The completed Bohr atom

Say we have a following formula

E - 1/(M_1c^2)^1/2

+ E - 1/(M_2c^2)^1/2 = N

Where

E = energy

M_1 and M_2 are rest masses with c^2 (celeritas squared of light speed)

N =some number related to Energy

Let us substitute

a = E - 1/(M_1c^2)^1/2

An

b = E - 1/(M_2c^2)^1/2

So we now have

a + E - 1/(M_1c^2)^1/2 = N

And

a + b = x

Using some algebra we find

(a -b)(a + b) = N ( a - b)

The left hand side becomes a difference of squares ie.

a^2 - b^2 = N(a - b)

This will now give

a^2 = E - 1/(M_1c^2)^1/2

And

b^2 = E - 1/(M_2c^2)^1/2

These solutions are important and can relationship to the golden ratio.

Now let's continue this. We can impose an extra relativistic correction to

(a^2 -b^2) = N/N_0

It is now a relativistic operator in the style of the ratio of energies it's a rough approximation without the need of the gravitational gamma function. This is highly unique because, it's said there are no relativistic operators but I have proven one exists.

Say the ratio represents a dimensionless set of numbers related to an energy solution

E(t)/E_0 = e^ λt

Now If

a^2 +b^2 = 9

as a number in an atom, it follows a cicular trajectory. If we plug in

a^2/9 +b^2/4 = 9

it travels the path in a hyperbolic trajectionary. Yet if N=O we get Bohrs "stable/special" trajectory where no radiation is given off:

N(a^2/9 - b^2/4) = 0

 

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Summary: I have been able to prove to first approximation where the nuclear charge when Z(e)=1 the mysterious reasons concerning why only certain orbits are allowed which do not give off radiation. While the wave function explains quite adequately why acceleration disappears in the stable orbit. This is the first model which predicts it without invoking any sense of waves, though the wave function certainly plays a strong theoretical consideration when concerning the laws waves in a space from debroglie theory. His extended reasons still hold. This model of mine is just a neat numerical solution for the simplest case of a Hydrogen atom.

 

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I heard Mary had a little lamb

It was delicious with mint sauce and dumplings

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Not really a lamb person, I'm a pescatarian. But I've heard they go well together

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If (a -b)(a + b) = N ( a - b),

then N=a +b, and obviously N = x.

Is that what you intended?


And then, if you make the subtitution a = E - 1/(M_1c²)^1/2,

and you claim to derive, as a "solution", that a^2 = E - 1/(M_1c^2)^1/2

Then a = a².

So that means that a = 1.

Is that what you intended?

 

N is notation for energy as some number in this case. I did say

a + b = x

But this is a bit vague because it looks like your simplifying everything to exact numbers when we really have orbitals following some set set rules which aren't defined by any number near 1.

Using some algebra we find

(a -b)(a + b) = N ( a - b)

The left hand side becomes a difference of squares ie.

a^2 - b^2 = N(a - b)

So we have two energy differences, those being


a^2 = E - 1/(M_1c^2)^1/2

b^2 = E - 1/(M_2c^2)^1/2

From geometry, we can set some rules, about whether the orbital gives a circular, hyperbolic or some trajectory which has no number assigned to it (as though it lacks a particle trajectory). I reality, it takes all those trajectories under a wave function but we have not defined this. From geometry we do find that

a^2 +b^2 = 9

as a number in an atom which follows a circular trajectory and gives off radiation.

a^2/9 +b^2/4 = 9

it travels the path in a hyperbolic trajectionary. It's non zero so it too will give off energy.

But if N=O we get Bohrs "stable/special" trajectory where no radiation is given off:

N(a^2/9 - b^2/4) = 0

It's very hypothetical.

 

Or maybe in the last case it is following mg a trajectory and Bohrs stable orbits hold, fir instance. That's why ot is a hypothetical model.

 

Ok let's do a quick post on the geometry bit!

 

The Cartesian graph of a circle is what we really want to work from I'm this hypothetical model. Any graph will link algebra with geometry by converting it into geometrical forms and operations fir easier survey. Without being able to post any picture of the graph, you may just have to accept the following algebra which describes the geometry. The equation for a circle on a graph is very simple. It is just

x^2_1 + y^2_1 = R^2

x^2_2 + y^2_2 = R^2

So the algebra for a circle is

x^2/R^2 + y^2/R^2 = 1

So yes the x= 1 is a bit vague, it's a bit more complicated in this form. But using this geometric picture, the rest follows. The big "if" comes in is when N = 0 for

a^2 - b^2 = N(a - b)

We ask what would this mean, if it has any implication to trajectories and energy given off by those trajectories, to be a case which satisfies Bohrs stable orbits. I worked on a much more interesting Bohr modelled maybe I should keep that for later as it's a bit more complicated.

 

I'm waiting

:EDIT:

Maybe that kinda number Pi will fuck you up.

 

Do we do have people who read opposing to what Origin claims. But not very patient.

It appears you posted this just after I posted a summary of the geometry part. I'm about to do a series of posts on some more interesting results which are less hypothetical but unifies Keplers amd Newton's equations with Bohrs atom model.

 

The biggest conclusions I reached from my papers during lockdown was that the ground state hydrogen atom does not radiate, not because of a wave function per se, but because the electron is in a state of free fall and that the wave is completely analogous to de Broglie's pilot wave theory - this was motivated largely from the fact Bohrs planatary model gives all the right results. Another one was an extension from the weak equivalence principle which drew on a peculiar resemblance of Einsteins work which states,” while matter tells space how to curve and curvature tells space and matter how to move,” to de Broglies principles corresponds by saying,”the wave tells the particle how to move whereas the particle tells the wave how to spread.” The extension could be a load of rubbish, but if true, it would mean the wave function is in fact microscopic gravitational waves. I gave some examples in which the absorption or emission of a gravitational wave at microscopic scales could explain quantum jumps inside the atom, either raising the electron or lowering it between energy levels. Another implication suggested showed that de Broglies ”empty wave" would be an in-phase gravitional wave. Maybe wild enough to be true or wild enough to be rubbish. Either way, my investigation led to a thorough examination of Bohr's orbit equation, linking scattering processes with the Bragg Condition and even came to 18 final propositions in the work elucidating on the nature of black holes as a special optik theory allowing me to extend the work of Satori (where photons are not indefinitely captured by black holes) and can he regarded from my own hypothesis as a special case of a prism where Snell's law and the equations of heat engines gives rise to a transition formula for internal reflection of light. I do not rule out the particle creation at the horizon but do conclude that treating black holes as a special prism plausible. There are far too many equations to cover so in next post, we shall look at only one result derived, that being the transition formula for the black hole entropy. The transition formula, while I apied it to black holes, Bohr did mention the transition for discrete quantum processes should be universal.

From my recent seven papers.... Which are still I'm the hands of my lawyer, the main equation for the transition of a black hole is given by the entropy transition equation for discrete quantum processes

S(ΔT) = k(ΔT)

= N⋅mc²/(1 - T (2)/T (1) (tan θ)²)

S is the entropy given in units which k which is the Boltzmann constant. N is the number of particles emitted by the black hole and the denominator comes from the laws of heat engines but the tan angle tells us about a geometric application of the radiation when leaving the cavity. It also came from applying Snells law and the index of refraction.

The idea behind using Snell's law was really based on the Brewster angle θ which is the angle of incidence. Snell's law will obey

tan θ = n(2)/n(1)

The transition formula was once again

S(ΔT) = k(ΔT)

= N⋅mc²/(1 - T (2)/T (1) (tan θ)²)

Whenever you see N⋅mc² it is proportional to the pressure and from relativity, an additive correction of density associated to the system. Basically

PV = N⋅mc²

Under normal convention and from variation using calculus we have also

PdV + dPV = dN⋅(m(2) - m(1))c²

A quick set of equations that feature in my paper which requires little format goes like this; we wish to derive the motion of the electron around the nucleus and see if it obeys Kepler's third law. We start off by writing down the general force equation in electromagnetic theory

k [e/R]^2 = B (e Z/R^2)

= Ze (v^2/R) = Ze (4π^2R^2/Rt)

Where Ze is the nuclear charge and k = B as the Boltzmann constant. We notice the nuclear charge cancels and by rearranging we obtain

e = 4π^2/k [R^3/t^2]

And viola! Its that simple, we retrieved Kepler's third law for the planetary motion in which it's acceleration is simply

a = k [Ze/R^2] = v^2/R

In the more complicated arguments of my papers I explain that this acceleration disappears in the ground state, again due to it being in a state of free fall. There are still wave mechanics in the theory, but I'm interested only in the case where the wave pilots the electron. That requires the simple de Broglie guiding equation.

It should be noted also that

R^3/t^2 = Gm/4π^2

Where Gm = R(s)c^2 as the gravitational parameter. Pulling the remaining constant to the LHS gives

ek/4π^2 = [R^3/t^2] => Gm/4π^2

As a conclusion by cancelling out factors we get

ek = 4π^2[R^3/t^2] => Gm

The gravitational parameter.

 

This really is only short snippet of work relatable to a theoretical Bohr model. The larger aspect of the work concentrated on black holes obeying optical laws. In fact Schrodinger believed his wave equation was something related by gravity being a quantum optical model.

 

To understand the post in question yoy need some idea of Kepler's third law, including how Newton finally derived it. While Kepler's could not and how they apply to Bohrs planetary model. To remove acceleration in shirt, the particle if following a real debroglie trajectory must in a way classical mean it us in a state of free fall.

 

OK, prove I'm dumb.

 

Oh yes, it does make me feel more dumb. But if you could make me feel more dumb, it would make me feel... dumber....

 

You feel dumb. As I should ask why?! Physics isn't easy so don't expect to understand everything. It's truly remarkable how dumb we can be made to feel... I won't mention his name, but an astrophysicist had a grad student who wrote a paper on black holes and he admitted aloud, that he only understood a fraction of it. It was thst complicated

 

But of course. This astrophysicist knew he wasn't dumb, it's just that we are all limited by our knowledge of certain models, ideas, and with it the ever evolving complexity of these theories.

 

If you want, I'll do a quick post on how Kepler's guessed his third law and how Newton derived it. Then maybe you won't feel so dumb afterall?

 

No, I want to know if Pi is a finite number.