To find out, let us look at how the first equation is derived in textbooks. We start with the constant acceleration equation,

2ad = v_f² - v_i²

Then substitute a = F/m into that

2Fd/m = v_f² - v_i²

If we let the initial velocity equal zero, and define work as force through a distance, we get

W = E = Fd = ½ mv_f²

Work is then defined as the change in kinetic energy, in the famous work-energy theorem.

The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will have a force. A car hitting you will apply a force, whether or not it is accelerating.

But can we derive the kinetic energy equation without a force? Can we achieve a square velocity without assuming an acceleration? Not according to classical mechanics. Classical mechanics answers this question by ignoring it completely. QED and string theory do not address it either: they are busy ignoring all sorts of new questions, and would not dream of addressing old ignorance.

Some quarters try to dodge this problem as Wikipedia does when it says, “Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.” But this is absurd. The equation is developed from the acceleration, as I just showed. The work-energy theorem requires a change in velocity, which is an acceleration. You cannot get work without a force and you cannot get a force without an acceleration. But the current kinetic energy equation has no change in velocity. A particle has kinetic energy with a constant velocity. If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated.

Wikipedia actually leads its page on kinetic energy with the assertion above, wanting to misdirect from the first utterance, proving once again that it is the primary playground of contemporary propaganda. Whereas most textbooks ignore the question completely, hoping to keep it in the dark, Wikipedia takes a more proactive approach in mind control. It sees the question coming and deflects you in a positive way from asking it.

But this is question begging of the first order, and it is amazing that 15 generations of students have failed to ask it.

Let me put it another way. If v_f = v_i , then the postulate equation becomes

2ad = v_f² - v_i² = 0

2Fd/m = 0

E = Fd = 0

You cannot postulate an acceleration in order to develop an equation, and then dump the acceleration. The equations that come after the first equation depend on the first equation. You cannot have different assumptions in the postulate equation and the derived equations. You cannot have variable motion in the first equation, and then derive constant motion from it! We see once again how our textbooks are riddled with gloriously negligent math.

In another paper, I showed that we can develop the equation E = ½mv² from the equation E = mc², by reworking Einstein’s equations and making a few corrections. But this brings us back to explaining E = mc². Einstein develops this equation mainly by assuming that E/c is the momentum of the photon. Then, by the equation E/c = mv, and substituting c for v, he gets his new equation. But E/c was found by experiment, not by theory, so the theory becomes circular at this point. We keep returning to the question, why c²? Einstein gives us the number but not the mechanical explanation. Why square the speed of light? Why should the energy depend on c²? Or, to extend the question, why should the energy of any moving particle, moving with a constant velocity, depend on the square of that velocity?

This reason this is all hidden is that the standard model has no answer for it. If it could derive the kinetic energy equation with good math, it wouldn’t need to derive it with bad math. If anyone could tell you why the kinetic energy is a function of the square velocity, they wouldn’t need to pin the equation to acceleration.

Even if I didn’t have an answer for this question, I think it would be important to show the hole. These holes should not be covered up, they should be put on the front page: that is the only hope of an answer. But I do have an answer. I can tell you why the kinetic energy is dependent on the square velocity.

In my paper on photon motion, I showed that the measured wavelength and the real wavelength of the photon differ by a factor of c². This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c². Energy, like velocity, is a relative measurement. A quantum with a certain energy has that energy only relative to us, since it has its velocity only relative to us. If the wavelength has to be multiplied by c² in order to match it to our measurements, then the mass or mass equivalence will also. Hence the equation E = mc². In this way, c² is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.

The same analysis can be applied to any object. The energy of any object is determined by summing the energies of its constituent atomic and quantum particles, and all these particles also have spins. The quanta will impart this spin energy in collision, so this spin energy must be included in the total kinetic energy.

You will say, “The object as a whole would have to be spinning to impart spin energy. The quanta cannot transmit this energy, since most of the quanta will be away from the surface of collision. How can quanta deep inside an object transmit their spin energy in collision?”

And I answer, how could they not? The quanta have this energy--no matter where they exist in the object--and the macro-object made up of these quanta is nothing but the summation of these quanta and their energies and interactions. A collision is not just an interaction of surfaces, it is an interaction of objects, through a surface. If a car hits you, you don’t just feel the forces from mass on the front bumper, you feel forces from the entire mass of the car. Any energy the car has, whether it is linear energy or spin energy or energy from mass, will sum to create the total effect upon you. If quanta are spinning, this spin must affect all energies and forces, at all levels. And this must apply to nucleons and electrons as well as photons.

So the short answer is that the kinetic energy equation, like the equation E = mc², always included the spin energy; but no one recognized that. Just as with the photon, all matter has a wavelength (see de Broglie), and the wavelength is determined by spin. The spin has a radius, and this radius is the local wavelength. Any linear velocity of the spinning particle will stretch our measurement of this wavelength, in a simple mechanical manner, as I showed in the photon paper. As the linear velocity increases, the spin velocity relative to the linear velocity decreases, by a factor of 1/v. This makes the difference between the linear velocity and the spin velocity v². The term v² transforms the local wavelength into the measured wavelength. This is why we find the term in the energy equation.

The only question remaining is why we have the term ½ in the kinetic energy equation. The reason is simple: we are basically multiplying a wavelength transform by a mass, in order to calculate an energy. So we have to look at how the mass and the wavelength interact. I have shown that the wavelength is caused by stacking several spins (at least two spins), so what we have is a material particle spinning end-over-end. If we look at this spin over any extended time interval, we find that half the time the material particle is moving in the reverse direction of the linear motion. Circular motion cannot follow linear motion, of course, and if we average the circular motion over time, only half the circular motion will match the linear vector. This means that half the effective mass will be lost, hence the equation we have.