Re: How an electrostatic gravity is possible

On Sep 18, 1:23 am, frankli...@xxxxxxxxx wrote:
On Sep 17, 10:33 am, Igor <thoov...@xxxxxxxxxx> wrote:





On Sep 17, 2:02 am, frankli...@xxxxxxxxx wrote:

I have been trying to figure out how to make a gravity force out of
the electrostatic force. Based on previous discussions, if we examine
the forced between 2 separated hydrogen atoms, the force between them
can be easily calulated as the sum of the forced between the separate
pairs of electrons and protons. The force is simply:

(Proton/Proton) + (Electron/Electron) - (Proton/Electron) - (Proton/
Electron)

So we have 2 pair that repel and 2 pair that attract. Normally, we
consider the distances and the charge forces to be equal, so that the
attractive and repelling forces exactly cancel, leaving no force
between the two hydrogen atoms.

But this is a massive simplification of the system because we know
that several differences exist between the electron and proton. The
proton is 1,836 times as massive as the electron. An electrostatic
solution to gravity could exist if this difference could be leveraged
to create an imbalance of forces between the particles of two hydrogen
atoms.

Such a difference could exist if the force generated by the (Proton/
Electron) opposites pair was slightly greater than the force generated
by similar charges. The math would work out:

Sum of similar charges
- Sum of disimilar charges
_____________________________
= Force of gravity?

So the question is, how could you justify a difference in force for
dissimilar charges? If such a difference could be justified, you could
find a force generated between neutrally charged matter that could
account for gravity on a purely electrostatic basis.

To try to justify this position, I made the presumption that since a
proton is so much more massive than an electron, it must have a larger
effective radius. I also presume that the charge of a proton is spread
out over this larger radius. Neither of these assumptions are terribly
unreasonable.

To represent this situation, I drew a simple picture of a 2 hydrogen
atom system. The electron was represented by a dot on the page, while
the proton was represented as a 3.25" circle surrounding the electron
dot. The circle represents the much larger radius of a proton. I drew
a similar electon/proton 6 inches away from the first electron. So now
we have 2 dots on 6 inch centers surrounded by 2 3.25" circles (get
the picture?)

To represent the fact that the charge is spread out over the
circumference of the circle, I drew 6 equally spaced dots on the
circumference of the circle to represent charge points and I assigned
them a charge value of 1/6 the value of the electron. This represents
the situation that the charge is spread out in the proton. Then I
simply took my ruler and measured the distances between the points and
completely calculated the force between the points. Force is
calculated using Coulomb's law F = KQ1Q2/R^2. For simplicity, I have
set K=1 and electron/proton charge = 1. Thus, each of the proton dots
have 1/6 the charge of the electron.

Between an electron and the 6 positive dots

Distance Force (1/6)/R^2
5.25 0.006046863
7 0.003401361
4.25 0.00922722
7.75 0.002774887
7 0.003401361
5.25 0.006046863

Net = 0.030898555

The measurement of the forces on the other electron to the other
proton is identical, so the total attractive force is double which is;

Total attractive force between (Proton/Electron) X 2 = 0.06179711

Next, I measured the repelling force between just the 2 electrons F =
1/R^2.

6 0.027777778

Repelling force between electrons = 0.027777778

The hard part was measuring the repelling force between all of the
positive dots which required measuring every possible combination
between the two protons and summing them: F = (1/6)*(1/6)/R^2

2.75 0.003673095
3.75 0.001975309
5.25 0.001007811
6 0.000771605
5.25 0.001007811
3.75 0.001975309
3.75 0.001975309
5 0.001111111
6.5 0.000657462
6.75 0.000609663
6 0.000771605
4.25 0.00153787
5.5 0.000918274
6.75 0.000609663
8.25 0.000408122
8.75 0.000362812
8 0.000434028
6 0.000771605
6 0.000771605
7 0.000566893
8.75 0.000362812
9.29 0.000321859
8.75 0.000362812
7 0.000566893
5.5 0.000918274
6.25 0.000711111
8 0.000434028
8.75 0.000362812
8.5 0.000384468
6.75 0.000609663
3.75 0.001975309
4.5 0.001371742
6.25 0.000711111
7 0.000566893
7 0.000566893
5.5 0.000918274

Net = 0.033061913

Now if you add up all the repelling forces, you get:

0.060839691

You will find that this is significantly less than the attractive
forces which is:

-0.06179711

This leaves an attractive force:

-0.000957419

Such an attractive force could account for the force of gravity
between neutrally charged objects and would be directly due to the
fact that the charge is more diffuse on a proton than an electron.

Now, this is hardly a proof that such a thing is possible. This is a
very crude calculation and protons aren't circles made up of 6 charge
centers. However, it does open the possibility that it is possible for
gravity to have an electrostatic origin and this is one possible way
of explaing how it might happen.

fhugravity

Congratulations! You just discovered the basic principles behind the
Van der Waals interactions, something that has been known to
physicists for centuries. These are the forces responsible for
cohesion.

And if you check the archives of usenet, someone posts this same
general "discovery" about every other month.- Hide quoted text -

- Show quoted text -

No, not exactly - if you look up what Van der Waals interactions are,
these are based on momentary dipoles. The calulations I have made do
not rely on any such dipoles. It relies on the assumption that the
charges are fixed in space and perfectly symmetric (unlike Van der
Waals). I have hardly seen anyone trying to equate electrostatics to
gravity beside myself - search the usenet yourself.

I don't need to search. I see postings like yours all the time. And
they're dead wrong in so many ways. Come back in twenty years and if
you still think you have a viable idea, we'll talk.







.

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