Re: Anti-gravitational effects demonstrated using a Van De Graaf generator

In article <1170745219.820301.244910@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
franklinhu@xxxxxxxxx wrote:

On Feb 5, 6:06 pm, Mitchell Jones <mjo...@xxxxxxxxxxxxxx> wrote:
In article <epr12a$c2...@xxxxxxxxxxxxxxxxxxx>,

carlip-nos...@xxxxxxxxxxxxxxxxxxx wrote:
frankli...@xxxxxxxxx wrote:

[...]
I was looking for something like "Why gravity can't be the
electrostatic force", but I find almost nothing in the literature
or the web.

The simplest reason is this: in gravity, all masses attract, while
in electrostatics, like charges repel.

For example, the Earth ands the Moon attract, so in an electrostatic
model, they must have opposite charges.

***{That's seems rather too strong, don't you think? In the particular
electrostatic model you are discussing, they do, in fact, have opposite
charges; but if you intend to claim that must be so in all possible
electrostatic models of gravity, then I must disagree. For example,
suppose that the attraction between unlike charges is greater than the
repulsion between like charges, in the very slight amount necessary to
account for the observed strength of gravity. In such an electrostatic
model, the Earth and Luna (the Moon) attract electrostatically with
gravitational force, yet have the same charges--to wit: essentially
zero.

Let me elaborate a bit.

According to Coulomb's law, the magnitude of the electrostatic force is
equal to a constant of proportionality times the product of the charges,
divided by the square of the distance between their centers.

By such a rule, the electrostatic force between Earth and Luna would
have four components:

(1) If K is the Coulomb constant, e1 is the charge of the electrons in
the Earth, e2 is the charge of the electrons in Luna, and r is the
distance from the center of the Earth to the center of Luna, we have:

F1 = Ke1e2/r^2

(2) If p1 is the charge of the protons in the Earth and p2 is the charge
of the protons in Luna, we have:

F2 = Kp1p2/r^2

(3) If k is the ratio, slightly greater than 1, of the absolute
magnitude of the attraction between like charges to that of the
repulsion between unlike charges, then we have:

F3 = Kke1p2/r^2

(4) And, similarly, we have:

F4 = Kke2p1/r^2

Therefore the force of gravity, Fg, would be such that

Fg = F1 + F2 + F3 + F4

Fg = Ke1e2/r^2 + Kp1p2/r^2 + Kke1p2/r^2 + Kke2p1/r^2

If M is the mass of Earth, m is the mass of Luna, and <e> is the mass of
the electron, then:

e1 = -(M/1836)/<e> = -M/1836<e>

p1 = -e1 = M/1836<e>

e2 = -(m/1836)/<e> = -m/1836<e>

p2 = -e2 = m/1836<e>

Substitution into the last force equation, above, gives:

Fg = K(-M/1836<e>)(-m/1836<e>)/r^2

+ K(M/1836<e>)(m/1836<e>)/r^2

+ Kk(-M/1836<e>)(m/1836<e>)/r^2

+ Kk(-m/1836<e>)(M/1836<e>)/r^2

= [2K/(1836<e>)^2][Mm/r^2] - k[2K/(1836<e>)^2][Mm/r^2]

= [2K/(1836<e>)^2][1 - k][Mm/r^2]

For simplicity, let z = 1 - k, so that the above becomes

Fg = [2zK/(1836<e>)^2][Mm/r^2]

By Newton's law of gravitation we have

Fg = GMm/r^2

And so we conclude that

G = [2zK/(1836<e>)^2]

Of course, 1836<e> = <p>, the mass of the proton, in the notation I'm
using, so we obtain

G = 2zK/<p>^2

In SI units, G = 6.6742x10^-11 Nm^2/kg^2, K = 8.988x10^9 Nm^2/C^2, and
<e> = 9.109x10^-31 kg. Hence <p> = (1836)(9.109x10^-31) = 1.672x10^-27.
Since in this problem we want to distinguish between attractive and
repulsive forces, and since the attractive ones traditionally get a
negative sign, I'm going to treat the value of G as a negative number
when I plug it in below. (Mass and radius are clearly positive, so G is
the only palusible way to enforce that distinction.) And so we have:

z = (-6.6742x10^-11)/{2(8.988x10^9)/[(1.672x10^-27)]^2}

z = -1.038x10^-74

What the above argument accomplishes, hopefully, is to demonstrate the
theoretical possibility that gravity is electromagnetic in nature, and
arises out of Coulomb's law. That's not to say that it would be
convenient to do gravitational calculations that way, of course. Such a
demonstration would be useful in the same sense that it is useful to
note that the relativistic kinetic energy formula, Ek = m0[1/(1 -
v^2/c^2)^.5 -1]c^2, could in principle be used to calculate the kinetic
energy involved in ordinary automobile collisions, despite the practical
inconvenience of doing so. The idea is to reveal a possible
connectedness that would in most circumstances remain hidden from view.

There are lots of counterarguments that can be directed at the above
idea, of course--so many, in fact, that I don't offer it as a necessary
truth, but merely as something worthy of consideration. I have been
thinking along these lines off and on for a long time, and there is
another theory of gravitation that I feel has more explanatory power
than this one, even though the issue between them is not settled by any
means.

What would be the point of trying to unify electromagnetics and
gravitation, after all? :-)

Any errors in the above are of course mine, but the credit for the basic
idea, for what it's worth, goes to Prof. Thomas Barnes. (See his book,
Space Medium: the key to unified physics, published in 1986.)

--Mitchell Jones}***





The Earth and the Sun
attract, so they, too, must have opposite charges. But that would
make the charge of the Moon the same sign as the charge of the Sun,
so they would repel. This does not happen -- the Moon's orbit
very clearly shows that it is attracted by both the Earth and the
Sun.

(Or for a more dramatic example, what "charge" would you ascribe to
the Apollo 11 lunar lander? It was attracted to the Earth -- they
didn't have to scrape it off the ceiling during assembly -- so it
must have had the opposite charge from the Earth. But the Moon is
also attracted to the Earth, and must also have the opposite charge
from the Earth. Why, then, wasn't the lunar lander repelled from
the Moon?)

Steve Carlip

*****************************************************************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ- Hide quoted text -

- Show quoted text -

Excellent work!

To restate what you are saying - the only thing you really need to
have a electrostatic based gravity is a tiny difference in strength
between the attraction of like opposite charges versus the repulsion
of similar charges. (I think you had this worded backwards in your
post

***{That's the sort of thing that could easily happen. I don't spend a
lot of time editing before I post, since this is, to put it mildly, an
informal group. However, when I glanced back over the material, I
didn't notice any instance of the sort you described. --MJ}***

). If the ratio is something like k = .038x10^-74, we can account
for the gravitational constant found in the formula Fg = GMm/r^2 which
in turn accounts for the Newtonian Orbit equation of v^2R =GM. That
ratio is exceedingly small and we have done experiments to determine
this type of ratio - is this below our current threshold of
sensitivity?

***{It's not really an experimental thing, once the other constants (G
and K) have been determined. All that is required is the basic idea--to
wit: that the attraction of unlike charges may slightly exceed the
repulsion of like charges. If it does, then a calculation along the
lines that I attempted in my post will reveal the value of z, which I
would call the unification constant. Thomas Barnes, as I noted, is the
first person, to my knowledge, who recognized the possibility that the
attractive force might slightly exceed the repulsive one. The book in
which he mentioned that idea, however, was perceived to have other
problems--i.e., he was an aether theorist--and so his book "fell
deadborn from the press," as they used to say.

That happened because while all modern physicists admit that a medium
exists which pervades all of space, few are aware that was all the term
"aether" ever meant, back in the days when it was in general usage. Of
those who are aware, the honest ones are like Barnes: they still use the
term today; and the dishonest ones rail at them, because they resent the
fact that they refuse to conform.

Why are they resented? Because physics is like a fraternity in that it
has initiation rites, and the purpose of the rites is exactly the same
in both cases: you have to prove that fitting in is more important to
you than the truth, in order to get in. To get into a fraternity you
might, for example, have to risk choking to death or becoming infested
with some horrible parasite by swallowing a pound of raw liver. And in
physics you will, in fact, have to swallow the nonexistence of the
aether, the relativity of simultaneity, things that magically pop into
and out of existence, and a lengthy parade of other ridiculous nonsense,
in order to fit in. The idea, in both cases, is for you to demonstrate
that you will do as you are told, that you will go along in order to get
along--which means: that you will take your cues from on high, and
support the goals of those in authority, as defined by the dominant
culture (even if those ideas will lead inexorably to the destruction of
your country and the enslavement of mankind, as, in fact, they will).

As far as the mathematical derivation that I posted, that was mine,
including any mistakes it may have contained. It has been 20 years since
I looked at Barnes' book, but I have no recollection of anything like
that being there. (Looking on Amazon.com, I see that several copies of
Barnes' book are currently available, if anyone would care to check me
on this.)

I disagreed with him in lots of places, by the way. The main thing I
considered memorable about the book was his notion that the attractive
Coulomb force might be slightly stronger than the repulsive one. Don't
take my reference as an endorsement of all of his opinions.

--Mitchell Jones}***

So we can keep the 1/r^2 force relationship rather than thinking it
may be a force like dielectorphoresis which seems rather problematic.
I am willing to accept any reasonable hypothesis that allows the
gravitational force to be composed of electrostatic forces.

So what to the nay sayers of gravity=electrostatic have to say now?
You cannot so easily toss this idea into the trash bin without some
real analysis. The difference in the force strength was the only
counter argument presented, and now even this can be explained. You
are tossing out the gravity=electrostatic possibility just like people
use to throw out the possibility that the Earth wasn't flat. By
looking around, it is obvious that the Earth is flat. Don't fall for
the obvious explanations without examining the alternatives.

***{You seem to be implying, in the above, that the original idea you
posted wasn't wrong. But, of course, it was. Steve Carlip's post
dissected your idea quite convincingly, and nothing in what I said was
intended to dispute that. I merely thought that he had stated his
argument in too general a form, that's all. --MJ}***

I think that the simplification and unification of the forces that you
would get with a gravity=electrostatic force pictures is so
appealling, that I think it has just got to be true.

***{It is a very powerful idea, but there are other theories of gravity
that have more explanatory power. The simplest theory that explains a
set of facts trumps more complex theories that explain the same facts,
but it does not necessarily trump more complex theories that explain
more facts. Thus in my mind this is a tough question, and still not
completely decided. --MJ}***

*****************************************************************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ
.

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