return to homepage (and a Refutation of Newton's Lemmae VI, VII & VIII) by Miles Mathis
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.
Newton used the equation a = v 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 will show 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 I do that, 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 = v ^{2}/rNewton 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 = v^{2}/2r then I must admit that d = v^{2}/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 = 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 I do below. In either case we find that over the ultimate interval, a = v. ^{2}/2rI want to highlight one thing before I move on. I quoted Newton above as 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. 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. ^{2}/rIn further analyzing this problem, I 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. I will prove that this is false. Beyond that, I would ask you to consider the very elementary fact that an arc is a curve. A curve 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 I will prove. 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} = 2arv ^{2}/2r = aAmazing, 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 I have copied here is the standard mathematical derivation of a = v rΔt 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 v_{t} ≠ 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, I 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/rThat 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 hamhanded 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. I am talking about 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, as I said. 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 I have simply taken a bad of example of the proof from a poor textbook. To answer this I will provide a different proof from a completely different book. I will critique a proof of the equation by Richard Feynman. Feynman is famous for explaining difficult problems lucidly and concisely, we are told. I take the proof from
We have seen three different failed derivations, from the likes of Newton and Feynman, no less (although I should probably not put Newton and Feynman in the same sentence). I will not critique Huygens derivation here, since I consider 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. I will do so now, by a very transparent method. 2 If we assume a positive motion around the circle, that reduces to a = √ v _{o}^{2} + r^{2}) - rFirst of all, notice that Δv = a. That vector is the centripetal acceleration. That vector is the number we are seeking. I followed the example of the book and Feynman in putting off a full description of what Δv applies to. Until now. 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 I have 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 my 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 my Δv. The only difference is that I point my 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 I have 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 I 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 I have shown that neither works in his proof.You may say that my Δv is a constant, too. Yes, it is a constant acceleration. But it is not zero, since I never differentiate it. As a final proof that my analysis of a = Δv 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 I have shown they are all dirty tricks.Someone may notice that my equation gives the wrong notation for an acceleration. Yes, that is true. In using the Pythagorean theorem on the lengths of the vectors, I lost their full notation. I only found the length of the acceleration vector, which is to say its number value. However, I will show below that this is not crucial. I 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 I should not be taken to task for it. The next thing to notice is that my new equation yields very similar proportions to the current equation between a, r, and v _{o}. If you think that my equation looks completely different from the current equation, I encourage you to put some numbers into it. Yes, it yields 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. I encourage you to test out the equation before you dismiss it out of hand. Now let's get back to my proof. My 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 my 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}/2ra = v_{orb}^{2}/2rΔv = √ v _{o}^{2} + r^{2}) - r = v_{orb}^{2}/2rv _{orb} = √[2r√ v_{o}^{2} + r^{2}) - 2r^{2}] As some may be aware, I just solved the problem using Newton's own idea of an ultimate ratio. I have applied the Pythagorean theorem over the last interval of the series—an interval which is not zero. I treat 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 again 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. 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 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. 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 = 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. *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.
We have seen that no matter which velocity you assign v to in the final equation—either orbital or tangential— (2πr/t) ^{2}/r = a_{orb}^{2}/2ra_{orb} = 2(√2)πr/t a + r = √(v _{o}^{2} + r^{2}) a ^{2} + 2ar = v_{o}^{2} v
_{o} = a√[1 + (2r/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 my 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 my 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. I assume this means 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. Mislabelling variables is not a semantic or metaphysical failure. Is it failure of science itself. All this is very big news, I hope you will agree. What makes it even bigger is that I will show in subsequent papers that this is far from the only basic equation that is fatally flawed. In putting physics under a microscope, I have found that the sloppy algebra and calculus exposed above is the rule, not the exception: one might say it is pandemic. It infects 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 my 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. I have led my attack with this short and shocking paper because I know that only a full-frontal assault has any hope of breaching the walls of science. Philosophical subtleties can always be dismissed as arbitrary or subjective or metaphysical, but I hope it is impossible to ignore simple math. I have now found other mathematical proof that the current equation is wrong, since I have shown in my paper on the virial 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, I am able to correct the equation 2K = -V, making it K = -V. This not only makes the virial logical, it acts to confirm my correction in this paper. The two corrected equations confirm one another.For more on this problem, go to my newer paper on pi, and my newest paper on a=v ^{2}/r.If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |