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Explaining The Electromagnetic Fields Of The Earth


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PDF file: http://www.scienceforums.com/topic/36455-gravitational-curvatures-might-be-a-vapor-of-liquid-space/

 

I read that Einstein wrote that he believed the electromagnetic fields of the earth were created from a charge imbalance in the particles that make up the earth.

 

I think I may have found a theory to explain that. First I will provide a possible voltage of space time which I will use to describe the energy of the elementary charge:

 

Vst = ((c^4)/(8πGε))^1/2

 

Where Vst is the voltage of space-time, c is the speed of light, G is the gravitational constant, ε is the permittivity of free space.

 

Next I will describe a gravitational relationship of energies. This is the gravitation that keeps the elementary charge from exploding in a particle:

 

8πG/c^4 * εV^2 * mc^2 = Vst * e

 

Where V is the internal voltage of particles that describes its matter charge and matter energy. m is the mass of the particle, e is the elementary charge.

 

Next I can describe the energy and momentum of any particle in an electromagnetic manner:

 

Vq = mc^2

 

B = V/c^2

 

p = mv = qvB

 

Where q is the secondary charge of the particle that provides the charge of the earth from its protons. B is the magnetic field of the particle, p is the momentum of the particle, v is the velocity.

 

Now one can easily calculate the proton charge contribution of the earth by obtaining the quantity of protons that make up the mass of the earth:

 

Q = M/mp * qp = 8.2E+05 Coulombs

 

Where Q is the charge of the earth, M is the mass of the earth, mp is the mass of the proton, qp is the matter charge of the proton described above as Vq = mc^2.

 

The matter charge of the proton calculates to: 2.3E-46 Coulombs

20191228WaveCurvatures.pdf

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The Internet is like 90% porn just look at the internet, we don't need girls. That being said are you saying that you know how to make a girl summoning device?

Here: https://en.wikipedia.org/wiki/Dynamo_theory   "Einstein believed that there might be an asymmetry between the charges of the electron and proton so that the Earth's magnetic field would be produ

How intriguing. I had no idea he thought that.   Thanks. 

I am considering that the elementary charge and other charge values might have an associated energy similar to E=mc^2. This energy is not an assembly energy like Gmm/r or Ke^2/r. It is like a charge substance energy. The elementary charge then has the energy:

 

E = Vst * e = 1.182E+08 Joules

 

Where Vst is the voltage of a black hole, and is simply a conversion of Coulombs to Joules:

 

Vst  = Ke/(2GKe^2/c^4)^1/2

 

Where K is the electric constant. And then it follows that the energy of any particle is:

 

E  = Vq = mc^2

 

Which I've already spelled out in the previous OP.

 

So then in the OP I've spelled out the gravitation that would contain the elementary charge substance energy. And the voltage of a particle can be obtained thru a black hole concept:

 

Vq = mc^2 = Keq/(2Gm/c^2 * (2GKe^2/c^4)^1/2)^1/2

 

Although I'm not sure actual black hole concepts apply, since c^2/G is the mass per meter of a black hole.

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Gravitation may be a magnetic mono-pole generated from a uniform velocity field of gravitation against a matter charge. I'll use the standard gravitational constant for simplicity:

 

v? = (8πG/c^2 * εV^2)^1/2

 

E = Gm1m2/r = 1/2 * K/c^2 * q1v1q2v2/r

 

 

Where E is the gravitational energy, m1 is a mass, m2 is a mass, r is the radius, v1 is the velocity field for q1 which is a matter charge, v2 is the same for a second matter charge also represented by v?.

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One of the properties of the particle equation is that of assembly energy balance:

 

1/2 * Ke^2/rK = Gm^2/rV

 

Where rV is the voltage radius or matter radius of the point particle, such that the voltage of the particle can be found:

 

Kq/rV = V

 

Then the elementary charge sits at rK and the mass and its associated charge sit at rV. Then there is a balance of energies.

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This is nonsense! There is no imbalance between charges otherwise places like Mars would have a magnetic field still which it doesn't.

 

First off Mars has a magnetic field: https://en.wikipedia.org/wiki/Terraforming_of_Mars

 

"Mars does not have an intrinsic global magnetic field, but the solar wind directly interacts with the atmosphere of Mars, leading to the formation of a magnetosphere from magnetic field tubes.[22] This poses challenges for mitigating solar radiation and retaining an atmosphere."

 

I would need to have more measurement information to actually decide if the above statement is correct. Since it appears to be hiding someones theory idea.

 

Second I'm not totally sure how to calculate the Tesla value for the earth. My theory explains the volts per meter at the surface of the earth. So that part checks out. But I think it would have to take an expert in magnetic fields to calculate the magnetic field from an intrinsic earth charge. I get a very small magnetic field value (undetectable) for the earth when I attempt to calculate it without using unit vectors. This is why I just stated the charge of the earth. For instance Mars magnetic field calculates to 2.5 micro Teslas using unit vectors.

 

Third, from my calculations the Mars magnetic field would be 1/10 that of earth. So did it go un-detected?

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Now one can easily calculate the proton charge contribution of the earth by obtaining the quantity of protons that make up the mass of the earth:

 

Q = M/mp * qp = 8.2E+05 Coulombs

 

Where Q is the charge of the earth, M is the mass of the earth, mp is the mass of the proton, qp is the matter charge of the proton described above as Vq = mc^2.

 

The matter charge of the proton calculates to: 2.3E-46 Coulombs

 

 

Did you forget about the electrons?

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Did you forget about the electrons?

 

No, the electrons only contribute a small portion of charge and it would be oppositely charged. So it would be not noticed. The charge of an electron might be 3e-51 Coulombs using my equations. So the total charge of the earth from electrons is about 10 Coulombs.

 

Or are you talking about atmospheric electrons?

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The gravitational constant has the energy per meter of a black hole of any mass. Therefore the energy per meter of the particle relates to the gravitational energy per meter in the following way:

 

c^4/8πG = εV^2 / rG^2 * rS^2

 

Where V is the voltage of the particle previously spelled out in this thread.

 

rK = (2GKe^2/c^4)^1/2

 

rS = 2Gm/c^2

 

rG = (rS * rK)^1/2

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The proton might be 3 orbiting electrons with an energy increase of a logarithmic strain of the electrons metric. The strain happens since the electric energy is too great that the charge separates and forms a charge radius to match or be less than the proton energy. The charge radius size of the proton may instead be explained via a wavelength idea of the mass relating to Planck's constant:

 

ln(8πG/c^4 * εV^2) * 8πG/c^4 * 3me^2c^4/rS = mpc^2 = 1.503 761E-10 Joules

 

Where V is the internal voltage of the electron as described earlier in this thread, me is the electron mass, rS is the Schwarzschild radius of the electron, mp is the proton mass. The logarithmic strain is also expressed as a radius factor:

 

ln(8πG/c^4 * εV^2) = ln(rK/rS)

 

The electron follows the maximum bend of space as a wave nature and therefore relates to the fine structure as a logarithmic strain. This allows the particle equation described earlier in this thread to describe transitions of particles from higher energy to lower energy through decay process until they reach a maximum log strain of the electron:

 

ln(3/10 * 8πG/c^4 * εV^2) = ln(2)/4π * (hc)/(Ke^2)

 

Which also relates the Bekenstein bound and gravitational binding energy.

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