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Thread: Ferromagnetism Relationship to Atomic Mass Defect

  1. #1 Ferromagnetism Relationship to Atomic Mass Defect 
    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    The most commonly accepted theory of ferromagnetism is the RKKY interaction source: https://en.wikipedia.org/wiki/RKKY_interaction

    I have observed that the only 3 chemical elements which are ferromagnetic above 70 degrees Fahrenheit -- iron, cobalt and nickel -- are also the 3 chemical elements with the lowest mass per atomic nucleon (mass defect) of all chemical elements (iron, cobalt and nickel).

    Can anyone explain whether or not it is merely coincidence that the only 3 room-temperature chemical elements which display ferromagnetism are the very same 3 elements which have the least mass per nucleon of all chemical elements? Does the RKKY interaction account for this mass defect?

    Source: Wiki https://en.wikipedia.org/wiki/Nickel-62:

    "Nickel-62 is an isotope of nickel having 28 protons and 34 neutrons.
    It is a stable isotope, with the highest binding energy per nucleon of any known nuclide (8.7945 MeV).[1] [2] It is often stated that 56Fe is the "most stable nucleus", but actually 56Fe merely has the lowest mass per nucleon (not binding energy per nucleon) of all nuclides. The lower mass per nucleon in Fe-56 is enhanced by the fact that 56Fe has 26/56 = 46.43% protons, while 62Ni has only 28/62 = 45.16% protons, and the relatively larger fraction of light protons in 56Fe lowers its mean mass-per-nucleon ratio in a way that has no effect on its binding energy."


    "most nickel is produced in supernovas in the r-process of neutron capture by nickel-56 immediately after the core-collapse, with any nickel-56 that escapes the supernova explosion rapidly decaying to cobalt-56 and then stable iron-56."

    Source: Wiki https://en.wikipedia.org/wiki/Nickel-62:

    I also note these are the 3 elements most predominantly produced in supernova explosions.









    For illustration, Iron-56, the element with the strongest room temperature ferromagnetism, has 56 protons and neutrons in the nucleus. The above 3 images are the result of my close packing of 56 magnetized spheres.


    Last edited by devin-m; January 12th, 2018 at 02:26 PM.
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    Quote Originally Posted by devin-m View Post
    The most commonly accepted theory of ferromagnetism is the RKKY interaction source: https://en.wikipedia.org/wiki/RKKY_interaction

    I have observed that the only 3 chemical elements which are ferromagnetic above 70 degrees Fahrenheit -- iron, cobalt and nickel -- are also the 3 chemical elements with the lowest mass per atomic nucleon (mass defect) of all chemical elements (iron, cobalt and nickel).

    Can anyone explain whether or not it is merely coincidence that the only 3 room-temperature chemical elements which display ferromagnetism are the very same 3 elements which have the least mass per nucleon of all chemical elements? Does the RKKY interaction account for this mass defect?





    For illustration, Iron-56 has 56 protons and neutrons in the nucleus. The above images are the result of close packing of 56 magnetized spheres.
    First, it is not my understanding that ferromagnetism is attributed to the RKKY effect. Your Wiki link in fact does not make such a claim and nor does the Wiki article on ferromagnetism, here: https://en.wikipedia.org/wiki/Ferromagnetism

    Second, there is no reason to link nuclear stability, which is what I think the mass per nucleon reflects, to ferromagnetism.


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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    The most commonly accepted theory of ferromagnetism is the RKKY interaction source: https://en.wikipedia.org/wiki/RKKY_interaction

    I have observed that the only 3 chemical elements which are ferromagnetic above 70 degrees Fahrenheit -- iron, cobalt and nickel -- are also the 3 chemical elements with the lowest mass per atomic nucleon (mass defect) of all chemical elements (iron, cobalt and nickel).

    Can anyone explain whether or not it is merely coincidence that the only 3 room-temperature chemical elements which display ferromagnetism are the very same 3 elements which have the least mass per nucleon of all chemical elements? Does the RKKY interaction account for this mass defect?





    For illustration, Iron-56 has 56 protons and neutrons in the nucleus. The above images are the result of close packing of 56 magnetized spheres.
    First, it is not my understanding that ferromagnetism is attributed to the RKKY effect. Your Wiki link in fact does not make such a claim and nor does the Wiki article on ferromagnetism, here: https://en.wikipedia.org/wiki/Ferromagnetism

    Second, there is no reason to link nuclear stability, which is what I think the mass per nucleon reflects, to ferromagnetism.
    Source: "https://en.wikipedia.org/wiki/Ferromagnetism"

    "There are different exchange interaction mechanisms which create the magnetism in different ferromagnetic, ferrimagnetic, and antiferromagnetic substances. These mechanisms include
    direct exchange, RKKY exchange, double exchange, and superexchange."

    Source: "
    https://en.wikipedia.org/wiki/Ferromagnetism"

    @exchemist I observe that the same 3 elements which display 70 degree fahrenheit ferromagnetism are also the 3 elements with greatest nuclear stability and lowest nuclear "excess energy." (and also the same 3 elements most predominantly produced in certain stellar supernova explosions.) Are we saying then this is a coincidence?

    https://en.m.wikipedia.org/wiki/Pair...lity_supernova

    "In addition to the immediate energy release, a large fraction of the star's core is transformed to nickel-56, a radioactive isotope which decays with a half-life of 6.1 days into cobalt-56. Cobalt-56 has a half-life of 77 days and then further decays to the stable isotope iron-56(see Supernova nucleosynthesis). For the hypernova SN 2006gy, studies indicate that perhaps 40 solar masses of the original star were released as Ni-56, almost the entire mass of the star's core regions.[5] Collision between the exploding star core and gas it ejected earlier, and radioactive decay, release most of the visible light."

    https://en.m.wikipedia.org/wiki/Pair...lity_supernova







    Wiki: https://en.wikipedia.org/wiki/SN_1572




    Image Caption: Close packing of 56 magnetized spheres
    Last edited by devin-m; January 12th, 2018 at 08:30 PM.
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Suppose I look at the following ferromagnetic-at-room-temperature STABLE isotopes:

    Iron-56 - 26 protons - https://en.wikipedia.org/wiki/Isotopes_of_iron
    Cobalt-59 - 27 protons - https://en.wikipedia.org/wiki/Isotopes_of_cobalt
    Nickel-62 - 28 protons - https://en.wikipedia.org/wiki/Isotopes_of_nickel

    Next I look at the following ferromagnetic-at-temperatures-below-room-temperature STABLE isotopes:

    Gadolinium-157 - 64 protons - https://en.wikipedia.org/wiki/Isotopes_of_gadolinium
    *Terbium-157 - 65 protons - https://en.wikipedia.org/wiki/Isotopes_of_terbium
    *Terbium-160 - 65 protons - https://en.wikipedia.org/wiki/Isotopes_of_terbium
    Dysprosium-160 - 66 protons - https://en.wikipedia.org/wiki/Isotopes_of_dysprosium
    Dysprosium-163 - 66 protons - https://en.wikipedia.org/wiki/Isotopes_of_dysprosium
    *Holmium-163 - 67 protons - https://en.wikipedia.org/wiki/Isotopes_of_holmium

    *Terbium-160 is not actually stable but instead beta decays to stable Dyprosium-160. Terbium-157 beta decays to stable Gadolinium-157. Holmium-163 is actually the most stable radioactive Holmium isotope with an extremely long 4570 year half life and decays with minuscule 0.003MeV decay energy via electron capture to stable Dysprosium-163.

    Notice the elements in each set of elements are directly adjacent to each other on the periodic table (they differ by one proton), and the chosen isotopes differ from their neighboring isotopes by 3 nucleons.



    If one attempts the most efficient close sphere packing of magnetized spheres with 56 (iron), 59 (cobalt), 62 (nickel), 157 (gadolinium), 160 (terbium/dysprosium) or 163 (holmium/dysprosium) spheres, the close sphere packing in each case results in what appears to be a fully rotationally-symmetrical truncated or non-truncated triangular bipyramidal solid.

    Wiki (Triangular Bipyramid) - https://en.wikipedia.org/wiki/Triangular_bipyramid



    56 Spheres - Truncated Triangular Bipyramid - Iron-56



    59 Spheres - Truncated Triangular Bipyramid - Cobalt-59



    62 Spheres - Triangular Bipyramid - Nickel-62



    157 Spheres - Truncated Triangular Bipyramid - Gadolinium-157



    160 Spheres - Truncated Triangular Bipyramid - Dysprosium-160 & Terbium-160



    163 Spheres - Triangular Bipyramid - Holmium-163 & Dysprosium-163



    All of the above solids can be constructed using the following "layers" composed entirely of interlocking equilateral triangles and hexagons.



    Strangely, if one counts the number of spheres in each layer, one witnesses the following pattern:

    3 -- First Layer

    3+4=7 -- Second Layer

    7+5=12 -- Third Layer

    12+6=18 -- Fourth Layer

    18+7=25 -- Fifth Layer

    25+8 = 33 -- Sixth Layer

    The number of additional spheres in each subsequent layer is one integer larger than the number of additional spheres in each previous layer...

    Also interesting to note is the comparison between the 62 & 163 sphere solids -- both are composed of (6) outer equilateral-triangle faces - 4x4x4 equilateral triangle side lengths in the case of 62 spheres (nickel-62) and 6x6x6 equilateral triangle side lengths in the case of 163 spheres (holmium-163).
    Last edited by devin-m; January 13th, 2018 at 02:23 AM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    First, it is not my understanding that ferromagnetism is attributed to the RKKY effect. Your Wiki link in fact does not make such a claim and nor does the Wiki article on ferromagnetism, here: https://en.wikipedia.org/wiki/Ferromagnetism

    Second, there is no reason to link nuclear stability, which is what I think the mass per nucleon reflects, to ferromagnetism.
    Source: "https://en.wikipedia.org/wiki/Ferromagnetism"

    "There are different exchange interaction mechanisms which create the magnetism in different ferromagnetic, ferrimagnetic, and antiferromagnetic substances. These mechanisms include
    direct exchange, RKKY exchange, double exchange, and superexchange."

    Source: "
    https://en.wikipedia.org/wiki/Ferromagnetism"

    @exchemist I observe that the same 3 elements which display 70 degree fahrenheit ferromagnetism are also the 3 elements with greatest nuclear stability and lowest nuclear "excess energy." (and also the same 3 elements most predominantly produced in certain stellar supernova explosions.) Are we saying then this is a coincidence?

    https://en.m.wikipedia.org/wiki/Pair...lity_supernova
    Yes.
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    @exchemist Would it be accurate to say that when a nucleus has lower “mass per nucleon” than another nucleus, that the particles including up quarks, down quarks, protons and neutrons composing the nucleus with lower mass per nucleon have less angular momentum (less energy) per particle than the nucleus with greater mass per nucleon?

    If the answer is yes... that lower mass per nucleon means essentially lower angular momentum per nucleon in substances such as Iron-56... since protons in the nucleus are attracted to electrons in the electron shells, if the nuclear protons and electrons are all below a certain energy level, ie “moving slowly” below the curie temperature, is it possible that an axis of rotational symmetry in the nucleus becomes geometrically correlated with a rotational axis of symmetry of the wavefunction of the 4 unpaired electron orbitals in the 3-d electron shell? In other words is it possible that above the curie temperature, the orientation of the nucleus and electron shell are not correlated but below the curie temperature, the physical orientation of the nucleus and the physical orientation of the wavefunction of the unpaired electron orbitals are correlated (this being the result of the protons in the Iron-56 nucleus having the least mass per proton and therefore lowest angular momentum per proton of all elemental protons)? simply is it possible that the electrons and protons in room temperature iron-56 are in a low enough energy state that the attractive forces between the protons and electrons leads to geometrical alignment between the nucleus and electron shell in ferromagnetic elements which are below their respective curie temperature?

    my basic understanding of electron shell hybridization theory leads me to believe the 4 unpaired electrons in the 3-d electron shell of an iron-56 would take on an “8-lobed” appearance — 2 “lobes” per electron. when these 8 lobes become hybridized, my understanding is they would take on the appearance of a cube composed of 8 spheroid lobes. when viewed corner-on, this cube of spheroids would appear to have “3 sides” which could correlate with the “3 sides” of the truncated triangular bipyramid formed by close-packing of 56 magnetic spheres to represent Iron-56.

    For illustration:

    8 “lobes” with “3-sided” rotational symmetry representing the hybridized 3-d electron shell of Iron-56 composed of 4 unpaired electrons



    56 magnetized close-packed spheres representing an Iron-56 nucleus with apparent “3-sided” rotational symmetry





    For example:



    Image Source:

    https://www.quizover.com/ocw/mirror/...HybrdOrbit.jpg



    Image Source: https://d2gne97vdumgn3.cloudfront.ne...CSoqT7jKG5ofA8

    Simply - could it be possible that when the rotational orientations of the nucleus and unpaired electron shell wavefunctions are geometrically aligned and correlated, we say the substance is ferromagnetic, and when there is no geometric correlation between the orientation of the electron shell wavefunctions and the nucleus, we say that the substance is not ferromagnetic?
    Last edited by devin-m; January 13th, 2018 at 12:04 PM.
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    Quote Originally Posted by devin-m View Post
    @exchemist Would it be accurate to say that when a nucleus has lower “mass per nucleon” than another nucleus, that the particles including up quarks, down quarks, protons and neutrons composing the nucleus with lower mass per nucleon have less angular momentum (less energy) per particle than the nucleus with greater mass per nucleon?
    No.
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Suppose we wind a copper coil (solenoid) around an Iron-56 core... if we apply a voltage to the coil, it takes much longer to reach the maximum current than if the iron-56 core was not present. this is inductance. if we substitute the iron core for a copper core, it takes almost the same amount of time as if a vacuum was used. what about the iron causes this delay in max current or inductance? if the unpaired electron orbitals in the iron-56 were geometrically correlated with the much heavier iron-56 nuclei, it would take additional time to rotate said electrons+nuclei “combination” due to the increased “correlated” rotational mass of the electron+nucleus combination versus changing the orientation of only the low mass electrons themselves... similarly to what is observed in circuits with a high inductance value versus low inductance value.



    Last edited by devin-m; January 13th, 2018 at 12:24 PM.
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    Quote Originally Posted by devin-m View Post
    Suppose we wind a copper coil (solenoid) around an Iron-56 core... if we apply a voltage to the coil, it takes much longer to reach the maximum current than if the iron-56 core was not present. this is inductance. if we substitute the iron core for a copper core, it takes almost the same amount of time as if a vacuum was used. what about the iron causes this delay in max current or inductance?


    The time and energy required to align the magnetic domains that exist in ferromagnetic materials.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Suppose we wind a copper coil (solenoid) around an Iron-56 core... if we apply a voltage to the coil, it takes much longer to reach the maximum current than if the iron-56 core was not present. this is inductance. if we substitute the iron core for a copper core, it takes almost the same amount of time as if a vacuum was used. what about the iron causes this delay in max current or inductance?


    The time and energy required to align the magnetic domains that exist in ferromagnetic materials.
    @exchemist so then at the atomic scale, why does it take so much additional time and energy to align the magnetic domains in iron-56 or nickel-62 (ferromagnetic) compared to, for example, copper-63 (non-ferromagnetic)?
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    The time and energy required to align the magnetic domains that exist in ferromagnetic materials.
    @exchemist so then at the atomic scale, why does it take so much additional time and energy to align the magnetic domains in iron-56 or nickel-62 (ferromagnetic) compared to, for example, copper-63 (non-ferromagnetic)?
    Domains are not atomic-scale entities. They are much larger regions of alignment of individual atoms.

    Copper has no domains as it is not ferromagnetic (it is diamagnetic).
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    The time and energy required to align the magnetic domains that exist in ferromagnetic materials.
    @exchemist so then at the atomic scale, why does it take so much additional time and energy to align the magnetic domains in iron-56 or nickel-62 (ferromagnetic) compared to, for example, copper-63 (non-ferromagnetic)?
    Domains are not atomic-scale entities. They are much larger regions of alignment of individual atoms.

    Copper has no domains as it is not ferromagnetic (it is diamagnetic).
    @exchemist... what is happening at the atomic scale which causes aggregations of iron-56, cobalt-59 & nickel-62 atoms to give rise to magnetic domains while aggregations of copper-63 atoms do not give rise to magnetic domains? what exactly happens at & above the curie temperature which causes the property of ferromagnetism to "switch off" in Iron-56, Cobalt-59, Nickel-62, Gadolinium-157, Terbium-160, Dysprosium-160 & Holmium-163? what factors determine this temperature?

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

    edit: since this has been moved to personal theory section, i'll admit I am convinced the rotational symmetry orientations of the low-mass electron orbitals and high-mass atomic nuclei are correlated/synchronized/aligned/"bound"/"tidally locked" in ferromagnetic elements, and not correlated in non-ferromagnetic elements... and I believe this is the correct explanation for the high measured inductance values of solenoids with iron, nickel & cobalt cores. i am also convinced the alignment of the nuclei with the electron orbitals explains the unusually high induced magnetic field strength of ferromagnetic substances when used as solenoid cores, such as those found in BLDC electric motors.




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

    also, it is not just a "personal theory" that the 3 elements which are ferromagnetic at room temperature are also the same 3 elements with the greatest atomic mass defect, highest binding energy, lowest mass per nucleon, and lowest "excess" nuclear energy of all the chemical elements.


    Last edited by devin-m; January 13th, 2018 at 05:29 PM.
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    STOP re-posting the same images over and over, and stop posting redundant images please!
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by Paleoichneum View Post
    STOP re-posting the same images over and over, and stop posting redundant images please!
    ok message received but the images are indeed all externally linked and hosted meaning they are not taking up any storage space (or bandwidth) from thescienceforum.com servers.
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    Its not a storage issue, its that they are already in the thread, and repeating them ad nauseum does not help your discussion and makes the thread a pain to read through, especially on mobile devices.
    If more of us valued food and cheer and song above hoarded gold, it would be a merrier world. -Thorin Oakenshield

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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    @exchemist Would it be accurate to say that when a nucleus has lower “mass per nucleon” than another nucleus, that the particles including up quarks, down quarks, protons and neutrons composing the nucleus with lower mass per nucleon have less angular momentum (less energy) per particle than the nucleus with greater mass per nucleon?
    No.
    can anyone explain like i'm 5 how the nucleus of say... plutonium-239... has greater "mass per nucleon" [E=MC^2] and greater "nuclear excess energy" than iron-56, without the plutonium-239's constituent nuclear particles (quarks/protons/neutrons) having additional angular momentum per particle in the nucleus, compared to iron-56? i previously believed this was necessary for conservation of momentum.

    Source: https://en.wikipedia.org/wiki/Momentum#Conservation

    "This law holds no matter how complicated the force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds up to zero, so the total change in momentum is zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces.[4] It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics.[6]"

    Source: https://en.wikipedia.org/wiki/Momentum#Conservation


    Last edited by devin-m; January 14th, 2018 at 04:35 PM.
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    Quote Originally Posted by devin-m View Post

    can anyone explain like i'm 5 how the nucleus of say... plutonium-239... has greater "mass per nucleon" [E=MC^2] and greater "nuclear excess energy" than iron-56, without the plutonium-239's constituent nuclear particles (quarks/protons/neutrons) having additional angular momentum per particle in the nucleus, compared to iron-56? i previously believed this was necessary for conservation of momentum.


    - Quantum theory: angular momentum is quantised and can not take just any value. The angular momenta of nuclei and their constituents take specific values, that are not connected with the mass in any particular way,

    and

    - general relativity: the stability of a nucleus affects the rest mass, as it is a reflection of the amount of energy present in the system, by E=mc. This has nothing to do with angular momentum, but with the strength of the nuclear binding forces.
    Last edited by exchemist; January 15th, 2018 at 06:07 AM.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    can anyone explain like i'm 5 how the nucleus of say... plutonium-239... has greater "mass per nucleon" [E=MC^2] and greater "nuclear excess energy" than iron-56, without the plutonium-239's constituent nuclear particles (quarks/protons/neutrons) having additional angular momentum per particle in the nucleus, compared to iron-56? i previously believed this was necessary for conservation of momentum.


    - Quantum theory: angular momentum is quantised and can not take just any value. The angular momenta of nuclei and their constituents take specific values, that are not connected with the mass in any particular way,

    and

    - general relativity: the stability of a nucleus affects the rest mass, as it is a reflection of the amount of energy present in the system, by E-mc. This has nothing to do with angular momentum, but with the strength of the nuclear binding forces.
    Suppose I swing a ball around on the end of a string like a game of tether ball. If the ball has too much angular momentum the string might be close to its breaking point (a result of centrifugal force) and I might say in this state say the system has comparatively "low binding energy" because a comparatively small amount of additional energy input added to the ball could "break the string." If the ball had comparatively lower angular momentum, it would take comparatively greater additional energy input to overcome the "binding energy" of the string (to get the ball spinning fast enough for its centrifugal force to break the string) -- I might say the system has greater "binding energy" when the ball has lower angular momentum.

    When comparing Pu-239 and Iron-56 we say the Pu-239 has greater "excess energy" but lower "binding energy," and the Iron-56 has lower "excess energy" but greater "binding energy" or greater "nuclear stability."

    Last edited by devin-m; January 14th, 2018 at 05:11 PM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    The time and energy required to align the magnetic domains that exist in ferromagnetic materials.
    @exchemist so then at the atomic scale, why does it take so much additional time and energy to align the magnetic domains in iron-56 or nickel-62 (ferromagnetic) compared to, for example, copper-63 (non-ferromagnetic)?
    Domains are not atomic-scale entities. They are much larger regions of alignment of individual atoms.

    Copper has no domains as it is not ferromagnetic (it is diamagnetic).
    @exchemist... what is happening at the atomic scale which causes aggregations of iron-56, cobalt-59 & nickel-62 atoms to give rise to magnetic domains while aggregations of copper-63 atoms do not give rise to magnetic domains? what exactly happens at & above the curie temperature which causes the property of ferromagnetism to "switch off" in Iron-56, Cobalt-59, Nickel-62, Gadolinium-157, Terbium-160, Dysprosium-160 & Holmium-163? what factors determine this temperature?
    Ferromagnetism is quite subtle and I do not know, nor can I easily find out, an exact description in detail. However it is to do with the differences between the atomic orbitals of the electrons in Fe and Cu. It is nothing to do with the nuclei. I can give you the following rough idea of how it works (I'm open to correct by an expert however, as I'm just going on what I have have read.)

    In Fe the valence shell structure is 3d64s2, i.e. 6 d electrons and 2 s electrons. So the s orbital is full and there are unpaired electrons in the d orbitals, whereas in Cu you have 3d104s1, i.e. the d orbitals are full and there is just one unpaired electron in the s. As you may know, the s orbitals are spherical, while the d orbitals are directional. Also, the d orbitals get pulled in tighter as you go across the period, due to increasing attractive charge on the nucleus. From what I understand, in the case of Fe, the arrangement of atoms in the crystal structure is such that the unpaired electrons in some of the d orbitals are not oriented in the correct way to participate in the conduction band of the metal, so they do not form part of the "sea" of delocalised electrons, but they interact with neighbours enough for exchange interactions to be significant - hence ferromagnetism. In Cu, the d is full, so there are no unpaired electrons. The only unpaired electron is in the conduction band and not available for creating a magnetic dipole on an individual atom.

    So my take on why we find ferromagnetism in the elements that we do is that it is due to the right combination of unpaired d electrons, the right degree of extension of the d orbitals out from the atom, and certain crystal structures that inhibit the participation of some orbitals in the metallic delocalisation.

    As I say, I am very much open to correction or clarification from a solid state physicist, but there is no reason at all to think the phenomenon is related to nuclear stability.
    Last edited by exchemist; January 15th, 2018 at 06:10 AM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    can anyone explain like i'm 5 how the nucleus of say... plutonium-239... has greater "mass per nucleon" [E=MC^2] and greater "nuclear excess energy" than iron-56, without the plutonium-239's constituent nuclear particles (quarks/protons/neutrons) having additional angular momentum per particle in the nucleus, compared to iron-56? i previously believed this was necessary for conservation of momentum.


    - Quantum theory: angular momentum is quantised and can not take just any value. The angular momenta of nuclei and their constituents take specific values, that are not connected with the mass in any particular way,

    and

    - general relativity: the stability of a nucleus affects the rest mass, as it is a reflection of the amount of energy present in the system, by E-mc. This has nothing to do with angular momentum, but with the strength of the nuclear binding forces.
    Suppose I swing a ball around on the end of a string like a game of tether ball. If the ball has too much angular momentum the string might be close to its breaking point and I might say in this state say the system has comparatively "low binding energy." If the ball had comparatively lower angular momentum, it would take comparatively greater additional energy input to overcome the binding force of the string (to get the ball spinning fast enough for its centrifugal force to overcome the "binding energy" of the string) -- I might say the system has more "binding energy" when the ball has less angular momentum.

    When comparing Pu-239 and Iron-56 we say the Pu-239 has greater "excess energy" but lower "binding energy," and the Iron-56 has lower "excess energy" but greater "binding energy" or greater "nuclear stability."
    Unfortunately this is ballocks when it comes to quantum systems such as those we are discussing. It looks as if you have not studied anything about how angular momentum is quantised in atomic-scale systems. I recommend that you do this, as it is very important if you want real understanding of this type of phenomenon.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    can anyone explain like i'm 5 how the nucleus of say... plutonium-239... has greater "mass per nucleon" [E=MC^2] and greater "nuclear excess energy" than iron-56, without the plutonium-239's constituent nuclear particles (quarks/protons/neutrons) having additional angular momentum per particle in the nucleus, compared to iron-56? i previously believed this was necessary for conservation of momentum.


    - Quantum theory: angular momentum is quantised and can not take just any value. The angular momenta of nuclei and their constituents take specific values, that are not connected with the mass in any particular way,

    and

    - general relativity: the stability of a nucleus affects the rest mass, as it is a reflection of the amount of energy present in the system, by E-mc. This has nothing to do with angular momentum, but with the strength of the nuclear binding forces.
    Suppose I swing a ball around on the end of a string like a game of tether ball. If the ball has too much angular momentum the string might be close to its breaking point and I might say in this state say the system has comparatively "low binding energy." If the ball had comparatively lower angular momentum, it would take comparatively greater additional energy input to overcome the binding force of the string (to get the ball spinning fast enough for its centrifugal force to overcome the "binding energy" of the string) -- I might say the system has more "binding energy" when the ball has less angular momentum.

    When comparing Pu-239 and Iron-56 we say the Pu-239 has greater "excess energy" but lower "binding energy," and the Iron-56 has lower "excess energy" but greater "binding energy" or greater "nuclear stability."
    Unfortunately this is ballocks when it comes to quantum systems such as those we are discussing. It looks as if you have not studied anything about how angular momentum is quantised in atomic-scale systems. I recommend that you do this, as it is very important if you want real understanding of this type of phenomenon.
    My understanding is there is generally an inverse relationship between "excess nuclear energy" and "binding energy." For example iron isotopes have nearly the greatest "binding energy" but lowest "excess energy" or "mass per nucleon" -- by comparison helium isotopes have nearly the lowest binding energy and nearly the greatest "excess energy."

    Notice the inverse nature of both graphs: (reposted for comparison)


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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Unfortunately this is ballocks when it comes to quantum systems such as those we are discussing. It looks as if you have not studied anything about how angular momentum is quantised in atomic-scale systems. I recommend that you do this, as it is very important if you want real understanding of this type of phenomenon.
    My understanding is there is generally an inverse relationship between "excess nuclear energy" and "binding energy." For example iron isotopes have nearly the greatest "binding energy" but lowest "excess energy" or "mass per nucleon" -- by comparison helium isotopes have nearly the lowest binding energy and nearly the greatest "excess energy."

    Notice the inverse nature of both graphs: (reposted for comparison)

    Yes, the "binding energy" of a nucleus is analogous to the "ionisation energy" of an atomic electron, i.e. the energy you need to put in, to separate the components. More here: https://en.wikipedia.org/wiki/Nuclear_binding_energy

    The binding energy is thus the energy the nucleus does not have, relative to free constituent particles.

    The nucleus has lower energy than the free particles that make it up and therefore has less mass than they do, by E=mc.


    Returning briefly to the issue of angular momentum, I invite you to think about a very simple case, namely the orbital angular momentum of the electron in the hydrogen atom. Do you know how much angular momentum it has? The answer is zero. And yet it does not fall into the nucleus. This finding is one of the reasons why quantum theory was originally developed. If we could simply use analogies of little girls and boys playing with balls on string, we would never have needed to develop QM.

    In fact we know the angular momentum of the nuclei of Fe and Cu. For 54Fe, 56Fe and 58Fe the spin is zero, while for 57Fe it is 1/2. For Cu, both 63Cu and 65Cu have nuclear spin of 3/2. More here: https://www.webelements.com/iron/isotopes.html That would seem to show pretty clearly that nuclear spin does not tell you anything about ferromagnetism.
    Last edited by exchemist; January 15th, 2018 at 06:12 AM.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Unfortunately this is ballocks when it comes to quantum systems such as those we are discussing. It looks as if you have not studied anything about how angular momentum is quantised in atomic-scale systems. I recommend that you do this, as it is very important if you want real understanding of this type of phenomenon.
    My understanding is there is generally an inverse relationship between "excess nuclear energy" and "binding energy." For example iron isotopes have nearly the greatest "binding energy" but lowest "excess energy" or "mass per nucleon" -- by comparison helium isotopes have nearly the lowest binding energy and nearly the greatest "excess energy."

    Notice the inverse nature of both graphs: (reposted for comparison)

    Yes, the "binding energy" of a nucleus is analogous to the "ionisation energy" of an atomic electron, i.e. the energy you need to put in, to separate the components. More here: https://en.wikipedia.org/wiki/Nuclear_binding_energy

    The binding energy is thus the energy the nucleus does not have, relative to free constituent particles.

    The nucleus has lower energy than the free particles that make it up and therefore has less mass than they do, by E=mc.


    Returning briefly to the issue of angular momentum, I invite you to think about a very simple case, namely the orbital angular momentum of the electron in the hydrogen atom. Do you know how much angular momentum it has? The answer is zero. And yet it does not fall into the nucleus. This finding is one of the reasons why quantum theory was originally developed. If we could simply use analogies of little girls and boys playing with balls on string, we would never have needed to develop QM.

    In fact we know the angular momentum of the nuclei of Fe and Cu. For 54Fe, 56Fe and 58Fe the spin is zero, while for 57Fe it is 1/2. For Cu, both 63Cu and 65Cu have nuclear spin of 3/2. More here: https://www.webelements.com/iron/isotopes.html That would seem to show pretty clearly that nuclear spin does not tell you anything about ferromagnetism.
    Suppose I look at a 3 dimensional depiction of the wavefunction of a P-Orbital electron. What stands out to my untrained eye is that it appears from the shape of the wavefunction that the electron does in fact "fall into the nucleus" -- but instead of stopping it just goes straight through without losing any momentum and out the other side before curving around again and falling back towards the nucleus in an unending "orbit." My understanding is that unlike a gravitational orbit, electrostatic repulsion between neighboring electrons and electrostatic attraction between the electrons and protons in the nucleus is what dictates the shape of the electron orbitals/wavefunctions.

    Now as an analogy suppose I simulate the path of a neutrino in the vicinity of earth. Neutrinos are able to pass through large amounts of matter without being deflected -- in fact they can pass through the entire Earth without being deflected or losing momentum. Since neutrinos have mass they are influenced by gravity. Suppose I follow the path of a neutrino which has just passed through the earth and the neutrino is emerging through the earth's surface towards outer space traveling 1000 meters per second slower than earth's escape velocity in the vicinity of earth's surface. My basic understanding is the neutrino would follow a ballistic trajectory into space, reach an apogee (maximum height) determined by it's "initial velocity," and then the neutrino would begin falling back towards the earth, ultimately reaching its maximum velocity in the vicinity of the center of the earth before passing straight through and out the other side again. My understanding is this cycle could continue for a very long time. Due to the oscillatory nature of the trajectory one might say the neutrino's trajectory is wavelike and one could write an equation describing the "wavefunction" of this trajectory. Since the neutrino would appear to "go out into space" then "turn around and fall back towards earth" then "pass straight through the planet and out the other side" before "falling back again" in an unending cycle, I presume that if I were to graph this trajectory over time it would take on an appearance quite similar to the shape of a P-orbital electron wave function.





    & about the electrons having zero angular momentum in their orbitals....

    If we take the number of copper free electrons in a typical BLDC electric motor winding, and send them on a straight line trajectory relative to the earth at their fermi velocity (1570km/s), how much kinetic energy is this?

    Source: http://hyperphysics.phy-astr.gsu.edu.../fermi.html#c1

    In theory, the kinetic energy equivalent of the angular momentum stored in the free electrons of 10 cubic centimeters of copper is equivalent to the kinetic energy at the muzzle of about 63 x .50 caliber rifle bullets or about 1688 x .45 caliber handgun bullets. This same amount of kinetic energy is also equivalent to about 264 watt hours, or about 5AH from a 50V battery pack.

    ——————————

    0.0000007725 kgs free electrons in 10 cubic centimeters copper (nearly a milligram or 1/1000th of a gram or roughly equivalent to a single grain of sugar)

    1570000 meters per second = V = fermi velocity of copper free electrons

    KE=(1/2)*(0.0000007725*(1570000*1570000))

    KE = 952067.625 joules

    —————

    Math:

    10 cubic centimeters copper

    1,000,000 cubic centimeters per cubic meter

    10 cubic centimeters is 1/100,000th of a cubic meter

    8.49x10^28 free electrons per m^3 copper

    (1/100000)*(8.49x10^28)=8.49x10^23 free electrons in 10 cubic centimeters copper

    8.49x10^23 free electrons * 9.10*10^-31 kg mass per electron

    (8.49*10^23)*(9.10*10^-31) = 7.725*10^(-7) kgs free electrons in 10cc’s copper

    7.725*10^(-7)= 0.0000007725 kgs free electrons in 10cc’s copper

    fermi velocity of copper free electrons = 1570000 meters per second = V

    M = 0.0000007725 kgs

    KE = 1/2 (M * (V * V))

    KE=(1/2)*(0.0000007725*(1570000*1570000))

    KE = 952067.625 joules

    ——————————

    952067 joules / 15037 joules per .50 cal rifle bullet = 63.31 x .50 cal rifle bullets equivalent Kinetic Energy

    952067 joules / 564 joules per .45 handgun bullet = 1688 x .45 bullets equivalent Kinetic Energy

    Source: https://en.wikipedia.org/wiki/Muzzle_energy

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



    Image Source: https://www.survivalmonkey.com/threa...-rounds.49556/

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

    *Also, the equivalent kinetic energy which is in theory stored in the roughly single-sugar-grain-sized mass of free electrons in 10 cubic centimeters of copper, 952067 joules, is the same amount of energy transferred by a human-sized 80kg object striking a brick wall at 345.10 miles per hour, 144.277 meters per second, or 555.39 kilometers per hour.

    Last edited by devin-m; January 15th, 2018 at 05:41 PM.
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    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    Last edited by exchemist; January 16th, 2018 at 05:05 AM.
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    Quote Originally Posted by exchemist View Post
    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    To the contrary... as described in my last post I believe just as electrons have a fermi velocity endowing them with momentum, that the nuclear particles (up quarks, down quarks, protons & neutrons) have momentum as well (unrelated to their 1/2 spin property)... and I believe elements with lower mass per nucleon have lower relative momentum inside the nucleus. by extension i believe the protons in Iron-56 have the lowest relative momenta of all chemical elements as they have the least mass per nucleon and mass is proportional to energy and energy is proportional to relative motion.. i also believe the iron-56 nucleus has an axis of rotational symmetry. since electrons are attracted to protons and vice versa, i believe when both the unpaired electrons are below a certain energy level (absence of phonons - https://en.m.wikipedia.org/wiki/Phonon ), and the protons are below a certain momentum level (lowered mass/energy per nucleon of iron-56) that the rotational orientation of the nucleus is corralated/bound with the orientation of the 4 unpaired d-orbital electrons. i believe the alignment of the nuclear magnetic moments with the alignment of the electron orbital magnetic moments is the reason the induced magnetic field strength of ferromagnetic substances is thousands of times stronger than non-ferromagnetic substances, and in the case of using an iron-56 core in a copper solenoid, i believe the high mass of the iron-56 nuclei being rotationally correlated with the orientation of the 4 d-orbital electrons leads to the extreme time delay to reach maximum current which is observed in inductive circuits with ferrite cores.

    (when i say the nuclear particles have angular momenum i am not talking about 1/2 spin, i am talking about relative motion of the particles inside the nucleus)

    ———————-

    Source: https://en.m.wikipedia.org/wiki/Angu...entum_operator

    “There are several angular momentum operators: total angular momentum (usually denoted J), orbital angular momentum(usually denoted L), and spin angular momentum (spin for short, usually denoted S).”

    ”Finally, there is total angular momentum J, which combines both the spin and orbital angular momentum of a particle or system:J=L+S.“

    “The orbital angular momentum operator L is mathematically defined as the cross product of a wave function's position operator (r) and momentum operator (p):L=rp
    This is analogous to the definition of angular momentum in classical physics

    Source: https://en.m.wikipedia.org/wiki/Angu...entum_operator

    —————————

    Source: https://en.m.wikipedia.org/wiki/Spin–orbit_interaction

    “In quantum physics, the spin–orbit interaction(also called spin–orbit effect or spin–orbit coupling) is a relativistic interaction of a particle's spin with its motion inside a potential. A key example of this phenomenon is the spin–orbit interaction leading to shifts in an electron's atomic energy levels, due to electromagnetic interaction between the electron's magnetic dipole, its orbital motion, and the electrostatic field of the positively charged nucleus.”

    Source: https://en.m.wikipedia.org/wiki/Spin–orbit_interaction

    ———————

    @exchemist — also iron-56 has 4 unpaired d-electrons not 6 as you stated previously — please note that many other elements have the same number of unpaired electrons as iron-56, but only iron, cobalt and nickel display ferromagnetism at room temperature (above 70 degrees fahrenheit)

    Image source: https://en.m.wikipedia.org/wiki/Unpaired_electron




    (reposted chart for comparison)
    Last edited by devin-m; January 16th, 2018 at 09:18 AM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    To the contrary...
    This in spite of the fact that the nucleus of all isotopes of Fe have zero angular momentum, save the 57 isotope, which has a spin of 1/2 but forms only 2% of naturally occurring iron?

    So you are saying that nuclear physics is wrong, then. Good luck with that proposal!
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    To the contrary...
    This in spite of the fact that the nucleus of all isotopes of Fe have zero angular momentum, save the 57 isotope, which has a spin of 1/2 but forms only 2% of naturally occurring iron?

    So you are saying that nuclear physics is wrong, then. Good luck with that proposal!
    @exchemist I am referring to the orbital angular momentum operator (L) and you are referring to the spin angular momentum operator (S)... these are 2 completely separate phenomena

    ———-

    Source: https://en.m.wikipedia.org/wiki/Angu...entum_operator

    “There are several angular momentum operators: total angular momentum (usually denoted J), orbital angular momentum(usually denoted L), and spin angular momentum (spin for short, usually denoted S).”

    ”Finally, there is total angular momentum J, which combines both the spin and orbital angular momentum of a particle or system:J=L+S.“

    “The orbital angular momentum operator L is mathematically defined as the cross product of a wave function's position operator(r) and momentum operator (p):L=rp
    This is analogous to the definition of angular momentum in classical physics

    Source: https://en.m.wikipedia.org/wiki/Angu...entum_operator
    Last edited by devin-m; January 16th, 2018 at 09:39 AM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    To the contrary...
    This in spite of the fact that the nucleus of all isotopes of Fe have zero angular momentum, save the 57 isotope, which has a spin of 1/2 but forms only 2% of naturally occurring iron?

    So you are saying that nuclear physics is wrong, then. Good luck with that proposal!
    @exchemist I am referring to the orbital angular momentum operator (L) and you are referring to the spin angular momentum operator (S)... these are 2 completely separate phenomena
    No they are not completely separate and in fact l and s are not well-defined in many cases, including the constituents of atomic nuclei, as it happens. In such cases one deals instead with the resultant j. The "spin" of an atomic nucleus is the resultant of all sources of angular momentum of the constituent nucleons and is what enables the magnetic moment of the nucleus to be determined: https://en.wikipedia.org/wiki/Nuclear_magnetic_moment.

    Also, a more concise description here: http://hyperphysics.phy-astr.gsu.edu...ear/nspin.html

    It is nuclear magnetic moment that you are looking for, is it not? Well, the magnetic moment of all isotopes of Fe, bar the 2% of isotope 57, is zero. That is a fact, like it or not.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    I take it we're now agreed there is no relation between ferromagnetism and atomic mass defect (the subject of the thread).
    To the contrary...
    This in spite of the fact that the nucleus of all isotopes of Fe have zero angular momentum, save the 57 isotope, which has a spin of 1/2 but forms only 2% of naturally occurring iron?

    So you are saying that nuclear physics is wrong, then. Good luck with that proposal!
    @exchemist I am referring to the orbital angular momentum operator (L) and you are referring to the spin angular momentum operator (S)... these are 2 completely separate phenomena
    No they are not completely separate and in fact l and s are not well-defined in many cases, including the constituents of atomic nuclei, as it happens. In such cases one deals instead with the resultant j. The "spin" of an atomic nucleus is the resultant of all sources of angular momentum of the constituent nucleons and is what enables the magnetic moment of the nucleus to be determined: https://en.wikipedia.org/wiki/Nuclear_magnetic_moment.

    Also, a more concise description here: Nuclear Spin

    It is nuclear magnetic moment that you are looking for, is it not? Well, the magnetic moment of all isotopes of Fe, bar the 2% of isotope 57, is zero. That is a fact, like it or not.
    @exchemist... if we look at the instantaneous value of L - orbital angular momentum operator for an individual proton constituent of an iron-56 nucleus, do you agree that the value is non-zero?

    in other words do you agree an individual proton within an iron-56 nucleus has momentum relative to the other protons and neutrons within said nucleus?
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    Quote Originally Posted by devin-m View Post

    @exchemist... if we look at the instantaneous value of L - orbital angular momentum operator for an individual proton constituent of an iron-56 nucleus, do you agree that the value is non-zero?

    in other words do you agree an individual proton within an iron-56 nucleus has momentum relative to the other protons and neutrons within said nucleus?
    I think it is a meaningless question, as l is not well-defined for a nucleon within the Fe nucleus. (This lack of definition of l and s is quite common in QM, as a result of what is in effect the uncertainty principle as applied to angular momentum. See also R-S and j-j coupling in electron angular momenta, for instance.)

    The resultant I for the nucleus is however well-defined and is zero, which is all that matters when assessing magnetic moment. The nucleus exerts no magnetic field. End of story.

    Look, I have spent a fair amount of time with you on this, but I am now starting to run out of patience. I am not going to pursue an endless debate with somebody who clearly has little understanding of this subject and who persists in silly ideas even after explanations have been provided.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    @exchemist... if we look at the instantaneous value of L - orbital angular momentum operator for an individual proton constituent of an iron-56 nucleus, do you agree that the value is non-zero?

    in other words do you agree an individual proton within an iron-56 nucleus has momentum relative to the other protons and neutrons within said nucleus?
    I think it is a meaningless question, as l is not well-defined for a nucleon within the Fe nucleus. (This lack of definition of l and s is quite common in QM, as a result of what is in effect the uncertainty principle as applied to angular momentum. See also R-S and j-j coupling in electron angular momenta, for instance.)

    The resultant I for the nucleus is however well-defined and is zero, which is all that matters when assessing magnetic moment. The nucleus exerts no magnetic field. End of story.

    Look, I have spent a fair amount of time with you on this, but I am now starting to run out of patience. I am not going to pursue an endless debate with somebody who clearly has little understanding of this subject and who persists in silly ideas even after explanations have been provided.
    @exchemist I will make the analogy for a stable helium atom with 2 protons and 2 electrons we say it has zero overall charge even though the 2 protons are each + 1 and the electrons are -1.

    by contrast if we had 4 neutrinos (each 0 charge, net 0 charge combined), with a small initial momentum they would fly apart

    in the case of 2 protons and 2 electrons (each + or - 1 charge, net 0 charge) they stick together

    in both cases the overall charge of the system is 0, but in the case of “overall zero charge” the summary of charges does nothing to predict the final behavior - whether it flies apart or sticks together

    similarly i argue referring to the overall angular momentum or magnetic moment of an entire iron-56 nucleus as a whole does nothing to inform a person about the hidden properties and relationships of the constituent particles - including the proper explanation for ferromagnetism.

    i would further argue that since the d-orbital electrons spend some of their time in close proximity to (or possibly within) the nucleus, the electrons could be influenced by the charges, spins, momenta and magnetic moments of the individual nuclear particles rather than the entire nucleus, averaged as a whole.
    Last edited by devin-m; January 16th, 2018 at 11:27 AM.
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    @exchemist... if we look at the instantaneous value of L - orbital angular momentum operator for an individual proton constituent of an iron-56 nucleus, do you agree that the value is non-zero?

    in other words do you agree an individual proton within an iron-56 nucleus has momentum relative to the other protons and neutrons within said nucleus?
    I think it is a meaningless question, as l is not well-defined for a nucleon within the Fe nucleus. (This lack of definition of l and s is quite common in QM, as a result of what is in effect the uncertainty principle as applied to angular momentum. See also R-S and j-j coupling in electron angular momenta, for instance.)

    The resultant I for the nucleus is however well-defined and is zero, which is all that matters when assessing magnetic moment. The nucleus exerts no magnetic field. End of story.

    Look, I have spent a fair amount of time with you on this, but I am now starting to run out of patience. I am not going to pursue an endless debate with somebody who clearly has little understanding of this subject and who persists in silly ideas even after explanations have been provided.
    @exchemist I will make the analogy for a stable helium atom with 2 protons and 2 electrons we say it has zero overall charge even though the 2 protons are each + 1 and the electrons are -1.

    by contrast if we had 4 neutrinos (each 0 charge, net 0 charge combined), with a small initial momentum they would fly apart

    in the case of 2 protons and 2 electrons (each + or - 1 charge, net 0 charge) they stick together

    in both cases the overall charge of the system is 0, but in the case of “overall zero charge” the summary of charges does nothing to predict the final behavior - whether it flies apart or sticks together

    similarly i argue referring to the overall angular momentum or magnetic moment of an entire iron-56 nucleus as a whole does nothing to inform a person about the hidden properties and relationships of the constituent particles - including the proper explanation for ferromagnetism.

    i would further argue that since the d-orbital electrons spend some of their time in close proximity to (or possibly within) the nucleus, the electrons could be influenced by the charges, spins, momenta and magnetic moments of the individual nuclear particles rather than the entire nucleus, averaged as a whole.
    The helium atom is neutral and exerts no electric or magnetic field.

    By the way, electrons in d orbitals spend zero time at the nucleus. There is a node (= zero electron density) in the wave function at the nucleus, for all orbital types except s orbitals.

    Why don't you study this stuff properly instead of just reading Wiki and making shit up? It's really rewarding, provided you have the stomach for the maths. And you clearly have the interest.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post

    @exchemist... if we look at the instantaneous value of L - orbital angular momentum operator for an individual proton constituent of an iron-56 nucleus, do you agree that the value is non-zero?

    in other words do you agree an individual proton within an iron-56 nucleus has momentum relative to the other protons and neutrons within said nucleus?
    I think it is a meaningless question, as l is not well-defined for a nucleon within the Fe nucleus. (This lack of definition of l and s is quite common in QM, as a result of what is in effect the uncertainty principle as applied to angular momentum. See also R-S and j-j coupling in electron angular momenta, for instance.)

    The resultant I for the nucleus is however well-defined and is zero, which is all that matters when assessing magnetic moment. The nucleus exerts no magnetic field. End of story.

    Look, I have spent a fair amount of time with you on this, but I am now starting to run out of patience. I am not going to pursue an endless debate with somebody who clearly has little understanding of this subject and who persists in silly ideas even after explanations have been provided.
    @exchemist I will make the analogy for a stable helium atom with 2 protons and 2 electrons we say it has zero overall charge even though the 2 protons are each + 1 and the electrons are -1.

    by contrast if we had 4 neutrinos (each 0 charge, net 0 charge combined), with a small initial momentum they would fly apart

    in the case of 2 protons and 2 electrons (each + or - 1 charge, net 0 charge) they stick together

    in both cases the overall charge of the system is 0, but in the case of “overall zero charge” the summary of charges does nothing to predict the final behavior - whether it flies apart or sticks together

    similarly i argue referring to the overall angular momentum or magnetic moment of an entire iron-56 nucleus as a whole does nothing to inform a person about the hidden properties and relationships of the constituent particles - including the proper explanation for ferromagnetism.

    i would further argue that since the d-orbital electrons spend some of their time in close proximity to (or possibly within) the nucleus, the electrons could be influenced by the charges, spins, momenta and magnetic moments of the individual nuclear particles rather than the entire nucleus, averaged as a whole.
    The helium atom is neutral and exerts no electric or magnetic field.

    By the way, electrons in d orbitals spend zero time at the nucleus. There is a node (= zero electron density) in the wave function at the nucleus, for all orbital types except s orbitals.

    Why don't you study this stuff properly instead of just reading Wiki and making shit up? It's really rewarding, provided you have the stomach for the maths. And you clearly have the interest.
    Suppose i use a cylinder of copper as the core of a solenoid, and I apply a current to the solenoid which in turn aligns the electron orbitals in the copper core while current is flowing (via inductance). As soon as I remove the current from the solenoid, the copper orbitals will “spontaneously” reorient themselves to random orientations. My understanding is “phonons” are responsible for reorienting the copper orbitals to random orientations... true or false?

    In the case of using an iron core, many of the iron’s d-orbitals will retain this induced directional preference... my understanding is below a certain temperature, the phonons are too “weak” to reorient the iron d-orbitals... and above the “curie temperature” the phonons have enough magnitude to reorient the d-orbitals... true or false?
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  35. #34  
    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by exchemist View Post
    The helium atom is neutral and exerts no electric or magnetic field.

    By the way, electrons in d orbitals spend zero time at the nucleus. There is a node (= zero electron density) in the wave function at the nucleus, for all orbital types except s orbitals.

    Why don't you study this stuff properly instead of just reading Wiki and making shit up? It's really rewarding, provided you have the stomach for the maths. And you clearly have the interest.
    @exchemist I’ve been doing some reading up on nodes and I found this:

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    [in reference to s-orbital electrons]

    “...antinode means the electron is most likely to be at the physical position of the nucleus (which it passes straight through without scattering or striking it), since it is moving (on average) most rapidly at that point, giving it maximal momentum.

    A mental "planetary orbit" picture closest to the behavior of electrons in s orbitals, all of which have no angular momentum, might perhaps be that of a Keplerian orbit with the orbital eccentricity of 1 but a finite major axis, not physically possible (because particles were to collide), but can be imagined as a limit of orbits with equal major axes but increasing eccentricity
    .”

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    In the above description, it says very similarly to my earlier neutrino analogy that in the case of S-orbitals they pass straight through the nucleus without strikng or being scattered by it. It also says the s electrons are moving most rapidly and have maximum momentum in the vicinity of the nucleus (also similarly to my earlier neutrino analogy). But then confusingly it goes on to say s-orbitals have no angular momentum. How does the electron have maximum momentum near the nucleus, but no angular momentum at the same time?

    The orbital angular momentum I was previously referring to (for both the iron-56 nuclear protons and iron-56 d-orbital electrons) is the type of momentum described in the article which is at maximum when the s-orbital electron is in the vicinity of the nucleus.

    In the case of Iron-56, since the 2 4s-orbital electrons are at a lower energy level than the 4 unpaired 3d-orbital electrons, and since the 3d orbitals are adjacent to the 4s orbitals, and since the 4s electrons spend time in the vicinity of the nucleus, and since the iron-56 protons have the least mass/energy of all elements, perhaps exchange interactions between the nucleus and the 4s orbitals combine with exchange interactions between the 4s and 3d orbitals, ultimately leading to a correlation between the physical orientation of the nucleus and the physical orientation of the unpaired 3d orbitals.

    one thing that i find troubling about the description of nodes, however, is the description that there is “zero” probability of finding an electron within a node. my undestanding is nodes separate the lobes of high electron probability. if it is impossible for an electron to be found within a node, how does an electron travel from one probability lobe to the next without passing directly through a node, leading to a non-zero probablility of finding the electron there? it made me think of a joke:

    Q: why did the electron cross the node?...

    A: to get to the other side...



    Last edited by devin-m; January 17th, 2018 at 11:00 AM.
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  36. #35  
    exchemist
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    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    The helium atom is neutral and exerts no electric or magnetic field.

    By the way, electrons in d orbitals spend zero time at the nucleus. There is a node (= zero electron density) in the wave function at the nucleus, for all orbital types except s orbitals.

    Why don't you study this stuff properly instead of just reading Wiki and making shit up? It's really rewarding, provided you have the stomach for the maths. And you clearly have the interest.
    @exchemist I’ve been doing some reading up on nodes and I found this:

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    [in reference to s-orbital electrons]

    “...antinode means the electron is most likely to be at the physical position of the nucleus (which it passes straight through without scattering or striking it), since it is moving (on average) most rapidly at that point, giving it maximal momentum.

    A mental "planetary orbit" picture closest to the behavior of electrons in s orbitals, all of which have no angular momentum, might perhaps be that of a Keplerian orbit with the orbital eccentricity of 1 but a finite major axis, not physically possible (because particles were to collide), but can be imagined as a limit of orbits with equal major axes but increasing eccentricity
    .”

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    In the above description, it says very similarly to my earlier neutrino analogy that in the case of S-orbitals they pass straight through the nucleus without strikng or being scattered by it. It also says the s electrons are moving most rapidly and have maximum momentum in the vicinity of the nucleus (also similarly to my earlier neutrino analogy). But then confusingly it goes on to say s-orbitals have no angular momentum. How does the electron have maximum momentum near the nucleus, but no angular momentum at the same time?

    Try learning the difference between momentum (=mv, in classical terms) and angular momentum (=Iω).
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    (6/5*Phi^2)-Pi=.000048132 devin-m's Avatar
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by devin-m View Post
    Quote Originally Posted by exchemist View Post
    The helium atom is neutral and exerts no electric or magnetic field.

    By the way, electrons in d orbitals spend zero time at the nucleus. There is a node (= zero electron density) in the wave function at the nucleus, for all orbital types except s orbitals.

    Why don't you study this stuff properly instead of just reading Wiki and making shit up? It's really rewarding, provided you have the stomach for the maths. And you clearly have the interest.
    @exchemist I’ve been doing some reading up on nodes and I found this:

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    [in reference to s-orbital electrons]

    “...antinode means the electron is most likely to be at the physical position of the nucleus (which it passes straight through without scattering or striking it), since it is moving (on average) most rapidly at that point, giving it maximal momentum.

    A mental "planetary orbit" picture closest to the behavior of electrons in s orbitals, all of which have no angular momentum, might perhaps be that of a Keplerian orbit with the orbital eccentricity of 1 but a finite major axis, not physically possible (because particles were to collide), but can be imagined as a limit of orbits with equal major axes but increasing eccentricity
    .”

    Source: https://en.m.wikipedia.org/wiki/Atomic_orbital

    In the above description, it says very similarly to my earlier neutrino analogy that in the case of S-orbitals they pass straight through the nucleus without strikng or being scattered by it. It also says the s electrons are moving most rapidly and have maximum momentum in the vicinity of the nucleus (also similarly to my earlier neutrino analogy). But then confusingly it goes on to say s-orbitals have no angular momentum. How does the electron have maximum momentum near the nucleus, but no angular momentum at the same time?

    Try learning the difference between momentum (=mv, in classical terms) and angular momentum (=Iω).
    @exchemist - my understanding is the classical “momentum” of an orbiting body is also considered “angular momentum” — this is the type of angular momentum i have been referencing in the case of the iron-56 protons and unpaired 3d shell electron orbitals and I believe the iron-56 nuclear protons have the lowest average value of this type of momentum when compared to the protons of all other chemical elements...

    Source: Planetary orbits

    ”Clearly, represents the angular momentum (per unit mass) of our planet around the Sun. Angular momentum is conserved (i.e., is constant) because the force of gravitational attraction between the planet and the Sun exerts zero torque on the planet.”

    Source: Planetary orbits
    Last edited by devin-m; January 17th, 2018 at 11:41 AM.
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  38. #37  
    exchemist
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    Quote Originally Posted by devin-m View Post

    @exchemist - my understanding is the classical “momentum” of an orbiting body is also considered “angular momentum” — this is the type of angular momentum i have been referencing in the case of the iron-56 protons and unpaired 3d shell electron orbitals and I believe the iron-56 nuclear protons have the lowest average value of this type of momentum when compared to the protons of all other chemical elements...

    Source: Planetary orbits

    ”Clearly, represents the angular momentum (per unit mass) of our planet around the Sun. Angular momentum is conserved (i.e., is constant) because the force of gravitational attraction between the planet and the Sun exerts zero torque on the planet.”

    Source: Planetary orbits
    Well you'd be wrong then. (This is what comes of trying to develop a "theory" when you do not understand the basics first.) In the s orbital, the electron is modelled as possessing kinetic and potential energy. Since it has rest mass, the possession of kinetic energy must imply the possession of momentum - linear momentum. But the angular momentum, which is what leads to magnetic moment, is zero.
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