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