A Correction to a Famous Equation
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.

Of course work continues on the mechanism of gravity, since it is still completely unknown, and no data has yet confirmed any competing theory. 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. I have 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. I show that all these derivations and computations are fatally compromised by this.

The equation is a = v²/r. 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 I would guess that 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.

I was led to examine this equation due to problems that have cropped up in several fields. I 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 my 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, I have discovered many errors that can only be called astonishing, I think. This paper relates one of them.

Newton used the equation a = v²/r to tie his famous equation of universal gravitation to Kepler’s Third Law. That is,

F = Gm₁m₂ / r² becomes

T²/r³ = 4π²/Gm₂ only by assuming that

a = v²/r

The full derivation is in all college textbooks and I will not repeat it here. I only mention it to show that a = v²/r has been a bedrock equation from the beginning. It was already dogma for Newton himself. What I will repeat here is the current mathematical derivation of a = v²/r. I have taken the steps below from a current college textbook, but they mirror Newton’s own steps, only making the conceptual mistakes more transparent.

The illustration on top I have taken from the textbook. The one below is my own illustration, correcting the mistakes in the current derivation.

This then is the accepted derivation of a = v²/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,

∆v/v = ∆L/r

∆v = v∆L/r

a = lim ∆v/ t→0 ∆t

= lim v∆L/ t→0 r∆t

And since the speed, v, of the object is lim ∆L/ t→0 ∆t

a = v²/r

The book says, "When ∆t is very small, ∆L and the angles are also very small, so v will be almost parallel to v_o, and ∆v will be essentially perpendicular to them. Thus ∆v 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 ∆L / t→0 ∆t."

But that is the orbital velocity. The variable v has switched! You can see that v in their first diagram is not lim∆L/∆t, since lim∆L/∆t is the curved velocity from A to B: v is only a component of that velocity. But the book substitutes one for the other. That is,

v∆L/r magically becomes v²/r

But, if v ≠ lim ∆L / t→0 ∆t

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, ∆L/∆t 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 = ∆L/∆t but v_t ≠ ∆L/∆t

So the equation a = v²/r should read a = v_t v_orb/r, if the book is following its own method very closely. v_t v_orb/r ≠ v²/r.

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

Very sloppy math, I must say. What we have above is a mess of poorly defined and shifting variables with some fake calculus to dress it up and obscure the mistakes. And I have only shown the most obvious error.

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 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. I will not take him to task for it. He shows how to do it right in the second diagram.]

He says, "The acceleration tangent to the path is of course just the change in length of the vector." I agree. But I point out 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 v₂, not v₁. 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 v₁, 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_┴ = vΔθ is a false equation. I hope you can see that it should be

Δv_┴ = vsinθ

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 v₂ shorter than v₁. Feynman’s triangle must be a right triangle for his last equation to work. But the lengths don’t add up. v₂ - Δv║ ≠ v₁ . 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_┴ = vΔθ/Δt. That should be a_┴ = vsinθ/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_┴ = v²/r

I hope you can see that 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 v₁. Therefore, vΔt describes a straight line length along that first vector v₁—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

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. 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? In order that the derivation seem to work. To trick the eye. Trompe l’oeil. 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?. I don’t know, but maybe Feynman knows. This proof must work, since I already know that a = v²/ r, so it is best just to pretend I 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.

Feynman’s proof fails. It fails in at least three places in a half-page derivation.

I am unsure what to think of this, frankly. I truly do not know whether 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 20^th century cannot see these errors. This sequence of steps in Feynman’s book is so obviously finessed that it can only mean that it was done purposefully, I think. You may think the arc length equation was left out because everyone knows it by heart. I think it was left out to help 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 quia adsurdum. I suspect 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?

In my 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 my 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. It points at the center if v_o is very long or if it is infinitesimal. Now all we need is the Pythagorean theorem.

v_o² + r² = (∆v + r) ²

∆v = √ v_o² + r²) - r

a = √ v_o² + r² ) - r

First of all, notice that ∆v = 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 a 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.

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 ∆v a velocity, and we would be calculating a change in that velocity, a ∆∆v, 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 d∆v/dt. Those equations would only yield a = 0. ∆v 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 ∆v.

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 a 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 ∆v) 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 = v²/ 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 v₁. 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 a 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 v₁ in his diagram (not Δv║), you get a║. You also get Δv║, since a║ = Δv║. In other words,

a║ = dv₁/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, ∆v, 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,

b² + ∆v² = v_o²

As t→0, b becomes the orbital velocity vector v_orb, which is what we seek.

v_orb² + ∆v² = v_o²

From above, v_o² + r² = (∆v + r) ²

So, by substitution, v_orb² + ∆v² + r² = ∆v² + 2∆vr + r²

2∆vr = v_orb²

∆v = v_orb²/2r

a = v_orb²/2r

∆v = √ v_o² + r²) - r = v_orb²/2r

v_orb = √[2r√ v_o² + r²) - 2r²]

I stress again that at the limit, the angle at B (between b and ∆v) 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.

Amazing how things turn out. A small correction, hidden all these years by bad math. No matter which velocity you assign v to in the final equation—either orbital or tangential a_┴ ≠ v²/r.

You may ask, "Well, which is it? Which of your new equations are you proposing as a replacement for a = v²/r ?" Notice that one factor in this decision might be supplied by Newton himself, for in the proof I mentioned in the beginning for deriving Kepler’s third law from his gravity equation, he uses this step:

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 = v²/r must be the orbital velocity. And according to his math, my correction would not affect his derivation of Kepler’s law. It only changes the constant:

t²/r³ = 2π²/GM instead of 4π²/GM

However, that leads us into the final embarrassment in this whole ignominious history of blunders. Not only is it now clear that ∆v 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 gives us an equation that devolves into this:

a_┴ = v²/r = a_orb²/r

That’s fine in one sense, since I do the same thing above. My equation is a_┴ = a_orb²/2r. The problem is not the form, the problem is that Newton and the textbooks substitute v²/r for a_┴ as if v is a velocity. Newton does it in the Kepler proof. He lets

ma = mv²/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.

The answer to the question, "which of my equations must replace the current one?" is therefore the first one: a = √ v_o² + r² ) - 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_┴ = a_orb²/2r.

A reader may well ask how this finding can be commensurable with current engineering and applied physics. First, consider this: gravity is not a force that is directly measurable. It is always calculated from other forces or motions or velocities. The number we now have for gravity was generated by this equation and other equations derived from it. In this sense the equation is not testable, at least as regards gravity. All that is testable is the relation between the variables. We know that as you increase r and v_o, a increases at a certain rate. But I have pointed out that my first equation give the same rates, although not the same numbers. Therefore my correction implies no conflict with experiment, it only implies a conflict with the number we have generated for gravity in given fields.

For example, when we put a satellite into orbit we are putting to the test not our number for g of the earth. We are putting to the test the relation of the variables in our equations. If these relations hold, in the same way they have done in the past, then we will achieve a safe orbit. Look at it this way: the first time we did an experiment to find the value of g, we created a stable orbit, by trial and error, and then calculated g from our current equation. The stability of the orbit (or the stability of the first test, whatever it may have been historically) was the raw data. This told us that some forces were in balance. Our theory at that time, which is still basically the same as theory now, gave us an equation to calculate g, given a balance of forces. I am saying we used the wrong equation the first time, and have been using the wrong equation ever since. But because the rates of change among the variables were almost exactly the same, we never saw the mistake. In each new orbit we create—each new experiment or project—the relationship between the variables yields stability. If we double the radius of the orbit, then g decreases some amount, and that requires a decrease in the velocity. You can see that g might be the wrong number at radius r and at radius 2r and at every other radius. It wouldn’t matter if g was changing at a rate that matched experiment. That is what I am saying has happened.

There are other ways to test the equation a = v²/r, it is true. It does not require a gravitational test. But my guess is that since the gravitational test was so successful, no one thought to test the equation closely by other means. The tolerances in circus rides, for example, are much greater than the tolerances of orbits, since an orbit is a balance of forces while circus rides are made of huge steel bars. A Tilt-a-Whirl is hardly a precision instrument. But it would require just such an instrument to really test the equation. The force on a single spoke or filament connected to a little whirling object would have to be measured directly, not calculated. And the velocity would also have to be measured to a high degree of accuracy. Which brings up another problem. Unlike Newton, Feynman implies that the velocity in the equation a = v²/r applies to the tangential velocity. If that is true, then 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.

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 let to stand 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.