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New Physics Based on Yoon's Universal Atomic Model



 
 
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  #71  
Old May 27th 05, 04:59 PM
Bjoern Feuerbacher
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newedana wrote:
May 26, 3:38 am show options

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Date: 26 May 2005 00:38:35 -0700
Local: Thurs,May 26 2005 3:38 am
Subject: New Physics Based on Yoon's Universal Atomic Model
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hi there Hansik Yoon, unrepentant crank.........




You have to know the stupid origin of equation, E=Mc^2.


I know it. You obviously don't


According to Dr.Yoon, it starts from the special theory of relativity for mass,


There is no "special theory of relativity for mass".

Thanks for confirming once again that you don't know what you are
talking about.


m=m'(1-v^2/c^2)^-1/2


That formula is *vaguely* right - but E=mc^2 does *not* come from it.


If we expand this as a poly-nominal series, it gives,
m=m'[1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . . .]


Congratulations, you got at least that right.


Since v/c is negligibly small as in usual, we can eliminate after third term.


Right again.


Thus the simplified equation becomes, (m-m')c^2 =1/2m'v^2=E, and E=Mc^2.


That is *not* the usual way of deriving E=mc^2. Try again.


Do you believe this equation can explain the atomic nuclear energy?


It can explain at least part of it. BTW, please make up your mind: do
you want to talk about atomic or about nuclear energy?


nonsense!


Why?

It is a *fact* that the energy released in nuclear reactions is equal
to the mass difference between the nuclei in the beginning and in the
end times c^2. Do you dispute that?


Bye,
Bjoern
  #72  
Old May 27th 05, 05:12 PM
Bjoern Feuerbacher
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newedana wrote:
Fission has nothing to do with electrons. You're an idiot.



hi there Hansik Yoon, unrepentant crank.........



I newedana posted on May 26, 3;38, you have to know the stupid origin of
equation, E=Mc^2.


And then you displayed nicely that you don't know its origin. Thanks
for playing.


According to Dr.Yoon, it starts from the special theory of relativity for mass,
m=m'(1-v^2/c^2)^-1/2. If we expand this to be a poly-nominal series, it gives
m=m'[1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . . .]. Since v/c is negligibly small as
in usual, we can eliminate after third term. Thus the simplified equation
becomes, (m-m')c^2 =1/2m'v^2=E, and E=Mc^2. Do you believe this equation can
explain the atomic nuclear energy? nonsense!


Then you replied we cannot cancel v/c.


No, he did not say that. Try again.




If so, you have to know another way of proving the stupidity of E=mc^2.


How could one prove the stupidity of an equation which isn't stupid?


Dr.Yoon ridiculed both deBroglei equation, λ=h/p,


Then how does he explain that it agrees with experiment?

and the key equation of your particle physicists, E=hν.


1) That's not our "key equation".
2) Then how does he explain that it agrees with experiment?


From these two stupid equations


What's stupid about them?


another stupid equation, E=mc^2 is straightly deduced.


Plain nonsense. E=mc^2 does in no way follow from the previous two
equations.

Thanks yet again for demonstrating that you have no clue what you are
talking about.


You see?


Yes, I see that you are an ignorant, arrogant idiot.



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


Wrong. We do not "define" that, we obtain that result from
experimental data.

Thanks yet again for demonstrating that you have no clue what you are
talking about.


so they defined arbitrarily, E=pc, pc=hν)


This is not an arbitrary definition, this follows straightforwardly
from E^2 = p^2 c^2 + m^2 c^4.

Thanks yet again for demonstrating that you have no clue what you are
talking about.


it becomes 1/ν=h/mc, where λ=1/ν, p=mc.


Plain utter *nonsense*. p=mc does *not* hold for photons.


So the E=mc^2 is established,
combining with E=hν. Right??


No, wrong. Utterly wrong. Plain nonsense. Bull****.

Thanks yet again for demonstrating that you have no clue what you are




Fission has nothing to do with electrons. You're an idiot.



You are quite free to believe such a stupid equation, E=Mc^2.


IT WORKS!


But you have to reconsider to use this equation E=mc^2, if you teach
younger generations of your science disciples who are innocently eager
to know what is the atomic nuclear energy.


Explaining what this energy *is* has little to do with that equation.

Thanks yet again for demonstrating that you have no clue what you are


If you explain atomic fission and fusion energy with the same, E=mc^2,
you have to realized that you become also stupid.


Indeed, someone who did do that would indeed be stupid. Fortunately,
no one ever did. That formula can be used to calculate the energy
which is released, but it is *not* used for explaining the *processes*
themselves!

Thanks yet again for demonstrating that you have no clue what you are



According to current physics, atomic fission and fusion are philosophically
opposit reactions,


Physics is not about "philosophically opposite".


in the former case, mass gain occurs, while in the later case, mass deficit or loss,


Wrong yet again.

Thanks yet again for demonstrating that you have no clue what you are


due to nuclear reaction. Despite that both reactions are the same exothermic.


Indeed, both are exothermic, and in *both* mass is lost. And that mass
which is lost is related to the released energy by E=mc^2. That is
an *experimental* *fact*. Live with it.


Gained and defisitted mass are transformed alike into energy.


Plain nonsense. Only *lost* mass is transformed into energy. According
to E=mc^2.


It is not a science but a kind of funny comics!


Yes, your straw men are indeed a kind of funny comics.

What about learning what physics *actually* says, for a change?


Dr. Yoon explains elegantly both nuclear reactions with atomic electron
rings and nuclear electron rings


And how does he explain that the lost mass and the released energy,
***ACCORDING TO OBSERVATIONS***, are related by E=mc^2, if he claims
that that formula is wrong?


without violating any natural laws, unlike your particle physicists do
desperately.


What natural laws do we violate, in your opinion?


Then you would rebut, how electrons can be in a nuclear structure, forming ypur
strange nuclear electron ring? Yes, it is quite possible.


1) It is not possible.
2) Experiments have shown clearly that it indeed does not happen. Read
up on "Hofstadter".


Evidence is the β-ray
electrons ejected out from radioactive atomic nuclei, carrying a huge energy.


In what way is that evidence for electrons in the nuclei?


Dr Yoon defined this nuclear electron ring to act as the nuclear strong force,


What on earth is "defined" supposed to mean here?


possible to bind a number of protons in atomic nuclei against their repulsions.


How does he explain the observed saturation?


I recommend you better read his textbook(www.yoonsatom.net)if you want to know
more details, what is the origin of the rest of α and γ rays. newedana wrote


Why should we read the book of someone who obviously has no clue of
what physics
actually says, and, worse, what observations actually say?


Bye,
Bjoern
  #73  
Old May 28th 05, 02:06 AM
newedana
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In order to get hydrogen spectrum in a usual way is to bombard orbital ele=
ctrons of hydrogen atoms with external electrons accelerated.
And how does the manage to get the Rydberg frequency?
Niels Bohr obtained Rydberg frequency theoretically based on his atomic mo=

del very precisely, but it is quite natural since Planck's constant h is or=
iginally estimated from observed Rydberg frequency, and ionizing energy E i=
s artificially modulated to be, E=3D hc=CE=BD=3D2.18x10^-18 J, where h: Pla=
nck constant, c: speed of light, =CE=BD: Rydberg frequency. Don't be cheate=
d so nicely!
Balmer equation is an empirically built equation, but Dr.Yoon's equation i=

s theoretically deduced one, based on the energy of orbital electron ring, =
E=3DE1 + E2, E: total energy, E1: energy for maintaining atomic structure, =
or circling energy of orbital electron around its nucleus keeping its radiu=
s constant until its energy capacity is saturated, E2: precessing energy by=
absorbing external radiation, which is extractable as radiations. In quant=
um atomic physics, as you may know, the energy of orbital electron is E=3D(=
1/2+n)hv, in which 1/2h=CE=BD is to maintain atomic structure,which is nice=
ly modulated by handwaving, while nh=CE=BD is extractable energy. How primi=
tive is this equation? newedana

  #74  
Old May 28th 05, 02:08 AM
newedana
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You have to know the stupid origin of equation, E=Mc^2. According to Dr.Yoon, it starts from the special theory of relativity for mass, m=m'(1-v^2/c^2)^-1/2. If we expand this as a poly-nominal series, it gives, m=m'[1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . . .] Since v/c is negligibly small as in usual, we can eliminate after third term. Thus the simplified equation becomes, (m-m')c^2 =1/2m'v^2=E, and E=Mc^2. Do you believe this equation can explain the atomic nuclear energy? nonsense! Newedana

  #75  
Old May 28th 05, 02:10 AM
newedana
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You are quite free to believe such a stupid equation, E=Mc^2. But you have to reconsider if you are going to teach younger generations of science students who are eager to know what is the atomic nuclear energy.
If you explain atomic fission and fusion energy you have to utiize this same stupid equation E=mc^2. However, according to current physics, atomic fission and fusion are philosopically opposit reaction, right???. But both reactions are exothermic. It is not a science but a comics. Yoon explains both nuclear reactions with atomic electron rings and nuclear electron rings without violating any natural law, unlike your particle physicists do. Newedana


  #76  
Old May 28th 05, 02:12 AM
newedana
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I newedana posted on May 26, 3;38, you have to know the stupid origin of e=
quation, E=3DMc^2.
According to Dr.Yoon, it starts from the special theory of relativity for =

mass, m=3Dm'(1-v^2/c^2)^-1/2. If we expand this to be a poly-nominal series=
, it gives, m=3Dm'[1+1/2(v/c)^2 + 3/8(v/c)^4 +. . . . .]. Since v/c is negl=
igibly small as in usual, we can eliminate after third term. Thus the simpl=
ified equation becomes, (m-m')c^2 =3D1/2m'v^2=3DE, and E=3DMc^2. Do you bel=
ieve this equation can explain the atomic nuclear energy? nonsense!
Then you replied we cannot cancel v/c.
If so, you have to know another way of proving the stupidity of E=3Dmc^2.
Dr.Yoon ridiculed both deBroglei equation, =CE=BB=3Dh/p, and the key equat=

ion of your particle physicists, E=3Dh=CE=BD. From these two stupid equatio=
ns another stupid equation, E=3Dmc^2 is straightly deduced. You see?

When deBroglei equation is applied to a photon( QM theorists defined

photon has zero mass, so they defined arbitrarily, E=3Dpc, pc=3Dh=CE=BD) it
becomes 1/=CE=BD=3Dh/mc, where =CE=BB=3D1/=CE=BD, p=3Dmc. So the E=3Dmc^2 i=
s established,
combining with E=3Dh=CE=BD. Right??

  #77  
Old May 28th 05, 02:16 AM
newedana
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You are quite free to believe such a stupid equation, E=3DMc^2. But you ha=
ve to reconsider to use this equation E=3Dmc^2, if you teach younger gener=
ations of your science disciples who are innocently eager to know what is t=
he atomic nuclear energy. If you explain atomic fission and fusion energy w=
ith the same, E=3Dmc^2, you have to realized that you become also stupid.
According to current physics, atomic fission and fusion are philosophicall=

y opposit reactions, in the former case, mass gain occurs, while in the lat=
er case, mass deficit or loss, due to nuclear reaction. Despite that both r=
eactions are the same exothermic. Gained and defisitted mass are transforme=
d alike into energy. It is not a science but a kind of funny comics! Dr. Yo=
on explains elegantly both nuclear reactions with atomic electron rings and=
nuclear electron rings without violating any natural laws, unlike your par=
ticle physicists do desperately. Then you would rebut, how electrons can be=
in a nuclear structure, forming ypur strange nuclear electron ring? Yes, i=
t is quite possible. Evidence is the =CE=B2-ray electrons ejected out from =
radioactive atomic nuclei, carrying a huge energy. Dr Yoon defined this nuc=
lear electron ring to act as the nuclear strong force, possible to bind a n=
umber of protons in atomic nuclei against their repulsions.
I recommend you better read his textbook(www.yoonsatom.net)if you want to =

know more details, what is the origin of the rest of =CE=B1 and =CE=B3 rays=
.. newedana wrote

  #78  
Old May 28th 05, 02:35 AM
newedana
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Posts: n/a
Default

You are quite free to believe such a stupid equation, E=3DMc^2. But you h=
ave to reconsider to use this equation E=3Dmc^2, if you teach younger gene=
rations of your science disciples who are innocently eager to know what is =
the atomic nuclear energy. If you explain atomic fission and fusion energy =
with the same, E=3Dmc^2, you have to realized that you become also stupid.
According to current physics, atomic fission and fusion are philosophicall=

y opposit reactions, in the former case, mass gain occurs, while in the lat=
er case, mass deficit or loss, due to nuclear reaction. Despite that both r=
eactions are the same exothermic. Gained and deficit mass are transformed a=
like into energy? bah! It is not a science but a kind of funny comics! Dr. =
Yoon explains elegantly both nuclear reactions with atomic electron rings a=
nd nuclear electron rings without violating any natural laws, unlike your p=
article physicists do desperately. Then you would rebut, how electrons can =
be in a nuclear structure, forming your strange nuclear electron ring? Yes,=
it is quite possible. Evidence is the =CE=B2-ray electrons ejected out fro=
m radioactive atomic nuclei, carrying a huge energy. Dr Yoon defined this n=
uclear electron ring to act as the nuclear strong force, possible to bind a=
number of protons in atomic nuclei against their repulsions.
I recommend you better read his textbook(www.yoonsatom.net)if you want to =

know more details, what is the origin of the rest of =CE=B1 and =CE=B3 rays=
.. newedana wrote

  #79  
Old May 28th 05, 03:15 AM
newedana
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You don't know, nuclear fission and fusion are philosophically opposite reaction, right? For example, meeting and departing are the same? You know only energy. Do you know energy is a macroscopic concept? Your people reversely interprets the natural law like that way? If a reaction is exothermic there must occur a mass loss. Nonsense! We know when particle physicists encounter a logical crises they hide themselves behind this energy shadow, under the name of quantitative calculation, done based on a stupid non-scientific postulation. newedana

  #80  
Old May 28th 05, 08:57 AM
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newedana wrote:
[snip a lot of non-sensical rant]

you are boring. go away Yoon.

 




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