Mon Feb 16, 2015 9:03 pm

In a discussion with my son recently, the subject of electric vs centripetal force came up, and it seemed logical that GP theory ought to hold for the simple case of Hydrogen.

\[F = K\frac{qq}{r^2} = \frac{mv^2}{r}\]

For the first expression we know the coulomb constant and the elementary charges, and I have looked up the Hydrogen radius to be 25 picometer. The Hydrogen radius seems to be the constant which is least accurately defined.

For the second expression we know theelectron mass and we use the same radius 25 pm, but we don't know the exact velocity of the electron.

My first rather naive attempt at solving this problem was to calculate the electron velocity using the formula for particle velocity from groundpotential;

\[v_e=c(\frac{\Delta\phi}{\Phi})\]

Where \(\Delta\phi\) is the difference between the electron potential and groundpotential, amd \(\Phi\) is the proton potential.

Well, I was out by a factor of 1000, and could not understand what was wrong, I rechecked and rechecked again because it just had to be right, and I suffered several sleepless nights.

Finally I realised that an electron in free flight is not the same as an electron in orbit, the electron in orbit is in constant accelleration because it is held at a potential other than it's natural (511 keV) potential. It is forcefully held at the surface potential of Hydrogen itself, far from 511 keV.

All we need to find the surface potential of Hydrogen is it's mass energy in MeV and divide that by the number of nucleons which is one, and it turns out that Hydrogen is actually a little heavier than protonium, and clocks in at 939 MeV.

So now I obtained a far more satisfactory solution to the problem.

\[F = K\frac{qq}{r^2} = 3.69 * 10^{-7}\] and

\[F = \frac{mv^2}{r} = 2.98 * 10^{-7}\]

The two forces agree to within 20%, and I guess that some further thinking might explain why there is the slight difference. For the above calculations I used the following numbers.

Proton Mass - 938.3 MeV

Electron Mass - 0.5 MeV

Hydrogen Mass - 939.2 MeV

Ground Potential - 930.4 MeV

Hydrogen Radius - 25 pico meter

So now that this has been sorted out, I look forward to a good nights sleep.

Not that the 20% difference doesn't bother me, but I am hoping that some of you readers out there will put your minds to the problem and help me get ground potential off the ground (if that makes any sense at all).

Ground potential works, it can solve problems with, it's rules are fewer and simpler than existing relativity theory, and it is this simplicity which will eventually take over physics.

Enjoy..

Steven Sesselmann

\[F = K\frac{qq}{r^2} = \frac{mv^2}{r}\]

For the first expression we know the coulomb constant and the elementary charges, and I have looked up the Hydrogen radius to be 25 picometer. The Hydrogen radius seems to be the constant which is least accurately defined.

For the second expression we know theelectron mass and we use the same radius 25 pm, but we don't know the exact velocity of the electron.

My first rather naive attempt at solving this problem was to calculate the electron velocity using the formula for particle velocity from groundpotential;

\[v_e=c(\frac{\Delta\phi}{\Phi})\]

Where \(\Delta\phi\) is the difference between the electron potential and groundpotential, amd \(\Phi\) is the proton potential.

Well, I was out by a factor of 1000, and could not understand what was wrong, I rechecked and rechecked again because it just had to be right, and I suffered several sleepless nights.

Finally I realised that an electron in free flight is not the same as an electron in orbit, the electron in orbit is in constant accelleration because it is held at a potential other than it's natural (511 keV) potential. It is forcefully held at the surface potential of Hydrogen itself, far from 511 keV.

All we need to find the surface potential of Hydrogen is it's mass energy in MeV and divide that by the number of nucleons which is one, and it turns out that Hydrogen is actually a little heavier than protonium, and clocks in at 939 MeV.

So now I obtained a far more satisfactory solution to the problem.

\[F = K\frac{qq}{r^2} = 3.69 * 10^{-7}\] and

\[F = \frac{mv^2}{r} = 2.98 * 10^{-7}\]

The two forces agree to within 20%, and I guess that some further thinking might explain why there is the slight difference. For the above calculations I used the following numbers.

Proton Mass - 938.3 MeV

Electron Mass - 0.5 MeV

Hydrogen Mass - 939.2 MeV

Ground Potential - 930.4 MeV

Hydrogen Radius - 25 pico meter

So now that this has been sorted out, I look forward to a good nights sleep.

Not that the 20% difference doesn't bother me, but I am hoping that some of you readers out there will put your minds to the problem and help me get ground potential off the ground (if that makes any sense at all).

Ground potential works, it can solve problems with, it's rules are fewer and simpler than existing relativity theory, and it is this simplicity which will eventually take over physics.

Enjoy..

Steven Sesselmann

Last edited by Steven Sesselmann on Tue Feb 17, 2015 8:44 am, edited 1 time in total.

**Reason:** *Minor grammatical corrections*

Tue Feb 24, 2015 10:47 am

Addendum:

Calculating the exact radius of the single Hydrogen atom using Ground Potential theory.

First assumption is that the electric force is equal to the centripetal force.

\[ K\frac{qq}{r^2} = \frac{mv^2}{r} \]

We then use GP theory to find the electrons velocity relative to the observer.

The orbital electron velocity is calculated by first finding \( \Delta \phi \), (Hydrogen Surface Potential - Ground Potential) = ∆Ø = 8.8 MV

We can now find \( \Delta v\), by the formula

\[ v_e = c * (\frac{\Delta\phi}{\Phi}) = c * (\frac{8.8 MV}{938.3 MV}) = 2.812 * 10^6 m/s \]

Where the constant \( \Phi \) is the surface potential of the proton.

Important to understand that this velocity is the relative velocity between the observer at ground potential and the electron

Now we can find the radius of the H atom by the following equation;

\[ r_H = K\frac{qq}{m_e * v_e ^2} = r = 32.02 pm \]

This result agrees 100% with the measured radius of the single H atom.

Steven

Calculating the exact radius of the single Hydrogen atom using Ground Potential theory.

First assumption is that the electric force is equal to the centripetal force.

\[ K\frac{qq}{r^2} = \frac{mv^2}{r} \]

We then use GP theory to find the electrons velocity relative to the observer.

The orbital electron velocity is calculated by first finding \( \Delta \phi \), (Hydrogen Surface Potential - Ground Potential) = ∆Ø = 8.8 MV

We can now find \( \Delta v\), by the formula

\[ v_e = c * (\frac{\Delta\phi}{\Phi}) = c * (\frac{8.8 MV}{938.3 MV}) = 2.812 * 10^6 m/s \]

Where the constant \( \Phi \) is the surface potential of the proton.

Important to understand that this velocity is the relative velocity between the observer at ground potential and the electron

Now we can find the radius of the H atom by the following equation;

\[ r_H = K\frac{qq}{m_e * v_e ^2} = r = 32.02 pm \]

This result agrees 100% with the measured radius of the single H atom.

Steven