the other founders of the calculus misunderstood the problem. The falsification of these lemmae is just another example of that.
Newton "proved" the equation in a very early part of The Principia (Section 2, Proposition 4). I say "proved" because the equation is 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. His 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." I read 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. I 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 (see my previous paper).
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:
versine = h2/2r (where h = rθ.)
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 arc2/2r. Furthermore, he says, the arc is equal to the velocity, so that a is proportional to v2/2r. But, the versine is only half the force at the limit, he says [see Prop. I, Corollary IV], so that the full acceleration becomes d = v2/r. You can see that the sagitta equation is the key to understanding Newton’s derivation. Newton gives away none of this in the Principia, but it is the only way to understand his comments on the versed sine.
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. I show that the tangent must be allowed to remain greater than the chord at the limit; only then can the problem be solved without contradiction [for a further critique of Lemma VII, see below].
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 my correction is that he finds the force over the interval from A to C, whereas I find 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. I agree that this force is, d = 2a = v2/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. If I admit that a = v2/2r then I must admit that d = v2/r. I do admit it. But there remains one very big problem. Newton has gone to the limit to find d. I have gone to the limit to find a. We are both supposed to be at the ultimate ratio. I have just shown, however, that he 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 = v2/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 I do below. In either case we find that over the ultimate interval, a = v2/2r.
In the previous paper, I showed that the triangle ABn must be taken to the limit to solve for vo and Δv. You can see that this triangle is half of Newton’s total vector problem, and so will give me the same value for a that I got above using the versine. In other words, I solved the problem using Newton’s own idea of an ultimate ratio. I applied the Pythagorean theorem over the last interval of the series—an interval which is not zero. I treated 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.
I stress here again, as I did before, that at the limit the angle at B (between b and ∆v) is 90o, but vo ≠ d. For vo to equal d or b, the angle at B would have to go past 90o. It would have to be slightly acute. But that implies a negative time interval. B therefore cannot go past 90o. 90o is the limit. And when the angle is at 90o, vo > d.
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. I have 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 my illustration above, taking that angle to zero. I have shown that the limit must apply first to the angle at B. That angle hits the limit at 90o 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.
For further clarification, go back to the first illustration above. Newton’s states that BC is compounded of (is the vector addition of) Bc and cC. 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, Lemma VI must also be false, since it concerns taking the angle θ to zero. I have shown that θ is not zero at the limit.
It helps me to think of it this way, when I am 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 = ∆v/vo ≠ 0. This is now generally understood, in some form or another. What is not understood, apparently, is that this dooms not only Lemmae VI and VII but all current derivations of circular motion and a = v2/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.
*For further clarification, see my paper A Redefinition of the Derivative—why the calculus works and why it doesn’t (2003).
I have solved this problem using Newton’s idea of the ultimate ratio, since it mirrors the current conception of the calculus in most ways. I have corrected Lemma VII, but this has not affected my ability to use the historical concept of the limit. In my 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 my paper on the foundation of the calculus, knowledge of which I could not take for granted in this paper. I believe that my arguments here are clear enough in the present form, making it unnecessary to include that method in this paper.