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The Mills Radius

[Transcript of the explanation given by Dr. Mills starting at 1.53]
Dr. Mills:  So, how does the orbit change with an excited state? 

Well it absorbs a photon of a discrete energy - the difference between the state it was in and the state it arrives at - transitions to.

So, the excited state is an electron, a photon, and a proton. Current theory [quantum mechanics] just treats it purely mathematically. It doesn't say where the photon is, or what the electron looks like, or what the photon looks like. But if you do it classically and actually solve what the electron is... it's a resonator cavity... a dynamic resonator cavity. It has discrete frequencies of excitation. 

An excited state is an electron, a photon, and proton - and you can solve all those. The photon is like a packet of electromagnetic field. You can solve exactly what it is - and all its characteristics, and the double slit, and everything about it. It has E [electric] and B [magnetic] fields according to classical electrodynamics. And you have angular momentum. It's calculated by the R cross E cross B conjugate. It turns out to be h-bar for the photon. And that h-bar is conserved when the electron goes from one state to another state. And that photon carries an electric field. And that electric field superimposes the proton's electric field, so that the electron in the n=1 state sees a field of an integer of e+ [ the positive fundamental electric charge of a proton ]. 

But when it absorbs a photon, the resultant field is reciprocal integer times e+. That is, the photon's electric field adds to the proton's electric field and the resultant electric field at the electron is reciprocal integer. That comes out of Maxwell's equations. So the field can be a half, a third, a fourth, a fifth, all the way up to zero in the ionized state, where the photon identically cancels the proton's field, and the electron moves further away, and has a higher energy, and the h-bar omega of energy is equal to the energy of that transition, and the omega-change [ angular velocity change ] of the electron's angular velocity equals the omega angular velocity of the h-bar omega, and you conserve angular momentum. But the photon, because its reciprocal integer radial field produces an instability, a radiative dipole, at the electron, and that's unstable to radiation, so it will decay back, the photon will be released, and it will go back to the original n=1 state. 

But in the case of the hydrino state, the hydrinos are again, an electron, a photon, and a proton - but rather than the photon field adding to the proton field and decreasing it, it adds to the proton field and increases it by an integer.

That's why you have to take away 27.2 [electronvolts], because if you make the field go from 1 to 2, the potential energy changes by 27.2. Thereby, an acceptor has to accept that amount of energy to conserve energy. So, you can have the field increase by an integer, the electron will then be pulled radially, and achieve a minimum energy at a reciprocal integer of the original hydrogen radius, and that has an integer field interacting with an integer charge, fundamental charge electron... and that is stable to radiation.

So a hydrino state is an electron, a photon, and a proton. The way it gets that state is you transfer energy out, the field becomes stronger, the electron goes close to the proton, and it's stable. 

Conversely, with an excited state, energy comes in, the field is decreased, and the electron goes further away from the proton. And with enough cancellation of field it becomes completely ionized. 

And you can actually calculate - again this is classical, just to give you an idea how this works.

So, if you have the electron as a current, and it's going from radius one Bohr radius to, say, two Bohr radius, so it's going radially further away - it's obviously moving going from state one to state two - which current theory doesn't deal with. They say it's instantaneous; it's not.

So you have a radial current and you can calculate from a current. You have a vector potential - you take the curl of A, you get the B field... you take the curl of B, you get the E field. E cross B conjugate is the power. You divide the power into the energy, h-bar omega, and that gives you the state lifetime. 

You take the reciprocal, you got the Einstein A coefficients. As a function of fundamental constants n - the integer initial final state - you plug in the initial final state, it gives you all Einstein A coefficients and matches all the NIST Einstein A coefficients. (...) 

I'm saying you can do very, very fundamental calculations. You can't do that with current theory. It has the electron going faster than the speed of light, you get these huge hypergeometric series. It takes years - one PhD grad student will spend his entire career doing just one calculation. And they don't even make sense physically. They're totally nonsensical. Faster than lightspeed, things like that.

This is an integer formula, it gets all of them, just plug in the integers. And it's like that for everything. Molecules, atomic excited states, you can solve them in exact analytical expression, integers only, because it's a physical theory, it gives you a physical result. 

But you want to know how it does it? It transfers energy out. The photon's created inside the resonator cavity, increases the field. The electron is pulled closer to the proton. And it's stable. 

And [while] it's moving radially you get the continuum radiation. So, you have a resonant energy transfer, that creates the photon, that increases the field by an integer, then the electron radiates by moving - accelerating - closer to the proton. And that's why we get the continuum radiation. It's not a resonant excitation like it is in the excited state, it's an energy transfer and then a radial collapse of the radius. And that emits continuum radiation. 

[End of Transcript]