Explaining The Electromagnetic Fields Of The Earth

[devin553344]

The logarithmic strain may relate to the neutron instead of a relativistic electron. Or both are correct. The following is 8 digits accurate:

 

mn = mp + me * ln(4π) = 1.674 927 505E-27 kg

 

Where mn is the mass of the neutron, mp is the mass of the proton and me is the mass of the electron. Which relates the logarithmic strain of the solid angle of a sphere.

Edited December 18, 2019 by devin553344

[devin553344]

Regarding post number 15 the proton charge radius could be due to log strain of the charge compression energy exceeding a log strain, such that a log strain happens instead of compressing the charge. Then the charge could expand its radius to the charge radius of the proton:

 

rC = 6 * rK^2/rS * ln(4π)/ln(rK/rS) = 8.783E-16 meters

 

Where rC is the charge radius of the proton, I've already defined rK and rS is the Schwarzschild radius of the electron.

 

I updated the description in post #15 to explain the logarithmic strain idea.

Edited December 18, 2019 by devin553344

[devin553344]

And if the logarithm of 4 pi relates to spheroid particles then it should relate to Planck's constant. I will use the Plane of charge solution with logarithms only. 6 pi relates the total solid angle for electromagnetic of 4pi plus 2 pi:

 

(e * ln(4π+2π))^2/(2εc) * ln(4π) * a = h/(2π)

 

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

 

In order for this to be correct, QED must be incorporated. Otherwise it's a close approximation.

 

The QED apparently relates to a 5 dimensional strong force:

 

a = 1 - 1/(8/3 * π^2 * exp(4))

 

I use 1.0 for the radius/radius which indicates it's happening at the wavelength.

Edited December 18, 2019 by devin553344

[OverUnityDeviceUAP]

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[devin553344]

And thus ends religion and there will be a hot thermonuclear war unlike any 0ther as the pagan god of abstinence and suffering is suffocated alas

 

I'm Christian ;)

[Vmedvil2]

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Edited December 18, 2019 by VictorMedvil

[OverUnityDeviceUAP]

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[devin553344]

I found a more accurate Planck solution that uses the 3 dimensional strong force:

 

(e * ln(4π+2π))^2/(2εc) * ln(4π) * a - e^2/(8εc) = h/(2π)

 

Where a is the 3 dimensional strong force at the wavelength:

 

a = 1 + 1/(4πexp(2))

 

I've subtracted the basic plane electric force using only half of each charge from the logarithmic strain portion.

Edited December 18, 2019 by devin553344

[OverUnityDeviceUAP]

I found a more accurate Planck solution that uses the 3 dimensional strong force:

 

(e * ln(4π+2π))^2/(2εc) * ln(4π) * a - e^2/(8εc) = h/(2π)

 

Where a is the 3 dimensional strong force at the wavelength:

 

a = 1 + 1/(4πexp(2))

 

I've subtracted the basic plane electric force using only half of each charge from the logarithmic strain portion.

You did good work.

[devin553344]

You did good work.

Thanks

[devin553344]

I found one more solution that uses the strong force, and this one matches the expected value exactly to 8 digits. I started thinking that perhaps the radius is reciprocal for hc and Ke^2, such that a Riemann zeta -5 appears on top and the ln(2π) appears on bottom. Now it has plane field and self energy:

 

h = -252e^2/(2εc * ln(2π)) + 1/2 * Ke^2/c * (1 - 1/(4πexp(2))) = -6.626 070 189E-34 Joules * seconds

 

I'm examining it and wondering if it is the solution I've been looking for?

Edited December 19, 2019 by devin553344

[OverUnityDeviceUAP]

I found one more solution that uses the strong force, and this one matches the expected value exactly to 8 digits. I started thinking that perhaps the radius is reciprocal for hc and Ke^2, such that a Riemann zeta -5 appears on top and the ln(2π) appears on bottom. Now it has plane field and self energy:

 

h = -252e^2/(2εc * ln(2π)) + 1/2 * Ke^2/c * (1 - 1/(4πexp(2))) = -6.626 070 186E-34 Joules * seconds

 

I'm examining it and wondering if it is the solution I've been looking for?

Solution for what? What exactly is this in reference to?

[OverUnityDeviceUAP]

The electromagnetic fields of the earth? You haven't graphed anything, I don't see a graph.

[OverUnityDeviceUAP]

I dont see coordinates anywhere. I dont see lines. What the ef are you even trying to do.

[devin553344]

I dont see coordinates anywhere. I dont see lines. What the ef are you even trying to do.

I don't need coordinates. I've already defined the charge of the earth based on the protons and the equation I've supplied. Perhaps you should take another look at it. I've defined the overall charge of the earth based on the number of protons that make up its mass. The charge then defines the volts per meter at the surface which is about 100 volts per meter. Check again please. It's in post #1.

Edited December 19, 2019 by devin553344

[devin553344]

Solution for what? What exactly is this in reference to?

 

This is in reference to the other equations I've supplied which link back to post #1 regarding the connection of logarithmic strain on the nature of wave particle duality. That particular reference is for Planck from electric charge. The idea is that a bend of space carries the elementary charge regardless of its characteristics. Then the elementary charge is a space bend quantum. And the bending also produces Planck's constant which links to mass energies.

 

And that would be why all particles have a DeBroglie wavelength and also carry the elementary charge. It also matches my strong force idea which also calculates accurately, defying the current Nobel prize winner for what I consider is a strange idea (particle exchange forces "Yukawa potential"). My strong force being an inverse to logarithmic strain.

 

The equation needs to be re-arranged to understand it correctly. But I've arranged it to solve for Planck's constant value.

Edited December 19, 2019 by devin553344

[devin553344]

The electromagnetic fields of the earth? You haven't graphed anything, I don't see a graph.

 

You might be wondering what I'm doing by adding additional equations for other areas? This all supports the original idea. The more poof I provide for my ideas that link back to calculating the electric field of the earth, the more I can say, perhaps it's correct.