Atomic radius.
max...@optusnet.com.au
proton radius evolved to be around 1e-10 meters, which is the
atomic radius. This of course has nothing whatever to do with
the proton in the fabled nuclear atom, but is the proton as
predicted by the zero origin concept.
Calculating the orbital speed of an electron required to
counteract its coulomb force interaction with the charge of its
parent proton could perhaps provide me with some kind of lead.
But that wouldn't be enough to accommodate all coulomb forces
involved. From the electron and proton mass difference, the
total coulomb force generators within a proton is 1833. So, 916
electrons and 917 positrons make up the entire proton content.
Assuming that the electron is drawn toward every charge within
the proton would be a more realistic approach. Or I could assume
that the electron plays a dual role, the second being that of a
positron which interacts with the internal e- charges. But how
this is done is irrelevant. The only necessity is that the entire
proton content be represented.
1.313e-10 meters emerges as the true proton radius.
Firstly, the force acting on the orbiting charge carrier at the
atomic radius has to be determined.
f = (Q1 * Q2) / (4 * pi * e0 * r ^ 2) does the trick.
f = 1.602e-19 *(1.602e-19 *1833)/(4*pi*8.8462e-12 * 1.313e-10 ^2)
f = 2.476e-5 newtons at a radius of 1.313e-10 meters.
Next is the orbital speed : v = SQR(f * r / m)
m is the mass held in orbit at the atomic radius (9.11e-31 kg).
v = 5.97e+7 m/sec.
The acceleration is : a = v ^ 2 / r : a = 2.71e+25 m/sec^2.
Or simply : N / m : 2.476e-5 / 9.11e-31 = 2.71e+25 m/sec^2
That figure represents the global acceleration rate applied
to dimension everywhere around the 1.313e-10 meter radius.
The entire three dimensions are enclosed by coulomb forces at
that radius if space-time is distorted inward at the speed of
light in all directions.
The acceleration rate given above is close to the speed of light
cubed (2.7e+25 m/sec), which accounts for the volume of dimension
housed within the atomic radius. Of all possible numbers in the
universe, why did it land so close to the one that sets the
radius of the enclosed dimension at the atomic radius? Such an
amazing coincidence doesn't come along every day, does it.
There is of course no electron orbiting the perimeter of the
proton. And if there is no thermal energy field present
(0 degrees K), there's no reason why it can't be stationary at
the proton event horizon because the proton charge would be seen
to be coming from all directions parallel with that horizon. The
electron would be driven equally in all directions in that plane.
The story can't end here though.
This paragraph is a necessary precursor to the rest.
When an electron and positron come together in the same point,
dimension encloses around them and they disappear. There are no
means of detecting them other than with the high energy E/M
radiation that brings them back into detectable existence. The
fact that this happens is evidence enough that the individual
charges had remained accessible to some degree to the world
outside their immediate association.
The black hole arising from coulomb forces is not the same as
that generated by gravity. Increasing the matter content of a
gravitationally induced black hole increases the black hole
radius, as does increasing the contained coulomb forces. But
the contained coulomb forces can fall to another level where
a second event horizon is established, which diminishes the
force setting the initial event horizon radius.
In a fusion environment, the resident electron can be hammered
into the proton body and the charge balance within becomes
symmetric. If interaction energy was not contained by the
enclosed dimension, there would be nothing to stop the entire
set of opposite charges from collapsing into individual points,
which would end up as one point when the initial event horizon
collapses because there are no coulomb forces left to sustain
it. But because dimension is closed, the gamma ray products of
partial or complete e-e+ collapses would cycle around and
continually revive the charges. That represents an enormous
thermal energy field trapped within the horizon. *And such a
field has mass*, which can account for the anomalous mass
increase when a proton converts to a neutron.
So what would be the radius of the neutron? Whatever it
is, there's little doubt that it's going to end up being
*enormously* smaller than the proton.
-----
Max Keon
Eric Gisse
> Something I've never before attempted to address is how the
> proton radius evolved to be around 1e-10 meters, which is the
> atomic radius.
The radius of the atom is much, much larger than the radius of the
nucleus. This has been known for nearly a century.
Go away until you learn some basic physics.
[snip rest, unread]
_...@jeff_relf.seattle.invalid
research and educational activities pertaining to
the North Slope and the adjacent portions of the Arctic Ocean.
Uncle Al
>
> Something I've never before attempted to address is how the
> proton radius evolved to be around 1e-10 meters, which is the
> atomic radius. This of course has nothing whatever to do with
> the proton in the fabled nuclear atom, but is the proton as
> predicted by the zero origin concept.
1) Rutherford
2) Gold foil and alpha decay
3) idiot
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
VMCM1905
<max...@optusnet.com.au> wrote in message
news:671fd896-199b-401f...@y10g2000prc.googlegroups.com...
> Something I've never before attempted to address is how the
> proton radius evolved to be around 1e-10 meters,
<SNIP>
Bzzt!
Full stop error. You are only off by six orders of magnitude... about how
much my IQ surpasses yours.
Y.Porat
-------------------
there are no elctrons orbiting in shells
around the nuc
2
one of my new (now quite old ...) findings is
that the
overall volume - difference (nuc plus elctrons orbitals)
between a small Atom and a big Atom
is much less than the existing assumptions
there are not as may' shells '
around the nuc as commonly assumes!
see table 2 and 3 in my abstract
http://sites.google.com/site/theyporatmodel/an-abstract
if it is not understood
i will explain its meaning
ATB
Y.Porat
----------------------------
PD
> Something I've never before attempted to address is how the
> proton radius evolved to be around 1e-10 meters, which is the
> atomic radius. This of course has nothing whatever to do with
> the proton in the fabled nuclear atom, but is the proton as
> predicted by the zero origin concept.
But the proton radius is not 1E-10 m.
Max Keon
Uncle Al wrote:
> max...@optusnet.com.au wrote:
>> Something I've never before attempted to address is how the
>> proton radius evolved to be around 1e-10 meters, which is the
>> atomic radius. This of course has nothing whatever to do with
>> the proton in the fabled nuclear atom, but is the proton as
>> predicted by the zero origin concept.
> [snip crap]
>
> 1) Rutherford
> 2) Gold foil and alpha decay
> 3) idiot
I calculate the orbital speed of an electron traveling around
the nucleus of a gold atom at the Bohr radius of 5.292e-11
meters, which is the same as the Schrodinger length constant,
thus;
f = (Q1 * Q2) / (4 * pi * e0 * r ^ 2)
f = 1.602e-19 *(1.602e-19 *79)/(4*pi*8.8462e-12 * 5.292e-11 ^2)
f = 6.52e-6 newtons acting at a radius of 5.292e-11 meters.
f/m = 7.16e+24 m/sec^2 acceleration rate. m is the electron mass.
v = (f / m * r)^.5
v = 1.947e+7 m/sec orbital speed.
f/m / c = 2.388e+16
The acceleration rate is 2.388e+16 times the speed of light.
Doesn't that enormous acceleration rate bother you?
The electron may not be moving relative to the nucleus, but
it's certainly turning corners at a hell of a rate in its
inertial frame.
Put your intuition back in the drivers seat because the whole
concept is absurd. Mathematical reality is based on whatever is
required to get the job done, and that's exactly how it should
be. But claiming that it represents physical reality is
ridiculous.
Oh what a tangled mess we make when
we base reality on math that's fake.
-----
Max Keon
Max Keon
Eric Gisse wrote:
> On May 17, 5:05am, maxk...@optusnet.com.au wrote:
>> Something I've never before attempted to address is how the
>> proton radius evolved to be around 1e-10 meters, which is the
>> atomic radius.
>
> The radius of the atom is much, much larger than the radius of the
> nucleus. This has been known for nearly a century.
It's not easy to understand how electrons are whizzing around at
enormous speeds at a lesser radius than the atomic radius when
the speed of light at the atomic radius is zero.
http://members.optusnet.com.au/maxkeon/atom-rad.html
-----
Max Keon
doug
Max Keon wrote:
Have you looked at any physics from the last century or so?
>
> The electron may not be moving relative to the nucleus, but
> it's certainly turning corners at a hell of a rate in its
> inertial frame.
Try looking at the physics research from about 1925.
>
> Put your intuition back in the drivers seat because the whole
> concept is absurd. Mathematical reality is based on whatever is
> required to get the job done, and that's exactly how it should
> be. But claiming that it represents physical reality is
> ridiculous.
>
> Oh what a tangled mess we make when
> we base reality on math that's fake.
Well then, you should not try to fake the math.
>
> -----
>
> Max Keon
>
>
>
Y.Porat
--------------
electron is the Atom
do not orbit
they vibrate
ATB
Y.Porat
---------------------
Eric Gisse
Then maybe the speed of light near an atom isn't zero?
You've been spewing random ideas for a few years now. Consider opening
a physics textbook to get the real answe.
tad...@comcast.net
> Something I've never before attempted to address is how the
> proton radius evolved to be around 1e-10 meters,
WRONG!
A "femtometer" is 10^-15 meters.
http://scienceworld.wolfram.com/physics/Proton.html
That makes the proton about 8.0x10-16 to 8.6x10^-16 meters.
I suspect the 'evolution" is in your changing eyesight. You appear to
have mistaken the "6" for a "0" in the exponent.
It happens to us all as we get older. See an optometrist and update
your prescription.
Tom Davidson
Richmond, VA
Max Keon
> On May 17, 9:05am, maxk...@optusnet.com.au wrote:
>> Something I've never before attempted to address is how the
>> proton radius evolved to be around 1e-10 meters,
>
> WRONG!
Your convincing argument overwhelms me. How can I possibly
possible counter such an argument?
**RIGHT!** should do the trick. But you could come back with a few
more stars and flatten my argument completely.
> A "femtometer" is 10^-15 meters.
> http://scienceworld.wolfram.com/physics/Proton.html
>
> That makes the proton about 8.0x10-16 to 8.6x10^-16 meters.
Does it! Did you ever see a proton?
Eric Gisse
Rutherford scattering, you clueless fucking moron. This was figured
out nearly a century ago.
tad...@comcast.net
> Tadchem wrote:
> > On May 17, 9:05am, maxk...@optusnet.com.au wrote:
> >> Something I've never before attempted to address is how the
> >> proton radius evolved to be around 1e-10 meters,
>
> > WRONG!
>
> Your convincing argument overwhelms me. How can I possibly
> possible counter such an argument?
>
> **RIGHT!** should do the trick. But you could come back with a few
> more stars and flatten my argument completely.
I cannot say you are right when your claim is inconcsistent with
empirical data.
The use of all caps (without stars) is simply 'emphasis,' not
shouting. It is a standard typographical technique.
Did you do *any* research into the size of the proton before you put
your pinkies to the keyboard for your post?
> Does it! Did you ever see a proton?
In the experimental sense of scattering experiments, the bare proton
has been 'seen' many times over the years.
I used to count them in my mass spectrometer while doing quantitative
analysis of gas mixtures.
Tom Davidson
Richmond, VA
Ahmed Ouahi, Architect
Atomic radius
http://www.answers.com/topic/atomic-radius
--
Ahmed Ouahi, Architect
Best Regards!
<tad...@comcast.net> kirjoitti
viestiss�:41f65560-ddcc-4876...@z7g2000vbh.googlegroups.com...
Max Keon
Eric Gisse wrote:
> Max Keon wrote:
>> Eric Gisse wrote:
>>> Max Keon wrote:
>>>> Something I've never before attempted to address is how the
>>>> proton radius evolved to be around 1e-10 meters, which is the
>>>> atomic radius.
>>>
>>> The radius of the atom is much, much larger than the radius of the
>>> nucleus. This has been known for nearly a century.
>>
>> It's not easy to understand how electrons are whizzing around at
>> enormous speeds at a lesser radius than the atomic radius when
>> the speed of light at the atomic radius is zero.
>>
>> http://members.optusnet.com.au/maxkeon/atom-rad.html
>
Near an atom, that would be true. But it's zero at the surface
of the proton that makes up the entire volume of the hydrogen
atom.
Perhaps I haven't properly explained how the charges within the
proton interact is such a way that they will enclose dimension
around them. Or at very least, enclose the coulomb force field
around them. Because the proton contents contain almost equal
quantities of opposite charges, there seems to be no reason why
they should be drawn together any more than they would be forced
apart. But geometry insists that they are.
In this configuration, the distance between the unlike charges
is .707 of the distance between the like charges. The unlike
charges are drawn together at, 1^2 / .707^2 = twice the rate that
the like charges are being forced apart.
e+
e- e-
e+
That configuration is the reason why hydrogen naturally forms
into a two part molecule. The charges are drawn together until
proton-proton contact is made. The atomic radius is of course
the proton radius.
Overlay a similar charge configuration to that shown in the
diagram, on top of that set, but indexed around by 90 degrees.
The attraction between the charges in the vertical plane is
unrestricted. There's really no doubt that the contents would
fall to a point if the interaction energy of the collapse is
not somehow contained.
Every charge within the proton is drawn toward all opposite
charges at twice the rate that they are forced away from the
like charges. That also applies for the electron on the proton
surface.
The numbers have been altered slightly, but the acceleration
rate was given as : f = (Q1*Q2)/(4*pi*e0*r^2)
f = 1.602e-19 *(1.602e-19 *1836)/(4*pi*8.8462e-12 * 1.314e-10 ^2)
f = 2.458e-5 newtons at a radius of 1.314e-10 meters.
Acceleration rate = newtons / electron mass
= 2.458e-5 / 9.11e-31
= 2.7e+25 m/sec^2, which is c^3.
Assuming that the electron is drawn to only half the contents
will account for the 50% imbalance between the attraction and
repulsion rates for all charges involved.
That sets the atomic radius at 9.3e-11 meters for a c^3 m/sec^2
acceleration rate. That's even better.
A black hole formed by coulomb forces will bend a light path
in exactly the same way as a matter black hole. A light ray
traveling around either black hole will behave in the same
manner. Why wouldn't it?
The above link demonstrates why the speed of light will be zero
at the event horizon of any kind of black hole.
> You've been spewing random ideas for a few years now. Consider
> opening a physics textbook to get the real answe.
Random ideas are derived when one aimlessly postulates, when
one has no direction other than a need to overcome an immediate
problem. It happens when theory fails.
My guiding light has always been the zero origin concept,
and there's nothing random about its predictions.
http://members.optusnet.com.au/maxkeon
-----
Max Keon
Max Keon
<tad...@comcast.net> wrote in message
news:41f65560-ddcc-4876...@z7g2000vbh.googlegroups.com...
> Max Keon wrote:
---
>> Does it! Did you ever see a proton?
>
> In the experimental sense of scattering experiments, the bare proton
> has been 'seen' many times over the years.
>
> I used to count them in my mass spectrometer while doing quantitative
> analysis of gas mixtures.
Did you think to catch one and measure it while you were doing
the counting?
PD
> <tadc...@comcast.net> wrote in message
The size and structure of individual protons have been measured
decades ago with deep inelastic scattering experiments at SLAC.
VMCM1905
"Max Keon" <max...@optusnet.com.au> wrote in message
news:4a1209dc$0$2602$afc3...@news.optusnet.com.au...
>
> Uncle Al wrote:
>> max...@optusnet.com.au wrote:
>>> Something I've never before attempted to address is how the
>>> proton radius evolved to be around 1e-10 meters, which is the
>>> atomic radius. This of course has nothing whatever to do with
>>> the proton in the fabled nuclear atom, but is the proton as
>>> predicted by the zero origin concept.
>> [snip crap]
>>
>> 1) Rutherford
>> 2) Gold foil and alpha decay
>> 3) idiot
> The acceleration rate is 2.388e+16 times the speed of light.
BZZT!
Acceleration is not the same as speed. Try at least to keep your units
straight.
VMCM1905
>
> Eric Gisse wrote:
>> On May 17, 5:05am, maxk...@optusnet.com.au wrote:
>>> Something I've never before attempted to address is how the
>>> proton radius evolved to be around 1e-10 meters, which is the
>>> atomic radius.
>>
>> The radius of the atom is much, much larger than the radius of the
>> nucleus. This has been known for nearly a century.
>
> It's not easy to understand how electrons are whizzing around at
> enormous speeds at a lesser radius than the atomic radius when
> the speed of light at the atomic radius is zero.
Dumbass.
VMCM1905
Wow, you have just disproved something fundamental with an amazing
logical proof. Now, go stand near your mailbox and we will send you your
Nobel Prize right away by certified mail!
Sam Wormley
> The acceleration rate is 2.388e+16 times the speed of light.
>
Acceleration is dv/dt
The speed of light is a defined physical constant of nature
with unit as the magnitude of velocity, i.e., |dx/dt|
Watch out, Max< or Dirk will archive an immortal fumble.
Sam Wormley
> Eric Gisse wrote:
>> On May 17, 5:05am, maxk...@optusnet.com.au wrote:
>>> Something I've never before attempted to address is how the
>>> proton radius evolved to be around 1e-10 meters, which is the
>>> atomic radius.
>> The radius of the atom is much, much larger than the radius of the
>> nucleus. This has been known for nearly a century.
>
> It's not easy to understand how electrons are whizzing around at
> enormous speeds at a lesser radius than the atomic radius when
> the speed of light at the atomic radius is zero.
Try not to think like a Newtonian.
Sam Wormley
>
> Near an atom, that would be true. But it's zero at the surface
> of the proton that makes up the entire volume of the hydrogen
> atom.
>
ILLUCID
Y.Porat
--------------------
even the proton is not a little
billiard ball
it is a chain of sub orbitals
orbitals are even smaller entities moving in orbits
so that orbit is not laways
all laong 'populated' by that
small entity
and that is why
youcan meet it *at a certian point of time
in one of its orbit location
and at another point of time
at another 'corner' of that orbital
iow
'it has the 'option' to be at any point of the orbital
but not all time at all points
did i made myself clear ??
and that is why it seem to be
'a ''point aprticle' but *practically*
you can find it in a much bigger volume than its real volume !!
TIA
Y.Porat
--------------------
Max Keon
VMCM1905 wrote:
> Max Keon wrote:
>> Tadchem wrote:
>>> A "femtometer" is 10^-15 meters.
>>> http://scienceworld.wolfram.com/physics/Proton.html
>>>
>>> That makes the proton about 8.0x10-16 to 8.6x10^-16 meters.
>>
>> Does it! Did you ever see a proton?
>
> Wow, you have just disproved something fundamental with an amazing
> logical proof. Now, go stand near your mailbox and we will send you your
> Nobel Prize right away by certified mail!
I would like to thank the Nobel committee for this gesture of support
in the quest for truth. I can't tell you how much it means to me.
Max Keon
VMCM1905 wrote:
> Max Keon wrote:
>> Tadchem wrote:
>>> http://scienceworld.wolfram.com/physics/Proton.html
>>>
>>> That makes the proton about 8.0x10-16 to 8.6x10^-16 meters.
>>
>> Does it! Did you ever see a proton?
>
> Wow, you have just disproved something fundamental with an amazing
> logical proof. Now, go stand near your mailbox and we will send you your
> Nobel Prize right away by certified mail!
I would like to thank the Nobel commity for this gesture of support
Max Keon
> Max Keon wrote:
>> The acceleration rate is 2.388e+16 times the speed of light.
>
> Acceleration is dv/dt
>
> The speed of light is a defined physical constant of nature
> with unit as the magnitude of velocity, i.e., |dx/dt|
http://members.optusnet.com.au/maxkeon/atom-rad.html
proves you wrong. The reasoning is so simple that almost anyone
will understand it.
How does this wavefront made up of photons turn a corner if the
speed of light is constant on the inner and outer edges of the
radius?
o
o
o--->
o
o
Perhaps I can help. The entire wavefront can be considered as
a moving carrier for the photons as it turns the corner, and a
photon's true speed can be the speed of light within that
carrier.
With the support of a self consistent math loop, such a postulate
could surely squeeze its way into reality, as has happened many
times before.
-----
Max Keon
Max Keon
PD wrote:
> On May 21, 6:03am, "Max Keon" <maxk...@optusnet.com.au> wrote:
>> <tadc...@comcast.net> wrote:
>>> Max Keon wrote:
>>>> Does it! Did you ever see a proton?
>>>
>>> In the experimental sense of scattering experiments, the bare proton
>>> has been 'seen' many times over the years.
>>>
>>> I used to count them in my mass spectrometer while doing quantitative
>>> analysis of gas mixtures.
>>
>> Did you think to catch one and measure it while you were doing
>> the counting?
>
> The size and structure of individual protons have been measured
> decades ago with deep inelastic scattering experiments at SLAC.
The proton size is "observed" at the point where the scatter
pattern remains constant while the momentum of the bombarding
electrons continues to increase. But what is actually observed?
e- -----------> _
----------> o o - _
-------> o 1 o - _ _
------> o Proton o 1e-15? meter radius. _ X _
-------> o o _ -
----------> o o _ -
--------------->
Assuming that the electron is drawn to the proton charge; the
maximum deflection angles per velocity for electrons grazing the
proton surface on each side of the proton will coincide at some
point X, through which they will continue on, expanding apart.
All scattered electron trajectories at any common angle on both
sides of the proton can be traced back to a source which is not
where the proton resides. For very small deflections the proton
radius could perhaps be determined in this way.
This is how I would explain such scattering assuming that the
proton radius is the event horizon of a black and that its radius
is the atomic radius (1e-10 meters).
_
e- -----------> _ -
--------x-> o o ==== - _
-------> o 1 o - _
------> o Proton o _
-------> o o _ -
--------x-> o o ==== - _
---------------> - _
The rising trajectory in the top set was caused by a deflection
off the event horizon, while the falling trajectory in that set
results from the electron following around the curved dimension
slightly above the horizon. In that set, the lines along the
trajectories at two common angles (rising and falling) extended
back to the source at point 'x' pass by the proton centerline
and still have some distance between them. At that location, the
proton radius is perhaps assumed to be measured. But the
measurement taken where the scattering occurs, around the event
horizon, is *far* less than the true proton radius. Why not
1e-15 meters? Why not whatever it is?
Regardless of what scatter angles are chosen, the upper set and
lower set of scatter angles shown will be separated by a constant
1e-10 meters, which can't be magnified with distance. They will
be overlaid on each other as one.
As for the proton content being revealed through deep inelastic
scattering, being a truly remarkable achievement in itself, what
has been revealed is the disruption of the interacting e- and
e+ components that made up the entire proton content. The so
called quarks that carry proportional charge, exist only in the
space between the proton collapse and the gamma ray emission that
quickly follows when all opposite charges have been paired off
and begin to fall into a multitude of individual points. Which
is where they were before they had grouped together in the
required numbers to enclose dimension around them in the first
place. The energy of the gamma ray represents the energy which
maintained their separation within the enclosed dimension.
There should be a spare positron floating about somewhere as
well. Although the bombarding electron would probably counteract
it.
-----
Max Keon
PD
Max, all you really need to do is to look at the DIS papers rather
than just guessing what they say.
PD
Sam Wormley
Single photons follow the geodesic of curved space about masses.
You confuse yourself with wavefronts.
Eric Gisse
Guessing is more fun.
Max Keon
PD wrote:
> Max Keon wrote:
>> PD wrote:
---
>>> The size and structure of individual protons have been measured
>>> decades ago with deep inelastic scattering experiments at SLAC.
>>
>> The proton size is "observed" at the point where the scatter
>> pattern remains constant while the momentum of the bombarding
>> electrons continues to increase. But what is actually observed?
> Max, all you really need to do is to look at the DIS papers rather
> than just guessing what they say.
Guessing was necessary because your claim that the size and
structure of *individual* protons had been measured with deep
inelastic scattering experiments led me to believe that an
isolated proton size had been measured, and since I couldn't find
any evidence to the contrary, I had no choice but to go along
for the ride.
It was necessary for me to anticipate how such a scattering
system might be designed. Firstly, the proton target is held in
position with repulsive electric fields. For obvious reasons,
the scattering electrons would need to be uniformly distributed
over a single wavefront and hit the target in unison.
This is the expected result. Note that the x's have been
relocated to the correct side of the proton, and the X
intersection has been added.
_
e- -----------> _ -
----------> o o =x== - _
-------> o 1 o - _ _
------> o Proton o _ X _
-------> o o _ -
----------> o o =x== - _
---------------> - _
The angle of the trajectories focused through X would be the
maximum, and if all other trajectories could be eliminated,
the true proton radius could be determined. Otherwise, the
proton size is measured at both, x and x. Not from x to x.
That's as detailed as it needs to be in the circumstances.
http://arxiv.org/abs/nucl-ex/0408020
describes a proposed deep inelastic scattering experiment which
uses 1.6 GEV electrons to expose the workings of a proton (2004).
Hydrogen, cooled by liquid nitrogen, was used as a target. The
scatter cell is 400mm long. The maximum gas density at the cell
center is 5e+13 atoms/cm3.
It's blatantly obvious that the scattering by a single proton
is far beyond the detection capability of a calorimeter, or
anything else.
You were obviously referring to the individual protons in a
target group and not an isolated proton? But "deep inelastic
scattering"? What does that have to do with measuring the proton
size?
So what you were really trying to say is that all experiments
which are extensions of the Rutherford scattering experiment give
much the same result, with improved accuracy. That's hardly
surprising is it.
If an electron probe was used to determine the size of a proton
through elastic scattering, it would still be shifted off course
by the influence of the approaching target to the same degree as
any other probe.
http://members.optusnet.com.au/maxkeon/ruthfd.html
But there were no guesses involved in this part of my post.
> As for the proton content being revealed through deep inelastic
> scattering, being a truly remarkable achievement in itself, what
> has been revealed is the disruption of the interacting e- and
> e+ components that made up the entire proton content. The so
> called quarks that carry proportional charge, exist only in the
> space between the proton collapse and the gamma ray emission that
> quickly follows when all opposite charges have been paired off
> and begin to fall into a multitude of individual points. Which
> is where they were before they had grouped together in the
> required numbers to enclose dimension around them in the first
> place. The energy of the gamma ray represents the energy which
> maintained their separation within the enclosed dimension.
>
> There should be a spare positron floating about somewhere as
> well. Although the bombarding electron would probably counteract
> it.
You may not agree with it, but I have a valid theory and you
don't.
-----
Max Keon
Eric Gisse
[...]
> You may not agree with it, but I have a valid theory and you
> don't.
>
> -----
>
> Max Keon
Stupid only ever wins because stupid simply grinds you down.
PD
> PD wrote:
> > Max Keon wrote:
> >> PD wrote:
> ---
> >>> The size and structure of individual protons have been measured
> >>> decades ago with deep inelastic scattering experiments at SLAC.
>
> >> The proton size is "observed" at the point where the scatter
> >> pattern remains constant while the momentum of the bombarding
> >> electrons continues to increase. But what is actually observed?
>
> ---
>
> > Max, all you really need to do is to look at the DIS papers rather
> > than just guessing what they say.
>
> Guessing was necessary because your claim that the size and
> structure of *individual* protons had been measured with deep
> inelastic scattering experiments led me to believe that an
> isolated proton size had been measured, and since I couldn't find
> any evidence to the contrary, I had no choice but to go along
> for the ride.
As I said, Max, the way to solve this is not to decide to go along
with the ride or not, but to look up the original references and read
them. This way you know what's actually been done, and what the
measurements show about the way nature actually behaves.
>
> It was necessary for me to anticipate how such a scattering
> system might be designed. Firstly, the proton target is held in
> position with repulsive electric fields.
No, Max. This is pointless. For every correct answer, there are
1,000,000,001 incorrect guesses. Why you would rather expend your
entire existence exploring the 1,000,000,001 incorrect guesses than
simply look up the correct answer is beyond me.
Yes, OK, that's a good example. Now, if you would, please note the
ratio of the size of the hydrogen atoms to the distance between the
hydrogen atoms in this target. Then tell me the protons are not
isolated.
>
> It's blatantly obvious that the scattering by a single proton
> is far beyond the detection capability of a calorimeter, or
> anything else.
I don't see where you got that idea at all.
You might also want to look up electron-proton scattering experiments
at HERA.
>
> You were obviously referring to the individual protons in a
> target group and not an isolated proton? But "deep inelastic
> scattering"? What does that have to do with measuring the proton
> size?
Read the articles, read a book. You will NOT get what you need by
reading abstracts. Get up off your butt and go to a library.
>
> So what you were really trying to say is that all experiments
> which are extensions of the Rutherford scattering experiment give
> much the same result, with improved accuracy. That's hardly
> surprising is it.
No, Rutherford scattering is *elastic* scattering, not deep inelastic
scattering. Completely different.
>
> If an electron probe was used to determine the size of a proton
> through elastic scattering, it would still be shifted off course
> by the influence of the approaching target to the same degree as
> any other probe.http://members.optusnet.com.au/maxkeon/ruthfd.html
>
> But there were no guesses involved in this part of my post.
>
>
>
> > As for the proton content being revealed through deep inelastic
> > scattering, being a truly remarkable achievement in itself, what
> > has been revealed is the disruption of the interacting e- and
> > e+ components that made up the entire proton content. The so
> > called quarks that carry proportional charge, exist only in the
> > space between the proton collapse and the gamma ray emission that
> > quickly follows when all opposite charges have been paired off
> > and begin to fall into a multitude of individual points. Which
> > is where they were before they had grouped together in the
> > required numbers to enclose dimension around them in the first
> > place. The energy of the gamma ray represents the energy which
> > maintained their separation within the enclosed dimension.
>
> > There should be a spare positron floating about somewhere as
> > well. Although the bombarding electron would probably counteract
> > it.
>
> You may not agree with it, but I have a valid theory and you
> don't.
First of all, Max, the validity of a theory is not based on whether it
coheres and looks pretty. An idea that coheres and looks pretty, plus
75 cents, will buy you a cup of coffee. Those are literally worthless.
Valid theories are ones that make quantitative predictions of
measurable quantities that distinguish themselves from competing
theories AND for which those predictions are borne out by measurement.
You want to be credited for having something you do not yet have. You
want what you have to be called a valid theory, but that is not what
you have.
In order for you to have a valid theory, you need to start by becoming
more acquainted with what is already measured, already unambiguously
settled.
PD
Max Keon
PD wrote:
> On May 27, 6:50pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
---
>>>> PD wrote:
>>>>> The size and structure of individual protons have been measured
>>>>> decades ago with deep inelastic scattering experiments at SLAC.
This argument is based on your above claim, which is wrong.
Apart from giving me the opportunity to describe the true nature
of the proton content, the entire argument has been fairly
pointless.
---
> As I said, Max, the way to solve this is not to decide to go along
> with the ride or not, but to look up the original references and read
> them. This way you know what's actually been done, and what the
> measurements show about the way nature actually behaves.
Yes, I certainly agree.
This is the link that I had previously misplaced. But it doesn't
seem to work any more. Perhaps you can find it somewhere else.
http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-19
13.html
You have caused me to properly address the problem though,
according to the *prediction* of the zero origin concept.
The next two paragraphs are extracts from my web page at this
address
http://members.optusnet.com.au/maxkeon/ruthfd.html
Note that protons and neutrons are predicted as being enclosed
dimensions which are made up entirely of electrons and positrons.
And the nucleus of the predicted atom is not a tiny fraction of
the atomic volume, it's the entire volume, up to the atomic
radius.
------------
Nothing but the individual charges inside the black holes of
the probe particle are included in the field which extends into
the past around them. The field becomes a global recording of
bygone coulomb force interactions at all points in space where
the sources once resided in the present. The fields generated
by readily apparent charges are known to exist in the past
around the source. The hidden fields are no different. A field
has potential mass which will be realized if it's changed in
any way. If any part of a field is altered, the light speed
expansion of the field at the location of the change has been
restrained to a speed less than that of light. The field is
being accelerated, which requires a force. The force is
applied between the probe and the scatter material atoms as
the probe approaches the scatter material and the fields of
each are increasingly restrained by the influence of the other.
The field around the probe will interact with the atoms of the
scatter material as would a massive object. And the scatter
material atoms would affect the probe atom in the same way. The
probe's forward motion would be resisted, while the scatter
material atoms would be forced off the path of the probe.
---------
Apart from being a simple and inaccurate fix to the problem, it
provides no tangible link between the fields and the particles
that could accelerate-decelerate the particles in any way because
the fields are generated by near enough to equal quantities of
opposite charges within protons and neutrons.
The very basis of the zero origin concept is that electrons and
positrons provide the foundation for all that exists, and they
are each the point origin of the dimension that they leave
behind in their pasts, creating dual past dimensions which are
perceived as one. Which may all seem rather bizarre, but it's
really not that hard to understand.
Each e- component within the probe is drawn toward all e+
components within the atoms of the scatter material, and vice
versa. The force direction is outward from each component.
Each e- component within the probe is forced away from all e-
components within the atoms of the scatter material, and vice
versa. The same applies for the e+ components. The force
direction is then pointed toward each component.
If the probe is moving at light speed toward the scatter material
atoms, the opposite charges are moving toward each other at the
rate at which the attractive forces are updated, and the force
will be zero. But the repulsive force will have doubled because
it points in the opposite direction. The resultant total force
is then equal to the combined coulomb forces of the whole system.
v^2 / c^2 * Q1 * Q2 etc. gives the resultant force.
If the probe is moving away at light speed, the attractive forces
would double and the repulsive forces would be zero.
Why else do you think the probe slows so dramatically in the
nuclear atom if there's virtually nothing in its way to slow it
down? A 100 atom thickness foil, with a volume of empty space
1.5e+12 times the volume enclosed by matter, couldn't do it,
surely? It's irrelevant though because the nuclear atom isn't
in question here. Not directly anyway.
The consequences of the above extract assumes that coulomb forces
acting from the probe would be sufficient to clear a straight
path through the scatter foil before it arrives. But the math
suggests otherwise if the hidden charges emanating from within a
proton or neutron are considered to be originating from the
center of the atom. Contact between the probe and a foil atom
would be made when the centers of each are 2e-10 meters apart.
If protons and neutrons are black holes as the theory predicts,
the event horizon is an absolute lateral distortion of dimension
through which nothing can penetrate (below an energy threshold).
The positive charge excess within a proton exist only at the
horizon. Beyond that, nothing exists. Assuming that the charge
is centered at a point which is almost infinitely removed from
the outside world cannot be correct because the charge at the
horizon could only be zero. In a sense the horizon is the point
origin of dimension around which the proton charge is
represented if full, over the zero thickness horizon. There's no
other way. The charge field of an electron embedded deep in the
horizon will spread around the horizon many times before any part
of it emerges into the outside world, along with its counterpart
field from within the proton.
A path can easily be cleared through the target material.
This image was generated from the attached Qbasic program. The
reason why it's attached is because it saves me giving a million
word explanation.
http://members.optusnet.com.au/maxkeon/probe.jpg
The image is the result of 21 steps through the program, with
each step halving the distance to the surface of the target
atom. The start point was 10 times the atomic radius. The
surfaces of the circles to the right were initially in contact,
with the probe trajectory pointed at where the lower circle
resided at the start. And the scale is reasonably accurate.
In the program, the probe trajectory continues to bend upward,
which is obviously wrong, but the very first jump tells the
story. If the restraining force applied by the repulsive fields
was directed at 90 degrees to the probe trajectory the sideways
acceleration in that step is certainly enough to clear the target
and direct it toward the top atom shown, from which it will be
directed back down again toward a point of balance between the
two fields. If there was no variation in the field, the probe
would not be deflected off course.
There's little doubt that the probe will quickly align with the
point of least resistance in the field. When it enters the target
material, its trajectory will be aligned exactly with the path
of least resistance. The program was stopped when the probe was
4.77e-16 meters from the target (if the target hadn't moved),
which is the point where the pathway was first opened up enough
for the probe to pass through. The path clearing process
continues on from one atom to the next beyond that point. You can
see how easily this is done by entering .1 at the radius prompt
in the program, which sets the initial deflection at the atomic
radius from one encounter to the next through the foil.
The pathway clearing process would obviously follow a crooked
path, but the momentum of the probe would maintain a reasonably
straight line on average through the foil. Once that pathway has
been opened up, the field at the point of entry will have been
weakened and will become the natural target for following probes.
The kinks in the path can be hammered out of the way and the
entire beam of probe particles could end up going straight
through a small number of holes spread out across the incident
beam width.
If a probe particle traveling a .1c is stopped completely in
a collision with the 100 gold atoms over the foil thickness, the
gold atoms would be accelerated up to 609137m/sec. The bond with
the surrounding atoms would first need to be broken. But there's
little doubt that a single gold atom could be beaten into the
surrounding structure by a passing probe.
If the trajectory of a probe approaching the foil is almost
exactly centered on a "high" part of the field between holes,
there will be no line of least resistance emerging until the
probe has almost reached the target. By the time the shift
direction has been determined, it could be too late for the
probe to align itself with the line of the chosen hole. It will
make its way down the hole bouncing off walls. The final bounce
will be the deflection angle.
If the possibility of encountering a "high" in the field is the
same as the possibility of encountering the chosen size of the
nucleus of a nuclear atom, then the scattering results will be
identical.
Copy-paste the program directly off this screen.
'-----Start-------------------------
'Escape exits program.
SCREEN 12
CLS
pi = 3.1416
c = 3E+08
e0 = 8.8462E-12
INPUT " Radius multplier (.1 = 1e-10 meter radius)"; mul
INPUT " Probe speed (c=1)"; v
IF v = 0 OR mul = 0 THEN END
v = v * c
e1 = 1.6029E-19 * 1836 'Proton total charge
e = e1 * 4 'Alpha particle total charge
ep = e1 * 197 'Gold atom total charge
m = 1.67E-27 * 4 'Alpha particle mass
ma = 1.67E-27 * 197 'Gold atom mass
r = 1E-09 'The graphics are based on this radius.
r = r * mul
mul = 1 / mul 'Graphics
b = 340 'Graphics
LOCATE 17, 21: PRINT "<--- "; r; "meters"
aa: ON ERROR GOTO ab
LOCATE 3, 1
PRINT " Step"; p
p = p + 1
PRINT r; "radius from impact zone. "
'f = v^2 / c^2 * Q1 * Q2 / (4 * pi * e0 * r ^ 2)
f = v ^ 2 / c ^ 2 * e * ep / (4 * pi * e0 * r ^ 2)
PRINT r / v; "seconds from impact (r / v) "
acc = f / m
PRINT acc; "m/sec^2 acceleration rate. "
time = r / v
shift = time ^ 2 * acc
PRINT shift; "meter shift at time of impact. "
PRINT " For a specific mass the shift"
PRINT " is constant for any speed"
PRINT " and from any radius."
r = r / 2 'Radius decrement per step
CIRCLE (390 + mul * 34, 250 - mul * 68 - u * mul), 34 * mul
CIRCLE (390 + mul * 34, 250 + u * mul), 34 * mul
CIRCLE (390 - b - mul * 34, 250 - a * mul), 34 * mul
CIRCLE (390 - b, 250 - a * mul), 2
LINE (390 - b, 250 - a * mul)-(390 - bb, 250 - aa * mul)
bb = b: aa = a
DO: s$ = INKEY$: LOOP UNTIL s$ <> ""
IF s$ = CHR$(27) THEN END
CIRCLE (390 + mul * 34, 250 - mul * 68 - u * mul), 34 * mul, 0
CIRCLE (390 + mul * 34, 250 + u * mul), 34 * mul, 0
CIRCLE (390 - b - mul * 34, 250 - a * mul), 34 * mul, 0
b = b / 2 'Graphics
a = a + shift / m * 5E-16 'Graphics
u = u + shift / ma * 5E-16 'Graphics
GOTO aa
ab: END
'----------end-------
-----
Max Keon
PD
> PD wrote:
> > On May 27, 6:50pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
> ---
> >>>> PD wrote:
> >>>>> The size and structure of individual protons have been measured
> >>>>> decades ago with deep inelastic scattering experiments at SLAC.
>
> ---
>
> This argument is based on your above claim, which is wrong.
> Apart from giving me the opportunity to describe the true nature
> of the proton content, the entire argument has been fairly
> pointless.
> ---
>
> > As I said, Max, the way to solve this is not to decide to go along
> > with the ride or not, but to look up the original references and read
> > them. This way you know what's actually been done, and what the
> > measurements show about the way nature actually behaves.
>
> Yes, I certainly agree.
>
> This is the link that I had previously misplaced. But it doesn't
> 13.html
Max, this is a link to a HIGH SCHOOL physics course. I'm suggesting
that you look up the papers in physics journals, published by the
professional physicists.
[rest snipped, further waste of time making blind guesses when better
information is readily available to you]
Max Keon
"PD" <TheDrap...@gmail.com> wrote:
> On May 31, 10:25pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
>> PD wrote:
---
>>> As I said, Max, the way to solve this is not to decide to go along
>>> with the ride or not, but to look up the original references and read
>>> them. This way you know what's actually been done, and what the
>>> measurements show about the way nature actually behaves.
>>
>> Yes, I certainly agree.
>>
>> This is the link that I had previously misplaced. But it doesn't
>> seem to work any more. Perhaps you can find it somewhere else.
>> http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/
>
> Max, this is a link to a HIGH SCHOOL physics course. I'm suggesting
> that you look up the papers in physics journals, published by the
> professional physicists.
I found the correct address with a Google search for
GeigerMarsden-1913. Good old Google aye. It's still in the same
place, but has been relocated since 2004.
http://www.chemteam.info/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.
html
Paragraph 3 in the final experiment listed gives a 1mm air
equivalent stopping power for a 2.1e-7 meter thick gold foil.
Table V11 in the experiment listed prior to that gives 5.5mm as
the stopping range in air for the probe particle when no mica
sheets are in place. So the probe is stopped completely by 5.5
gold foils, 2.1e-7 meters thick.
How can you explain that when chances of a collision is so
extremely remote through 1 foil? You can't blame coulomb forces
once the probe has entered the atomic environment either because,
in your universe, there are just as many such forces pulling it
forward as there are holding it back.
In my previous post, my description right up to the point where
the probe enters the scatter foil was exactly according to
theory. There's little doubt that the probe's trajectory would
be shifted to align with the weakest part of the coulomb force
field at the foil surface before it arrives there.
I then suggested that a series of holes would be opened up
through the foil in the early stages of the bombardment and that
following probes would be directed toward them because that's
where the coulomb force fields would be weakest. But it's a
doubtful scenario because the tensile strength in the bond over
the area of an (square) atom is 6.3e-13kg for 10 ton per square
inch tensile strength gold foil. I imagine that it would be at
least that much. The probe would not be capable of breaking just
one gold atom from its bond with rest of the atoms unless it was
moving at relativistic speed.
There are alternatives of course.
In this case, the bond between the gold atoms is not broken but
is elastically distorted around the probe as it passes through
the foil.
If an atom is assumed to be spherical, the minimum distance
between gaps in contacting atoms is the atomic radius. In that
case the probe need only shift course by that amount on its
journey toward the foil surface. Which is certainly more feasible
than having to shift the distance between a small number of holes
spread out across the incident beam width, as suggested. So long
as the probe enters the foil at the depression in the field,
being a gap between contacting atoms, its trajectory will remain
unchanged. And if coulomb forces are the only means of contact
between the probe and its surroundings, as per my previous post,
its movement through the foil will not be affected much.
The rolling process involved in the manufacture of scatter foils
demands that the atoms on both surfaces all be oriented in the
plane of the surface, and the atoms that bridge the distance
between the two surfaces would be expected to be compressed
together as much as possible.
The arrangement of atoms shown in this animation is about as
close together as they could get. And this is not supposed to be
an accurate representation of the elastic movement. The atomic
structure will shift wherever it can.
http://members.optusnet.com.au/maxkeon/foil.gif
When the probe has negotiated the first layer, it's confronted
by an atom in the second layer that's directly in its path. It
will deflect in the direction where the repulsive forces are
weakest. Chances of an exact alignment with the centers of mass
along the trajectory would be zero in that environment. It's
certainly not stable because the probe just crashed through the
surface. The rest of the journey is obvious. The restraining
force applied as the probe pushes forward is almost the same as
the driving force applied as the structure closes back in behind
it.
As the probe approaches the final layer, it deflects downward
and then along the plane of the original trajectory. The gold
atom configuration on both sides of the foil is the same. How it
goes in is how it comes out.
From my previous post, this equation was used to determine the
the coulomb forces generated by the probe's motion toward the
target foil. f = v^2/c^2*Q1*Q2/(4*pi*e0*r^2) The probe's motion
is restrained to some degree as soon it leaves the source. As it
nears the target, the nearest depression in the field around the
gold atoms to which it's heading will begin to become apparent
and the probe will be shifted off course accordingly. The
possibility of an atom in the target surface being exactly
aligned through the center of mass along the trajectory of the
approaching probe is extremely remote, but it is possible because
the target is not being disturbed as in the previous scenario.
If the alignment is exact, the scatter angle is 180 degrees.
If the depression in the field only becomes apparent when the
probe is close to the target, it will driven to that depression
at an angle and will enter the surface, colliding with the atom
on the far side of the entrance. That bounce will resonate right
through the foil, and wherever it's heading at the time of
departure will be the scatter angle.
According to the animation, there is a second channel through
which a probe could travel, which is at 45 degrees to the foil
surface. But it probably wouldn't be accessible to a probe
entering the surface at 45 degrees because the probe would be
deflected straight into the foil in the initial encounter.
If the foil is tilted at i.e. 20 degrees to the incident particle
beam, a probe's initial deflection into the foil at the point of
entry would set up a resonant oscillation that's added to its
normal wobbly path and that would manifest itself at the exit
point, deflecting at 20 degrees to the surface. But depending on
the phase of the oscillation cycle, it can go either way in the
plane of the tilt. The scatter pattern would show an enormous
spike at 40 degrees to the incident beam in the plane of the
tilt. ????
> [rest snipped, further waste of time making blind guesses when
> better information is readily available to you]
You really don't want to know about this do you. What I'm
attempting to do is understand the consequences of direct
*predictions* stemming from the zero origin concept. I can only
anticipate those consequences to the best of my ability, but what
is observed in nature must always set the boundaries. Just like
you do, but you have no valid prediction to guide you, so your
initial parameters are boundless. The nuclear atom "theory" was
simply designed to fit the evidence.
The problem now is that it has become so entrenched in the system
over the past century that it can't be removed without taking
down everything that has been built up around it. 100 years of
"progress" comes tumbling down. But we're all going to have to
face reality one day. A sinking barge overloaded with garbage
can't be kept afloat by piling more garbage on it. Tell me that
statement isn't true.
If this pixel '.' represents one hydrogen atom nucleus in a jar
of liquid hydrogen, its nearest neighboring nucleus is 22000
pixels away, or around 100 meters depending on screen size. And
there are only two electrons that will pass through the void
between them. That's what is proposed as the foundation of
solids.
-----
Max Keon
PD
> "PD" <TheDraperFam...@gmail.com> wrote:
> > On May 31, 10:25pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
> >> PD wrote:
> ---
> >>> As I said, Max, the way to solve this is not to decide to go along
> >>> with the ride or not, but to look up the original references and read
> >>> them. This way you know what's actually been done, and what the
> >>> measurements show about the way nature actually behaves.
>
> >> Yes, I certainly agree.
>
> >> This is the link that I had previously misplaced. But it doesn't
> >> seem to work any more. Perhaps you can find it somewhere else.
> >>http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/
> >> GeigerMarsden-1913.html
>
> > Max, this is a link to a HIGH SCHOOL physics course. I'm suggesting
> > that you look up the papers in physics journals, published by the
> > professional physicists.
>
> I found the correct address with a Google search for
> GeigerMarsden-1913. Good old Google aye. It's still in the same
> html
>
No, this isn't the deep inelastic scattering experiment at SLAC I was
referring to. If you need a hint, you can check the Nobel Prize list
in physics. The work was done in the late 60's and early 70's, and the
Nobel was given in 1990.
Max Keon
"PD" <TheDrap...@gmail.com> wrote:
> On Jun 3, 10:48pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
>> "PD" <TheDraperFam...@gmail.com> wrote:
>>> On May 31, 10:25pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
>>>> This is the link that I had previously misplaced. But it doesn't
>>>> seem to work any more. Perhaps you can find it somewhere else.
>>>>http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/
>>>> GeigerMarsden-1913.html
>>>
>>> Max, this is a link to a HIGH SCHOOL physics course. I'm suggesting
>>> that you look up the papers in physics journals, published by the
>>> professional physicists.
>>
>> I found the correct address with a Google search for
>> GeigerMarsden-1913. Good old Google aye. It's still in the same
>> place, but has been relocated since 2004.
html
html is usually wordwrapped, but that's easy fixed.
> No, this isn't the deep inelastic scattering experiment at SLAC I was
> referring to.
Of course not. But what has that to do with elastic scattering?
That's how the atomic radius was presumed to have been
determined wasn't it?
This paragraph from my previous post really needs to be addressed.
>> If this pixel '.' represents one hydrogen atom nucleus in a jar
>> of liquid hydrogen, its nearest neighboring nucleus is 22000
>> pixels away, or around 100 meters depending on screen size. And
>> there are only two electrons that will pass through the void
>> between them. That's what is proposed as the foundation of
>> solids.
100 meters was supposed to be 10. It's closer to 20 on my screen
though because the dot measures almost 1*1mm. But, how would you
make that paragraph seem more realistic?
As for this paragraph:
>> If the foil is tilted at i.e. 20 degrees to the incident particle
>> beam, a probe's initial deflection into the foil at the point of
>> entry would set up a resonant oscillation that's added to its
>> normal wobbly path and that would manifest itself at the exit
>> point, deflecting at 20 degrees to the surface. But depending on
>> the phase of the oscillation cycle, it can go either way in the
>> plane of the tilt. The scatter pattern would show an enormous
>> spike at 40 degrees to the incident beam in the plane of the
>> tilt. ????
That wouldn't work. The configuration of atoms shown in the
animation http://members.optusnet.com.au/maxkeon/foil.gif
appears to open up only one path, being perpendicular to the
surface, but when the 3D picture is considered on the actual
incident face, that's not the case at all. The probe can be
driven down any path it chooses.
This could test your nuclear atom theory though.
Tilting the foil by at least 45 degrees relative to the probe
trajectory could give the result shown in this diagram. Nothing
passes through the scatter material. That would be impossible to
explain with the nuclear atom.
Source l
. . l Scintillation
. . l detector.
. . l
.
0000000
Nothing here.
I think that amount of shift from the perpendicular is a bit of
an overkill though. Much smaller deviations would probably give
the same result.
-----
Max Keon
glird
< 1.313e-10 meters emerges as the true proton radius.
Firstly, the force acting on the orbiting charge carrier at the
atomic radius has to be determined.
f = (Q1 * Q2)/(4 * pi * e0 * r ^ 2) does the trick.
f = 1.602e-19 * (1.602e-19 *1833)/(4*pi*8.8462e-12 * 1.313e-10 ^2)f =
2.476e-5 newtons at a radius of 1.313e-10 meters.
Next is the orbital speed : v = SQR(f * r / m)
m is the mass held in orbit at the atomic radius (9.11e-31 kg). v =
5.97e+7 m/sec.>
Let m = 9.11^-27 grams be the mass of an orbiting electron and let the
length of its path be
2pi r = 3.366709 -9 cm.
The orbital speed of an electron is
c' = cFs = 2.187768 cm/sec.
> The acceleration is : a = v ^ 2 / r : a = 2.71e+25 m/sec^2.
Or simply: N/m : 2.476e-5 / 9.11e-31 = 2.71e+25 m/sec^2.
That figure represents the global acceleration rate applied to
dimension everywhere around the 1.313e-10 meter radius.
The entire three dimensions are enclosed by coulomb forces at that
radius if space-time is distorted inward at the speed of light in all
directions.
The acceleration rate given above is close to the speed of light
cubed (2.7e+25 m/sec), which accounts for the volume of dimension
housed within the atomic radius. Of all possible numbers in the
universe, why did it land so close to the one that sets the
radius of the enclosed dimension at the atomic radius? Such an amazing
coincidence doesn't come along every day, does it. >
Here is another such "coincidence": Using the numerical values I
wrote above,
2 pi r m c' = 6.624^-27.
That is exactly the value of Plank's quantum of action.
> There is of course no electron orbiting the >perimeter of the proton.
The orbit is at the perimeter of the atom, not of its nucleus.
>And if there is no thermal energy field present
(0 degrees K), there's no reason why it can't be stationary at the
proton event horizon because the proton charge would be seen to be
coming from all directions parallel with that horizon. The electron
would be driven equally in all directions in that plane.
The story can't end here though. >
Although it's the wrong story [;-)] it doesn't end there.
A proton is a bodily spinning portion of continuous matter. As such,
it is a particle. Because a vector representing the quantity of
pressure against the surface is aslant relative to the proton, there
will be a decrease in the pressure against that surface.
Impelled by that grad p, the surrounding continuous resistively-
compressible matter will flow toward that interface from both sides.
Because there is a small amount of material inside the proton, the
density gradient within it will be very steep. However, there is an
unlimited amount of matter outside it, and the grad d will decrease by
the square of the distance, overall, but at a series of such patterns
when examined closely.
In an H atom, whose spinning nucleus in one proton,
there is only one such circumnuclear layer. Inside it, a nodal wave
system circulates, at c'. THAT is the "electron". When the frequency
of a "photon" is such that it matches the grad s-d pattern of such a
layer, the entire layer will escape. When it does, the local grad d in
that atom will have decreased and the proton will AUTOMATICALLY weigh
less, even though an electron has no weight of its own once it
escaped.
Btw, a gram is a unit of pressure and one of the units of measure of
force and weight and energy.
Since a "mass" is a quantity of matter, a gram is the wrong unit of
measure. THAT's why today's physicists misunderstand the physical
meaning of many of their own equations. Example:
In e = mc2 and m = m_r/(c2-v2).5, the m is measured in grams,
therefore denotes the WEIGHT of a given amount of atomic matter. (Raw
matter has no weight.)
Being unaware that continuous matter fills all spaces within and
around particles such as atoms, and thinking that a gram is a quantity
of matter rather than just its weight, physicists think that the above
two equations mean that MATTER and Energy are interconvertible. (They
aren't! Matter is one of the six basic items of which the universe is
made. Energy is something that matter does.)
< The black hole arising from coulomb forces is not the same as that
generated by gravity. Increasing the matter content of a
gravitationally induced black hole increases the black hole radius, as
does increasing the contained coulomb forces. But the contained
coulomb forces can fall to another level where a second event horizon
is established, which diminishes the force setting the initial event
horizon radius. >
Carried to its logical conclusion, perhaps that's why there is a
staggered series of density gradients around the nucleus of atoms. The
dinsity gradients are the seat and cause of gravitational force. The
mechanism is fully explained elsewhere.
Btw, since a gravitational FIELD is a density gradient, and since
that grad d is staggered - when examined closely each layer is about
double the thickness of the prior one, that's why the force of gravity
doesn't fall off by 1/r^2. It has nothing to do with a "fifth or sixth
primary force" at all.
Indeed, since a force is a net pressure, there is only one such
basic entity in nature.
< If interaction energy was not contained by the
enclosed dimension, there would be nothing to stop the entire set of
opposite charges from collapsing into individual points, which would
end up as one point when the initial event horizon collapses because
there are no coulomb forces left to sustain
it. >
A point has zero volume, and does not physically exist other than as
an imaginary tool of mathematics.
Matter does exist, and has several basic properties.
Here is one of them:
Ontropy: "Ontropy" denotes the property of matter such that it takes
increasingly greater increments of pressure to gain unit-volume
condensation per unit increase in dinsity, wherefore matter
increasingly rapidly increases its expansive pressure ["sorce"] as the
dinsity increases. It is an escalating curve. Ontropy includes the
converse of this; which is that the lower the dinsity the weaker the
expansive pressure and the more easily the material can be compressed.
One of the consequences of ontropy is that as the local dinsity
increases it will reach a level where there isn't enough pressure
available anywhere to increase it by meaningful amounts. Therefore,
regardless of how much a given portion of matter may be compressed it
will still have a volume; the more matter per portion the greater its
final size will be.
Regards,
glird
PD
> "PD" <TheDraperFam...@gmail.com> wrote:
> > On Jun 3, 10:48pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
> >> "PD" <TheDraperFam...@gmail.com> wrote:
> >>> On May 31, 10:25pm, "Max Keon" <maxk...@optusnet.com.au> wrote:
> ---
> >>>> This is the link that I had previously misplaced. But it doesn't
> >>>> seem to work any more. Perhaps you can find it somewhere else.
> >>>>http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/
> >>>> GeigerMarsden-1913.html
>
> >>> Max, this is a link to a HIGH SCHOOL physics course. I'm suggesting
> >>> that you look up the papers in physics journals, published by the
> >>> professional physicists.
>
> >> I found the correct address with a Google search for
> >> GeigerMarsden-1913. Good old Google aye. It's still in the same
> >> place, but has been relocated since 2004.
>
> html
>
> html is usually wordwrapped, but that's easy fixed.
>
> > No, this isn't the deep inelastic scattering experiment at SLAC I was
> > referring to.
>
> Of course not. But what has that to do with elastic scattering?
> That's how the atomic radius was presumed to have been
> determined wasn't it?
Not at all. Where are you getting your information?
Besides, atomic size has nothing to do with proton size. The latter
was found by deep inelastic scattering.
>
> This paragraph from my previous post really needs to be addressed.
>
> >> If this pixel '.' represents one hydrogen atom nucleus in a jar
> >> of liquid hydrogen, its nearest neighboring nucleus is 22000
> >> pixels away, or around 100 meters depending on screen size. And
> >> there are only two electrons that will pass through the void
> >> between them. That's what is proposed as the foundation of
> >> solids.
>
> 100 meters was supposed to be 10. It's closer to 20 on my screen
> though because the dot measures almost 1*1mm. But, how would you
> make that paragraph seem more realistic?
>
> As for this paragraph:
>
> >> If the foil is tilted at i.e. 20 degrees to the incident particle
> >> beam, a probe's initial deflection into the foil at the point of
> >> entry would set up a resonant oscillation that's added to its
> >> normal wobbly path and that would manifest itself at the exit
> >> point, deflecting at 20 degrees to the surface. But depending on
> >> the phase of the oscillation cycle, it can go either way in the
> >> plane of the tilt. The scatter pattern would show an enormous
> >> spike at 40 degrees to the incident beam in the plane of the
> >> tilt. ????
>
> That wouldn't work. The configuration of atoms shown in the
> animationhttp://members.optusnet.com.au/maxkeon/foil.gif