One error I noticed.
Fmax is not the force calculated between proton and electron at ground
state. 29.05N is the force at the coulombic barrier, even with
proton/electron.
Food for thought, last night I was messing with the numbers and realized:
q^2/(8pi e0 Fmax Rc) = classical electron radius
You can switch this equation around (because classical electron radius =
2Rc), and then solve for Fmax, producing 29.05 (and the same equation you
mentioned, just factored differently).
Fmax = q^2/(16pi e0 Rc^2)
Also you can solve for e0 yielding:
e0 = q^2/(16pi Fmax Rc^2)
But the really cool thing I was able to do tonight was formulate Planck's
constant from Frank's orbital radii equations... And the answer it gave me
was the same Planck's constant as in the brackets of the probability of
transition equation.
I had it typed up in latex so you can copy pasta the latex code if you want
to see what I did.
I'll also explain it in part 23 of my series. In part 22 I explain how to
calculate the orbital radii of both standard hydrogen and muonic hydrogen
with frank's equations. And I rapped the whole thing.
Anywhere here's the latex code.
Start with standard equation for orbital levels:
a_{o}=\frac{h}{2\pi m_{-e}c\alpha}
And we know that:
\alpha=\frac{2v_{t}}{c}
So:
a_{o}=\frac{h}{2\pi m_{-e}c\left(\frac{2v_{t}}{c}\right)}
The c's cancel leaving:
a_{o}=\frac{h}{2\pi m_{-e}2v_{t}}
Solve for h yielding:
h=4\pi m_{-e}v_{t}a_{0}
And it's been proven that:
\frac{F_{max}r_{c}^{2}}{v_{t}^{2}m_{-e}}=a_{0}
So we can say:
h=4\pi m_{-e}v_{t}\left(\frac{F_{max}r_{c}^{2}}{v_{t}^{2}m_{-e}}\right)
Reducing yields:
h=\frac{4\pi F_{max}r_{c}^{2}}{v_{t}}
Like I said, this is the same version of Planck's constant as formulated by
Frank in the probability of transition equations. I just came to it from a
different way (by cheating LOL).
Peace.
On Thu, Dec 2, 2010 at 3:16 AM, Craig Haynie <[email protected]>wrote:
> Frank, I find your idea interesting. I've worked through your basic
> equations and have included them simply because I spent so much time on
> them, I figured I should do something with them. :)
>
> In the palladium lattice, when the molecules are stimulated such that
> they are vibrating near the transitional frequency, I understand from
> your theory that the coulombic barrier opens up. Do you have a way to
> calculate the size of the coulombic barrier at this point?
>
> Thanks,
>
> Craig
>
> ---------------------------
>
> The theory postulates that for energy to travel from space into matter,
> an impedance match must occur. Frank calculates the speed of transition
> to be equal to 1,094,000 meters / second, which is, essentially, the
> speed of sound within the nucleus of an atom. Once he calculates this
> number, he notices a lot of little interesting things. For instance:
> this speed can be translated into a vibrational frequency in the
> nucleus, and all electron orbitals are at integer multiples
> of a wavelength calculated from the frequency and the speed.
>
> To calculate the speed of transition, (Vt)
>
> 1) Newton's Law
> F=ma
>
> Now, what we're going to do is use classical equations to solve for the
> speed of sound in the nucleus, from the vibrational frequency in the
> nucleus.
>
> 2) Coulomb's Law
> Calculate the maximum force between 2 protons. This is also the force
> between the proton and electron in a hydrogen atom at the ground state.
> Maximum force occurs at the Coulombic Barrier and can be calculated from
> Coulomb's law.
>
> Fmax = Q^2 / (4 * pi * e0) * (2Rc)^2)
> Q = charge of a proton = 1.602176487*10^-19 Coulombs
> e0 = permittivity of free space = 8.854187817*10^-12
> (http://en.wikipedia.org/wiki/Vacuum_permittivity)
> Rc = the radius of the Coulombic barrier. This is also known as the
> classical radius of a proton.
>
> Fmax = Q^2 / ( 4 * pi * e0 (2*1.409 x 10-15 )^2 ) = 29.053 Newtons
> Fmax = 29.053 Newtons
>
> 3) The equation for simple harmonic motion as applied to a simple
> vibrating nucleus.
> f = (1/(2 * pi)) * sqrt (k/m)
> f = frequency
> m = mass = average mass of nucleons
> k = spring constant = Fmax / Rn, where Rn = displacement, from Hooke's
> Law.
> Rn = 1.36 * 10^-15 = radius of a proton
>
> 4) Frequency (f) can be turned into a speed by multiplying both sides of
> the equation by the distance covered during a vibration. This is 2 *
> displacement.
> Vt = (1/(2*pi)) * sqrt (k/m) * 2Rn
> k = Fmax / Rn
> Vt = (1/(2*pi)) * sqrt ((Fmax / Rn) / m) * 2Rn
> m = mass of proton = 1.67*10^-27 kg
> Rn = radius of a proton = 1.36*10^-15 meters
> Vt = (1/(2*pi)) * sqrt((29.053 / (2*1.36e-15)) / 1.67e-27) *
> (2*1.36e-15) = 1,094,817.78
>
> Vt = 1,094,817 m/s
>
> This is the speed of transition, and the number Frank wants to call
> Znidarsic's Constant. It represents the speed of sound in a nucleus.
>
> Since we're talking about a vibrational speed, we can go back to a
> frequency and a wavelength.
> 5) Vt = f*w
> f = frequency
> w = wavelength
>
> w = Vt / f
> This is the wavelength of a photon inside of the nucleus, not the
> emitted photon.
>
> 6) This is the equation for capacitance.
> C = e0 * A / D
> C = Capacitance
> e0 = Permittivity of Free Space
> A = Area between
> D = Distance
>
> Let's assume that the wavelength of a photon in the nucleus carries a
> capacitance. Twice the wavelength would be the area,
> and 1/2 the wavelength would be used instead of the distance between the
> plates of a capacitor, in the equation.
> C = e0 * w^2 / 0.5 * w
> C = 2*e0*w
>
> Substituting for wavelength:
> C = 2*e0*Vt / f
>
> This is the capacitance of energy in the transitional state.
>
> 7) E = Q^2 / 2 * C
> Q = Charge
> C = Capacitance
> E = Energy
>
> Substituting
>
> E = (Q^2 / 4 * e0 * Vt) * f
> E = h * f (This is Einstein's Photo-Electric Equation)
>
> h = Planck's Constant
>
> >
> >
> >
> >
>
>
>
