Bohr's FirstBig Mistake by Miles Mathis First published October 21, 2005 In this paper I will reveal the first major mathematical error of Quantum Mechanics. As promised in other papers, I have gone back to the very beginning, to rerun all the equations. Before the Heisenberg Uncertainty Principle, before the problem of superposition, before the problem of the two-slit experiment and all the other theoretical problems, there existed the foundational math of Quantum Mechanics. This was the basis and expression of the theory, and still is. The math I will analyze here is still taught to this day as the bedrock of QED. All physics students learn this math when they are first taught the Bohr Model.       Of course the Bohr Model has long been superceded, but the math below has never been corrected. It still infects QED at the foundational level, since this very math is used to obtain all the existing maths of Quantum Theory. This math underlies and infects Schrodinger's equations and all subsequent maths and theories. Bohr's Math In determining the allowed orbits, Bohr first equated the centripetal electrostatic force on the electron to the centripetal acceleration, by this equation:F = ma = mv2/rmv2/r = ke2/r2mv2 = ke2/rBohr next uses the following equations to express the angular momentum of the electron:L = rmvL = nh/2π       Where h is Planck's constantrmv = nh/2π ke2/r = m[nh/2πmr]2r = h2n2/4π2mke2r1 = h2/4π2mke2 This is the Bohr radius, which we are told is in agreement with the observed size of hydrogen atoms. And we can go back and find the equation for v:v = [h/2πm][4π2mke2 /h2] = 2πke2/h = 2.18 x 106 m/sNext, we find the energy of the orbit:E = K + U = mv2/2 + U mv2 = kZe2/r      (where Z is the atomic number)E = kZe2/2r – kZe2/r E = -kZe2 /2r E1 = -13.6 eV The Problem The problem is that for some reason Bohr has used the rotational momentum but not the rotational kinetic energy. He assumes that the momentum of the electron will be expressed by an equation that includes the radius (L = mrv), so that we have an angular momentum. But when he finds the kinetic energy of the electron, he uses a straight translational equation—one that does not include the radius. You will say that it includes the radius after we make the substitution, but that does not count. The first equation is a translational equation that contains the variable v, not the variable ω. The substitution is made on the variable v, which is not thought to be an angular velocity in either equation. In the equation mv2 = kZe2/r, v is supposed to be the instantaneous tangential velocity. In the equation E = mv2/2 + U, v is a straight-line velocity. We can make the substitution precisely because they are both assumed to be translational variables.       Current theory always assures us that rotational motion must be expressed by rotational variables. In the chapters in physics books on rotational kinetic energy, we are given this equation for rotational kinetic energy:K = mv2/2v = rω K = m(rω)2/2 = Iω2/2      Why didn't Bohr use this equation? Because he couldn’t figure out how to make it yield the right experimental numbers.       You will say, "You can use either equation, since, as you just showed, they are equivalent: m(rω)2/2 = mv2/2."      I answer, "Are they? Are you telling me that in the equation K = mv2/2, v = 2πr/t?" You will no doubt throw up your hands in frustration, and say, "Yes, just do the substitution, you fool!" But you wilfully miss my point. I know they are transferable--I just showed that--but they are not the same. You can't measure a straight-line velocity with pi and a radius, since you aren't given pi and a radius. And you can't measure a rotational velocity with a distance, since you require pi and a radius. I have said before that mainstream physics has lost sight of Newton's original notation somewhere along the way, and we see it again here. For instance, the Lagrangian can be written F = – GMm/r2 + mv2/2r Newtons equation from Proposition LXVI in the Principia is F(r) = (-GM0m0/r2) + m0ω2r Newton explicitly wrote the velocity as an angular or orbital velocity, but somewhere along the way that has been replaced by a tangential or linear velocity, compromising the math. A compensation of errors has saved Bohr in all the equations above. He has achieved the right experimental numbers only by finessing the math to fix the conceptual mistakes. The fundamental conceptual mistake is in assuming that the v variable in v = 2πr/t is an instantaneous tangential velocity. It is not. It is an orbital velocity. This velocity describes an arc of the circle; it curves; it cannot be a tangential velocity. The tangential velocity vector is a straight-line vector with its tail at the tangent. It does not follow the curve of the arc over any interval, even an infinitesimal interval. Newton never claimed that the v variable in his equation was the tangential velocity. In my paper on the equation a = v2/r, I show exhaustively that this is so, by quoting Newton directly from The Principia, and by rerunning his versine derivation. Even at the ultimate interval, or at the limit, the tangential velocity and the orbital velocity are not equal.       The orbital velocity and the angular velocity are actually the same thing. The orbital velocity is just the angular velocity at a given radius. They both curve. The only difference is that the angular velocity is measured in radians and the orbital velocity is measured in meters. That is why the difference between them is just r. The variable v, as used in most places above, is no more a tangential velocity than the variable ω is. I suggest a new letter to denote orbital velocity, and I suppose for this paper w is good enough. v = x/t ω = 2π/tw = 2πr/t a = w2/r L = rmwThe last two equations show how Bohr was saved from his first mistake. The variable is an orbital velocity in both places, which saves his substitution. Neither velocity variable is a translational velocity nor a tangential velocity, although current textbooks still tell us they both are. His energy equations are saved in a different way. You can see that he is trying to do a straight substitution of w for v, which cannot work. But he should be seeking a rotational kinetic energy for his electron, not a translational kinetic energy. That is to say, he should be writing his energy equations in terms of w, not v. So the equations should run like this:mw2 = kZe2/r E = mw2/2 + U       In which case the equations work as before. I know that most physicists will see my corrections as caviling. They will not care about these subtleties, since the equations yield the right numbers, whether they are in the old form or my new form. They will dismiss my points as semantics or metaphysics, or some other easy epithet of contemporary science.       But my correction here is strictly mathematical. Part of applied mathematics is variable assignment. Sloppy variable assignment must be fudged over later with sloppy equation assignment, so that you end up with proofs like those above. The misassigned variables were misassigned in the same way in Bohr’s derivation of the radius, so that he didn’t have to do anything else to make the equations work. But with the kinetic energy, he had to make a further misassignment to make them work out. He had to misassign an entire equation, giving his electron translational kinetic energy instead of rotational kinetic energy.       No one seems to have been embarrassed by this since then, but it is more than a cosmetic failure. As I will show in subsequent papers, it is just this sort of error that has brought us to the theoretical impasses we now face. Bohr’s compensations were minor. In the equations above he needed only very small mathematical finesses. But very soon these finesses snowballed on QED. Before long the math was requiring huge finesses like renormalization. Most contemporary physicists seem not to be embarrassed by renormalization. But they should be. I will show that if you are rigorous in your math at all points, you don’t need any renormalizing. The trick is to keep the equations "normal" from the start.       In my paper on Newton linked above, I predicted a "kinetic energy meltdown" due to the form of F = ma = mv2/r. I said that an improper substitution was being begged by not properly differentiating between the orbital velocity and the tangential velocity. The term mv2/r is a strictly rotational term, but there is nothing but subtle conceptual theory to keep someone from applying it to a translational kinetic energy situation. Bohr has only just avoided this catastrophe, since his electron really is in circular motion. But what if an electron in orbit ejected a photon, and that photon was assumed to be ejected from the tangent? The photon would now be in linear motion, not circular motion. What is its kinetic energy? You can see the problem.      Furthermore, there is actually not even subtle conceptual theory to keep anyone from making this improper substitution, since current theory does not theoretically disallow it. Only my theory disallows it. The subtle conceptual theory is so far mine alone.       Current theory believes that the velocity variable in a = v2/r is an instantaneous tangential velocity. That would make it the same as a translational velocity. That is why current theory uses the same variables for both. And that is why the proofs of Bohr followed the form they did and why they have never been corrected. That is why all modern physics textbooks conflate tangential velocity and orbital velocity.       And that is why I repeat my prediction: this improper substitution will be found to be at the heart of some mathematical impasse in contemporary physics. If I do not root it out of QED very soon, I will be quite surprised. Update: I have now reworked Bohr's equation completely, finding a new value for the Bohr radius. Coulomb's constant is shown to be an expression of the Bohr radius. See my paper The Bohr Magneton. 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 noms de plume. If you are a Paypal user, there is no fee; so it might be worth your while to become one. 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