Re: Einstein's Mistakes

On Thu, 24 Oct 2024 21:42:00 +0000, J. J. Lodder wrote:

Paul B. Andersen <relativity@paulba.no> wrote:
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Den 23.10.2024 18:38, skrev rhertz:

On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:

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Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.

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This is not disputed!
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So why do you act as it is?
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And you know that this _confirms_ E = mc? because:

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  we know:
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Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
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All physicists knew that in 1939, obviously.
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https://www.atomicarchive.com/resources/documents/beginnings/nature_meitner

..html

Quote:
"It seems therefore possible that the uranium nucleus has only small
   stability of form, and may, after neutron capture, divide itself
   into two nuclei of roughly equal size (the precise ratio of sizes
   depending on finer structural features and perhaps partly on chance).
   These two nuclei will repel each other and should gain a total
kinetic
   energy of c. 200 Mev., as calculated from nuclear radius and charge."
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Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.

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Because this was the simplest way to estimate the released energy.
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Quote:
   "This amount of energy may actually be expected to be available
   from the difference in packing fraction between uranium and the
   elements in the middle of the periodic system."
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When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc?.

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So Meitner, like all physicists, took E = mc? for granted.
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You know this, because I told you 5 years ago.
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So why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc? for granted?

>

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These are excerpts from Serber's 1992, "Los Alamos Primer":
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Somehow the popular notion took hold long ago that Einstein's theory of
relativity, in particular his famous equation E = mc?, plays some
essential role in the theory of fission. Albert Einstein had a part in
alerting the United States government to the possibility of building an
atomic bomb, but his theory of relativity is not required in discussing
fission. The theory of fission is what physicists call a nonrelativistic
theory, meaning that relativistic effects are too small to affect the
dynamics of the fission process significantly.
Section 2 of the Primer gives a more exact calculation of the ratio of
the
energy released by the fission of a gram of uranium to the energy
released by the explosion of a gram of TNT.

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Even if the atom bomb could have been made without E = mc?,
the statement above shows that Serber, as all physicists,
knew E = mc?, they all took it for granted.
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Serber doesn't say that E = mc? is not a valid theory,
he says that E = mc? wasn't much help in making the atom bomb.
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So I ask you again:
Why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc? for granted?

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It's hopeless. RH is completely clueless when it comes to real physics.
(and he is unwilling to learn)

Are physicists willing to learn that the MMI actually shows that light
speed varies with the speed of the emitter if the Earth is moving in
space?
Woof-woof
What fools these apes be!
Bertietaylor

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E = mc? is now thoroughly experimentally verified, and the atom bomb
is part of the experimental evidence.

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Of course, but not really needed.
Mass spectroscopy was invented by J. J. Thomson in 1913,
and refined by his student, F. W. Aston. (discovering lots of isotopes)
In 1932, Kenneth Bainbridge pushed the accuracy of it to about 10^-4,
which was good enough to verify E = mc^2 directly, for atomic nuclei.
So the mass excess of the Uranium nucleus of about 200 MeV
was well known to 'everybody', well before WWII got started,
>
Jan