Unification Of The Forces

[devin553344]

PDF File: 20200415WaveCurvatures.pdf

 

I have finished my theory and unified the forces thru logarithmic strain. I found that there are two types of fields similar to what Einstein described for gravitation. There are space-time hills and space-time valleys. One repels while the other attracts. Both are curvatures of space-time.

 

I have based most of the theory on logarithmic strain energies using the electron and proton wavelengths. Which are involved in the creation of protons-antiprotons and electron-positrons. I've described all the stable particles and left the unstable particles as higher energies of stable particles.

 

I have included my strong force idea and a boiling point idea, the boiling point idea may relate to DoubleOSix's previous work. Contact him for his theory please.

 

Everything  calculates and I have provided the values as they calculate to. I  will provide one equation here and the rest I have included in the attached PDF file. Here is the calculation for the proton, it describes a proton composed of two supercharged positrons and one supercharged electron:

 

Logarithmic strain energy is defined as (https://en.wikipedia.org/wiki/Strain_energy):

 

U = 1/2 * V * E * s^2

 

Where U is the logarithmic strain energy, V is the volume, E is Young's Modulus, s is the strain.

 

I'm attempting to establish that Young's modulus relates to any  base energy that is strained (see PDF file for the 4 calculations that demonstrate that idea). The proton calculation is:

 

mpc^2 = 1/2 * 3mec^2 * (4ln(rp/re) + ln((2πKe^2)/(hc)))^2

 

Where mp is the mass of the proton, c is the  speed of light, me is the mass of the electron, ln is the natural logarithm, rp is the wavelength of the proton, re is the wavelength of the electron, K is the electric constant, e is the elementary  charge, h is Planck's constant.

 

I should also  provide the calculation for the  electron. It is created during interaction with the nucleus and strains  into formation while interacting with a proton and perhaps neutron curvature:

 

mec^2 = 1/2 * 2Ke^2/re * (4ln(re/rp))^2 * (1-g^2)^1/2

 

The electron is a vacuum curvature which makes it require a small vacuum adjustment similar to the proton gravitation (see PDF file) and therefore uses g^2 which is the coupling constant of  electromagnetic:

 

g = (((8π^2Ke^2)/(hc)))^1/2

 

See (https://en.wikipedia.org/wiki/Coupling_constant) for more information on the electromagnetic coupling constant. I will provide more of the equations in this thread later. But they are included in the PDF file attached here.

 

Basically in this theory there are two types of logarithmic strains. One is wavelength strains which create matter curvatures. Those create vacuum curvatures (valleys), the other include the fine structure and represent pressure curvatures (hills), if it includes the wavelength strain then it creates a matter curvature (but in the case of defining Planck's constant, see PDF file, then no wavelength strain and no matter curvature).

 

For definition of wavelength see: https://en.wikipedia.org/wiki/Compton_wavelength

Edited July 25, 2020 by devin553344

[Vmedvil2]

PDF File: 20200415WaveCurvatures.pdf

 

I have finished my theory and unified the forces thru logarithmic strain. I found that there are two types of fields similar to what Einstein described for gravitation. There are space-time hills and space-time valleys. One repels while the other attracts. Both are curvatures of space-time.

 

I have based most of the theory on logarithmic strain energies using the electron and proton wavelengths. Which are involved in the creation of protons-antiprotons and electron-positrons. I've described all the stable particles and left the unstable particles as higher energies of stable particles.

 

I have included my strong force idea and a boiling point idea which may relate to DoubleOSix's previous work. Contact him for his theory please.

 

Everything  calculates and I have provided the values as they calculate to. I  will provide one equation here and the rest I have included in the attached PDF file. Here is the calculation for the proton, it describes a proton composed of two supercharged positrons and one supercharged electron:

 

Logarithmic strain energy is defined as (https://en.wikipedia.org/wiki/Strain_energy):

 

U = 1/2 * V * E * s^2

 

Where U is the logarithmic strain energy, V is the volume, E is Young's Modulus, s is the strain.

 

I'm attempting to establish that Young's modulus relates to any  base energy that is strained (see PDF file for the 4 calculations that demonstrate that idea). The proton calculation is:

 

mpc^2 = 1/2 * 3mec^2 * (4ln(rp/re) + ln((2πKe^2)/(hc)))^2

 

Where mp is the mass of the proton, c is the  speed of light, me is the mass of the electron, ln is the natural logarithm, rp is the wavelength of the proton, re is the wavelength of the electron, K is the electric constant, e is the elementary  charge, h is Planck's constant.

 

I should also  provide the calculation for the  electron. It is created during interaction with the nucleus and strains  into formation while interacting with a proton and perhaps neutron curvature:

 

mec^2 = 1/2 * 2Ke^2/re * (4ln(re/rp))^2 * (1-g^2)^1/2

 

The electron is a vacuum curvature which makes it require a vacuum adjustment similar to the proton gravitation (see PDF file) and therefore uses g^2 which is the coupling constant of  electromagnetic:

 

g = (((8π^2Ke^2)/(hc)))^2

 

See (https://en.wikipedia.org/wiki/Coupling_constant) for more information on the electromagnetic coupling constant. I will provide more of the equations in this thread later. But they are included in the PDF file attached here.

 

Basically in this theory there are two types of logarithmic strains. One is wavelength strains which create matter curvatures. Those create vacuum curvatures (valleys), the other include the fine structure and represent pressure curvatures (hills), if it includes the wavelength strain then it creates a matter curvature (but in the case of defining Planck's constant, see PDF file, then no wavelength strain and no matter curvature).

Well, this is wrong all of your equations fail dimensional Analysis.

Edited May 25, 2020 by VictorMedvil

[devin553344]

Well, this is wrong all of your equations fail dimensional Analysis.

Can you demonstrate how they fail mathematically? The strain uses four times the log strain, which relates to hc/r^4 which is more of a 3 dimensional pressure.

Edited May 25, 2020 by devin553344

[devin553344]

The gravitation of the particles is simple. The logarithmic strain generates a change in distance that must be accounted for. And the proper field used. I've defined the vacuum field as a strain of the wavelengths, and a pressure field as a strain of the wavelengths plus the fine structure which relates it to charge.

 

The proton is a pressure field (which gives it a boundary) surrounded by a vacuum field that has a strong force falloff and a gravitation. The gravitation is defined as:

 

Gmp^2/rp = mpc^2 * exp^3(4ln(rp/re))

 

Where G is the gravitational constant, mp is the proton mass and rp is the proton wavelength, c is the speed of light, re is the electron wavelength. I've used the natural exponent to the third to account for each mass (including wavelength) on the left side of the equation.

 

The electron gravitation is the same but instead of a pressure field surrounded by a vacuum field like the proton, it is a vacuum field surrounded by a pressure field:

 

Gme^2/re = mec^2 * exp^3(4ln(rp/re) + ln((2πKe^2)/(hc)))

 

Where me is the electron mass, K is the electric constant, e is the elementary charge, h is Planck's constant.

 

The approximations above are fairly precise, but small vacuum corrections must be made which I've included in the PDF file in the beginning of the OP. Cheers!

Edited May 25, 2020 by devin553344

[devin553344]

Planck's constant may be a curvature of space-time and relate to planes of charge as electromagnetic waves. It is not matter and therefore has no wavelength strain. I have defined it as it relates to a pressure field so it uses something like the fine structure:

 

g = ((8π^2Ke^2)/(hc))^1/2

 

Where g is the electromagnetic coupling constant described earlier in this thread. Using this we find the electromagnetic planes of charge solution and must include the magnetic portion. It has planes of charge separated by half the wavelength and uses the radius to the fourth like any pressure so I've used a factor of 16:

 

Recalling the strain energy equation: U = 1/2 * VEs^2

 

Where U is the energy, V is the volume, E is Young's modulus and s is the strain.

 

Planck's constant is then:

 

hc = 1/2 * 3 * 16 * e^2/(2ε) * ln^2(g^2)

 

Where h is Planck's constant, c is the speed of light, e is the elementary charge, ε is the permittivity of free space.

 

Above 3 defines the charge planes as electric plus magnetic. Cheers!

Edited May 25, 2020 by devin553344

[Vmedvil2]

Planck's constant may be a curvature of space-time and relate to planes of charge as electromagnetic waves. It is not matter and therefore has no wavelength strain. I have defined it as it relates to a pressure field so it uses something like the fine structure:

 

g = ((8π^2Ke^2)/(hc))^1/2

 

Where g is the electromagnetic coupling constant described earlier in this thread. Using this we find the electromagnetic planes of charge solution and must include the magnetic portion. It has planes of charge separated by half the wavelength and uses the radius to the fourth like any pressure so I've used a factor of 16:

 

Recalling the strain energy equation: U = 1/2 * VEs^2

 

Where U is the energy, V is the volume, E is Young's modulus and s is the strain.

 

Planck's constant is then:

 

hc = 1/2 * 3 * 16 * e^2/(2ε) * ln^2(g^2)

 

Where h is Planck's constant, c is the speed of light, e is the elementary charge, ε is the permittivity of free space.

 

Above 3 defines the charge planes as electric plus magnetic. Cheers!

Ya except when you take those in units do their cancel out on both sides, if they don't then the equation is incorrect and fails dimensional analysis as these do fail dimensional analysis. So basically as I have been saying you are just spouting nonsense.

 

 

Some properties and their units

Edited May 25, 2020 by VictorMedvil

[devin553344]

Ya except when you take those in units do their cancel out on both sides, if they don't then the equation is incorrect and fails dimensional analysis as these do fail dimensional analysis. So basically as I have been saying you are just spouting nonsense.

The units are correct. I'm not sure what you're referring to. Could you be more specific please.

[Vmedvil2]

The units are correct. I'm not sure what you're referring to. Could you be more specific please.

No, they aren't however anyone who knows anything about physics can instantly spot this so I will just leave it as it is.

 

as a example look at this equation

 

U = 1/2 * VEs^2 , This doesn't equal energy in units which is newton * meters, you have like Newtons³ / meters³ = Newton * meters as I said it fails dimensional Analysis.

Edited May 25, 2020 by VictorMedvil

[devin553344]

No, they aren't however anyone who knows anything about physics can instantly spot this so I will just leave it as it is.

 

as a example look at this equation

 

U = 1/2 * VEs^2 , This doesn't equal energy in units which is newton * meters, you have like Newtons⁵ / meters = Newton * meters as I said it fails dimensional Analysis.

Thanks for the clarification, perhaps you are confusing something. I just copied that from wikipedia: https://en.wikipedia.org/wiki/Strain_energy

 

V is meters cubed, E is energy per meter cubed, and s or strain is dimensionless.

[Vmedvil2]

Thanks for the clarification, perhaps you are confusing something. I just copied that from wikipedia: https://en.wikipedia.org/wiki/Strain_energy

 

V is meters cubed, E is energy per meter cubed, and s or strain is dimensionless.

Oh okay you were not using σ and using ε properly name your units next time because it made that equation looks dimensionally wrong, it makes all the difference in the world what letter of the alphabet you use in physics for your variables.

 

However I know this is wrong show the dimensions on this one.

 

hc = 1/2 * 3 * 16 * e^2/(2ε) * ln^2(g^2)

Edited May 25, 2020 by VictorMedvil

[devin553344]

Oh okay you were not using σ and using ε properly name your units next time because it made that equation looks dimensionally wrong, it makes all the difference in the world what letter of the alphabet you use in physics for your variables.

 

However I know this is wrong show the dimensions on this one.

 

hc = 1/2 * 3 * 16 * e^2/(2ε) * ln^2(g^2)

Thanks Victor, I described the character I used below the equation, sorry for the confusion. Guess I'm limited on characters.

 

The equation you're mentioning here solves to energy times meters on each side of the equation. it is similar to:

 

hc=Ke^2 * s^2

 

where s is strain and therefore dimensionless. On the left side is Plank's constant times speed of light. Right side is electric. Both are energy times meters.

[Vmedvil2]

Thanks Victor, I described the character I used below the equation, sorry for the confusion. Guess I'm limited on characters.

 

The equation you're mentioning here solves to energy times meters on each side of the equation. it is similar to:

 

hc=Ke^2 * s^2

 

where s is strain and therefore dimensionless. On the left side is Plank's constant times speed of light. Right side is electric. Both are energy times meters.

Whatever name your variables properly, it makes a huge difference when trying to understand what is written, use standard variable notations.

Edited May 25, 2020 by VictorMedvil

[devin553344]

Whatever name your variables properly, it makes a huge difference when trying to understand what is written, use standard variable notations.

I did that time I think. For planes of charge see: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

[devin553344]

I'm modifying the electron equation slightly to the following:

 

mec^2 = 1/2 * 3Ke^2/(re * exp(3/2 * g)) * (4ln(re/rp))^2 = 8.183 961E-14 Joules

 

Where g is the electromagnetic coupling constant again as previously described. This gives the electron a charge radius and also the proton follows the negative version since it is opposite charged and the proton is a pressure instead of a vacuum. The proton is:

 

rC = rp * exp(-3/2 * g) = 8.40E-16 meters.

 

Where rC is the charge radius derived from the wavelength. I updated the PDF file in the OP. That matches current measurements for the proton, for the proton charge radius see: https://en.wikipedia.org/wiki/Proton

 

Basically the charge radius extends to the pressure region of the particle. Either inside or outside the wavelength.

Edited May 25, 2020 by devin553344

[devin553344]

I've updated the OP pdf file to slightly modify the gravitation of the proton and electron, they both relate to a charge radius condition which allows for accurate calculations, which is precise considering the exponential expanse of the logarithmic strain of the fields. Basically the exponential value is large and accurate. The proton gravitation is now:

 

3/5 * Gmp^2/rp = 1/3 * mpc^2/exp^2(-3/2 * g) * exp^3(4ln(rp/re))

 

Where G is the gravitational constant, mp is the proton mass, rp is the proton wavelength, c is the speed of light, g is the electromagnetic coupling constant, previously described in this thread and related here: (https://en.wikipedia.org/wiki/Coupling_constant), re is the wavelength of the electron.

 

The electron gravitation is now defined as (which already uses a charge radius factor so it's considered squared here and matches the proton gravitation math):

 

3/5 * Gme^2/re = mec^2/exp(3/2 * g) * exp^3(4ln(rp/re) + ln(a0))

 

Where me is the electron mass, a0 is the fine structure constant.

 

I'm using a gravitational binding energy factor of 3/5, and that is explained on wickipedia here:https://en.wikipedia.org/wiki/Gravitational_binding_energy

 

The equations were already pretty accurate. But to split some more hairs, now the gravitational equations are about 3 digits accurate, which I think is half a percent.

Edited June 7, 2020 by devin553344

[Vmedvil2]

I see hills of stupidity and valleys of ignorance in your mathematics. 

You know I would alert the admins to ban you but I totally agree with this statement.

[Vmedvil2]

Well-met I extend the proverbial olive branch

Na, they are gonna ban you nothing I can do about however I find the raw determination of you always trying to sneak on here inspiring.