On Mon, 21 Oct 2024 16:50:30 +0000, LaurenceClarkCrossen wrote:
E= mc^2 has never been the formula to calculate the energy released in
atomic explosions.
Serber and Oppenheimer taught (1943) several waves of scientists that
joined the Manhattan Project with lectures prior joining any team, and
Serber wrote the booklet "Los Alamos Primer", where they taught that
energy liberated in the fission of the atom was of electrostatic nature,
due to the repulsion of the byproducts of the fission, plus radiation.
The numbers given in the booklet are almost a match with E=mc^2
calculations of that epoch, but have NOTHING TO DO with relativity, as
Serber wrote in his 1992 book, once some data of the project was cleared
for publishing.
But relativists, since 1945 (Time Magazine cover) hyper-hyped the figure
of Einstein and relativism, and the cult succeeded in re-writing
history.
E= mc^2 has never been the formula to calculate the energy released in
atomic explosions.
Dr. Serber, 2nd. in command after Oppenheimer, CLEARLY AFFIRMED (in his
1992 book, published once he got a security clearance) that the atomic
bomb WAS COMPLETELY UNRELATED WITH RELATIVITY AND E=mc^2.
He WROTE "Los Alamos Primer" in 1943, to teach recruited scientists
about the power behind atom fission, and MADE CALCULATIONS that were
present in the booklet, giving results close to 170 MeV of KE plus some radiation. And the calculations were done by using COULOMB'S FORCES in electrostatic repulsion, once two (or more) new heavy elements were
produced in the atom fission.
splitting of the uranium nucleus has the same origins as in the case of
the [alpha-]decay of a radium nucleus, i.e. mainly from the
electrostatic repulsion of the two parts into which the atomic nucleus
is split. The energy released by an atomic explosion thus comes directly
from this source and does not derive from a conversion of mass into
energy.'- Heisenberg" - "Catalogue of Errors for Both Theories of
Relativity"
Starmaker: The mass-energy relation was already well known long before Einstein imagined he thought it up all by himself.
On Mon, 21 Oct 2024 16:50:30 +0000, LaurenceClarkCrossen wrote:
E= mc^2 has never been the formula to calculate the energy released in atomic explosions.
Serber and Oppenheimer taught (1943) several waves of scientists that
joined the Manhattan Project with lectures prior joining any team, and
Serber wrote the booklet "Los Alamos Primer", where they taught that
energy liberated in the fission of the atom was of electrostatic nature,
due to the repulsion of the byproducts of the fission, plus radiation.
The numbers given in the booklet are almost a match with E=mc^2
calculations of that epoch, but have NOTHING TO DO with relativity, as
Serber wrote in his 1992 book, once some data of the project was cleared
for publishing.
But relativists, since 1945 (Time Magazine cover) hyper-hyped the figure
of Einstein and relativism, and the cult succeeded in re-writing
history.
rhertz <hertz778@gmail.com> wrote:
On Mon, 21 Oct 2024 16:50:30 +0000, LaurenceClarkCrossen wrote:
E= mc^2 has never been the formula to calculate the energy released in
atomic explosions.
Serber and Oppenheimer taught (1943) several waves of scientists that
joined the Manhattan Project with lectures prior joining any team, and
Serber wrote the booklet "Los Alamos Primer", where they taught that
energy liberated in the fission of the atom was of electrostatic nature,
due to the repulsion of the byproducts of the fission, plus radiation.
The numbers given in the booklet are almost a match with E=mc^2
calculations of that epoch, but have NOTHING TO DO with relativity, as
Serber wrote in his 1992 book, once some data of the project was cleared
for publishing.
But relativists, since 1945 (Time Magazine cover) hyper-hyped the figure
of Einstein and relativism, and the cult succeeded in re-writing
history.
Your lack of basic understanding of physics is showing again.
Conservation laws are about the overall result.
Equations of motion are about what forms
the (conserved!) energy appears in, in the reaction products.
Serber and Oppenheimer of course talked to people who would understand,
(so not to you)
Jan
Paul Anderson: You seem to be assuming what you wish to conclude because Heisenberg says this fission process is not a conversion of matter into energy:
"'The energy associated with the
splitting of the uranium nucleus has the same origins as in the case of
the [alpha-]decay of a radium nucleus, i.e. mainly from the
electrostatic repulsion of the two parts into which the atomic nucleus
is split. The energy released by an atomic explosion thus comes directly
from this source and does not derive from a conversion of mass into
energy.'- Heisenberg" - "Catalogue of Errors for Both Theories of
Relativity"
On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:
Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.
And you know that this _confirms_ E = mc² because:
Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
All physicists knew that in 1939, obviously.
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."
Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.
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."
When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc².
You know this, because I told you 5 years ago.
So why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc² for granted?
These are excerpts from Serber's 1992, "Los Alamos Primer":
Somehow the popular notion took hold long ago that Einstein’s theory of relativity, in particular his famous equation E = mc², 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.
Den 23.10.2024 18:38, skrev rhertz:.html
On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:
Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.
This is not disputed!
So why do you act as it is?
And you know that this _confirms_ E = mc? because:
we know:
Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
All physicists knew that in 1939, obviously.
https://www.atomicarchive.com/resources/documents/beginnings/nature_meitner
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." >>
Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.
Because this was the simplest way to estimate the released energy.
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."
When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc?.
So Meitner, like all physicists, took E = mc? for granted.
You know this, because I told you 5 years ago.
So why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc? for granted?
These are excerpts from Serber's 1992, "Los Alamos Primer":
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.
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.
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.
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?
E = mc? is now thoroughly experimentally verified, and the atom bomb
is part of the experimental evidence.
Paul B. Andersen <relativity@paulba.no> wrote:
Den 23.10.2024 18:38, skrev rhertz:..html
On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:
Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.
This is not disputed!
So why do you act as it is?
And you know that this _confirms_ E = mc? because:
we know:
Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
All physicists knew that in 1939, obviously.
https://www.atomicarchive.com/resources/documents/beginnings/nature_meitner
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." >>>>
Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.
Because this was the simplest way to estimate the released energy.
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."
When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc?.
So Meitner, like all physicists, took E = mc? for granted.
You know this, because I told you 5 years ago.
So why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc? for granted?
These are excerpts from Serber's 1992, "Los Alamos Primer":
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.
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.
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.
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?
It's hopeless. RH is completely clueless when it comes to real physics.
(and he is unwilling to learn)
E = mc? is now thoroughly experimentally verified, and the atom bomb
is part of the experimental evidence.
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
Paul B. Andersen <relativity@paulba.no> wrote:
Den 23.10.2024 18:38, skrev rhertz:.html
On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:
Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.
This is not disputed!
So why do you act as it is?
And you know that this _confirms_ E = mc? because:
we know:
Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
All physicists knew that in 1939, obviously.
https://www.atomicarchive.com/resources/documents/beginnings/nature_meitner
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."
Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.
Because this was the simplest way to estimate the released energy.
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."
When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc?.
So Meitner, like all physicists, took E = mc? for granted.
You know this, because I told you 5 years ago.
So why do you pretend to be ignorant of the fact that all physicists
(and chemists) at the time took E = mc? for granted?
These are excerpts from Serber's 1992, "Los Alamos Primer":
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.
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.
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.
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?
It's hopeless. RH is completely clueless when it comes to real physics.
(and he is unwilling to learn)
E = mc? is now thoroughly experimentally verified, and the atom bomb
is part of the experimental evidence.
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
Den 22.10.2024 22:45, skrev LaurenceClarkCrossen:
Paul Anderson: You seem to be assuming what you wish to conclude because
Heisenberg says this fission process is not a conversion of matter into
energy:
"'The energy associated with the
splitting of the uranium nucleus has the same origins as in the case of
the [alpha-]decay of a radium nucleus, i.e. mainly from the
electrostatic repulsion of the two parts into which the atomic nucleus
is split. The energy released by an atomic explosion thus comes directly >>> from this source and does not derive from a conversion of mass into
energy.'- Heisenberg" - "Catalogue of Errors for Both Theories of
Relativity"
Heisenberg's statement shows that he knew E = mc².
An atom isn't spitted because some mass decides to convert
itself into energy. It is not the _cause_.
When an atom is spitted, the kinetic energy comes from
the repulsion of the protons, obviously.
But as a _consequence_, the mass of the constituents will
be less than the mass of of the atom before it was spitted
according to E = mc².
--------------
If you heat a body with a bunsen burner, the body
will be hotter, it will contain more heat energy.
No energy is _converted_ into something else, the heat
energy is still energy (excited electrons).
As a _consequence_ the mass of the hot body will increase
according to m = E/c². Some of the mass _is_ energy.
E = mc² will always me fulfilled.
It is a fundamental law of nature.
And of course all the physicists at Los Alamos knew that.
On 2024-10-24 21:42:00 +0000, J. J. Lodder said:er
Paul B. Andersen <relativity@paulba.no> wrote:
Den 23.10.2024 18:38, skrev rhertz:
On Tue, 22 Oct 2024 12:41:19 +0000, Paul.B.Andersen wrote:
Of course there are Coulomb forces that accelerate the parts of
the atom in a fission.
This is not disputed!
So why do you act as it is?
And you know that this _confirms_ E = mc? because:
we know:
Generally:
In a fission the mass of the constituents is less than
the mass of the fissioned atom.
------------------
All physicists knew that in 1939, obviously.
https://www.atomicarchive.com/resources/documents/beginnings/nature_meitn
.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." >>>>
Meitner calculated from the electrostatic repulsion that
the kinetic energy of the constituents would be ca 200 Mev.
Because this was the simplest way to estimate the released energy.
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."
When Meitner found that this mass difference was equivalent to
ca.200 Mev it could only be through E = mc?.
So Meitner, like all physicists, took E = mc? for granted.
You know this, because I told you 5 years ago.
So why do you pretend to be ignorant of the fact that all physicists >>>> (and chemists) at the time took E = mc? for granted?
These are excerpts from Serber's 1992, "Los Alamos Primer":
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.
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.
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.
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?
It's hopeless. RH is completely clueless when it comes to real physics. (and he is unwilling to learn)
The problem with using just initials that two crackpots can have the
same ones. At first reading I thought you meant "Dr" Hachel, who is
indeed completely clueless about many things, but I was puzzled as I
didn't think he had contributed to this thread.
...
If someone tells you "Doctor Hachel is perfectly ignorant in physics", weigh his
words on a scale, before swallowing them.
Athel Cornish-Bowden <me@yahoo.com> wrote:
On 2024-10-24 21:42:00 +0000, J. J. Lodder said:
Paul B. Andersen <relativity@paulba.no> wrote:
The problem with using just initials that two crackpots can have the
same ones. At first reading I thought you meant "Dr" Hachel, who is
indeed completely clueless about many things, but I was puzzled as I
didn't think he had contributed to this thread.
Sorry, I didn't pay attention. I try to focus on content, if any.
To the point: do you have grounds for supposing one RH to be more
clueless about the general state of physics than the other one is?
Jan
Le 26/10/2024 à 13:35, M.D. Richard "Hachel" Lengrand a écrit :
...
If someone tells you "Doctor Hachel is perfectly ignorant in physics", weigh his
words on a scale, before swallowing them.
It is quite difficult to find anything more true that this statement.
Le 26/10/2024 à 13:47, Python a écrit :
Le 26/10/2024 à 13:35, M.D. Richard "Hachel" Lengrand a écrit :
...
If someone tells you "Doctor Hachel is perfectly ignorant in physics", weigh his
words on a scale, before swallowing them.
It is quite difficult to find anything more true that this statement.
By the way, you asked me to place the hyperplanes of simultaneity of Terrence and Stella on a drawing.
I posted the two drawings for you.
This is unprecedented in the history of humanity, because no one draws relationships as precise, as clear, and as beautiful as me, with horizontal hyperplanes, as logic would require.
I then explained to you that for Stella, it was necessary to take into account
the change of inertial frame of reference, and to adapt it to her own point-of-reference, that is to say her rocket, always at the origin of her frame
of reference (it is space that accelerates and deforms around her, but her own
rocket does not vary for her).
It is therefore necessary to take into account the space-zoom predicted by Poincaré on the x-axis. Let D'=D.sqrt(1-Vo²/c²)/(1+cosµ.Vo/c)
I then explained to you that in this example, there was a hint of reasoning (like in chess the winning move that only Kasparov will find, here it is the good
doctor Hachel), and that to find the date on which Stella will receive her answer
(she sends her message at six years old and receives her answer at 14 years old,
on the way back), you have to use the diagram as follows.
<http://nemoweb.net/jntp?B_UDIY4o1RGc2Z3c3FLtDlnEdpg@jntp/Data.Media:1>
However, despite the beauty, clarity and logic of my relativistic thinking, it
would seem that you still haven't understood anything, like a few other morons
here, who don't understand anything about the theory of relativity, but spit on
Hachel.
All this is horribly stupid, tinged with jealousy and imbecilic narcissism.
We're just spitting for the sake of spitting.
R.H. (suivi sci.physics.relativity)
Le 26/10/2024 à 14:17, Richard Hachel a écrit :
Le 26/10/2024 à 13:47, Python a écrit :
Le 26/10/2024 à 13:35, M.D. Richard "Hachel" Lengrand a écrit :
...
If someone tells you "Doctor Hachel is perfectly ignorant in
physics", weigh his words on a scale, before swallowing them.
It is quite difficult to find anything more true that this statement.
By the way, you asked me to place the hyperplanes of simultaneity of
Terrence and Stella on a drawing.
I posted the two drawings for you.
This is unprecedented in the history of humanity, because no one draws
relationships as precise, as clear, and as beautiful as me, with
horizontal hyperplanes, as logic would require.
I then explained to you that for Stella, it was necessary to take into
account the change of inertial frame of reference, and to adapt it to
her own point-of-reference, that is to say her rocket, always at the
origin of her frame of reference (it is space that accelerates and
deforms around her, but her own rocket does not vary for her).
It is therefore necessary to take into account the space-zoom
predicted by Poincaré on the x-axis. Let D'=D.sqrt(1-Vo²/c²)/
(1+cosµ.Vo/c)
Poincaré never predicted such a stupidity. You are a liar.
By the way, you asked me to place the hyperplanes of simultaneity of
Terrence and Stella on a drawing.
I posted the two drawings for you.
This is unprecedented in the history of humanity, because no one draws relationships as precise, as clear, and as beautiful as me, with
horizontal hyperplanes, as logic would require.
Den 26.10.2024 14:17, skrev Richard Hachel:
Can you please explain the difference between a horizontal hyperplane
and a vertical hyperplane?
Is a horizontal hyperplane parallel to the ocean?
In that case, at what latitude?
Le 26/10/2024 à 19:36, "Paul.B.Andersen" a écrit :
Den 26.10.2024 14:17, skrev Richard Hachel:
Can you please explain the difference between a horizontal hyperplane
and a vertical hyperplane?
Is a horizontal hyperplane parallel to the ocean?
In that case, at what latitude?
Je ne suis pas sûr que votre intervention soit si humoristique.
Je vous signale, avec délicatesse; que vos pdf recèlent des erreurs de concepts,
comme par exemple des intégrations de Leibniz mêlant des carottes et des navets.
Vous n'avez pas rectifié.
Ne me reprochez pas de ne pas vous l'avoir dit, et encore moins de trouver vos
réponses humoristiques et hors-sujet.
R.H.
Le 26/10/2024 à 19:36, "Paul.B.Andersen" a écrit :
Den 26.10.2024 14:17, skrev Richard Hachel:
Can you please explain the difference between a horizontal hyperplane
and a vertical hyperplane?
Is a horizontal hyperplane parallel to the ocean?
In that case, at what latitude?
Je ne suis pas sûr que votre intervention soit si humoristique.
Je vous signale, avec délicatesse; que vos pdf recèlent des erreurs de concepts,
comme par exemple des intégrations de Leibniz mêlant des carottes et des navets.
Vous n'avez pas rectifié.
Ne me reprochez pas de ne pas vous l'avoir dit, et encore moins de
trouver vos réponses humoristiques et hors-sujet.
R.H.
The point is obviously:
What is the meaning of the word "horizontal" in the context:
"with horizontal hyperplanes, as logic would require."
Why do logic require that hyperplanes must be "horizontal" as opposed
to "vertical"?
Paul
Starmaker: Did he really say that? Was he pretending to have contributed
to that great accomplishment?
Starmaker: It was already in the marketplace for a long time.
Le 27/10/2024 à 19:53, "Paul.B.Andersen" a écrit :
The point is obviously:
What is the meaning of the word "horizontal" in the context:
"with horizontal hyperplanes, as logic would require."
Why do logic require that hyperplanes must be "horizontal" as opposed
to "vertical"?
Paul
Paul! Paul! Please!!!
It is perfectly normal, perfectly legitimate and completely scientific
to propose, for a given individual, a representation of the space-time
that is his.
This has four dimensions. Three of space, and one of time.
LaurenceClarkCrossen wrote:
Starmaker: Did he really say that? Was he pretending to have contributed
to that great accomplishment?
"If I had foreseen Hiroshima and Nagasaki, I would have torn up my
formula in 1905." - Albert Einstein
p. 112 - Einstein the Poet (1983)
https://libquotes.com/albert-einstein/quote/lbr9f3o
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
https://libquotes.com/albert-einstein/quote/lbr9f3o
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Le 29/10/2024 à 07:27, Thomas Heger a écrit :
https://libquotes.com/albert-einstein/quote/lbr9f3o
Fake citation.
Non seulement la formule n'est pas d'Einstein, mais de Poincaré, mais encore, ce n'est pas le fait de l'avoir popularisée qui a fait mettre au point les bombes atomiques.
On savait bien avant Einstein que la matière pouvait dégager des quantités considérable d'énergie, et les physiciens de Los Alamos se moquaient pas mal du niveau d'Albert Einstein en physique nucléaire.
Le reste est du pipeau médiatique visant à attribuer à Einstein :
- la relativité restreinte de Poincaré
- la relativité générale de Grosmann et Hilbert
- Les recherches d'Oppenheimer et de ses collaborateurs.
Je ne comprendrais jamais ce battage médiatique incroyable fait sur un obscur copiste du bureau des brevets de Berne.
Aujourd'hui encore on retrouve des tas de citations signées Albert Einstein, citations qu'il n'a jamais écrite, mais qu'on lui attribue.
Tout ceci tourne à l'hystérie collective.
R.H.
Starmaker: So he was taking credit for it.
Thomas Heger wrote:
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Physicists there were held against their own will. They did not
volunteer for it, they were...drafted.
Thomas Heger wrote:
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Physicists there were held against their own will. They did not
volunteer for it, they were...drafted.
LaurenceClarkCrossen wrote:
Starmaker: So he was taking credit for it.
Credit for what?
It's just a book of Einstein's quotes like so many.
https://books.google.com/books?id=ZQsFEAAAQBAJ&pg=PA238&dq=%22If+I+had+foreseen+Hiroshima+and+Nagasaki,+I+would+have+torn+up+my+formula+in+1905
Am Dienstag000029, 29.10.2024 um 15:35 schrieb The Starmaker:
Thomas Heger wrote:
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Physicists there were held against their own will. They did not
volunteer for it, they were...drafted.
Sure: Los Alamos looked like barracks, which were placed remotely in the desert.
But my assumption was, that the atomic bomb was much older than the
Manhattan project, which had the purpose, to 'gag' the physicists and
sort out those, which were not willing to comply
(e.g. proponents of 'free energy', abiogenic oil theory, homeopathy and similar heresies).
The Soviets did something quite similar and regarded e.g. Andre Sacharov ('father' of the Soviet bomb) as 'enemy of the people' (for what reason
I don't know).
The atomic bomb itself was most likely already known befor WWII.
A reason to think so:
Einstein and Szillard patented their 'Einstein fridge' in 1930 in Berlin.
The only known use of that device is as part of a fast breeding reactor (those reactors that produce plutonium).
But for which purpose would you like to breed plutonium, if you had no
atomic bomb?
The Einstein fridge doesn't coll (this was at least found out by a bunch
of students, who rebuilt the device).
It is kind of 'three substances absorption cooler' (with ammonia, water
and butan in liquid form).
(To me this 'fridge' does not look like a device, which could eventually cool. Therefore I would support the claim of those students.)
A theory of my taste goes like this:
the bomb was already known in 1930 and Szillard and Einstein played a
role in its development (undertanken long befor 1930 and certainly not
at Los Alamos).
To reward Einstein and Szillard secretly, that patent was used, even if
the device does not cool.
...
TH
Thomas Heger wrote:
Am Dienstag000029, 29.10.2024 um 15:35 schrieb The Starmaker:
Thomas Heger wrote:
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Physicists there were held against their own will. They did not
volunteer for it, they were...drafted.
Sure: Los Alamos looked like barracks, which were placed remotely in the desert.
But my assumption was, that the atomic bomb was much older than the Manhattan project, which had the purpose, to 'gag' the physicists and
sort out those, which were not willing to comply
(e.g. proponents of 'free energy', abiogenic oil theory, homeopathy and similar heresies).
Yes, it was much older. Albert Einstein invented the atomic bomb in 1905...that is when it first came to him in his mind.
From then on it became his passion and quest to make it happen.
Einstein gave How To Build an Atomic Bomb classes in the 1920's...Enrico Fermi was a student. (Leo Zilgard also his main student)
The Starmaker <starmaker@ix.netcom.com> wrote:
Thomas Heger wrote:
Am Dienstag000029, 29.10.2024 um 15:35 schrieb The Starmaker:
Thomas Heger wrote:
Am Montag000028, 28.10.2024 um 06:26 schrieb The Starmaker:
LaurenceClarkCrossen wrote:
Starmaker: It was already in the marketplace for a long time.
sorry, i only know of only one Manhattan Project..
Sure, but what were they doing there in Los Alamos?
To me it looked like a concentration camp for physicists.
TH
Physicists there were held against their own will. They did not volunteer for it, they were...drafted.
Sure: Los Alamos looked like barracks, which were placed remotely in the desert.
But my assumption was, that the atomic bomb was much older than the Manhattan project, which had the purpose, to 'gag' the physicists and sort out those, which were not willing to comply
(e.g. proponents of 'free energy', abiogenic oil theory, homeopathy and similar heresies).
Yes, it was much older. Albert Einstein invented the atomic bomb in 1905...that is when it first came to him in his mind.
From then on it became his passion and quest to make it happen.
Einstein gave How To Build an Atomic Bomb classes in the 1920's...Enrico Fermi was a student. (Leo Zilgard also his main student)
Really absolutely incredibly clever of him,
considering that the Uranium 235 isotope wasn't discovered until 1936.
Jan
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