14th Experiments proving HYASYS; Correlations of nuclear sizes
Archimedes Plutonium
hahn@newshost (Karl Hahn) writes:
> In article <45n35g$s...@dartvax.dartmouth.edu> Archimedes...@dartmouth.edu (Archimedes Plutonium) writes:
>
> [deletia]
>
> [Feynman quote:]
> > vacuum). Now the point is that the potential energy is reduced if A
> > gets smaller, but the smaller A is, the higher the momentum required,
> > because of the uncertainty principle, and therefore the higher the
> > kinetic energy. The total energy is
> > E = h^2/2mA^2 - e^2/A. (38.10)
> >
> > We do not know what A is, but we know that the atom is going to arrange
> > itself to make some kind of compromise so that the energy is as little
> > as possible. In order to minimize E, we differentiate with respect to
> > A, set the derivative equal to zero, and solve for A. The derivative of
> > E is
> > dE/dA = -h^2/mA^3 + e^2/A^2, (38.11)
> >
> > and setting dE/dA = 0 gives for A the value
> >
> > A_0 = h^2/me^2 = 0.528 angstrom
> > = 0.528 x 10^-10 meter. (38.12)
> >
> > This particular distance is called the Bohr radius, and we have thus
> > learned that atomic dimensions are of the order of angstroms, which is
> > right: This is pretty good-- in fact, it is amazing, since until now we
> > have had no basis for understanding the size of atoms! Atoms are
> > completely impossible from the classical point of view, since the
> > electrons would spiral into the nucleus.
> > Now if we put the value (38.12) for A_0 into (38.10) to find the
> > energy, it comes out
> > E_0 = -e^2/2A_0 = -me^4/2h^2 = -13.6ev. (38.13)
> >
> [deletia]
>
> > --- end of quoting of Feynman Lectures, vol 1, page 38-6 ---
> >
> > So, momentum P, where P = mc
> >
> > then Uncertainty Principle (UP) we have h/P = h/mc = Compton
> > wavelength
> >
> > Thus, to get around UP, or better yet, put UP to work. We can
> > calculate what the mass of a nuclear electron is, in order for it to
> > hold together say the 4 protons of helium 4@2 by the 2 nuclear
> > electrons. Would the mass of the 2 nuclear electrons be muon masses or
> > would they be tau masses?
> >
> > This is all pretty for the UP predicts what the masses of nuclear
> > electrons must be in order to be inside the nucleus, or inside the
> > individual protons moving very rapidly from one proton to another in
> > order to strong force hold them together. Understand that the Coulombic
> > nuclear strong force of nuclear electrons is 83 times stronger than
> > normal Coulomb force.
>
> You mean you cannot do this yourself, AP, with all your math expertise??
> Allow me to assist:
>
> A neutron size is on the order of 1e-15 meters. Substitute that into
> the right hand side of equation 38.12. Since both h and e are
> constants, m must change. In fact it must grow by a factor of
> 0.528e-10 meters / 1e-15 meters, or in other words, by a factor of
> 52800. That would place its mass higher by an order of magnitude
> than a tau and higher by two orders of magnitude than a muon. So
> tell me, AP, is your conjecture predicting the existence of a new
> lepton of rest energy in the order of 26000 MeV (nearly 30 proton
> masses)? And if so, why does a neutron weigh far less than 30 proton
> masses, since according to material that you posted above (and I
> agree with) the electron (or whatever you claim is the proton's
> partner in your model of a neutron would have to have that amount
> of mass?
The 14th experiments is very much math oriented. Correlate all the
isotopes as to size and shape of the nucleus. We do have modern
techniques to give sizes of nucleuses.
The above shows that isotopes containing 30 hadrons should have a
radius of 0.528e-10 meters assuming all neutrons are of the same size
of 1e-15 meters.
Using the same calibration scheme then the size of the stable
deuterium nucleus and the stable helium nucleus is calculable and then
confirmed against observations. And most importantly the same
calibration scheme should agree with the nuclear size of the alpha
particle containing 4 hadrons, thus the factor should yield approx 4000
MEV and correlate with the size of the alpha particle.
This correlation scheme should agree with the nuclear geometry of
all isotopes assuming neutrons have the same size.
Archimedes Plutonium
alt.sci.physics.plutonium-- ceases and desists-- I have no choice but
just repost all and every one of them so as to keep my train of
thought.
In article <466gvs$j...@dartvax.dartmouth.edu>
Karl Hahn
> In article <466gvs$j...@dartvax.dartmouth.edu>
> Archimedes...@dartmouth.edu (Archimedes Plutonium) writes:
>
> > In article <951017...@are107.lds.loral.com>
> > hahn@newshost (Karl Hahn) writes:
> >
> > > In article <45n35g$s...@dartvax.dartmouth.edu> Archimedes...@dartmouth.edu (Archimedes Plutonium) writes:
> > >
> > > [deletia]
> > >
> > > [Feynman quote:]
[Feynman quote, some AP stuff and some of my stuff deleted]
> > > A neutron size is on the order of 1e-15 meters. Substitute that into
> > > the right hand side of equation 38.12. Since both h and e are
> > > constants, m must change. In fact it must grow by a factor of
> > > 0.528e-10 meters / 1e-15 meters, or in other words, by a factor of
> > > 52800. That would place its mass higher by an order of magnitude
> > > than a tau and higher by two orders of magnitude than a muon. So
> > > tell me, AP, is your conjecture predicting the existence of a new
> > > lepton of rest energy in the order of 26000 MeV (nearly 30 proton
> > > masses)? And if so, why does a neutron weigh far less than 30 proton
> > > masses, since according to material that you posted above (and I
> > > agree with) the electron (or whatever you claim is the proton's
> > > partner in your model of a neutron would have to have that amount
> > > of mass?
> >
> >
> > The 14th experiments is very much math oriented. Correlate all the
> > isotopes as to size and shape of the nucleus. We do have modern
> > techniques to give sizes of nucleuses.
> > The above shows that isotopes containing 30 hadrons should have a
> > radius of 0.528e-10 meters assuming all neutrons are of the same size
> > of 1e-15 meters.
Where do you arrive at this? Experiments of nuclear scattering of electrons
and protons have yielded the following formula for the size of a nucleus:
r = k * A^(1/3)
or equvalently:
A = (r/k)^3
where r is the radius, A is the atomic mass number (i.e. the sum of the
number of neutrons and protons), and k is on the order of 1e-15 meters.
You can find this in _Elementary Modern Physics_ by Weidner & Sells.
So a nucleus of 30 nucleons would have a size of about 3e-15 meters.
To get a nuclear radius of 0.528e-10 meters you would need 1.5e14 nucleons.
I have yet to see evidence of nucleus over 300 nucleons.
[deletia]
--
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| (^ (`> | Man gotta sit and wonder why, why, why.
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| \< ) |
| ( / | Kurt Vonnegut Jr.
| | |
| ^ ha...@lds.loral.com my opinions need not be Loral's
Archimedes Plutonium
hahn@newshost (Karl Hahn) writes:
> > > The above shows that isotopes containing 30 hadrons should have a
> > > radius of 0.528e-10 meters assuming all neutrons are of the same size
> > > of 1e-15 meters.
>
> Where do you arrive at this? Experiments of nuclear scattering of electrons
> and protons have yielded the following formula for the size of a nucleus:
>
> r = k * A^(1/3)
>
> or equvalently:
>
> A = (r/k)^3
>
> where r is the radius, A is the atomic mass number (i.e. the sum of the
> number of neutrons and protons), and k is on the order of 1e-15 meters.
> You can find this in _Elementary Modern Physics_ by Weidner & Sells.
> So a nucleus of 30 nucleons would have a size of about 3e-15 meters.
> To get a nuclear radius of 0.528e-10 meters you would need 1.5e14 nucleons.
> I have yet to see evidence of nucleus over 300 nucleons.
How accurate is that formula Karl? How accurately has the radius of
the neutron and muon been measured?
Archimedes Plutonium
Archimedes...@dartmouth.edu (Archimedes Plutonium) writes:
> In article <951025...@are107.lds.loral.com>
> hahn@newshost (Karl Hahn) writes:
>
> > > > The above shows that isotopes containing 30 hadrons should have a
> > > > radius of 0.528e-10 meters assuming all neutrons are of the same size
> > > > of 1e-15 meters.
> >
> > Where do you arrive at this? Experiments of nuclear scattering of electrons
> > and protons have yielded the following formula for the size of a nucleus:
> >
> > r = k * A^(1/3)
> >
> > or equvalently:
> >
> > A = (r/k)^3
> >
> > where r is the radius, A is the atomic mass number (i.e. the sum of the
> > number of neutrons and protons), and k is on the order of 1e-15 meters.
> > You can find this in _Elementary Modern Physics_ by Weidner & Sells.
> > So a nucleus of 30 nucleons would have a size of about 3e-15 meters.
> > To get a nuclear radius of 0.528e-10 meters you would need 1.5e14 nucleons.
> > I have yet to see evidence of nucleus over 300 nucleons.
>
> How accurate is that formula Karl? How accurately has the radius of
> the neutron and muon been measured?
That formula is merely a volume of sphere x density is proportional
to A. But it assumes nuclei are packed marbles. Any formula of the
nuclear radius must contend with the strong nuclear force. At what
point does the strong nuclear force apply?
How to define the collapsed nuclear radius? What is the radius of a
neutron where the neutron has no magnetic field? Trouble with these
nuclear measurements is to use the Uncertainty Principle so that it
aids in a successful measurement of the nuclear radii instead of UP
hindering and destroying the measurement. Find some confinement or cage
of the nuclear radii. Take for example the space of an electron
governed by UP then you can confine the radius to saying where 90% of
the probability density distribution is.
I propose such a confinement or cage of the nucleus. The s orbitals.
Karl, is there a formula for 90% density distribution of the electrons
in s orbitals. I would think that the 7s orbital radius of 90% would be
much larger than the 1s. I would think no nucleus would be larger than
its s orbitals and if it is, for a very brief time.
Archimedes Plutonium
alch...@pacificnet.net (John Milligan) writes:
> I would assume that the neutron has been measured fairly accurately. The
> muon, however, has such a short lifetime that any "measurements" of its radius
> are probably estimates.
>
> The value that A.P. has quoted here is the Bohr radius. It is not the radius
> of the nucleus but of the 1s orbital in the hydrogen atom. An atom with 30
> nucleons would have the radius of say phosphorus which actually has 31
> nucleons in its most abundant isotope. Phosphorus has an atomic radius of
> 0.93 angstroms in one allotrope and 1.15 angstroms in the other. This is
> about 30% larger than the atomic radius of hydrogen which is measured to be
> 0.78 angstroms. And this is again about 50% larger than the value quoted by
> A.P.
>
> If the nucleus of an atom with 30 nucleons, like P, had a radius of 0.528
> angstroms, the atomic radius would be on the order of 5 microns, since the
> atomic radius is fairly accurately measured to be about 5 orders of magnitude
> larger than the nuclear radius. This atoms would be visible in a simple light
> microscope. Thus another blow has been struck against HYASYS, which is a
> stupid name anyway.
It is a beautiful name, a fantastically beautiful name and was meant
to be. In fact it will be around as long as the name 'quantum' is
around. Who was that early scientist who said all things ultimately
were made up of water? Was it Thales? Water is hydr and where we get
the name hydrogen. HYASYS is the abbreviation for Hydrogen Atom
Systems. And HYASYS is the strong nuclear force. Hydrogen thus in
effect builds the entire universe from the lowly proton and electron of
hydrogen. HYASYS builds all atoms and it is the reason of the Strong
Nuclear force. A beautiful name indeed for it has some of the heritage
and history of science inbodied within it.
Every time I see Hyasys I am reminded of the beautiful and gorgeous
hyacinth flower and that family of flowers. It is your name John
Milligan which is an eye sore to me and many others.
Karl Hahn
> In article <951025...@are107.lds.loral.com>
> hahn@newshost (Karl Hahn) writes:
>
> > > > The above shows that isotopes containing 30 hadrons should have a
> > > > radius of 0.528e-10 meters assuming all neutrons are of the same size
> > > > of 1e-15 meters.
> >
> > Where do you arrive at this? Experiments of nuclear scattering of electrons
> > and protons have yielded the following formula for the size of a nucleus:
> >
> > r = k * A^(1/3)
> >
> > or equvalently:
> >
> > A = (r/k)^3
> >
> > where r is the radius, A is the atomic mass number (i.e. the sum of the
> > number of neutrons and protons), and k is on the order of 1e-15 meters.
> > You can find this in _Elementary Modern Physics_ by Weidner & Sells.
> > So a nucleus of 30 nucleons would have a size of about 3e-15 meters.
> > To get a nuclear radius of 0.528e-10 meters you would need 1.5e14 nucleons.
> > I have yet to see evidence of nucleus over 300 nucleons.
>
> How accurate is that formula Karl? How accurately has the radius of
> the neutron and muon been measured?
The library at Dartmouth ought to have a copy of Weidner & Sells. It is
quite a popular textbook. I don't have my copy with me, but here's what
I recollect from the discussion of this. That the measurements were made
with several different projectile particles (i.e. electrons and protons).
The 1/3 power rule worked out to within 10% or better. The value of k
varied slightly among the different projectile particles -- with a variation
of about +/-20%.
All the forumula really says is that the volume of the nucleus is the
sum of the volumes of its constituents.
As far as the radius of a bare neutron -- that is a more difficult experiment.
It's hard to assemble a large array of bare neutrons to use as targets.
So I don't know if it has been measured.
The muon is a lepton, and therefore ought to behave like an electron, which
is to say that it is a point source of charge.