New Physics Based on Yoon's Universal Atomic Model

newedana wrote:
You don't know, nuclear fission and fusion are philosophically opposite reaction,

Nuclear fission and fusion are *physically* opposite reactions. They
have nothing to do with philosophy.


right? For example, meeting and departing are the same?

No.


You know only energy.

Why do you think so?



Do you know energy is a macroscopic concept?

It originally was, but it works also nicely in the microscopic world.


Your people reversely interprets the natural law like that way?

Huh?


If a reaction is exothermic there must occur a mass loss. Nonsense!

Why is that nonsense?


BTW: you have *still* not explained why in nuclear reactions, the lost
mass m is related to the released energy E by E=mc^2, *ACCORDING TO
OBSERVATIONS*.


We know when particle physicists encounter a logical crises they hide
themselves behind this energy shadow,

Huh?


under the name of quantitative calculation,

Hint: that are these things which Yoon is apparently unable to do.


done based on a stupid non-scientific postulation.

This "postulation" *AGREES WITH OBSERVATIONS*. So how could it be
stupid and non-scientific?

Why do you keep fleeing the experimental evidence?


Bye,
Bjoern

newedana wrote:

[snip]

When deBroglei equation is applied to a photon( QM theorists defined
photon has zero mass,

Wrong, we did not "define" that, we determined that experimentally.

so they defined arbitrarily, E=pc,

Wrong, we did not define that, we derived that mathematically from the
general equation E^2 = p^2 c^2 + m^2 c^4.


pc=hν) it
becomes 1/ν=h/mc, where λ=1/ν,

Right.


p=mc.

Utter nonsense.


So the E=mc^2 is established,

Utter nonsense.


combining with E=hν. Right??

No, wrong.

Thanks for showing yet again that you have not the faintest clue what
you are talking about.


Bye,
Bjoern

newedana wrote:
wrote:

why don't you just admit that you are Yoon himself, instead of
constantly using the referral 'according to Dr. Yoon'...



If you say so, I'm very flattered by you because Dr. Yoon's IQ is 250
or so according to Syd, the first author of the thread, "Yoonatom vs
Standard Model". om

And what makes you think that "Syd" is right on that?


[snip]


and what is M? be clear when you write something. and hey stupid,


Don't you know M? stupid! It's alphabet letter,....... L, "M", N, ....
There are other things too, like A, B, C, D, .......

If you were in my class, I'll present you F triple plus. You don't know
what F is, either? Don't ask me, do your own homework by yourself.
idiot.

Hint: asking you to define your terms is quite sensible. Your reaction
is childish.


why must i take v/c neglible? the equation E=mc^2 holds for a mass at any
speed, not only when vc. it is an exact relation.... go back to
Halliday & Resnick (a proper textbook!) .....



Are you still worshiping unscientific E=mc^2?

***********************************
****** I T W O R K S ! ! ! ******
***********************************

Explain that!


[snip]

Bye,
Bjoern

newedana wrote:
The energy emission taking place when orbital electron rings expand, can be observed in the case when chemical explosives such as TNT (trinitrotoluen) explodes. The outermost orbital electron rings of their component atoms contributing to combine them, expand only a little bit in this case of explosion, due to dissociation of TNT to form various kinds of gas molecules, such as H2O, CO2, and NO2 etc.
It is well known that the explosion of only about 7 kgs of uranium 235 produces an enormous energy equivalent to that emitted by explosion of TNT 20,000 metric tons. The mass ratio of these two explosive materials is about, 1 : 2.86x10^6.


If the orbital electron rings in K shell of uranium atom with radial parameter, say, γ=1/100, expands to be the orbital electron rings in K shell of newly created atoms, such as Pb that has radial parameter, say, γ=1/99.28, then the ratio of energy capacity of these two orbital electron rings becomes identical to the mass ratio, 2.86x10^6, as shown above when we estimate it with Eq.△E=E'[1/r^2 -1/(r+n)^2]^2. The difference of radial parameter between these two electron rings is negligibly small, or Δγ=1/99.28-1/100=1/13,789, but the ratio of their energy capacity is such enormous, as shown above. However, this energy emission comes only from the expansion of orbital electron rings in K shell of uranium 235. Other orbital electron rings in L, M, N,. . . .shells of uranium 235 should also have to expand their orbital radii emitting huge energies also as in the case of electron rings of K shell. Thus the explosion of only 7 kg of uranium 235 gives rise to producing such
a tremendous energy. newedanna wrote

Explain why the energy E released in uranium fission is related to the
mass lost in such a fission process, ***ACCORDING TO OBSERVATIONS***,
by E = mc^2.

Or shut up, you brain-dead idiot who doesn't care at all for
experimental evidence and can't face the real world.


Bye,
Bjoern

newedana wrote:
newedana wrote:

The energy emission taking place when orbital electron rings expand, can be observed in the case when chemical explosives such as TNT (trinitrotoluen) explodes. The outermost orbital electron rings of their component atoms contributing to combine them, expand only a little bit in this case of explosion, due to dissociation of TNT to form various kinds of gas molecules, such as H2O, CO2, and NO2 etc.

It is well known that the explosion of only about 7 kgs of uranium 235 produces such an enormous energy equivalent to that emitted by explosion of TNT 20,000 metric tons. The mass ratio of these two explosive materials is about, 1 : 2.86x10^6. If the orbital electron rings in K shell of uranium atom with radial parameter, say, γ=1/100, expands to be the orbital electron rings in K shell of newly created atoms that has radial parameter, say, γ=1/99.28, then the ratio of energy capacity of these two orbital electron rings becomes the same as the mass ratio, 2.86x10^6 , as shown above when we estimate it with Eq.△E=E'(1/r^4, previously posted. The difference of radial parameter between these two electron rings is negligibly small, or Δγ=1/99.28-1/100=1/13,789, but the ratio of their energy capacity is enormous, as shown above. However, this energy emission comes only from the expansion of orbital electron rings in K shell of uranium 235. Other orbital electron rings i
n L, M, N,. . . .shells of uranium 235 would also have to expand their orbital radii emitting huge energies also as in the case of electron rings of K shell. Thus the explosion of only 7 kg of uranium 235 gives rise to producing such a tremendous energy.

The fundamental mechanism of emitting energy from nuclear fusion of deuterons is exactly the same as that of nuclear fission of uranium 235. It is also the expanding energy of electron rings. In the case of nuclear fission atomic electron rings expand, while in the case of nuclear fusion nuclear electron rings associated in the structure of two deuterons expand, emitting nuclear energy. When nuclear electron rings of two deterons combine to build a unified nuclear electron ring with pair electrons, they have to expand their orbital radii emitting energy, in order to bind four protons to build two neutrons and two protons in a helium nucleus. This is the nuclear fusion energy. It is the same as that when two hydrogen atoms combine to form a hydrogen molecule having molecular electron rings carrying pair electrons, with their two single atomic electron rings, emitting energy.
A single nuclear electron ring that binds two protons in constructing a neuteron, can emit γ-rays at the nearest distance to its two nuclear protons when it breaks. Since this single nuclear electron ring can emit γ-rays with wavelength 0.005 Å, its radial parameter must be, γ=1/430, when we estimate it with the same equation, △E=E'(1/r^4) I posted above. If these single nuclear electron rings expand their orbital radii and emit energy equivalent to that energy given by explosion of 7 kgs of uranium 235, their radial parameter has to expand from γ=1/430 to γ=1/429.991. The distinction between them is, △γ=1/429.991-1/430=1/20,544,014. It is an awfully small expansion compared to that in the case of nuclear fission. However the energy emission in the process of nuclear fusion is the same as that in the case of nuclear fission. The ratio of atomic volume of deuterium and helium is, D : He = 14.1 : 31.8. The larger atomic volume of helium than deuterium is attrib
uted to the reason that the helium nucleus is stabilized in the lowest energy level than deuteron. Nuclear fusion energy includes also the energy of orbital electron rings of deuterium when they expand their orbital radii to be those of helium. newedana

Posting the same nonsense 10 times doesn't make it better.

p6 wrote:
Congrats Newedana. You made Dr. Yoon richer by $180.
I received the book yesterday.

Idiot.


Before you go on to so many topics. First justify that when electrons
move. The electric field at the front move back to the back. Dr. Yoon
said the field concept is wrong

He is free to explain then why Maxwell's equations work so well.


[snip]

Bye,
Bjoern

I ask all of your particle physicists who believe blindly the QM theory as God's theory. With that theory, can you explain what are gas, liquid and solid with amorphous and crystal structures, in the level of orbital electrons? Can you calculate and estimate the evaporating potential energy 529 cal/gr of H2O, and freezing potential energy 80 cal/gr, as well as its its freezing and boiling temperature and the highest density at temperature 4 degree C? Can you calculate why molecules or atoms build a characteristic crystal structure, and what is the essential force to make them arrange in a three dimensional order, excluding very effectively their different kinds? Can you explain with your quantitative calculation, what is the reason liquid is as incompressible as solid, but have fluidity, easy to change its shape? Can you explain the essence of gas and liquid viscosity, with your calculation? Can you explain what is the critical temperature and pressure when gas changes into liquids in atomic orbital electron level? I believe QM theory can explain nothing for this plain problems in physics. newedana

In article ,
Bjoern Feuerbacher wrote:
newedana wrote:

[snip]

When deBroglei equation is applied to a photon( QM theorists defined
photon has zero mass,

Wrong, we did not "define" that, we determined that experimentally.

so they defined arbitrarily, E=pc,

Wrong, we did not define that, we derived that mathematically from the
general equation E^2 = p^2 c^2 + m^2 c^4.

Rather, that was another way to get to the same result known from
classical electromagnetism. Either way, a theoretical conclusion with
experimental validation, not a definition.


--
"I fart for joy and I laugh more than if I had cast my old age, as a
serpent does its skin." -- Aristophanes, Peace, 421 BC

no need to lie p6 -- Zhayne -- Syd -- A.S. -- Q-on --
Qi-on -- What is Qi -- Landle -- Caltechdude -- Cinquirer , you
are not interested in good physics with explaining power, you are only
interested in non-quantitative and high-school-maths based theories, so
you don't need to work hard to understand. Do you even know what
divergence or flux is?

general equation E^2 = p^2 c^2 + m^2 c^4.

You do not know this equation came out from a fraudulent bases. Newedana