Is gravity the same as the electrostatic force?
FrankH
electrostatic charge because the protons are fixed and the electrons
can move far enough away from the nucleus to allow the positive
nucleus to be unshielded some of the time. If this is true, then
couldn't gravity just be this residual positive electrostatic force?
All of the tiny positive forces of all the atoms in the earth add up
to a big gravity force (which is nothing more than a positive
electrostatic force). What is the evidence that gravity and the
electrostatic force are different? The force equations are similar
except in magnitude. What else is similar or different?
Joseph Legris
I used to think this too. Dick Tracy comics had Diet Smith's
electromagnetic space coupe that somehow repelled gravity.
The shielding of the nucleus by electrons may be significant at very
close range (atomic distances) but out here in the macroscopic world the
electrostatic forces all cancel because neutral atoms have a net charge
of zero.
--
Joe Legris
Timo Nieminen
> It has been suggested that all matter has a slight positive
> electrostatic charge because the protons are fixed and the electrons
> can move far enough away from the nucleus to allow the positive
> nucleus to be unshielded some of the time. If this is true, then
> couldn't gravity just be this residual positive electrostatic force?
Consider the gravitional acceleration of a proton and an electron - the
acceleration is the same for each, and their masses are very different, so
the force on each is very different (proportional to the mass). The
magnitude of their charges are the same - can't be electrostatic.
Also one sees that neutrons, photons, etc are affected by gravity.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
FrankH
Perhaps the word 'shielding' is too strong. How do we know that
neutral atoms have a net charge of zero? Did we go out and measure it
experimentally, or did someone just say that a proton has this much
charge and an electron has this much charge and so they should cancel
each other out? I think the only way a electron and proton could fully
cancel themselves out all the time is if the proton and electron took
up exactly the same space and they didn't move. This would insure that
the fields would overlap perfectly to cause a zero net charge.
Otherwise, if the electron is allowed to move about the proton, you
would see that for any particular point in space, the charge was not
zero because the electron was either closer or farther away than the
proton. If the electron got stripped away from the proton, then you
would definitely see a positive charge, and I would think this would
have to happen in a sufficient quantity of normal earth-like matter.
So for a large quantity of so-called neutral atoms, the overall charge
an any particular volume of space may still be zero, but this doesn't
automatically mean zero fields eminating from this space becuase the
electrons still move relative to the fixed protons. Or perhaps most
normal matter is missing some electrons. They are not tightly bound on
the outer shells, leaving enough atoms to produce a net positive
field.
FrankH
> On Sat, 25 Jul 2003, FrankH wrote:
>
> > It has been suggested that all matter has a slight positive
> > electrostatic charge because the protons are fixed and the electrons
> > can move far enough away from the nucleus to allow the positive
> > nucleus to be unshielded some of the time. If this is true, then
> > couldn't gravity just be this residual positive electrostatic force?
>
> Consider the gravitional acceleration of a proton and an electron - the
> acceleration is the same for each, and their masses are very different, so
> the force on each is very different (proportional to the mass). The
> magnitude of their charges are the same - can't be electrostatic.
>
> Also one sees that neutrons, photons, etc are affected by gravity.
Have we done experiments to verify that the gravitaional acceleration
is the same for a proton and an electron, or did you just assume that.
Experiments done at Stanford suggest that electrons are not effected
by gravity. No acceleration in a gravitational field was found in the
experiment. This is a weird result which may have been caused by other
factors, but my point is that you may be making assumptions about
things we haven't verified experimentally. Neutrons are also effected
by electrostatic fields since there appears to be evidence that they
have dielectric properties which can only be revealed by the
manipulation of electrostatic fields. Whether photons are affected by
gravity is questionable since we have only been able to measure this
effect around large sun-like bodies which obviously have a large
atmosphere around them that would cause light to bend through a normal
density-lensing effect. If you could show light bending around a large
airless asteroid, that might be convincing. Rather than using a proton
and electron as an example, we know that in an airless environment, a
hammer will drop as fast as a feather. It would be an interesting
experiment to show that hammer would be accelerated to a source of
electrostatic charge as fast as a feather in a similar airless
environment.
Timo Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.030726...@kolmogorov.physics.uq.edu.au>...
> > On Sat, 25 Jul 2003, FrankH wrote:
> >
> > > It has been suggested that all matter has a slight positive
> > > electrostatic charge because the protons are fixed and the electrons
> > > can move far enough away from the nucleus to allow the positive
> > > nucleus to be unshielded some of the time. If this is true, then
> > > couldn't gravity just be this residual positive electrostatic force?
> >
> > Consider the gravitional acceleration of a proton and an electron - the
> > acceleration is the same for each, and their masses are very different, so
> > the force on each is very different (proportional to the mass). The
> > magnitude of their charges are the same - can't be electrostatic.
> >
> > Also one sees that neutrons, photons, etc are affected by gravity.
>
> Have we done experiments to verify that the gravitaional acceleration
> is the same for a proton and an electron, or did you just assume that.
Don't know of any electron experiments off-hand, but the experiments some
years ago (late '80s?) which suggested that G might not be constant led to
a whole rash of gravity experiments, including some trying to measure any
difference between the gravitational acceleration of protons, neutrons,
and anti-protons (one theory predicted that anti-protons should accelerate
at 2g). No differences found.
To a reasonable degree of accuracy, this is easily tested. Compare the
weights of water and heavy water - do those extra neutrons make a
difference?
Millikan's famous oil-drop experiment also tells much about neutral and
charged matter in uniform fields. Various other EM levitation and trapping
experiments tell you exactly how large a field or field gradient is needed
to produce a force equal to the gravitational force.
Gravitational acceleration of neutral atoms and ions is the same, again
experimentally well-tested.
Finally, the strength of electrostatic forces between neutral atoms can
be, and has been, measured - do some reading about van der Waals forces.
> Neutrons are also effected
> by electrostatic fields since there appears to be evidence that they
> have dielectric properties which can only be revealed by the
> manipulation of electrostatic fields.
How large is the effect? Gravitational acceleration is the same as for a
proton.
> Whether photons are affected by
> gravity is questionable since we have only been able to measure this
> effect around large sun-like bodies which obviously have a large
> atmosphere around them that would cause light to bend through a normal
> density-lensing effect. If you could show light bending around a large
> airless asteroid, that might be convincing.
Not questionable at all. Multiple-frequency measurements, including RF
measurements as well as optical are very convincing.
> Rather than using a proton
> and electron as an example, we know that in an airless environment, a
> hammer will drop as fast as a feather. It would be an interesting
> experiment to show that hammer would be accelerated to a source of
> electrostatic charge as fast as a feather in a similar airless
> environment.
Try it and see. You don't even need an airless environment if you are
content to measure force rather than acceleration (no motion required, so
air resistance is not a problem). Check out Coulomb's measurements.
Richard Herring
<frank...@yahoo.com> writes
>It has been suggested that all matter has a slight positive
>electrostatic charge because the protons are fixed and the electrons
>can move far enough away from the nucleus to allow the positive
>nucleus to be unshielded some of the time. If this is true, then
>couldn't gravity just be this residual positive electrostatic force?
Wouldn't that be _repulsive_?
--
Richard Herring
Joseph.D.Warner
FrankH wrote:
> It has been suggested that all matter has a slight positive
> electrostatic charge because the protons are fixed and the electrons
> can move far enough away from the nucleus to allow the positive
> nucleus to be unshielded some of the time.
This effect is called the Van der Waal forces. It falls off as 1/r^6.
This is the wrong form for gravity; therefore, the answer to your
question that I deleted is no.
FrankH
If all matter had a positive charge, then you would think that
everything should repel each other. However, the positive charge I'm
talking about is tiny compared with the covalent bonds formed by
electron sharing between atoms. So neutral atoms left to themselves
may repel, but once they're stuck together, the positive charge still
remains, but isn't enough to break the covalent bonds and cause the
atoms to repel.
FrankH
gravity force to have the same equation form, I would think that you
would expect that particles with different masses would have exactly
the same acceleration under an electrostatic field. A=F/M and the
F=Force created by residual positive charges would be proportional to
M=Mass (the more mass, then more force) which would make
A=Acceleration a constant regardless of mass just like gravity.
Gravity and electrostaic interaction are so similar that some students
understand electrostatics by understanding how objects behave under
gravitational fields.
> Gravitational acceleration of neutral atoms and ions is the same, again
> experimentally well-tested.
>
> Finally, the strength of electrostatic forces between neutral atoms can
> be, and has been, measured - do some reading about van der Waals forces.
>
> > Neutrons are also effected
> > by electrostatic fields since there appears to be evidence that they
> > have dielectric properties which can only be revealed by the
> > manipulation of electrostatic fields.
>
> How large is the effect? Gravitational acceleration is the same as for a
> proton.
>
> > Whether photons are affected by
> > gravity is questionable since we have only been able to measure this
> > effect around large sun-like bodies which obviously have a large
> > atmosphere around them that would cause light to bend through a normal
> > density-lensing effect. If you could show light bending around a large
> > airless asteroid, that might be convincing.
>
> Not questionable at all. Multiple-frequency measurements, including RF
> measurements as well as optical are very convincing.
I would think that all electromagnetic radition would show some
lensing effect when travelling through an atmosphere. When RF or light
waves hit a dense atmosphere, it slows, and when it exits, it speeds
back up and exits at a different angle. This is why it important to
show the effect around an object which has no atmosphere to interfere
with the results.
>
> > Rather than using a proton
> > and electron as an example, we know that in an airless environment, a
> > hammer will drop as fast as a feather. It would be an interesting
> > experiment to show that hammer would be accelerated to a source of
> > electrostatic charge as fast as a feather in a similar airless
> > environment.
>
> Try it and see. You don't even need an airless environment if you are
> content to measure force rather than acceleration (no motion required, so
> air resistance is not a problem). Check out Coulomb's measurements.
Measuring force is not sufficient. It is the identical acceleration
that you're after and according to my new research, the hammer and the
feather should be accelerated at exactly the same rate towards a
charge source, just like gravity. So gravity is looking a lot like the
electrostatic force.
But if it is this simple, why hasn't anyone figured it out. There
still must be some big obvious objection that I don't understand.
FrankH
This isn't the Van der Waalh forces which has to do with the
attraction of dipoles within atoms. I am referring to just the
plain-old electrostatic force you get when charges within an atom are
not completely balanced. For example, you have 100 atoms of iron and
one of the atoms is missing an electron. The overall effect is that
the 100 atoms is slightly positively charged. Multiplied billions of
times, the bar of iron appears neutrally charged to the casual
observer, but actually has a slight positive charge. The reason why we
don't see a balancing negative charge is because it is the asymetry
between the proton and the electron whereby the proton is not allowed
to stray away from the atom.
Timo Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.030728...@kolmogorov.physics.uq.edu.au>...
> > On Sun, 27 Jul 2003, FrankH wrote:
> >
> In doing more research on what it means for the electrostatic and
> gravity force to have the same equation form, I would think that you
> would expect that particles with different masses would have exactly
> the same acceleration under an electrostatic field. A=F/M and the
> F=Force created by residual positive charges would be proportional to
> M=Mass (the more mass, then more force) which would make
> A=Acceleration a constant regardless of mass just like gravity.
If you want to develop a working theory of electromagnetic gravitation,
you need to explain many things:
Compare different isotopes. The charges are the same. Gravitational force
is different. Compare gravitational acceleration of different elements,
using isotopes of (approximately) the same mass.
Sure, you can say the electric/gravitational force acting on neutrons and
protons is (almost) the same, but then you have to try to explain why the
electric forces on them are so very different.
Next, you need to explain why gravity is attractive and not repulsive.
An electromagnetic explanation of gravity isn't a stupid idea; it's been
tried (some references below). It just hasn't succeeded. The search goes
on, in the modified form of trying to unite gravitational forces with
electromagnetic forces and weak and strong nuclear forces.
It's not a big mystery that electric and gravitational forces have similar
equations - Gauss's law and simple geometry.
> > > Whether photons are affected by
> > > gravity is questionable since we have only been able to measure this
> > > effect around large sun-like bodies which obviously have a large
> > > atmosphere around them that would cause light to bend through a normal
> > > density-lensing effect. If you could show light bending around a large
> > > airless asteroid, that might be convincing.
> >
> > Not questionable at all. Multiple-frequency measurements, including RF
> > measurements as well as optical are very convincing.
> I would think that all electromagnetic radition would show some
> lensing effect when travelling through an atmosphere. When RF or light
> waves hit a dense atmosphere, it slows, and when it exits, it speeds
> back up and exits at a different angle. This is why it important to
> show the effect around an object which has no atmosphere to interfere
> with the results.
Propagation of EM waves in low density plasma is well understood. The
frequency dependence is well-known. The effect of the atmosphere can be
separated from gravitational effects. Note that the atmosphere of the sun
at those distances is what we would call a "good vacuum" here on earth.
> > > Rather than using a proton
> > > and electron as an example, we know that in an airless environment, a
> > > hammer will drop as fast as a feather. It would be an interesting
> > > experiment to show that hammer would be accelerated to a source of
> > > electrostatic charge as fast as a feather in a similar airless
> > > environment.
> >
> > Try it and see. You don't even need an airless environment if you are
> > content to measure force rather than acceleration (no motion required, so
> > air resistance is not a problem). Check out Coulomb's measurements.
>
> Measuring force is not sufficient. It is the identical acceleration
> that you're after and according to my new research, the hammer and the
> feather should be accelerated at exactly the same rate towards a
> charge source, just like gravity. So gravity is looking a lot like the
> electrostatic force.
So try it and see. Get a good enough vacuum in a bell jar so that a lump
of steel and a feather fall at the same rate. Stick a nice big static
charge next to the jar, see what happens when you drop them then. If you
can attract the feather and not the lump of steel, it's a clear negative
result. If you always attract both or neither, it's a positive result,
limited by how well you can control the amount of charge (ie would some
charge in between the highest non-attracting charge, and the lowest
attracting-both charge attract the feather only?).
> But if it is this simple, why hasn't anyone figured it out. There
> still must be some big obvious objection that I don't understand.
It was a big thing, about a hundred years ago. Weber in the 1870s, Lorentz
and others in about 1900, did a lot of work on electromagnetic
explanations of gravity. While the idea that gravity is due to electrical
forces might be simple, the details were far from simple. If you're
interested, the best place to start might be the Collected Works of H. A.
Lorentz. IIRC, Lorentz gave up on the idea, and started looking at gravity
as a different force, but propagated through the same ether as EM forces,
also at the speed of light. The discovery of the electron, and other such
discoveries, pretty much killed off the electrical theories of gravity.
The big obvious objection is the experimental evidence that we have now,
some of which I listed in the previous post. Just the fact that ions and
neutral atoms accelerate at the same rate due to gravity is pretty
conclusive.
Phaedrus
"FrankH" <frank...@yahoo.com> wrote in message
news:46484c9f.03072...@posting.google.com...
Supposing that this were the case, then two separated bodies, say planets
would according to your idea contain net positive electrostatic charges,
this would then lead to repulsion. As we know the gravitational force is
attractive not repulsive. The should pretty conclusively consign your
attempts to argue this point to the scrap heap sadly.
GR even does away with the concept of a gravitational force and explains
gravitational attraction as caused by bodies following geodesics in
positively curved space-time. To get gravitational repulsion would require
negative space-time curvature, which would require anti-mass. Anti-matter
(anti-particles) still has mass not anti-mass.
Joe
FrankH
>
> Supposing that this were the case, then two separated bodies, say planets
> would according to your idea contain net positive electrostatic charges,
> this would then lead to repulsion. As we know the gravitational force is
> attractive not repulsive. The should pretty conclusively consign your
> attempts to argue this point to the scrap heap sadly.
>
> GR even does away with the concept of a gravitational force and explains
> gravitational attraction as caused by bodies following geodesics in
> positively curved space-time. To get gravitational repulsion would require
> negative space-time curvature, which would require anti-mass. Anti-matter
> (anti-particles) still has mass not anti-mass.
>
> Joe
Yes, there would be some repulsion due to the normal electrostatic
repulsion, but I neglected to say that the attraction is actually
caused by the attraction of neutral dilectric material in a diverging
electrostatic field. This is the same force that allows you to pick up
neutrally charged pieces of paper with a charged comb. The charge of
the comb causes charges to separate (become dilectric) in the
molecules of the paper. In a diverging field (it gets weaker as it
goes out) the dilectric is attracted to the area of greater field
strength. (See http://rognerud.com/physics/html/sec_4.html). A rock on
the ground, is neutrally charged and subject to this dilectric effect,
so it is strongly attracted to the Earth. The electrostaic replusion
is neglegible compared to the dilectric effect generated by the Earth.
On a larger scale, the Earth generates a large positive electrostatic
field, but doesn't actually contain the electron deficit you would
think would be associated with it. The mobile electron allows a tiny
positive field without an actual charge deficit. Since there is no
actual charge to repel against, the individual atoms in a planet-like
bodies appear as neutral objects, also subject to the dilectric
effect, so they (along with the entire planet) are attracted to other
large planets despite the same polarity electrostatic field.
Joseph Legris
*** WARNING ***
The above paragraph violates Gauss's law. If you continue with this line
of reasoning you risk being charged and confined by a closed surface.
--
Joe Legris
Borun Dev Chowdhury
plausible hypothesis but has it been proved by experimentation ?
Borun Dev Chowdhury
1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA
Phone 1-614-292-0067
Borun Dev Chowdhury
the effect of shileding given. You can see it cant work out if gravity
obey's inverse square. Besides, your whole argument falls apart since the
excess positive effect would be repulsive... unless you usually fly, you'd
know gravity is attractive.
Borun Dev Chowdhury
1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA
Phone 1-614-292-0067
Borun Dev Chowdhury
finally a word of wisdom, I thought I was the only one who saw this...
Borun Dev Chowdhury
1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA
Phone 1-614-292-0067
Borun Dev Chowdhury
I dont beleive this is Van der Wall effect since that has to do with
molecules and I beleive the original author is referring to atoms.
Borun Dev Chowdhury
1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA
Phone 1-614-292-0067
Borun Dev Chowdhury
what on earth is the matter with you man ? think a little wider, maybe two
atoms wont repel each other because of covalent bonds, but certainly two
macroscopic objects would which do not have a covalent bonding. And try to
punch in some numbers to your theory and give us some solid argument.
Borun Dev Chowdhury
1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA
Phone 1-614-292-0067
Timo Nieminen
> On Sat, 26 Jul 2003, Timo Nieminen wrote:
>
> > On Sat, 25 Jul 2003, FrankH wrote:
> >
> > > It has been suggested that all matter has a slight positive
> > > electrostatic charge because the protons are fixed and the electrons
> > > can move far enough away from the nucleus to allow the positive
> > > nucleus to be unshielded some of the time. If this is true, then
> > > couldn't gravity just be this residual positive electrostatic force?
> >
> > Consider the gravitional acceleration of a proton and an electron - the
> > acceleration is the same for each, and their masses are very different, so
> > the force on each is very different (proportional to the mass). The
> > magnitude of their charges are the same - can't be electrostatic.
> >
> > Also one sees that neutrons, photons, etc are affected by gravity.
>
[top-posting corrected]
> what do you mean one sees neutrons being affected by gravity ? It the most
> plausible hypothesis but has it been proved by experimentation ?
Gravitational deflection of neutron beams has been measured. Don't do that
kind of stuff myself, so no references at hand.
I don't know if anybody has done measurements on single neutrons just
being dropped. But at the most basic level, one needs only consider the
dependence of gravitational acceleration of atoms with differing numbers
of neutrons. For example, compare acceleration of hydrogen vs deuterium.
FrankH
> you would do good to look up Jackson chapter 1 problems. One of them has
> the effect of shileding given. You can see it cant work out if gravity
> obey's inverse square. Besides, your whole argument falls apart since the
> excess positive effect would be repulsive... unless you usually fly, you'd
> know gravity is attractive.
>
question, the excess positive effect would be attractive because of
the dilectric effect. A charged comb is capable of attracting a
neutral piece of paper. I have found it attracts dirt, rocks, metals,
plastics and just about anything you can find just like gravity. So
while the source of the positive electrostatic charge is up for grabs,
if it did exist, it would tend to pull everything down like a huge
comb.
FrankH
> The above paragraph violates Gauss's law. If you continue with this line
> of reasoning you risk being charged and confined by a closed surface.
Yep, what I am suggesting would violate Guass's law. I am suggesting
that the field from an area of enclosed space doesn't equal the sum of
the charges. But who says that Guass's Law is really the law? If you
take a single hydrogen atom and measured the force at any given point,
you'd find that the charge was not zero because either the proton or
the electron would be closer to the point you were measuring. The
effect I am looking for would put the proton closer to the outside of
a macro object. Think of 4 hydrogen atoms where all of the electrons
are facing inwards. The net charge is still zero, but anything around
the 4 atoms would see a net positive charge. The overall effect would
be to produce a positively charged electrostatic field from a
completely neutrally charged mass. The big problem I see to this
theory based on the posts so far is that protons should fly upward
away from the positive electrostatic gravity source, and they do not.
But otherwise, an electrostatic force would act the same as gravity.
FrankH
> On Tue, 28 Jul 2003, FrankH wrote:
>
>
> Compare different isotopes. The charges are the same. Gravitational force
> is different. Compare gravitational acceleration of different elements,
> using isotopes of (approximately) the same mass.
due to the dilectric effect which is the same effect that allows you
to pick up neutrally charged paper with a comb. Isotopes with more
neutrons would have a larger dilectric effect, and would therefore
weigh more.
>
> Sure, you can say the electric/gravitational force acting on neutrons and
> protons is (almost) the same, but then you have to try to explain why the
> electric forces on them are so very different.
>
You make a good point about protons and neutrons both having the same
G constant in your previous post. A theory like the one I proposed
would mean that protons should fly upward, electrons downward, and
neutrons downward (due to dilectric effect). But we do not see this
experimentally. This is the biggest hole I can see in the theory so
far. The only thing I can suggest is that maybe the experiments were
flawed.
> Next, you need to explain why gravity is attractive and not repulsive.
>
A sufficiently strong positive electrostatic gravity will attract
everything like a charged comb attracts everything. (See my other
posts on why this isn't a repulsive force). Any tiny positive
electrostatic charge generated by a rock I am holding in my hand is
overwhelmed by the dilectric force created by the trillions upon
trillions of atoms in the earth. So if the rock I'm holding is
repelling with the force of 100 protons, but is being dilectrically
attracted by zillions of Earth atoms, the Earth wins.
Timo Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> wrote:
> >
> > Compare different isotopes. The charges are the same. Gravitational force
> > is different. Compare gravitational acceleration of different elements,
> > using isotopes of (approximately) the same mass.
> The attraction is not due to difference in charges, the attraction is
> due to the dilectric effect which is the same effect that allows you
> to pick up neutrally charged paper with a comb.
Now, for this to result in a force, you need an electric field gradient.
For an electric field gradient, you need an electric field. This is the
whole point of comparing gravitational acceleration of ions with that of
neutral atoms.
Note that the polarisability of atoms is reasonably well-known, so you
could go ahead and calculate what the field gradient, and therefore the
mininum electric field, must be to provide an acceleration of g. At this
point, it might be an interesting exercise for you to run the numbers, and
see if the required fields and field gradients are reasonable, or
nonsensical.
The observation that ions accelerate at the same rate as neutral atoms
also gives the field. Is this electric field compatible with the field
gradient that gives the correct acceleration for neutral atoms? Again,
time to do some calculations.
> Isotopes with more
> neutrons would have a larger dilectric effect, and would therefore
> weigh more.
Why? The number of electrons is still the same. The charge on the nucleus
of the atom is still the same. Why should the polarisability be different?
Specifically, for the case of deuterium vs hydrogen, why should it be
double? Perhaps these polarisabilities have been measured; this is
something for you to search for.
> You make a good point about protons and neutrons both having the same
> G constant in your previous post. A theory like the one I proposed
> would mean that protons should fly upward, electrons downward, and
> neutrons downward (due to dilectric effect). But we do not see this
> experimentally. This is the biggest hole I can see in the theory so
> far. The only thing I can suggest is that maybe the experiments were
> flawed.
A more likely explanation is that the electromagnetic-gravity theory is
flawed.
> > Next, you need to explain why gravity is attractive and not repulsive.
> >
> A sufficiently strong positive electrostatic gravity will attract
> everything like a charged comb attracts everything. (See my other
> posts on why this isn't a repulsive force).
In which case, you need to explain why gravity is an inverse-square force.
The electric force due to polarisation between two neutral atoms is known
to be 1/r^6 (van der Waals forces). Why the difference?
Anyway, time for you to look up some polarisabilities for single atoms
(start with nice simple things first), and see if you can get sensible
results for gravitational acceleration of single atoms.
The other thing to note is that the dielectric polarisability of a
(macroscopic) uniform dielectric ellipsoid is a tensor, which gives rise
to torques acting to align them with the applied fields. How well do these
torques compare with torques due to tidal forces? Non-agreement here might
well be a complete non-issue, though, since the macroscopic dielectric
polarisability of such bodies is small enough so that extremely large
fields would be needed to account for the gravitational acceleration of,
say, a glass marble.
FrankH
> On Tue, 4 Aug 2003, FrankH wrote:
>
> > Timo Nieminen <ti...@physics.uq.edu.au> wrote:
> > >
> > > Compare different isotopes. The charges are the same. Gravitational force
> > > is different. Compare gravitational acceleration of different elements,
> > > using isotopes of (approximately) the same mass.
> > The attraction is not due to difference in charges, the attraction is
> > due to the dilectric effect which is the same effect that allows you
> > to pick up neutrally charged paper with a comb.
>
> Now, for this to result in a force, you need an electric field gradient.
> For an electric field gradient, you need an electric field. This is the
> whole point of comparing gravitational acceleration of ions with that of
> neutral atoms.
>
> Note that the polarisability of atoms is reasonably well-known, so you
> could go ahead and calculate what the field gradient, and therefore the
> mininum electric field, must be to provide an acceleration of g. At this
> point, it might be an interesting exercise for you to run the numbers, and
> see if the required fields and field gradients are reasonable, or
> nonsensical.
>
> The observation that ions accelerate at the same rate as neutral atoms
> also gives the field. Is this electric field compatible with the field
> gradient that gives the correct acceleration for neutral atoms? Again,
> time to do some calculations.
crazy after all? I don't have a degree in theoretical physics, so the
math is beyond me. I do know that the electrostatic force as measured
on the earth is quite high at around 100 Volts/meter. I understand
this is equivalent of saying that it takes 100 Joules to move a charge
of 1 columb 1 meter. A Joule is the amount of energy it takes to
accelerate 1 KG to 1 meter/second. If you had 1 columb of charge, it
would be accelerated towards the earth at 100 meters/second (I think).
Gravity is 9.8 m/s which is much less, but the force is due to
dilectric, not direct electrostatic effects, but there is certainly
lots of electrostatic force avaliable. Of course, my math could be
completely screwed up.
>
> > Isotopes with more
> > neutrons would have a larger dilectric effect, and would therefore
> > weigh more.
>
> Why? The number of electrons is still the same. The charge on the nucleus
> of the atom is still the same. Why should the polarisability be different?
> Specifically, for the case of deuterium vs hydrogen, why should it be
> double? Perhaps these polarisabilities have been measured; this is
> something for you to search for.
>
The number of electrons/protons doesn't make any difference. The atom
is effectively neutral. The dilectric force may act on the individual
neutrons, so if there are 2 of them, it pulls down on the entire atom
with twice the force than if there were only 1 neutron.
> > You make a good point about protons and neutrons both having the same
> > G constant in your previous post. A theory like the one I proposed
> > would mean that protons should fly upward, electrons downward, and
> > neutrons downward (due to dilectric effect). But we do not see this
> > experimentally. This is the biggest hole I can see in the theory so
> > far. The only thing I can suggest is that maybe the experiments were
> > flawed.
>
> A more likely explanation is that the electromagnetic-gravity theory is
> flawed.
Looking at the possible structure of protons being the combination of
quarks with partial charges, it may be that the proton exhibits
dilectric properties and it too may be attracted to sufficiently high
electrostatic field gradients. The dilectric force may be able to
overcome the repulsive positive force of a proton. Although I can't
help but think that positively charged objects should be lighter than
their neutral counterparts - any evidence for this?
>
> > > Next, you need to explain why gravity is attractive and not repulsive.
> > >
> > A sufficiently strong positive electrostatic gravity will attract
> > everything like a charged comb attracts everything. (See my other
> > posts on why this isn't a repulsive force).
>
> In which case, you need to explain why gravity is an inverse-square force.
> The electric force due to polarisation between two neutral atoms is known
> to be 1/r^6 (van der Waals forces). Why the difference?
The force isn't between two neutral atoms. It is between the
collection of all the atoms in the earth, and the atoms for anything
sitting on the earth. Naturally, the force generated by the earth
overwhelms any force for any atom sitting on the earth. The force is
purely electrostatic using the inverse-square law.
>
> Anyway, time for you to look up some polarisabilities for single atoms
> (start with nice simple things first), and see if you can get sensible
> results for gravitational acceleration of single atoms.
>
> The other thing to note is that the dielectric polarisability of a
> (macroscopic) uniform dielectric ellipsoid is a tensor, which gives rise
> to torques acting to align them with the applied fields. How well do these
> torques compare with torques due to tidal forces? Non-agreement here might
> well be a complete non-issue, though, since the macroscopic dielectric
> polarisability of such bodies is small enough so that extremely large
> fields would be needed to account for the gravitational acceleration of,
> say, a glass marble.
As I mentioned above, the electrostatic field measured at ground level
is huge at 100 V/M.
Timo Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> wrote:
> >
> > Now, for this to result in a force, you need an electric field gradient.
> > For an electric field gradient, you need an electric field. This is the
> > whole point of comparing gravitational acceleration of ions with that of
> > neutral atoms.
> >
> > Note that the polarisability of atoms is reasonably well-known, so you
> > could go ahead and calculate what the field gradient, and therefore the
> > mininum electric field, must be to provide an acceleration of g. At this
> > point, it might be an interesting exercise for you to run the numbers, and
> > see if the required fields and field gradients are reasonable, or
> > nonsensical.
> >
> > The observation that ions accelerate at the same rate as neutral atoms
> > also gives the field. Is this electric field compatible with the field
> > gradient that gives the correct acceleration for neutral atoms? Again,
> > time to do some calculations.
> Does this mean that this gravity=electrostatic force isn't completely
> crazy after all?
As I said earlier, it was very serious physics about 100 years ago.
> I don't have a degree in theoretical physics, so the
> math is beyond me.
Just algebra is needed.
> I do know that the electrostatic force as measured
> on the earth is quite high at around 100 Volts/meter. I understand
> this is equivalent of saying that it takes 100 Joules to move a charge
> of 1 columb 1 meter. A Joule is the amount of energy it takes to
> accelerate 1 KG to 1 meter/second. If you had 1 columb of charge, it
> would be accelerated towards the earth at 100 meters/second (I think).
> Gravity is 9.8 m/s which is much less, but the force is due to
> dilectric, not direct electrostatic effects, but there is certainly
> lots of electrostatic force avaliable. Of course, my math could be
> completely screwed up.
Time to read some books. Note the difference between force and
acceleration. Put an electron and a proton and a neutron into a static
field, and the neutron experiences no observable electric force, the
electron and the proton experience electric forces equal in magnitude, and
opposite in direction. Due to their very different masses, their
accelerations are very different.
Anyway, charge on a proton is 1.6e-19 C, so in a field on 100 V/m, the
force will be 1.6e-17 N. Since the mass is 1.7e-27 kg, the acceleration
should be about 1e10 m/s^2; much larger than g = 9.8 m/s^2.
By the way, it's worth checking that 100 V/m figure.
Maths for forces due to dielectric polarisation will be harder, since the
force depends on the gradient of the field, not the field directly.
> > > Isotopes with more
> > > neutrons would have a larger dilectric effect, and would therefore
> > > weigh more.
> >
> > Why? The number of electrons is still the same. The charge on the nucleus
> > of the atom is still the same. Why should the polarisability be different?
> > Specifically, for the case of deuterium vs hydrogen, why should it be
> > double? Perhaps these polarisabilities have been measured; this is
> > something for you to search for.
> >
> The number of electrons/protons doesn't make any difference. The atom
> is effectively neutral. The dilectric force may act on the individual
> neutrons, so if there are 2 of them, it pulls down on the entire atom
> with twice the force than if there were only 1 neutron.
Like I said, the dielectric polarisability of hydrogen atoms is known, see
if you can find measurements for deuterium. If it isn't double, then
they'll accelerate at different rates in the same electric field.
It's reasonable enough to assume that the polarisability of two neutrons
should be double that of a single one, but the big question is why should
the polarisability of a single neutron be the same as proton+electron, or
neutron+proton+electron have double the polarisability of proton+electron.
Time to bite the bullet and check the numbers. If experimental observation
doesn't support the theory, discard the theory.
> > > > Next, you need to explain why gravity is attractive and not repulsive.
> > > >
> > > A sufficiently strong positive electrostatic gravity will attract
> > > everything like a charged comb attracts everything. (See my other
> > > posts on why this isn't a repulsive force).
> >
> > In which case, you need to explain why gravity is an inverse-square force.
> > The electric force due to polarisation between two neutral atoms is known
> > to be 1/r^6 (van der Waals forces). Why the difference?
> The force isn't between two neutral atoms. It is between the
> collection of all the atoms in the earth, and the atoms for anything
> sitting on the earth. Naturally, the force generated by the earth
> overwhelms any force for any atom sitting on the earth. The force is
> purely electrostatic using the inverse-square law.
So? The important thing is that electrostatic forces due to dielectric
polarisation are _not_ inverse square laws. Check this by using Coulomb's
law to calculate the total force acting on a neutral, but polarised,
object.
> > Anyway, time for you to look up some polarisabilities for single atoms
> > (start with nice simple things first), and see if you can get sensible
> > results for gravitational acceleration of single atoms.
> >
> > The other thing to note is that the dielectric polarisability of a
> > (macroscopic) uniform dielectric ellipsoid is a tensor, which gives rise
> > to torques acting to align them with the applied fields. How well do these
> > torques compare with torques due to tidal forces? Non-agreement here might
> > well be a complete non-issue, though, since the macroscopic dielectric
> > polarisability of such bodies is small enough so that extremely large
> > fields would be needed to account for the gravitational acceleration of,
> > say, a glass marble.
>
> As I mentioned above, the electrostatic field measured at ground level
> is huge at 100 V/M.
And the field gradient is?
Check this number.
What is the electric field and electric field gradient inside a closed
metal box? What is the gravitational accleration inside the box?
FrankH
> On Thu, 13 Aug 2003, FrankH wrote:
>
> > Timo Nieminen <ti...@physics.uq.edu.au> wrote:
> > >
> > > Now, for this to result in a force, you need an electric field gradient.
> > > For an electric field gradient, you need an electric field. This is the
> > > whole point of comparing gravitational acceleration of ions with that of
> > > neutral atoms.
> > >
> > > Note that the polarisability of atoms is reasonably well-known, so you
> > > could go ahead and calculate what the field gradient, and therefore the
> > > mininum electric field, must be to provide an acceleration of g. At this
> > > point, it might be an interesting exercise for you to run the numbers, and
> > > see if the required fields and field gradients are reasonable, or
> > > nonsensical.
The formula for calulating the force due to the divergent
electrostatic field can be found at:
http://www.blazelabs.com/f-efield.htm
If I just ignore the things I don't understand and start plugging in
numbers, I get an interesting result. The density of graphite is 1.46
g/cm3, so it takes a 8.8 cm cube to make 1 kg of graphite. This is
equal to .00681 cubic meters in volume. The polarizsability of
graphite is 1.6 A^3. If I just plug this in to the formula, I get
F = (1.6 * .000681 * 120^2)/2 = 7.8 (joules ?)
The 7.8 joules means that there is enough force to cause a 1 kg mass
to accelerate to 7.8 meters/second (I think). This is fairly close to
the 9.8 meters/second required for gravity. Curiously, the term
.000681 * 120^2 is exactly equal to 9.8 and if the polarisability is
1, the result would be exactly 9.8 Joules (exactly what is needed) -
is this just coincidence? The force increases in proportion to the
mass, so this means that the acceleration will be the same for a 1 kg
mass and a 2 kg mass - just like gravity.
If this calculation isn't completely crazy, then I'd say we found the
source of gravity, because there is enough force from just the
divergent electric field to cause the affects we attribute to gravity.
Timo Nieminen
> The formula for calulating the force due to the divergent
> electrostatic field can be found at:
> http://www.blazelabs.com/f-efield.htm
>
> If I just ignore the things I don't understand and start plugging in
> numbers, I get an interesting result. The density of graphite is 1.46
> g/cm3, so it takes a 8.8 cm cube to make 1 kg of graphite. This is
> equal to .00681 cubic meters in volume. The polarizsability of
> graphite is 1.6 A^3. If I just plug this in to the formula, I get
>
> F = (1.6 * .000681 * 120^2)/2 = 7.8 (joules ?)
Force will be in newtons. OK, so with grad (E^2) = 120^2, then the force
is similar to gravitational force.
(Is this value for the polarisability of graphite for a static field, RF,
or optical?)
So, this means that the electric field near the earth's surface needs to
change by 120 V/m per metre of change in height. Next step: measure
electric field and rate of change.
Next step after that, consider that the formula for polarisability that
you used is for a spherical particle. What about a carbon rod? This will
align along the electric field. Maybe assume that the object is a
spheroid; this gives an analytical result for the polarisability.
See Stratton, Electromagnetic theory for a thorough overview.
For a prolate spheroid with aspect ratio of 2 (ie twice as long as it is
wide), and putting some fairly arbitrary numbers in (2m x 1m, glass), the
polarisability along the long axis is about double that along the short
axis.
Assuming that a similar ratio of polarisabilities can be achieved for a
graphite spheroid (and it will just be a matter of choosing the correct
aspect ratio), the torque should be about 10 N.m if the field is 100 V/m.
Since the 1 kg spheroid will be about 20cm long, you would need to apply
5N of force with each hand to force it into a horizontal orientation, and
it will happily stand upright due to this torque.
This also leads to the interesting result that if gravity is due to
electric fields and induced polarisation, the gravitational acceleration
of an elongated object depends strongly on its orientation. A simple thing
to test, just by dropping two identical pens, one upright, one horizontal.
> If this calculation isn't completely crazy, then I'd say we found the
> source of gravity, because there is enough force from just the
> divergent electric field to cause the affects we attribute to gravity.
Check the strength of the electric field needed. Your calculation shows
that grad(E^2) = 120^2 gives about the right answer. Since the field must
be spherically symmetric, it must be an inverse-square field.
At the surface of the earth, grad(E^2) = 120^2 = d/dr E^2 = 2E dE/dr
For an inverse square field, dE/dr = 2E/r, so
4 E^2/r = 120^2
and
E = 60 * sqrt(r)
where r is the radius of the earth, sqrt(r) = approx 2500 m^(1/2), giving
E = 150,000 V/m at the surface of the earth.
(So multiply the torque above by 1000).
Jacques Lavau
Thank for the link !
http://www.blazelabs.com/f-efield.htm
Interesting.
--
Ce message a ete poste via la plateforme Web club-Internet.fr
This message has been posted by the Web platform club-Internet.fr
FrankH
> where r is the radius of the earth, sqrt(r) = approx 2500 m^(1/2), giving
> E = 150,000 V/m at the surface of the earth.
>
I have been having another conversation on this subject on the
http://groups.yahoo.com/group/blazelabs discussion board. I shared
your calcualation of E at the surface of the Earth. This is the
reponse I got:
"
From: "famille-bondar" <henribondar@y...>
Date: Thu Sep 4, 2003 12:01 am
Subject: RE : [blazelabs] Need help calculating gravity
dielectrophoresis effect or : How to rise the hair on my head.
ADVERTISEMENT
Guys, what the hell are you doing ?
If you want to compute the electric field that is needed at the
surface of the hearth to compensate gravity there is an easiest and
correct way instead of the awful comments I have seen in your mails
below.
First it is easy to understand that the charges cannot be in the
atmosphere itself because through electrostatic repulsion they will be
expelled to the upper atmosphere. and consequently there will be no
field in the lower levels of the atmosphere and then on the surface.
This is one of the first electrostatic laws that has been discovered:
The charges always migrate (by auto repulsion) to the surface of a
conductive body and there is no field inside the body but only
outside.
This is called Faraday's effect (in France "cage de Faraday") and is
used in every electronic item and in special "embassy rooms" to
protect the external field to go inside the device (and the opposite
for embassy protection).
Then in your case you can only imagine to charge the earth in order to
obtain an electrostatic pressure on it surface and then a force equal
to the gravity one.
The earth is then the conductive body whereas the atmosphere has to be
insulating.
The right formula is then: Ax(Electrostatic Pressure) = mg (where A is
the surface of the body in contact with the earth).
We then have: Electrostatic Pressure = mg / A
And then: 1/2xEpsilonxE^2 = mg/A
Finally: E^2 = 2mg / (AxEpsilon)
For a standard standing human body we have approximately m = 60 Kg,
A = 0.16 m2
And then E = sqrt(2x60x9.81 / 0.16x9.9x10^-12)
Then E = 750 MV / m !!!!!
This is well above the atmosphere can assume before becoming a
conductor by field effect (the one used in corona glows).
Sorry to destroy your hopes but the maximum pressure that can be
obtained in earth can only expelled a piece of paper from a conductive
rod or rise my hair on my head (what you have done without the help of
electricity guys) but will unfortunately never expelled you to space
and this is known since the ancient Greeks.
Henri.
-----Message d'origine-----
De : Saviour [mailto:sav...@blazelabs.com]
Envoyé : jeudi 4 septembre 2003 00:55
À : blaz...@yahoogroups.com
Objet : [blazelabs] Re: Need help calculating gravity
dielectrophoresis effect
Hello,
Just a small correction to Timo's statement. Timo is working with the
inverse square law INSIDE the earth 'sphere'. This not true. Electric
field (and surprisingly similarly gravity) under earth's crust does
not obey the inverse square law. The law has to be applied from
r1=radius of earth to r2=radius of ionosphere=r1+60km, starting with
zero charge on earth with a net positive charge at the ionosphere,
or starting with a zero charge at ionosphere with a net negative
charge on earth (which is more realistic). If this results in E= -
300kV on earth relative to the ionosphere than everything will fit in
nicely. You can also use the known values for E and r1 & r2 to get
grad(E^2).
Regards
Saviour
"