by Miles Mathis
I get emails from spotty readers, asking me,
“Where is this charge field of yours?
I can’t seem to find it in the field equations or in the data.”
So, I will take this opportunity to rerun for you some of the sweetest math I have ever done.
This will show you where the charge field is hidden, both in the math and the data.
These are the equations which, if I live long enough, I will write on some big blackboard with dramatic flourishes, in a grand auditorium (I would like to suggest the Vienna State Opera House), in front of a bank of cameras.
I would also like to suggest that the world not wait too long to figure this out, since it will be so much more cinematic if I have a full head of “scientific” hair when this happens.
The clock is definitely ticking.
But seriously, these two proofs of the charge field have been separated so far, buried in longish papers that few may be able to get through.
I have linked them, but that probably isn’t enough.
And so, I thought it best to put them here side by side, in a short paper that may be readable.
In short, this is how I developed the number .009545m/s2 for the charge field of the Earth, in two different papers, in different years, on different continents, working on separate problems, and with unrelated math.
I will try to insert flourishes, and if we are imagining a time too far in the future, you can imagine me in a wig.
Don’t forget to applaud at the end.
The first calculation came about after I had been staring at the Moon for 72 hours straight (not really, that was a flourish).
It occurred to me that perhaps Newton’s equation for gravity—the famous F=GMm/r2—was already a unified field equation.
If it was true, how could I tease the E/M field out of that equation, without going to a lot of trouble?
I am lazy that way.
I thought, why not just compare the current numbers for the Earth and the Moon, with the added assumption that gravity by itself was proportional to radius?
In other words, if we take E/M out of Newton’s equation, we are left with gravity by itself.
What is gravity by itself?
Well, it might be a force that was dependent only on size.
Acceleration is dependent only on distance, gravity is an acceleration, and radius is a distance.
So, if gravity is an acceleration, it should be dependent only on distance.
The mass dependence was always fishy, and we accepted it only because Newton told us to.
What I mean is, we know that a=v2/r isn’t dependent on mass, so why should F be dependent on mass?
You will say it is because F=ma.
There is the mass dependence.
But if F=ma is also a unified field equation, and also includes E/M, then my point remains.
Perhaps the m goes with the E/M field part of the equation, and it has nothing to do with gravity?
To test this, I just calculated the corrections we would have to make to the Earth and Moon to make it work.
If I were wrong, I would find out very quickly, because I wouldn’t be able to make the numbers work.
You can’t really push equations like this:
they either work or they don’t.
If I were right, then the gravity of the Moon should be:
gE / gM = 3.672 9.8 m/s2 / gM = 3.672 gM = 2.669 m/s2
That’s much higher than the current number, but if I could show that the E/M field of the Moon offset that, bringing that number back down to the current number, I would be in high cotton.
So, what I did is assume that the E/M field fell off by the fourth power.
Again, that assumption will either work or it won’t.
There is nothing in between.
You can’t push this kind of math, because the first assumption— that gravity increased with radius—gave me a definite gap to fill.
The number 2.669 is solid, and it can’t be moved.
My second assumption also will give me a definite number.
Either those two numbers offset, or they don’t.
I have postulated that the gravitational part of this acceleration should be proportional to the radii. gE / gM = 3.672 gM = .2723 gE
And I have just postulated that the E/M field is proportional to 1/r4. EE /EM = 1/3.6724 = .0055 EM = 181.81 EE
But that last equation is assuming that the Earth and Moon have the same density.
So, I must now correct for density.
DE /DM = 5.52/3.344 = 1.6507 = 1/.6057 EM = 110.12 EE
So, we just substitute:
.2723 gE – 110.12 EE = 1.62 m/s2 gE – EE = 9.8 m/s2 .2723gE – .2723EE = 2.6685 m/s2 [subtract the two equations] -109.85EE = -1.0485 m/s2 EE = .009545 m/s2 EM = 1.051 m/s2 gM – EM = 1.62 m/s2 gM = 2.671 m/s2
The proof is complete.
It worked.
And I have a number for the charge field of the Earth.
It’s pretty small, which explains why it has remained hidden in most cases.
Now, in the years after I wrote this paper, I discovered that this number could be found in other places.
For instance, it explained the gap in the Bohr magneton.
The Bohr magneton is just the calculation of the magnetism of the electron from field equations.
But the magnetic moment of the real electron is found to be a slightly different value.
In other words, the math didn’t match the data.
And, as it turns out, the gap between the math and the data was my number, .009545.
The mathematicians forget to include the charge of the Earth, since they didn’t know about it.
It was hidden inside the field equations.
Soon after that, I found the number in another place, this time with some new math that had never been done before.
In a paper on atmospheric pressure, I needed to calculate the total weight of the atmosphere as a fraction of g.
To do this, I assumed that the atmosphere was contained by the ionosphere above.
I let the ionosphere start at 50km.
Then I calculated the lateral freedom of the atmosphere compared to the vertical freedom.
If r is the radius of the Earth, and a is the radius of the atmosphere at 50 km, then we can find the third leg of the right triangle very easily.
It is 800 km, and we double it to find the full width:
1600 km.
This means that the atmosphere has 32 times {1600/50} more lateral freedom than vertical freedom.
But that is only in one plane.
If we make our diagram three dimensional, we must square the lateral freedom, giving us 322.
If we assume this allows the gas to express only 1/322 of its weight, then we find a downward field of about .009550 m/s2 {9.78/322}.
At first, I was astonished to find that number.
I didn’t expect it, I was just playing with an idea, as usual.
I expected the problem to be more difficult than that.
But of course, that number has to match my number, since the weight of the atmosphere down has to equal the charge force up.
Otherwise, the sky would fall!
This way we have equilibrium, and a sky that appears to weightless—because it doesn’t fall.
In this way, I had confirmation of my number from a problem that was unrelated to my first proof.
The Moon and the atmosphere are two separate problems, as I hope you will agree.
And there you have it, proof of the charge field, both in the math and in the data.
I used current data, and I used current math, just updating it with a little linking math of my own.
I hope that got through to you.
If it did, please call the Vienna State Opera House and make reservations.
I should think 2021 would be a good date.
I shall start perusing perukes.