### S. Sesselmann (26/03/2015) First Law

Posted:

**Thu Mar 26, 2015 10:57 am**Ground Potential theory starts with one basic postulate,

This may appear inconsistent with current theory, as it is widely assumed that particle pairs created in a single event, must have opposite charge and equal mass (matter/antimatter). This assumption turns out to be wrong.

We shall see that the difference in mass between the electron and the proton is due to mass defect, the same kind of mass defect as seen in heavy nuclei which is well understood. The electron is trapped in it's own negative electric potential and dispite having the same mass as the proton, appears to the observer as a smaller particle.

It was obvious from these assumptions that the observers potential (Ground Potential) had to lie somewhere between the electrical potential of the proton and the potential of the electron, and the reason is simply that we know of no charge more positive than the proton and likewise no charge more negative than the electron, ergo everything else must lie in-between.

This inspired me to go after the equation describing the relationship between electron-potential, ground-potential and proton-potential, it was tricky, but the solution presented itself to me within a couple of days of working on it.

\[ a (\gamma) = \frac{1}{2} (c-b) \]

Where 'a' is electon-potential, 'b' is ground-potential and 'c' is proton-potential.

The problem here is \(\gamma\) being the same kind of \(\gamma\) as in Einstein's relativity, being velocity dependent, so how does one define gamma?

After some further thinking it became apparent that four-velocity was a function of potential and since the proton represented the absolute maximum potential, it could be concidered as a physical constant in the same way as the speed of light, so the full equation now became;

\[ a = \frac{(c-b)}{2} \sqrt{1-\frac{b^2}{c^2}} \]

So here we have the first law of ground potential, a very simple and elegant statement for the first time defining ground potential.

Solving the above equation with known values for electron and proton potential gives a ground potential of 930 million volts.

I have chosen volts as the ideal unit to work in, because it is the SI unit which moves one elementary charge 1 meter in 1 second, so it is the perfect unit to define the potentials of electrons and protons.

A proton which has a mass of \(\frac{938 MeV}{c^2}\) contains 938 MeV of energy which divided by one elementary charge gives a potential of 938 million volts. Likewise the electron can be said to have 0.511 million volts potential.

This first law of GP shall be the law upon which all other laws shall rest.

Steven

**that the electron and the proton are a particle pair**This may appear inconsistent with current theory, as it is widely assumed that particle pairs created in a single event, must have opposite charge and equal mass (matter/antimatter). This assumption turns out to be wrong.

We shall see that the difference in mass between the electron and the proton is due to mass defect, the same kind of mass defect as seen in heavy nuclei which is well understood. The electron is trapped in it's own negative electric potential and dispite having the same mass as the proton, appears to the observer as a smaller particle.

It was obvious from these assumptions that the observers potential (Ground Potential) had to lie somewhere between the electrical potential of the proton and the potential of the electron, and the reason is simply that we know of no charge more positive than the proton and likewise no charge more negative than the electron, ergo everything else must lie in-between.

This inspired me to go after the equation describing the relationship between electron-potential, ground-potential and proton-potential, it was tricky, but the solution presented itself to me within a couple of days of working on it.

\[ a (\gamma) = \frac{1}{2} (c-b) \]

Where 'a' is electon-potential, 'b' is ground-potential and 'c' is proton-potential.

The problem here is \(\gamma\) being the same kind of \(\gamma\) as in Einstein's relativity, being velocity dependent, so how does one define gamma?

After some further thinking it became apparent that four-velocity was a function of potential and since the proton represented the absolute maximum potential, it could be concidered as a physical constant in the same way as the speed of light, so the full equation now became;

\[ a = \frac{(c-b)}{2} \sqrt{1-\frac{b^2}{c^2}} \]

So here we have the first law of ground potential, a very simple and elegant statement for the first time defining ground potential.

Solving the above equation with known values for electron and proton potential gives a ground potential of 930 million volts.

I have chosen volts as the ideal unit to work in, because it is the SI unit which moves one elementary charge 1 meter in 1 second, so it is the perfect unit to define the potentials of electrons and protons.

A proton which has a mass of \(\frac{938 MeV}{c^2}\) contains 938 MeV of energy which divided by one elementary charge gives a potential of 938 million volts. Likewise the electron can be said to have 0.511 million volts potential.

This first law of GP shall be the law upon which all other laws shall rest.

Steven