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Electric Potential Replaces Gravity

Posted: Tue Aug 19, 2014 9:22 pm
by Steven Sesselmann
Anyone who has been following my recent posts should have given up all ambitions of understanding gravity....

Right, forget gravity, instead it appears that bodies move according to their surface potential, or more correctly stated, according the their difference in electric surface potential.

[ignore this:
This electric potential decreases at the rate of 1/r^2 with height above ground. From this we can now work out the potential gradient above ground, with a two step process;]

The Ground Potential law for relative velocity is:

\[\Delta v = c(\frac{\Delta V}{\Phi})\]

Where \(\Phi\) is the surface potential of the proton and \(\Delta V\) is the difference in surface potential between two bodies. The classical law for velocity under free fall is:

\[v = \sqrt{2gh}\]

Where g is accelleration due to gravity and h is the height in meters.

By combining these two equations we can find the potential difference at one meter above ground.

\[\sqrt{2gh} = c (\frac{\Delta V}{\Phi})\]

\[4.429 mps = c (\frac{\Delta V}{\Phi})\]

If we now solve this equation for height = 1 meter, we get \(\Delta V = 13.86\) volts, ie. the potential between ground and a body 1 meter above ground is about 13.9 volts.

Benjamin Franklin was way ahead of his time...


Re: Electric Potential Replaces Gravity

Posted: Wed Dec 02, 2015 12:48 pm
by Steven Sesselmann
Christopher F wrote:So.., I have a question(s)?
Christopher, welcome to the forum and thanks for the questions.
Christopher F wrote:If I'm understanding correctly, which I'm probably not, Ground Potential is saying that, an apple on a branch is at higher potential in relationship to ground. For this reason, the apple will move towards ground if it is disconnected from the branch.
Yes, GPT shows there is a direct and simple linear relationship between a bodies surface potential and its four velocity. Which gives rise to the relative velocity between two bodies being a function of relative surface potential.

\[\Delta v = c(\frac{\Delta V}{\Phi})\]

Note \(\Phi\) is the surface potential of a proton, which in GPT is a constant like the speed of light. I could also write it like this as the ratio of two constants:

\[\Delta v = \frac{c}{\Phi} * \Delta V\]
Christopher F wrote:So, if we were then to hold a plate above the apple, and allow it to have a lower potential in respect to ground. Would the apple then, if disconnected from the branch, move towards the plate instead of the ground?
You are contradicting GPT here. "..holding a plate above the apple.." implies the plate is not moving with respect to the apple its surface potential must therefore be the same as the apple. The GPT equation I just wrote above says there is a rigid linear relation between velocity and potential, therefore if you are hoding the plate still its at the same potential.
Christopher F wrote:I thought somewhere in your paper you mentioned not to think of "Potential" as electrical potential, or did I imagine that?
Yes, essentially it is electrical potential, but the electric potential energy \(E=mc^2\) is stored within the atom between the electron and the proton, so when you set a body in motion or elevate it in a feld you are changing the radius of every atom in the body and by doing so changing the stored energy and therefore the surface potential.
Christopher F wrote:If a person could walk into a room where the ceiling was a plate with a negative potential in respect to ground (without any arcing, let's pretend it's safe) would we be able to measure them weighing less or floating towards the ceiling?
Well you might see the persons hair stand up.
Christopher F wrote:Perhaps I've completely misunderstood in which case my questions probably seem silly. If so, I do apologize.
Not in the slightest, questions often help me fine tune the theory.
Christopher F wrote:What about the huge negative charge clouds build up before lighting strikes? What effect does that have on "Gravity"
Always remember, GPT says ground potential is 930,000,000 Volts so when you say huge charge build up in the clouds, it isn't huge at all, maybe half a million volts at the most 0.5/930 = 0.00053 not very huge at all. Same reason why you dont stick to the ceiling if it is charged to a few thousand volts.

We have lived our entire life at 930 million volts, and just like birds on a wire we don't really notice it.

Feel free to ask more questions if there is something you don't understand, it's the reason I set up the forum.


Re: Electric Potential Replaces Gravity

Posted: Fri Dec 04, 2015 7:42 am
by Steven Sesselmann
Christopher F wrote:Thank you for your reply, that has helped my understanding a little bit. I'm still trying to understand completely though.
Me to..
Christopher F wrote: So for two bodies to have a difference in potential they need to have some velocity in respect to each other?
We have to trust the GPT equation unless we can prove it wrong.
Christopher F wrote:Does an object only have potential in respect to ground if it is moving then (Falling?)
Once again I can't see any wiggle room in the equation it clearly says ∆v = ∆ø, so what is potential? it is a just a fancy word for energy density, and two objects with the same energy density are inertial with respect to each other. If you then go and transfer some more energy to one object or remove some energy from the other object you create motion (this sounds almost Newtonian).
Christopher F wrote:So in the example of the apple on the branch, while attached to the branch the Apple would be at the same potential as ground, because it is being held above ground in a constant position?
Good question, and I had to drink a cup of coffee before answering this question. Newton's coffee probably wasn't as strong as mine, because it suddenly became obvious to me what's going on.

When the apple becomes detached from the branch there is no change whatsoever in the apple's potential, let's face it we haven't even touched it, it just dropped. So we must assume the apples energy density/potential hasn't changed.

How about the earth?

Yes, the earth is a ball of decaying matter, the nuclear decay processes, Uranium, Thorium, Potassium and all those elements are constantly emitting gamma rays and falling to lower potentials, this is happening at a steady rate and is also the main cause of heat inside the earth.

So once again GPT turns physics upside down, it's not the apple falling off the branch which is changing potential it is the Earth, this also explains why all objects fall with the same velocity,

This definition of gravity completely eliminates the need for any field to be present, so please ignore any reference I might have made to fields in the past. It is obvious to me now that a field is just another "fairy" something we invented in order to explain something we didn't understand.
Christopher F wrote:or am I still misunderstanding? I know you mentioned potential increase as elevation increases. Didn't you say there should be a gradient in potential if we measured at the base and top of a mountain? (has this ever been done?) If so, I'm a little bit confused. If this is the case, how come when the plate is raised above the Apple it has the same potential as the Apple? Also, do you mean that they have the same potential in respect to each other? (Does relative potential between two object increase with distance?) or that they have the same potential in respect to ground?
Yes electrical potential does increase with height, but it is difficult to measure and there isn't very much literature on it. Late american president Benjamin Franklin was actually interested in the phenomenon and conducted experiments to measure it.

But remember if something doesn't have relative movement then it doesn't have a different potential, so as you go higher up above the earth one would expect the air molecules to have higher velocities, and as we know the density of air is lower it would be reasonable to assume air molecules moving faster. Just think of the jet stream in the upper atmosphere.
Christopher F wrote:Also, probably should be in a different thread, but I'm running late!, what becomes of the Positron in your theory? (Since the proton and electron are now particle pears)
Another good question. The best definition of matter is a wave, the Hydrogen atom is a wave with energy 938 MeV, observed asymmetrically by us here on earth. It just happens that the electron has a mass of 511keV/c^2 at the this point in time.

Clearly there is something special about Hydrogen because there is so much of it, and I suspect we will find some underlying law of nature which explains this soon.

That said, we know it is possible to generate electromagnetic waves with any amplitude (remember amplitude is potential). Most of our electromagnetic wave generating equipment is designed with one pole at ground potential (light, radio waves, x-rays etc) so these waves have zero surface potential, the trough and the peak have equal and opposite potentials and rotate so the net surface potential is zero, so when you apply the GPT equation to a body with zero potential it must travel at the local speed of light.

\[\Delta v = c (\frac{930-0}{938})\]

As you can see the local speed of light on earth is slightly less than the absolute speed of light in vacuum.

We have also learned how to make asymmetric waves by smashing atoms together, and 99% of the time these collisions produce short lived particles, and scientists have fun giving silly names to them.

Occasionally and with 1022 keV or more we can produce a pair, where one of the particles happen to have the same energy as the electron with an anti particle of the same mass. For all intents and purposes it's identical to an electron, but it's not a constituent of Hydrogen, it's a constituent of positronium which is just another short lived particle.

Think of it as a half inch long bolt with a half inch nut, vs a ten inch long bolt with a half inch nut, both have the same nut.

Hope I haven't confused you even more.


Re: Electric Potential Replaces Gravity

Posted: Sat Dec 05, 2015 9:12 am
by Steven Sesselmann
I am still not 100% happy that we understand the apple problem, yes the earth's potential is falling, and a quick back of an envelope calculation puts the decay rate at around 0.5 millivolt per year. I calculate this based on the difference between the proton potential and ground potential (around 8 million volts) divided by the current estimated age of the Universe and it works out to around 0.5 mV per year. Somehow I don't think it is a linear decay, I actually think the decay is speeding up, so the current rate of fall may well be more than that.

Yet it is not immediately obvious how this rate of decay can explain the falling apple.

An apple in a very tall apple tree is hanging on a branch 10 m above ground, and as it has zero relative velocity to the ground GPT says it must have the same potential as ground.

At some point the apple becomes detached from the branch and at that very instant it has zero relative movement to the ground so we conclude it still has the same potential as ground.

With the passage of time the relative velocity between the apple and ground increases and after one second the relative velocity is 10 m/s

\[\frac{v}{c} = \frac{\Delta V}{\Phi}\]

\(\Delta\phi\) works out close enough to 30 volts
Surface potential shells
shells.png (63.3 KiB) Viewed 9293 times
The dilemma to be solved here is, nothing is in contact or interacts with the apple after it is detached from the branch, and as far as we know nothing physically interacts with the earth either.

So what then causes a potential difference of 30 Volts between the apple's potential and ground potential after 1 second?

In the image above we can see there are virtual surfaces above ground where the energy contained within such a virtual surface is constant. this would give rise to a \(1/r^2\) rule.

This problem needs more thought...

Ideas and opinions welcome..


Re: Electric Potential Replaces Gravity

Posted: Fri Dec 11, 2015 7:01 pm
by Steven Sesselmann
Christopher F wrote:So if I think about motion being caused by difference in potential. When the apple is held above ground via the branch on the tree it is physically connected to ground, and therefore shares the same potential. Since it shares the same potential as ground, like your equation says, it cannot be in motion, and it isn't as it is held in a fixed position due to the tree. (Ignoring wind and other external forces) When the apple becomes disconnected from the branch it is no longer tied to ground, and therefore no longer tied to ground potential.
Yes that's right.
Christopher F wrote:..this immediately set me down the following train of thought. The apple stays at whatever potential it was at when it was connected to ground, but ground potential is decaying, therefore there is a potential difference, therefore motion. The problem here is that the rate of decay you predicted seems to be much to small for the 30V potential difference you predicted. Am I completely off track?
So far so good..
Christopher F wrote:The other thought I then had was. (When you say decay, ground potential is decreasing right?) What if when the apple becomes disconnected from the branch, it loses it's potential. I say this thinking about applying the same bias voltage between two plates. If both plates have the same charge we wouldn't be able to measure a potential difference between them.. for our purposes we could say that the two plates represent ground and earth and are connected to the same electrode via wires. (Relating the wire to the branch of the apple) If we then cut the wire to one of the plates it would no longer have the same charge as the plate still connected to the electrode. We should now be able to measure some difference in potential.., right?
I have also considered the possibility of the apple changing it's potential, but the issue there is how? Once detached from a branch the apple does not physically interact with the earth, we know gravity works just as well in vacuum, so air is not a factor either, yes there may be some nuclear decay going on inside the apple but if this was the cause of change then we should be able to measure a different rate of fall between radioactive and non radioactive objects, and I am not aware of any such difference, at least not large enough to measure.

The earth on the other hand is decaying at a measurable rate, I make gamma spectrometers and I can measure between 10-30 µSv background anywhere, with some spots measurably higher, and who knows what the radiation is like a few km's down below the crust. Ultimately this energy radiates away into space, that said, ground potential isn't falling at 30 Volts per second, else it would all be over in a year. Could a much smaller rate of decay, say in the order of 0.5 to 1 mV cause a 30 V gradient over a distance of 10 meters, especially considering how space is a poor electrical conductor as in the permittivity of free space \(\epsilon_0\).
Christopher F wrote:In the example of the apple and the tree. The apple would be the plate whose wire gets cut, whereas the ground is the plate always connected to the electrode (The earth I guess). Once the branch breaks (The wire being cut) the apple loses its potential. The ground now has a huge positive potential in respect to the apple's new potential, therefore motion.?
If the apple takes on it's new potential immediately once the stem severs, GPT would demand instant relative velocity, not the accelerating motion we observe, so it seems reasonable that the apple starts off at ground potential with no relative velocity.

We also know from the days of Galileo, objects of different mass and composition fall at the same rate.

Obviously time is not governed exclusively by the decay of the earth, every star in the visible universe is decaying, so the rate of change is a constant throughout all matter.

Just for my own amusement I tried multiplying the permittivity of free space (a very small number) by -3 volts per meter, which gave me a result in electric surface charge density, maybe this method is heading down the right track.
Permittivity of free space * -3 V/m
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Re: Electric Potential Replaces Gravity

Posted: Sat Dec 12, 2015 2:40 pm
by Steven Sesselmann
Christopher F wrote:Just to make sure I understand how you got 30V: (Electric Potential)=((G)(T)(Phi))/(C) Where G is acceleration due to Gravity, T is 1 second, PHI is surface potential of Proton, and C is speed of light?
That's correct, note you can use LaTex on this site it is easy you just have to backslash bracket to start and end expressions.
Square bracket for centred expressions
centered.png (11.11 KiB) Viewed 9256 times
You can use normal round brackets for expressions inline with text, \(\phi_e=gt/\Phi\) , like this. Right click on any equation to see the code.
Christopher F wrote:What if the Apple changes potential because it has a mechanism similar to capacitance? (Maybe the Apple could be one plate, the Ground the other, and the air the dielectric) Then the Apples changing potential could be related to a capacitor discharging. The time it takes to discharge could also explain the gradual increase in acceleration. Perhaps the equation for discharging a capacitor could be used to figure out what the capacitance is?
The problem with this suggestion is we should see different acceleration for matter falling through different gases and more problematic, objects would fall slower in vacuum, and we know the opposite is true.
Christopher F wrote: (Capacitance)=(-T)/((R)(LN(Vfinal/Vinitial)) Where T would be 1 second, Vfinal 30V, Vinitial 980MV, R could maybe be defined as: (R)=(((Q)(C))/((G(T)(PHI))(4PI(permittivity of free space)) Where G is acceleration due to Gravity, T is Time, PHI is the surface Potential of Proton, Q is Charge, and C is the speed of light..., I'm not sure how the value of Q would be determined in this scenario though..,Just a thought.

Edit: Realized I was referring to Surface Potential of Proton, as just Surface Potential
Edit again: On second thought, in the scenario I described above I think Vinitial should be zero, because there is no potential difference between the apple and the ground to begin with. In which case, I guess the equation wouldn't work.., I'm still curious about if the apple could have a similar mechanism to capacitance.

I'm especially curious because you mention the permittivity of free space. There are several capacitance equations that use permittivity of dielectric, so..., I don't know.., there is a lot of food for thought on this forum.
I couldn't quite follow your equations here, the difference between \(V_i\) and \(V_f\) is \(\Delta V\) is just 30V, unless you want to work in absolute units. In this case ground potential is say 930,000,000 Volts and potential at 10 meter elevation would be 930,000,030 Volts.

In this case the velocity fraction would be \(\frac{930000000V-930000030V}{938000000V}*c = 9.6\) m/s

I agree it would not be surprising to see \(\epsilon_0\) turn up in the equations, especially since we are working in terms of electrical potential and space. The low permittivity of space is what prevents time running away at an uncontrollable rate, the low permittivity of space prevents the sun blowing up like a nuclear bomb. The sun burns it's nuclear fuel at the maximum rate space can carry the energy away.

Knowing the relative velocity between earth's surface and the surface of the sun, we can easily work out the surface potential of the sun.

\[\Delta\phi = \Phi(\frac{\Delta V}{c})\]
\[\Delta\phi = 938 MV(\frac{-29,800 mps}{c})\]
\[\Delta\phi = -93,239 V\]
Assume somewhat more than that as we should take the rotation into account as well, so 100 kV is a pretty good estimate. I don't think it is surprising to find the sun's surface is negatively charged, after all it is a fusion reactor suspended in empty space and as such it is shooting out positively charged alpha particles. The positive particles moving isotropically out from the surface will render the surface more and more negative until the charged particles just lack enough velocity to climb out of the well.

Let's give it some more thought...

Re: Electric Potential Replaces Gravity

Posted: Mon Dec 14, 2015 1:53 pm
by Steven Sesselmann
After afew days of deep thought, the solution seems most likely to be a geometric one.

I start off with a square hydrogen atom like this, it's just a simplistic way to represent the atom (sperical cow principle).
Square Hydrogen
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What you see in the representation above is an electron bound to the heavier proton in a mutual orbit around an imaginary center of rotation which we shall call ground potential.

The relative velocities for the electron and the proton with respect to an observer at rest is defined by GPT as;
\[ v_e = c(\frac{930.377-0.511MV}{938.272 MV})=2.971 * 10^8 mps\]
\[ v_p = c(\frac{930.377-938.272MV}{938.272 MV})= -2.523 * 10^6mps\]
As we see the electron velocity is a fast prograde motion and the proton velocity is a much slower retrograde motion.

These velocities must hold for all bodies at rest with respect to the observer, but not for a body in motion or in free fall.

Let us look at the following example of two bodies separated by some distance (click on image to expand). Let us imagine the atom (body-1) being inside an apple which has just become detached from a tree branch, and the other atom (body-2) being inside the observer at ground potential.
stretch.png (46.57 KiB) Viewed 8595 times
As the electron and proton in body-1 make a full rotation we see the electron is approaching the observer and the proton is receding, as a result, in order to satisfy the velocity function, the electron must take a slightly longer path, and the proton must take a slightly shorter path, these two inevitable path changes cause the centre of rotation (pivot) to move towards the proton resulting in a higher overall potential, this change in potential causes the relative motion.

Repeated orbits result in incremental changes to the apple's potential, which in turn results in acceleration of free fall. A falling apple therefore gains potential energy while falling to the ground, but instantly gives it up as it comes to rest on the ground.

In principle all free falling bodies are attracted towards you the observer, unless their potential exceeds some limit such that it escapes. We know from classical physics this orbital velocity is \(\sqrt{Gm/r}\) and escape velocity is \(\sqrt{2Gm/r}\), but now we need to find and explain these limits in terms of GPT.

Re: Electric Potential Replaces Gravity

Posted: Thu Dec 17, 2015 1:05 pm
by Steven Sesselmann

Sadly not much about physics is painfully obvious, I hope I will be able to formulate an excact equation for this effect similar to Maxwell's equations, for I believe the proof is very similar to that of Maxwells.

To understand the hydrogen atom we first we need to understand that we are observing it from GP, this effectively means the observer sits between the electron and the proton (obviously not physically but on the potential axis).

Accepting this, we also know that the ultimate faith of the electron and proton is to come back together and annihilate, which means both particles are moving in helical paths towards the observer at a velocity determined by GP.

So when an apple is attached to the branch it's atoms are held at ground potential forcing the charged particles to hold a given radius and velocity, but as soon as the stem breaks, the electron which has direction and momentum from it's last contact with GP, must now take a longer path and the proton which spins the opposite way must take a shorter path.

The consequence is a gradual increase in potential with respect to the observer.

Why does the electron take a longer path?

It does so because it has to retain it's momentum so whenever it moves towards GP it takes a longer path due special relativity and likewise when it moves away from GP it needs to take a shorter path. The consequense is an increase in the electrons orbital radius and a decrease in the protons orbital radius, which simply means higher potential between the two.

The confusing part here is that nothing has physically changed in the apple, all this is happening with respect to the observer. Only when the apple comes to a sudden stop on the ground will there be a measurable change to the apple.
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