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Definite Proof, From Robert Serber: Manhattan Project Unrelated To E=Mc2


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#1 rhertz

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Posted 12 June 2019 - 10:39 PM

This is my final post on E=mc2 and the myth that such a formula was related to the development of the atomic bombs

used in 1945, and developed in the Manhattan Project.

 

I quote here the words of Robert Serber, a Manhattan's physicist, who wrote his memoirs in a small book by 1992,

currently sold by Amazon. The book, which contain several documents declassified and memories from 1943-1945,

was edited by Richard Rhodes. I strongly recommend to read the book (12 chapters), which tell about obstacles,

fears, discussions, doubts and many anecdotes for the whole Manhattan Project (2 1/2 years).

 

The Los Alamos Primer
The first lectures on how to build an Atomic Bomb

 University of California Press
1992 by Robert Serber (edited by Richard Rhodes)

 

and can be read at this link: https://www.academia...s_Alamos_Primer

 

From the American Heritage Foundation:  https://www.atomiche...e/robert-serber

 

I quote:

 

Robert Serber (1909-1997) was an American physicist. He was recruited by J. Robert Oppenheimer to work on

the Manhattan Project. Serber was tasked with explaining the basic principles and goals of the project to all

incoming scientific staff. Moving to Los Alamos in 1943, he gave lectures to members of the Manhattan Project

about the design and construction of atomic bombs known as the “Los Alamos Primer."

 

While at Los Alamos, Serber developed the first good theory of bomb disassembly hydrodynamics.

...............

After the Trinity Test in July 1945, Serber traveled to Tinian Island to assist in the construction of the bomb, as a

special consultant to Project Alberta. While there, he reassured Col. Paul Tibbets, pilot of the Enola Gay, that the

bomb's blast would not harm the plane. In early September 1945, Serber was part of the first American team to enter

Hiroshima and Nagasaki, assessing the bombs' damage and collecting debris for testing on their five-week mission.

 

--------------------------------------------------------------

 

From the Chapter 2: Energy of Fission Process, I take some excerpts here, related with energy per gram of atom

and Serber's comment about the complete disconnexion between atomic energy and the "myth" of Einstein's E=mc2.

 

In my words:

 

1) The fission's energy release per gram of U-235 (as kinetic energy due to electrostatic repulsion) is 7.1017 erg/gram.

2) The energy conversion of 1 gram of mass, in Einstein's E=mc2, is 9.1020 erg/gram.

3) Einstein's formula E=mc2 has a difference 1285.71 times higher and is unrelated to nuclear fission (Serber words). 

 

In Robert Serber words (Chapter 2: Energy of Fission Process):

 

The direct energy release in the fission process is of the order of 170 Mev per atom.
...................

To start with a simpler particle than an atom, let’s look at two electrons pushed together. If you
released them, they would fly apart with an amount of energy equal to the work that went into pushing
them together. That energy E is given by the formula

E = e2/R      (1)

where e is the electron charge,e2 is e multiplied by itself, and R is the distance between the particles.
The electrostatic energy thus ends up as kinetic energy, the energy of motion.

.....................

Now let’s consider the electrostatic energy in the uranium nucleus.

The uranium nucleus contains 92 protons, each of which has the same charge as an electron, though of opposite
sign — particles of opposite sign attract each other,those of the same sign repel. So the uranium nucleus has a
charge 92 times as great as an electron; it’s positive rather than negative, + rather than -, but since only the
square of the charge is involved
, that difference doesn’t matter in equation (1). The numerator of (1) is thus 92
times bigger than for a chemical reaction. For our purposes, 922 is close enough to call 1002. So the numerator for
a uranium atom would be greater by a factor of 1002, 100 times 100, or 10,000 (104).

The uranium nucleus is also much smaller than an atom. In an atom, the distance R is 10-8 cm (cm meaning centimeters).
The radius of the uranium nucleus is 10-12 cm, which is 104 times smaller. The electrostatic energy for a uranium nucleus
is therefore 104 for the numerator and another 104 for the denominator, for a total of 108 times greater than the
electrostatic energy between atoms or molecules. When a uranium nucleus fissions, much of this energy is released as
kinetic energy in the two fission fragments that fly apart. Suppose that the uranium nucleus broke in half. Each fragment
would have half the charge. The numerator of equation (1) would be a quarter as big—a half times a half. Since the volume
is proportional to the cube of the radius, the radius would be smaller by a factor of

2-3/2 = 1/1.26

So each fragment would have an electrostatic energy of about a third of the total and the two fragments about two-thirds.

That leaves a third left over for the reaction energy. Thus we see that the energy of fission is about 108 —one hundred
million times — greater than the energy of a chemical reaction, confirming the statement that it’s “considerably more than 107.”

This is 170·106 · 4.8·10-10/300 = 2.7·10-4 erg/nucleus. Since the weight of 1 nucleus of 25 is 3.88·10-22 gram/nucleus the energy

release is  7·1017 erg/gram.

The energy release in TNT is 4·1010 erg/gram or 3.6·1016 erg/ton. Hence 1 kg of 25 ≈ 20000 tons of TNT

 

..............................................................

 

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 non relativistic 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
.


---------------- End of quotes ----------------------------------------------------
 

Finally, this is a copy of the original page Serber prepared by 1943 to give 5 lectures to every physicist (at Los Alamos facilities),

in April 1943, before they were assigned to different (compartmentalized) tasks. The charge of the electron 4.8.10-10 was given

by then in Electrostatic Units, and due this the division by 300 is to obtain ergs from eV (that is what physicists used by then in US).

 

 

 

27-6e7ecf777e.jpg


Edited by rhertz, 12 June 2019 - 10:53 PM.


#2 rhertz

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Posted 13 June 2019 - 10:06 AM

Continuing from the abandoned thread:

 

http://www.sciencefo...ein-emc2/page-2

 

here are performed the calculations made by Robert Serber (in modern units), which prove that

U-235 fission energy is of electrostatic nature and unrelated to E=mc2, being more than 1,000 times lower.

 

First, some historical facts behind the calculations (from the abandoned thread):

 

--------------------------------------------------------------------------------------------------------------

 

Following Robert Serber's 5 lectures in April 1943 to every physicist that went into Los Alamos,

whom had to attend Serber's Los Alamos Primer (the name of the lecture), this is HOW the

amount of 170 MeV was released as Kinetic Energy at the fission of one U-235 atom.

 

Each one of the more than 100 physicists (and chemists) who worked at the Manhattan Project

received also a mimeographed copy of the lecture.

 

The origin of the energy released at the fission of a U-235 atom is of electrostatic nature, and

follows the Coulomb's Law: The force between two electrostatic charges q1 and q2, separated by

a distance r is:

 

F = ke.q1.q2/r2

 

The energy content prior to any repulsive movement between charges of equal sign is:

 

E = F.r = ke.q1.q2/r

 

Now is when what was taught by Server is presented, as a summary:

 

1) The U-235 atom contains 92 protons and 143 neutrons (146 for the common U-238).

2) It's assumed that the U-235 atom split in halves of equal size, aproximately. It gives 2 x 46 protons to

    exert a repulsive electrostatic force (Coulomb).

3) At the moment of the fission, there is an electrostatic energy E, given by Coulomb's Law.

4) When a U-235 atom split in halves (1943 estimation), the total electrostatic energy split in three parts.

5) Each of the two fragments fly away with 1/3 of the energy each.

6) Two aditional fast neutrons are produced, besides the original one which caused the fission.

    To get a critical chain reaction, it's needed approximately 80 fissions per original neutron to produce 280

    neutrons "rain" within the critical mass. This occurs in approximately 1 microsecond, so there is an explosive

    generation of energy, which can produce 20 Kilotons of energy by fissioning 1 Kg of U-235.

 

7) The third part of the energy is released into the air as Kinetic Energy of electrostatic origin.

 

    There is not the slightest relationship between this energy per nucleus with Einstein's 1905 E=mc2 paper.

----------------------------------------------------------------------------------------------------

 

Some values needed for the calculations (Robert Serber numbers and magnitudes are in the SI MKS system of units).

 

1) Serber's radius of the U-235 nucleus: rU235 = 10-14 meter.

2) Serber's charge of the electron: qelec = 1.6 10-19 Coulombs.

3) Serber's radius of each half of U-235 nucleus: rUHalf = 7.94 10-15 meters

4) Serber's charge of each half of U-235 nucleus: qUHalf = 7.37 10-18 Coulombs.

5) Serber's electrostatic energy between two halves of U-235 nucleus: EU235-split = 7.41 10-11 Joules = 462.17 MeV

 

    This is by using EU235-split = FUHalves.rUHalf = ke.qUHalf2/rUHalf

 

6) Serber's 1/3 of electrostatic energy liberated after U-235 atom's fission: EKE-Free = 1/3   EU235-split = 154.1 MeV

 

This value of liberated electrostatic energy under the form of Kinetic Energy is CLOSE to the ESTIMATED 170 MeV/U-235 atom.

 

As some values have changed a little since 1942, and 170 MeV was an approximation by then, the value is IN THE RANGE of

Serber's estimations at his lectures.

 

Final Note: Each one of these calculations were written at the document that Serber used to instruct scientists at his 5 lectures.

None of them belong to me, except that I used the SI MKS system of units directly (instead of cgs) and used current values

for the charge of the electron.

 

So, it serves as a complete proof that E=mc2 is UNRELATED to the energy liberated by the fission of an U-235 atom.

 

Anyone who dislike these values has to refer to Serber's book or perform a time travel back to 1942, to instruct the primary

scientists involved in the nuclear power. Please, include Fermi and Lawrence, among others.

 

This is, by far, the most profound and valuable source about this MYSTERY available at the Internet.


Edited by rhertz, 13 June 2019 - 10:24 AM.


#3 Dubbelosix

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Posted 13 June 2019 - 11:49 AM

Of course E=mc^2 is related to atom bombs, Einstein tried ever so hard to hide this fact by stating ''we will never be able to split an atom.''

 

He wasn't silly, he knew fine well they could, but he deeply regretted that ''his'' work was the underlying basis of the atom bomb, even though the formula had been predicted by a few people even before him... it was only a matter of time before we realized that using the right frequency of a particle, could interact with an atom as such to split it apart and release a large amount of energy.

 

The sun is also a nuclear object, it too follows the same principles. It is in principle, a giant atom bomb.



#4 exchemist

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Posted 13 June 2019 - 03:16 PM

To show whether or not E=mc² has anything to do with the scenario outlined in this thread, one would obviously need to determine the total mass of the system before and after the nuclear reaction.

 

Since the above analysis does not attempt to do this, it is incapable of demonstrating whether E=mc² applies or not.

 

In fact, using real nuclear equations and real values for the binding energies of the species involved, it can easily be shown that the immediate fission generally gives about 170MeV of energy, with a further 20-30MeV from subsequent emission from the products, making about 200MeV for the whole process. (A typical scenario is given here: https://www.nuclear-...m-235-fission/)

 

These binding energies are, as I have pointed out on other threads, determined by measuring the masses of of the species and applying E=mc² to them.

 

The energy release certainly comes quite largely in the form of acceleration of the daughter nuclei, due to electrostatic repulsion. However discussion of the nature of the mechanism of energy release does not address the question of whether or not mass is lost due to the release of energy, as predicted by E=mc².  As I say, this demonstrably does occur.   



#5 VictorMedvil

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Posted 13 June 2019 - 06:26 PM

Rhertz, what you propose with the atomic bomb would be like genetic researchers in the modern era not realizing that Genetic Engineering Methods such as CRISPR not leading to the possibility of Synthetic Biological weapons. E = MCis like that for the physics without E = MCthe atomic bomb's creation would have been much more difficult having to guess the amount of Mass that is converted into energy. I guess what I am trying to say is the Atomic Bomb nor Genetic Warfare Agents were discovered blindly without things that predict its possibility and Einstein's methods for the atomic bomb were used just like Jennifer Doudna's methods are going to be used for new weaponry someday if it hasn't already, if it is not CRISPR then it will be something else history repeatedly repeats itself. The Discovery of Atomic weapons just like Genetic Weapons is inevitable each piece of the puzzle being necessary. To say Einstein's theories had no part to play in the physics of the atomic bomb would be like saying CRISPR had no part in the creation of Insect allies using a CRISPR, the atomic bomb used E=MC, thus denying its part would be no different than denying CRISPR's part in these weapons.

 

insectallies-1.jpg

mass-energy-relationship.jpg


Edited by VictorMedvil, 13 June 2019 - 06:47 PM.


#6 Dubbelosix

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Posted 13 June 2019 - 07:41 PM

To show whether or not E=mc² has anything to do with the scenario outlined in this thread, one would obviously need to determine the total mass of the system before and after the nuclear reaction.

 

Since the above analysis does not attempt to do this, it is incapable of demonstrating whether E=mc² applies or not.

 

In fact, using real nuclear equations and real values for the binding energies of the species involved, it can easily be shown that the immediate fission generally gives about 170MeV of energy, with a further 20-30MeV from subsequent emission from the products, making about 200MeV for the whole process. (A typical scenario is given here: https://www.nuclear-...m-235-fission/)

 

These binding energies are, as I have pointed out on other threads, determined by measuring the masses of of the species and applying E=mc² to them.

 

The energy release certainly comes quite largely in the form of acceleration of the daughter nuclei, due to electrostatic repulsion. However discussion of the nature of the mechanism of energy release does not address the question of whether or not mass is lost due to the release of energy, as predicted by E=mc².  As I say, this demonstrably does occur.   

 

The binding of a particle has less energy than that which put into to make an atom, this energy is converted to gluon energy, which is maybe what you are thinking about. In fact, I think wiki may be able to articulate slightly better, as I quote:
 

''The mass of an atomic nucleus is less than the sum of the individual masses of the free constituent protons and neutrons, according to Einstein's equation E=mc2. This 'missing mass' is known as the mass defect, and represents the energy that was released when the nucleus was formed.''



#7 Dubbelosix

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Posted 13 June 2019 - 09:02 PM

Look, stop reading my material then trying to re-word things, then adding a dash of insanity. You are intentionally winding people up and winding yourself up. No one says, E=mc^2 is perfect, it only applies to a rest system - the true formula isn't like this, the relativistic formula has an extension. Let's see if you understand that?

 

Secondly, while c may not be a constant in a gravitational field, it is a constant in free space. Likewise, it is not a constant moving through any medium - the catch here is that the speed of light does in fact always move at the speed of light, but because everything is relative, you can also argue it isn't. Since a photon has no frame of reference, the aether remains the frame of reference in which we defer that the speed of light is not necessarily a constant.

 

E=mc^2 is perfectly reasonable, well-tested and well-calculated concerning nuclear energy from matter into energy and energy into matter. You've confused yourself so much you can't even follow the right path. You keep contending something is wrong, without fully appreciating why it is right. This is why many of us, cannot be bothered to deal with you, because you won't help yourself in the long run... unless you take my advice, stop coming here asking for the answers, use your brain matter and get to the truth through some work. We are not here at your beckon call just to give you an education in something you disrespect with the highest impunity, while offending posters here concerning their own intelligences.

 

Time to grow up don't you think?



#8 Dubbelosix

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Posted 13 June 2019 - 10:36 PM

As I have compassion and commiseration, I refrain myself to answer in ways that could hurt sensible feeling of some

forists here.

 

I only wanted to say that and also say that I expect that those who feel triggered may find peace in their souls.

 

 

It's not me you should be worried about, the fact is you keep wasting other peoples time and not learning for yourself... that is your problem.



#9 Dubbelosix

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Posted 13 June 2019 - 11:40 PM

Look, you jump from one line to another, I am not replying to this any more. If you could keep focused on one subject and stop illogically tying things together, we might get somewhere. If you want a clue on how to logically construct a set of arguments, take a look at my UFO thread under a scientific analysis. It is pretty clear and concise, you write way too fast for even your brain to coherently put things down properly.

 

Now take my advice or forget it, I won't be entertaining this any more!



#10 rhertz

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Posted Today, 01:13 PM

Look, you jump from one line to another, I am not replying to this any more. If you could keep focused on one subject and stop illogically tying things together, we might get somewhere. If you want a clue on how to logically construct a set of arguments, take a look at my UFO thread under a scientific analysis. It is pretty clear and concise, you write way too fast for even your brain to coherently put things down properly.

 

Now take my advice or forget it, I won't be entertaining this any more!

 

Don't post here, then. It's not mandatory for you to do so.