E = h nu

E_e = 510 998.9027 eV

nu_e = E / h = 1.235589976E20 Hz

Lambda_e = c / nu = 2.426310213-12 m

We see that the wavelength of a free moving photon of same energy as
the electron is equal to the electron Compton wavelength.

De Broglie posed the hypothesis in 1924 that in the Bohr model,
the length of the ground orbit would be equal to

Lambda_o = h / (m_e v)

and this wavelength was found to exactly match the length
of the Bohr ground orbital.

Knowing the long established mass of the electron (9.10938188E-31 kg)
and the velocity (c) of the photon, I found it an interesting
exercise to calculate what the orbital "wavelength" of a decoupling
pair could be if the velocity of the photon was preserved in the
process (v = c = 299 792 458 m/s) by applying de Broglie's
equation to the case of a 1.021997805 MeV decoupling photon as it
grazed a heavy nucleus, interacting with the Coulomb field of said
nucleus (CI = Coulomb interaction).

Lambda_e = h / (m c) = 2.426310215E-12 m

We observe that we obtain the exact same orbital "wavelength" as
the linear "wavelength" of a free moving photon of same energy
as the electron.

So we have lambda_e = c / nu = h / (m c)

from which we draw

h / (m c) = c / nu

and finally

E = h nu = mc^2

Hence

E_[1.021886795 MeV] = (h nu) + CI = 2K / (alpha_0 alpha)^2 = 2 mc^2

K = 1.220852596E-38 J.m^2 (fundamental electrostatic induction constant)
alpha_0 = Bohr radius
alpha = fine structure constant

(alpha_0 alpha) = radius of orbital Compton wavelength

which is the decoupling radius of the pair in electrostatic space
(orthogonal to normal space)

Knowing that E = gamma m c^2 for moving electrons,

we thus have

E_[1.022 MeV to 211.316 MeV] = h nu + CI = 2 gamma m_e c^2

and since photons in excess of 211.316 seem to decouple as
muon antimuon pairs, we also have

E_[211.316 MeV] = h nu + CI = 2 gamma m_mu c^2

The "Compton orbital wavelenth" turns out to be the smallest
(and only) orbital wavelenth on which energy can reach the speed
of light, which constitutes the escape velocity of the de Broglie
half-photons in electrostatic space.

Conversely, the 1.021997805 MeV photon is the smallest amount of energy
that is sufficient to cause its half-photons to reach escape velocity
upon destabilization through Coulomb interaction with a heavy particle.

Q.E.D.

André Michaud

Very astucious this André Michaud, he derivates
E = mc² from E= mc².

His second formula:
E_electron = 0.511 Mev
would not be, (quite from hasard, of course)
m_electron*c2 = .511Mev ?

Would not that be an autoreferent derivation ?
hello.

--
cisfranrey

Le fait d'apprendre beaucoup n'intruit pas l'intelligence

The fact much to learn does not teach the intelligence



http//perso.wanadoo.fr/jeanfranraymond.rey
"Andr? Michaud" a écrit dans le message de
om...
E = h nu

E_e = 510 998.9027 eV

nu_e = E / h = 1.235589976E20 Hz

Lambda_e = c / nu = 2.426310213-12 m

We see that the wavelength of a free moving photon of same energy as
the electron is equal to the electron Compton wavelength.

De Broglie posed the hypothesis in 1924 that in the Bohr model,
the length of the ground orbit would be equal to

Lambda_o = h / (m_e v)

and this wavelength was found to exactly match the length
of the Bohr ground orbital.

Knowing the long established mass of the electron (9.10938188E-31 kg)
and the velocity (c) of the photon, I found it an interesting
exercise to calculate what the orbital "wavelength" of a decoupling
pair could be if the velocity of the photon was preserved in the
process (v = c = 299 792 458 m/s) by applying de Broglie's
equation to the case of a 1.021997805 MeV decoupling photon as it
grazed a heavy nucleus, interacting with the Coulomb field of said
nucleus (CI = Coulomb interaction).

Lambda_e = h / (m c) = 2.426310215E-12 m

We observe that we obtain the exact same orbital "wavelength" as
the linear "wavelength" of a free moving photon of same energy
as the electron.

So we have lambda_e = c / nu = h / (m c)

from which we draw

h / (m c) = c / nu

and finally

E = h nu = mc^2

Hence

E_[1.021886795 MeV] = (h nu) + CI = 2K / (alpha_0 alpha)^2 = 2 mc^2

K = 1.220852596E-38 J.m^2 (fundamental electrostatic induction constant)
alpha_0 = Bohr radius
alpha = fine structure constant

(alpha_0 alpha) = radius of orbital Compton wavelength

which is the decoupling radius of the pair in electrostatic space
(orthogonal to normal space)

Knowing that E = gamma m c^2 for moving electrons,

we thus have

E_[1.022 MeV to 211.316 MeV] = h nu + CI = 2 gamma m_e c^2

and since photons in excess of 211.316 seem to decouple as
muon antimuon pairs, we also have

E_[211.316 MeV] = h nu + CI = 2 gamma m_mu c^2

The "Compton orbital wavelenth" turns out to be the smallest
(and only) orbital wavelenth on which energy can reach the speed
of light, which constitutes the escape velocity of the de Broglie
half-photons in electrostatic space.

Conversely, the 1.021997805 MeV photon is the smallest amount of energy
that is sufficient to cause its half-photons to reach escape velocity
upon destabilization through Coulomb interaction with a heavy particle.

Q.E.D.

André Michaud