 # BLP update

• Let me explain further, for the electron we have the relationship modolu some constants

k^2 = (E - C/r)^2 - m^2c^4

the non radiation condition ground state us kr=constant so it is wise to study

constant = (rE-C)^2 - r^2m^2c^4

Now essenially rE = rmc^2 + 2C and if rE->nrE=r'E', then

nr'mc^2 + nC = rmc^2 + 2C or

(r-nr')mc^2=2(n-1)C/(mc^2)

r/n - 2(n-1)C/n/(mc^2) = r'

or r' = r/n aproximately

Now E'r'=E'r/n = nEr => E'=n^2E

Which you can find in GUTCP.

No charge need to be changed, no mass need to be changed.

Note that most likely there are two modulated standing EM waves in the inside (photons) in the QM approach, one related to the mass and one related to the trapped photon and we would then demand that

E_photon r ~n1

E_mass r ~n2

where n1 = 1 for non hydrinos, this as the energy of the mass is of order 1MeV and the energy of the trapped photon is of the order eV. I suspect that the QM approach will never be able to contract the shell to zero width but very thin sop that practically it can be taken such. This explains the higher accuracy of the QM's energy levels and that we can probably deduce a correction that makes the classical approach as exact as the QM's approach. A thing that has bothered me as I do not fancy infinitely thin thingies and QM people bash Mills theory for not being as exact as QM although it is super exact (and QM is super super exact). It seams that via E=mc^2 the mass decides how thin the shell is.

Edited once, last by stefan ().

• No the mass and charge i constant in this argument.. why does it loose mass and charge? energy is also in the fields.

A deeper orbit violates the force equation! The Bohr radius is a saddle point!

Note that most likely there are two modulated standing EM waves in the inside (photons) in the QM approach,

In fact you can model - as I did in SO(4) - the binding as two EM flux torus. But Mills logic of a trapped photon is weird as in reality you emit a photon to fall back in ground state.

Further the E=mc2 is the biggest blunder in physics history and total nonsense for a proton. It partly works for an electron as it is close to a photon. Only the original Poincaré equation is correct. dm =E/c2

The Dirac nonsense started with the claim that e+- annihilation leads to two photons. This is nonsense too shown by countless experiments.

• No if you like at a higher order photon it will oscillat and if you squeeze that to a disk (with a boost to c) you get something that is not a density but a more generalized mathematical object that most likely will be rejected by nature. Hence these hydrino states cannot radiate. The ground state does not go between 1,-1 and hence converges to a density when compressed to a disk at a boost to c. That explain why you will not be able to radiate hydrinos.you really need a mass field in proximity e.g. chemistry.

• You are ignoring that Dirac has a very good agreement with experimental data. It just can't be bunk, but possible a mathematical complexity that inspired a lot of crazy philosophical

ideas like entanglement. Shrödingers cat, many world interpretation and so on and so on. It is just a tool that spit out energy levels correctly. You have yourself found a structure, I find it very interesting, but the true physical meaning is hidden, mathematically correct yes but structure can be hidden in many ways, this is at least my take away from studying mathematics.

• To evaluate non perfect information is difficult. Scientists are used to have perfect information and any new theory (under development) that is not a small refinement of the old one will be ignored because scientists read those like the devil reads the bible.

• You are ignoring that Dirac has a very good agreement with experimental data.

I said that it closely works for the electron because two waves cover it for 99.8% . But 99.8% agreement is nothing than a bad joke or as I usually say: Engineering physics.

I do not say that the Dirac math is wrong. It just does not represent the physics of dense matter. As long as you stay on EM ground it works fine.

Everybody that models physics with just a Hamiltonian ignores reality. H is the cover of all possible solution. Next step is to find the proper working Lagrangian's! And here the Hydrino Fantasy ends.

In the CERN case you can ignore the Lagrangian because you never have a stable state! Read it again. Mills assumes the central charge does increase due to a locked in photon ...... .

• There is a photon ar a EM field inside, I get the same math as mills by exciting that photon. Normaly it's a standing wave with two nodes at the ends, the first excited one has one a node in the middle as well, and so on. The radial distribution of a EM standing wave is j_0(kr) = sin(kr)/(kr). My calculation shows that you get hydrino states doing this all while the charge and mass is the same, actually by forcing it to be the same you get the formula Mills produce in his book.

• Now you have to show how the wave produces charge...If the whole electron is a wave then no charge exists! In fact the e-p bond is a 3 rotation bond. So its 3 waves. This is also confirmed when you solve the n=1 state that is a two wave bond between e-p.

But for the Lagrangian you have to solved the radial rotating mass equation, that must hold for all points of the mean radius = charge radius shell. You also have to solve Maxwell equations. If you halve the radius then the magnetic energy increases dramatically. This energy is positive so you must provide it. The magnetic energy is >>> relativistic mass increase!

So Hydrinos, deep electron states simply are a waste of time.

• Now you have to show how the wave produces charge...If the whole electron is a wave then no charge exists! In fact the e-p bond is a 3 rotation bond. So its 3 waves. This is also confirmed when you solve the n=1 state that is a two wave bond between e-p.

But for the Lagrangian you have to solved the radial rotating mass equation, that must hold for all points of the mean radius = charge radius shell. You also have to solve Maxwell equations. If you halve the radius then the magnetic energy increases dramatically. This energy is positive so you must provide it. The magnetic energy is >>> relativistic mass increase!

So Hydrinos, deep electron states simply are a waste of time.

No, charge is a disccontinuity in the first derivative so essentially when you sepcify a boundary with nothing at the outside but and phi(r0)=0 at the boundary, typically

\phi'(r0) is not zero, but we have nothing to the right and then there will be a jump in values depending if we move from the iniside to r0 or from the outside to r0. Now

taking the derivative once more give you a delta meassure e.g charge. If you noe specify that mass and charge should be the same things conspire so that indeed you

get the rydberg series and also mills hydrinos.

• New

taking the derivative once more give you a delta meassure e.g charge.

Please solve once the force equation.... then you see the problem. Adding mass = flux to the proton needs energy. It does not produce energy !!! The corrections to the classic solution are all repulsive!

One more thing: Charge can not be given for a static flux solution !!!

• New

Form QED and Klein Gordon, if we assume that the radius is constant at r.

Bohr,

E_e = E_ph + E'

For the photon

Removing the mass part from w,

w_ph = sqrt((mc^2+E_ph)^2 - m^2c^4)/hbar = sqrt(2mE_ph)c/hbar

k_ph = w_ph/c = sqrt(2mE_ph)/hbar = 2 pi

==================================

for the electron see analysis

k_e = sqrt((mc^2+V-E_e)^2-m^2c^4)/c/hbar => k_e = k_ = sqrt(2mE_ph)/hbar

So V-E_e = E_ph =>E_e = V - E_ph, assume (huinted by Bohr) E_ph = V/2 and note sqrt(mV)/hbar = 0.997, missing 2pi !! So indeed E_ph must be V/2 = K.

Mills explain why 2pi i-> 1, not sure personally. My guess is that if you boost the trapped photon to the speed of light you get a factor of 2 pi when you compare energies, we know how radiative energy are with respect to frequency, but when we bottle it things will change most likely with the factor 2 pi.

• energy is also in the fields.

If a photon is emitted then you remove energy from a field not the opposite way... The field contains less energy. Free energy is always positive. To free the electron you will need more energy... The first order mass metric is 1 - alpha/ 2*phi. The QED basic factor or 2FC as I called it in SOP. This also explains why m^2c^4 is wrong for dense mass. The proper metric is missing. The classic Einstein metric fails. In fact alpha is the exact Einstein metric at R-Bohr. Do once the calcs! Then you see the missing 2phi. So after R-Bohr or R-dbr (De Broglie radisu) the metric changes!

• I did some refined work and the quantum model using a photon and reduced mass and using all term (ground state),

E = 13.59838eV

measured is,

E = 13.59844eV

• E = 13.59844eV

Lock at SOP! With the added wave I get all 10 digits (not just 4). The same with the EM resonance formula that only uses SOP metric.

First correction you need to do is the added magnetic energy! This you find in Mills!

• Lock at SOP! With the added wave I get all 10 digits (not just 4). The same with the EM resonance formula that only uses SOP metric.

First correction you need to do is the added magnetic energy! This you find in Mills!

this was the best meassured value 5 years ago and there was eesentially just the last digit that was off. Anyyhow for you that want to know where the 2pi comes from, let's consider the photon, yep we can descrbive the photon explicitly. see ....

"

In the photon paper below we argue that the Klein Gordon and the Dirac Equation is not as optimal as it can. What we do is to show that for the same frequency a free photon has 2pi the energy of a trapped photon (standing wave) for the ground frequency. This explains the missing 2\pi we need to model with QED hydrogen as a trapped photon and the electron density at the outer shell. SO this should mean that modelling the photon as a standing wave and Dirac we need to do the modification $\hbar \to 2\pi \hbar$. Then the new quantization condition are $j_0(w\pi k_{photon} r) = j_0(w\pi k_{electron} r) = i1,2,3,4,...$. This is hence a try to explain Mills mysterious 2\pi factor. Everything is heavily inspired of GUTCP.

"

photon paper

• I wonder why these people talk of exactness when it has not changed during the last 100 years....

These folks are hopeless and ridiculous. Almost everything in the standard model is 180 degree opposite to reality.

Did you ever think which force can bind a wave to a nucleus ?? OK Schrödinger still did believe he models an electron/charge where in reality it is magnetic flux but what binds the flux? What happens to the charge ? The charge bound energy?

Mills makes the same silly error with his excitation current model. Currents cannot cross!!! So all fail because they do not get the toroidal nature of any dense mass object. By the may the CT torus norm is 4*phi2. The other reason for the 2phi difference!

Edited once, last by Wyttenbach ().

• Guys, this may be relevant to the discussion, posted today (4 Dec.) on BrLP site:

"FREE ELECTRON PHOTON ABSORPTION MECHANISM"

A key part in this post is this:

"The photon in free space comprises a partial two-dimensional covering of a spherical shell by great circle electric (E) field lines and orthogonal great circle magnetic (B) field lines wherein the sphere has the size of the photon wavelength. The photon field lines move as an ensemble at light speed and appear in the sub-light speed laboratory frame as one of linear or right or left-handed circularly polarized E and B fields that vary harmonically in time and space over the period corresponding to the wavelength as the photon propagates past an observer.

The free electron comprises a two-dimensional planar membrane disc having a maximum charge density at the origin and transitioning to zero density at the disc radius given by de Broglie wavelength divided by 2Pi wherein the linear velocity of the free electron determines the be Broglie wavelength and radius. The photon carrying h bar of angular momentum in its E and B field lines impacts the supercurrent of free electron causing a reaction current that in turn causes the photon to be captured on the inner surface of the electron current as it forms a transient spherical bound state.

The resulting two-dimensional covering of a spherical shell by great circle current loops comprises the uniform current density pattern that gives rise to electron spin as well as a standing wave being the superposition of a Fourier series of time and spherical harmonic surface charge density waves.

The current pattern of the transient electronic state is formed in the same manner as the free electron current when it binds to a nucleus to form a bound atomic electron (https://brilliantlightpower.co…nd-ionization-simulation/) wherein the transiently bound photon is the source of the central field as in the case of an atomic electronic excited state (https://brilliantlightpower.com/photon-absorption-mechanism/).

The spherical radius of the bound transient decreases causing the spherical current and the linear current to increase as the transient bound state electron ionizes to a free electron with an increased velocity and decreased radius and de Broglie wavelength wherein the energy of the absorbed photon is conserved in the corresponding increased rotational and translational kinetic energies."

• "The photon in free space comprises a partial two-dimensional covering of a spherical shell by great circle electric (E) field lines and orthogonal great circle magnetic (B) field lines wherein the sphere has the size of the photon wavelength.

This is Kindergarten physics. You cannot postulate charge or a current. You have to show how charge/current is produced using strict Maxwell physics from the magnetic flux. So current/charge must evolve in a self sustaining structure. This first time works in SO(4) 6D topology.

• Korrection,

|E_photon |= |K| = 1/2 V

then

V-E = E

Nothing changes in the conclusion

• I modified the small math calculation in the link cited here, old link work as well. not so much math errors, but maybe still a bit handwavy.