devin553344 Posted July 28, 2020 Report Posted July 28, 2020 (edited) PDF file: 20200726WaveCurvatures.pdf I have made some adjustments to the theory, as I found the strain equations in my last set of equations to not match between energy of particles and their gravitational curvature leak. I've fixed them in the equations in this pdf file. I'll write out the equations over the next few days, but they can be found in the above pdf file. Edited August 23, 2020 by devin553344 Quote
devin553344 Posted July 28, 2020 Author Report Posted July 28, 2020 (edited) The electron mass energy from strain energy is: ϵ = (hc/(εK^2e^2))^1/2 mc^2 = εK^2e^2/r * ϵ^2 Where ϵ is the logarithmic strain, h is Planck's constant, c is the speed of light, ε is the permittivity of free space, K is the electric constant, e is the elementary charge, m is the mass of the electron, r is the Compton wavelength of the electron. The gravitation then is a leak of the curvature of the matter energy and relates to the electron-positron pair energy: 3/5 * 8Gm^2/r = 2mc^2 * exp(-ϵ) Where G is the gravitational constant. This follows gravitational binding energy: https://en.wikipedia.org/wiki/Gravitational_binding_energy Edited July 28, 2020 by devin553344 Quote
devin553344 Posted July 28, 2020 Author Report Posted July 28, 2020 (edited) The proton mass energy from strain energy is: ϵ = (9/2 * ђc/(εK^2e^2))^1/2 mc^2 = 2/9 * εK^2e^2/r * ϵ^2 Where ϵ is the logarithmic strain, ђ is the reduced Planck's constant, c is the speed of light, ε is the permittivity of free space, K is the electric constant, e is the elementary charge, m is the mass of the proton, r is the "reduced" Compton wavelength of the proton. The gravitation then is a leak of the curvature of the matter energy: Gm^2/r = mc^2 * exp(-ϵ) Where G is the gravitational constant. Edited July 28, 2020 by devin553344 Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 (edited) Still working on this: To unify charge I find that the elementary charge is the same for mass valleys (Planck's constant) as it is for charge hills (electric) in value. Since I have already defined the fine structure types as logarithmic strains for the electron and proton, then I must define the base strain for the charge to Planck relationship. I will do that now. For any cross section of a wave, a distance of 4 gets deformed into two pi leaving the deformation as two pi minus 4. The original length is 1.0: ђc = e^2/ε * (4ln((2π - 4)/1))^2 Where ђ is the reduced Planck constant, c is the speed of light, ε is the permittivity of free space, e is the elementary charge. Edited July 30, 2020 by devin553344 Quote
Vmedvil2 Posted July 29, 2020 Report Posted July 29, 2020 (edited) Your dimensions are wrong.Agreed, I told him this 2 months ago, he's a crank there is no arguing with a crank. Edited July 29, 2020 by VictorMedvil Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 Your dimensions are wrong.There not wrong, but if you would like to give me an example of what you think is wrong I will explain it for you. Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 If the dimensional analysis is wrong, it is wrong period. There is no two sides to this statement. Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 (edited) If the dimensional analysis is wrong, it is wrong period. There is no two sides to this statement.Then you have to point out what is wrong, unless your trolling ;) I already showed VictorMedvil on my other theory that it was correct, so he is definitely trolling! Edited July 29, 2020 by devin553344 Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 (edited) e^2/r is the energy for instance but you have one numerator of strain weighted by a factor of the squared strain, meaning that the dimensions cannot be justified. Edited July 29, 2020 by Dubbelosix devin553344 1 Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 Then you have to point out what is wrong, unless your trolling ;) I already showed VictorMedvil on my other theory that it was correct, so he is definitely trolling!I just did. Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 (edited) e^2/r is the energy for instance but you have one numerator of strain weighted by a factor of the squared strain, meaning that the dimensions cannot be justified.Thanks for the response, I welcome the chance to check my equations. I'm assuming you're talking about the electrical to Planck equation since you mentioned e^2/r, but it's not e^2/r it's e^2/ε which is energy times meters, since ε is the permittivity of free space and the reciprocal of the electrical constant. And ђc is energy times meters also. Also that is not a strain energy equation, it references a curvature adjustment where I've used exponential growth of factors: π + exp(1-2*2^1/2) Where each charge is adjusted by multiplying pi and an exponential factor based on the height of the wave: 1.0, and the distance between heights: 2*2^1/2 Edited July 29, 2020 by devin553344 Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 Not what I am reading, anyway you write your strain, the dimensions are not possible to construct the rest energy. Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 You also identified epsilon with the strain ... Start writing the equations clear and we can start a proper discussion,because even if it was permittivity, the dimensions are still not making sense. Quote
Dubbelosix Posted July 29, 2020 Report Posted July 29, 2020 E= mc^2 = e^2/R Because Mc^2 * R = e^2 Here the dimensions are explained. Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 Not what I am reading, anyway you write your strain, the dimensions are not possible to construct the rest energy. OK now I see what you're reading... I only used strain in the strain energy equations for the electron and proton rest energy and also for the length the curvature must escape for gravitation. Anyways the dimensions for the following are energy: mc^2 = εK^2e^2/r * ϵ^2 This is the same as energy times meters divided by meters. εK^2e^2 is energy times meters since εK = 1/(4pi) then the remainder is Ke^2 and the strain is squared since it is allowed in strain energy equations: E = 1/2 * V * E * ϵ^2 Where V is volume, E is Young's modulus, and ϵ is the strain which is dimensionless. See the strain energy equation here: https://en.wikipedia.org/wiki/Strain_energy And I've used the square root of the fine structure basically for the dimensionless strain. Quote
devin553344 Posted July 29, 2020 Author Report Posted July 29, 2020 You also identified epsilon with the strain ... Start writing the equations clear and we can start a proper discussion,because even if it was permittivity, the dimensions are still not making sense.Those are two different symbols, one is commonly used for permittivity while the other is used for strain. Sorry for any confusion. Quote
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