Lazarus Posted August 16, 2013 Author Posted August 16, 2013 Swansont said: The claim appears to be that the 6 neutrons are really 6 protons + 6 electrons, but now we have a problem with e.g. C-14 which would have 14 protons and 8 electrons, and undergoes beta decay. You have the electron emitted from the nucleus, but now have to explain where the antineutrino came from. Worse, you have neutron-deficient nuclei which undergo beta-plus decay. Reply: The various 3d configurations would have some stronger and some weaker bonds. With 2D models the "pretty" ones seem to coorespond to the more stable isotopes. The cause of radiation has been consideted to be neutino collisions with nuclei. The configurations that are not nice and neat would be more suceptable to ejection of particles. In the early 1900's, when attempts were being made to come up with a physical model, the reason they gave up was for the same 2 reasons you gave, the electrons should radiate and they had no explanation as to why the electron orbits were discrete. I am sure they would have or did come up with a reason that the electon in orbit didn't radiate since the photon is so much bigger than the atom that a photon leaving could be offset by one in the opposite direction. Not a smoking gun. The reason they couldn't come up with an explanation for the discrete orbits is that they had no clue that there could be a rotationg magnetic field around the nuculeus. A real structure would be important, because from that you can calculate if the nucleus has any multipole moments and compare that to experiment. But the bigger issue is how this fits in with decay, as I have mentioned, or the complete contradiction with the Heisenberg Uncertainty Principle I think I brought up before. The Heisenberg Uncertainty Principle is usefull but doesn't preclude there being something real that we are being uncertain about.
swansont Posted August 17, 2013 Posted August 17, 2013 Swansont said: The claim appears to be that the 6 neutrons are really 6 protons + 6 electrons, but now we have a problem with e.g. C-14 which would have 14 protons and 8 electrons, and undergoes beta decay. You have the electron emitted from the nucleus, but now have to explain where the antineutrino came from. Worse, you have neutron-deficient nuclei which undergo beta-plus decay. Reply: The various 3d configurations would have some stronger and some weaker bonds. With 2D models the "pretty" ones seem to coorespond to the more stable isotopes. The cause of radiation has been consideted to be neutino collisions with nuclei. The configurations that are not nice and neat would be more suceptable to ejection of particles. This explains nothing. Neutrino oscillations has an actual meaning in physics, and it doesn't apply here, as it is a change of one type of neutrino to another. The problem with your model is that there is no neutrino. How do you account for the appearance of the neutrino in decay? Where do the positrons come from in beta+ decay? In the early 1900's, when attempts were being made to come up with a physical model, the reason they gave up was for the same 2 reasons you gave, the electrons should radiate and they had no explanation as to why the electron orbits were discrete. I am sure they would have or did come up with a reason that the electon in orbit didn't radiate since the photon is so much bigger than the atom that a photon leaving could be offset by one in the opposite direction. Not a smoking gun. Then go find that reason. You can't just hand-wave the problem away by saying there must be a reason. The reason they couldn't come up with an explanation for the discrete orbits is that they had no clue that there could be a rotationg magnetic field around the nuculeus. Appealing to the stupidity of other scientists is probably not the best argument to use, but it doesn't matter because you don't actually have a model that uses a rotating magnetic field to explain anything. You can't just say "rotating magnetic field" as some sort of magic invocation. How does this actually solve anything?
Lazarus Posted August 18, 2013 Author Posted August 18, 2013 Swansont said: This explains nothing. Neutrino oscillations has an actual meaning in physics, and it doesn't apply here, as it is a change of one type of neutrino to another. The problem with your model is that there is no neutrino. How do you account for the appearance of the neutrino in decay? Where do the positrons come from in beta+ decay?Reply:The neutrino consists of an electron and a positron. The protons are full of electrons and positrons. ************************************************************************************************************************************************* Swansont said: Then go find that reason. You can't just hand-wave the problem away by saying there must be a reason. Reply: The reason for the distinct orbits of electrons about the nucleus is that the nucleus has a rotating magnetic field. The 2 lowest energy electrons orbit the nucleus in sync with the rotation of the magnetic field. The rotation time of the magnetic field is half the time of orbit of one of the 2 shell 1 electrons. In shell 2 with 8 electrons, an electron takes 12.5 times as long to complete 1 orbit as and electron in shell 1. That means when a shell 2 electron reaches its perigee the 2 shell 1 electrons will be exactly reversed in their positions. Each of the 8 shell 2 electrons arrives at its perigee when the 2 shell 1 electrons are in their reversed positions during the 200 cycles of a shell 1 electron. The rest of the shells have different numbers but behave in a similar manner. **************************************************************************************************************** Swansont said: Appealing to the stupidity of other scientists is probably not the best argument to use, but it doesn't matter because you don't actually have a model that uses a rotating magnetic field to explain anything. You can't just say "rotating magnetic field" as some sort of magic invocation. How does this actually solve anything? There is nothing stupid about not finding a solution because of missing or unknown data. The discription of the effect of a rotating magnetic field is above and also in the summary. ********************************************************************************************************************************************* SUMMARY OF THE MODEL The Red Shift of light is caused by electrons changing orbits about the nucleus. The reason for the distinct orbits of electrons about the nucleus is that the nucleus has a rotating magnetic field. The 2 lowest energy electrons orbit the nucleus in sync with the rotation of the magnetic field. The rotation time of the magnetic field is half the time of orbit of one of the 2 shell 1 electrons. In shell 2 with 8 electrons, an electron takes 12.5 times as long to complete 1 orbit as and electron in shell 1. That means when a shell 2 electron reaches its perigee the 2 shell 1 electrons will be exactly reversed in their positions. Each of the 8 shell 2 electrons arrives at its perigee when the 2 shell 1 electrons are in their reversed positions during the 200 cycles of a shell 1 electron. The rest of the shells have different numbers but behave in a similar manner. The reason clocks slow in gravity is that the gravity affects the orbit of the electrons about the nucleus. The reason clocks slow at high velocities is that the limitation of the speed of light causes the path of the electron to change. The reason particles and photons are limited to the speed of light is that all particles and photons consist of small entities that can be represented as positive and negative vectors which always travel at the speed of light. That implies that force on them can only be effective perpendicular to the direction of travel. There is no Nuclear Binding Force. The protons in a nucleus are held together by electrons. The neutron consists of 1 proton, 2 electrons and a positron. The proton and the neutron consist of electrons and positrons. The neutrino consists of an electron and a positron. Light slows in a medium because the particles cause the photon to change directions to go around them. The reason that an electron has an electromagnetic charge greater than a charge of 1 rotating on the surface of an electron is that there are multiple positive and negative charged entities in the electron. The reason there is no dispersion of the colors when light passes the sun is that the particles around the sun are far enough apart that the photons don't always hit feet first as they do with a prism.
swansont Posted August 18, 2013 Posted August 18, 2013 Swansont said: This explains nothing. Neutrino oscillations has an actual meaning in physics, and it doesn't apply here, as it is a change of one type of neutrino to another. The problem with your model is that there is no neutrino. How do you account for the appearance of the neutrino in decay? Where do the positrons come from in beta+ decay? Reply: The neutrino consists of an electron and a positron. The protons are full of electrons and positrons. Doesn't work. These are spin 1/2 particles. There is no way to take 2 spin 1/2 particles and combine them to get a spin 1/2 particle. Swansont said: Then go find that reason. You can't just hand-wave the problem away by saying there must be a reason. Reply: The reason for the distinct orbits of electrons about the nucleus is that the nucleus has a rotating magnetic field. The 2 lowest energy electrons orbit the nucleus in sync with the rotation of the magnetic field. The rotation time of the magnetic field is half the time of orbit of one of the 2 shell 1 electrons. In shell 2 with 8 electrons, an electron takes 12.5 times as long to complete 1 orbit as and electron in shell 1. That means when a shell 2 electron reaches its perigee the 2 shell 1 electrons will be exactly reversed in their positions. Each of the 8 shell 2 electrons arrives at its perigee when the 2 shell 1 electrons are in their reversed positions during the 200 cycles of a shell 1 electron. The rest of the shells have different numbers but behave in a similar manner. Too hand-wavy. You need an actual model that shows all of this. Electrons have an interaction energy; you need to write down and show how this matches up with the spectra we observe. **************************************************************************************************************** Swansont said: Appealing to the stupidity of other scientists is probably not the best argument to use, but it doesn't matter because you don't actually have a model that uses a rotating magnetic field to explain anything. You can't just say "rotating magnetic field" as some sort of magic invocation. How does this actually solve anything? There is nothing stupid about not finding a solution because of missing or unknown data. The discription of the effect of a rotating magnetic field is above and also in the summary. What missing or unknown facts? The effect of the nucleus's field is part of the current model. It's up to you to show that a rotating field actually works as a model. SUMMARY OF THE MODEL The Red Shift of light is caused by electrons changing orbits about the nucleus. The reason for the distinct orbits of electrons about the nucleus is that the nucleus has a rotating magnetic field. The 2 lowest energy electrons orbit the nucleus in sync with the rotation of the magnetic field. The rotation time of the magnetic field is half the time of orbit of one of the 2 shell 1 electrons. In shell 2 with 8 electrons, an electron takes 12.5 times as long to complete 1 orbit as and electron in shell 1. That means when a shell 2 electron reaches its perigee the 2 shell 1 electrons will be exactly reversed in their positions. Each of the 8 shell 2 electrons arrives at its perigee when the 2 shell 1 electrons are in their reversed positions during the 200 cycles of a shell 1 electron. The rest of the shells have different numbers but behave in a similar manner. The reason clocks slow in gravity is that the gravity affects the orbit of the electrons about the nucleus. The reason clocks slow at high velocities is that the limitation of the speed of light causes the path of the electron to change. The reason particles and photons are limited to the speed of light is that all particles and photons consist of small entities that can be represented as positive and negative vectors which always travel at the speed of light. That implies that force on them can only be effective perpendicular to the direction of travel. There is no Nuclear Binding Force. The protons in a nucleus are held together by electrons. The neutron consists of 1 proton, 2 electrons and a positron. The proton and the neutron consist of electrons and positrons. The neutrino consists of an electron and a positron. Light slows in a medium because the particles cause the photon to change directions to go around them. The reason that an electron has an electromagnetic charge greater than a charge of 1 rotating on the surface of an electron is that there are multiple positive and negative charged entities in the electron. The reason there is no dispersion of the colors when light passes the sun is that the particles around the sun are far enough apart that the photons don't always hit feet first as they do with a prism. You need math to back this up. How exactly does a nucleus with net positive charge stay together with only an electrostatic force?
Lazarus Posted August 19, 2013 Author Posted August 19, 2013 Swansont said: You need math to back this up. How exactly does a nucleus with net positive charge stay together with only an electrostatic force? Reply: To test the ability of the electron's force, calculate the net force vector on one of the most vunerable protons. The protons furthest from the origin are most vunerable. Or if you wish test any proton. The test proton has the greatest x value and is at coordinates (a,b). q is the charge of a particle. (1 or -1) d is the distance from the test particle. d = sqr((a-x)^2 + (b-y)^2) The force vector on the test particle from another particle is: q(a-x)i q(b-y)j F = ------- + ------- d^3 d^3 The sum of all the force vectors is: gi + hj The net force, f is; f = sqr(g^2 + h^2) The attraction or repulsion is dependent on the sign of g. Place the center of the nucleus at (0,0) of the coordinate system, Then rotate the nucleus until one of the protons closest to the origin is on the positive x axis. The proton with the smallest angle from it in the counter clockwise direction is the test proton. If the net force vector on the test proton is positive the proton will be expelled, if it is negative the proton will be held in place. The calculation of the force vector between two particles is straight forward. Then you have to add up all the force vectors to find the net force and the direction of the force. An example of of a way electrons could hold protons together. In this case we would have p1 on the x axis and p7 as the test proton. The x axis goes through p4 and p1, the y axis goes through p8 and p11.
swansont Posted August 19, 2013 Posted August 19, 2013 And? What is the result of your calculation? Am I to understand you have done the calculation for your ring shape, and gotten a net attraction for all protons?
Lazarus Posted August 20, 2013 Author Posted August 20, 2013 Swansont said: And? What is the result of your calculation? Am I to understand you have done the calculation for your ring shape, and gotten a net attraction for all protons? Reply: Here are the forces one proton from all the other particles. The arithmetic may not be correct but the values from my program are here. If you wish, I can tell the program to choose other particles or other nuclei. Lithium 6 p6 p2 e7 p5 e8 p1 e9 p3 p6 Net force vector (nfx,nfy) on proton number 4 Atomic number is 3 Proton 4 at x= 1.000 y= 1.732 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 1.732 0.333 -0.000 0.333 1 2 -0.500 0.866 1.732 0.333 0.289 0.500 1 3 -0.500 -0.866 3.000 0.111 0.344 0.596 1 5 -2.000 -0.000 3.464 0.083 0.416 0.638 1 6 1.000 -1.732 3.464 0.083 0.416 0.721 1 7 0.500 0.866 1.000 1.000 -0.084 -0.145-1 8 -1.000 -0.000 2.646 0.143 -0.192 -0.238-1 9 0.500 -0.866 2.646 0.143 -0.219 -0.379-1 Attractive force electron to proton -1.000 Total repulsive force on proton -0.437 ----------------------------------------------------- Net force vector (nfx,nfy) on proton number 7 Atomic number is 6 Proton 7 at x= 1.500 y= 0.866 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 1.000 1.000 0.500 0.866 1 2 0.500 0.866 1.000 1.000 1.500 0.866 1 3 -0.500 0.866 2.000 0.250 1.750 0.866 1 4 -1.000 -0.000 2.646 0.143 1.885 0.913 1 5 -0.500 -0.866 2.646 0.143 1.993 1.006 1 6 0.500 -0.866 2.000 0.250 2.118 1.223 1 8 -0.000 1.732 1.732 0.333 2.407 1.056 1 9 -1.500 0.866 3.000 0.111 2.518 1.056 1 10 -1.500 -0.866 3.464 0.083 2.590 1.098 1 11 0.000 -1.732 3.000 0.111 2.646 1.194 1 12 1.500 -0.866 1.732 0.333 2.646 1.527 1 13 1.000 0.577 0.577 3.000 0.047 0.027-1 14 -0.000 1.155 1.528 0.429 -0.374 0.108-1 15 -1.000 0.577 2.517 0.158 -0.530 0.090-1 16 -1.000 -0.577 2.887 0.120 -0.634 0.030-1 17 0.000 -1.155 2.517 0.158 -0.728 -0.097-1 18 1.000 -0.577 1.527 0.429 -0.869 -0.502-1 Attractive force electron to proton -3.000 Total repulsive force on proton -1.003 -------------------------------------------------------- Uranium 270 O Net force vector (nfx,nfy) on proton number 93 Atomic number is 92 Proton 93 at x= 1.058 y= 0.036 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 0.068 214.487 181.980 113.525 1 2 0.998 0.068 0.068 214.487 371.283 12.684 1 3 0.991 0.136 0.121 68.835 409.688 -44.442 1 91 0.991 -0.136 0.185 29.234 489.730 -41.700 1 92 0.998 -0.068 0.121 68.850 524.158 17.925 1 94 1.053 0.108 0.072 191.408 537.220 -173.037 1 274 1.004 -0.173 0.216 21.424 -215.954 166.582-1 275 1.014 -0.104 0.147 46.198 -229.816 122.512-1 276 1.019 -0.035 0.081 151.935 -303.600 -10.304-1 Attractive force electron to proton -643.469 Total repulsive force on proton -303.775
swansont Posted August 20, 2013 Posted August 20, 2013 f is the force, and nfx, nfy are the components of the net force?
Lazarus Posted August 20, 2013 Author Posted August 20, 2013 f is the force, and nfx, nfy are the components of the net force? f is the scalar force between the test proton and the current particle. nfx,nfy is the vector sum as we go dow the list. at the bottom or the list nfx,nfy is the vector of the net force on the test proton.
swansont Posted August 20, 2013 Posted August 20, 2013 f is the scalar force between the test proton and the current particle. nfx,nfy is the vector sum as we go dow the list. at the bottom or the list nfx,nfy is the vector of the net force on the test proton. So you have an unstable equilibrium. What happens when the system is perturbed? It should collapse in on itself. And, the next step: calculate the energy of the system, and compare it to the known binding energy of C-12. Also, what if you lack the symmetry you have here? What of Li-7, or C-13? What of Deuterium? Shouldn't there be a nuclear electric dipole moment in deuterium, according to your model?
Lazarus Posted August 20, 2013 Author Posted August 20, 2013 For synchrotron radiation magnetic fields are used to curve the path because it's most convenient. The power radiated depends on the square of the acceleration, so for a circle it depends on v4. Particles in an accelerator are moving at ~108 m/s, while particles in a wire are moving at ~10-2 m/s. So, all else being equal, the amount of radiation is down by 10-40 . That's why it would be hard to detect the radiation from electrons in a normal current loop. That is sensible and is probably the strongest objection to this model as the discrete orbits could be caused by the rotating magnetic field of the nucleus. However, that may not mean it is impossible that an electron in orbit that is the size of an atom to have some mechanism that defeats the radiation. The photon is so much longer than the atom's diameter that the electron can complete one or more cycles before the the photon is gone. If the photon is leaving in the east direction on one side it should be leaving in the west direction on the opposite side. Nether photon will have completely left the electron during a cycle. A physical situation that has a little similarlarity is: Attach a bucket to a horizonal wheel. Put a coil of rope in the bucket with a couple of feet hanging over the side. Spinning the wheel will cause the rope to be pulled out of the bucket and go off into space. Now attach a vertical pole to the bucket so it extends a bit above the bucket.. Spinning the wheel will cause the rope to wrap around the pole and not fly off into space. Since we do not fully understand the mechanism of an electron firing a photon, it may be a bit of a stretch to say that it is not possible for an electron to orbit a nucleus without radiating.
swansont Posted August 21, 2013 Posted August 21, 2013 Since we do not fully understand the mechanism of an electron firing a photon, it may be a bit of a stretch to say that it is not possible for an electron to orbit a nucleus without radiating. We also don't accept things as correct just because there is no evidence against it. You need to have predictions that can be tested, and your model has to match up with what we've already observed. I pointed out several problems in my last post.
Lazarus Posted August 22, 2013 Author Posted August 22, 2013 Swansont said: We also don't accept things as correct just because there is no evidence against it. You need to have predictions that can be tested, and your model has to match up with what we've already observed. I pointed out several problems in my last post. Reply: I apologize for missing your post. ***************************************** Swansont said: So you have an unstable equilibrium. What happens when the system is perturbed? It should collapse in on itself. Reply: This arrangement of electrons is just to demonstrate that electrons can hold more protons together than the number of electrons. To get strong enough forces to build satisfactory models it is necessary to consider the building blocks of protons, namely electrons and positrons. The proton could be modeled in the following manner as one possibility. Someone said the scattering pattern of the proton is like 3 point masses. May be 3 humps would do instead of point masses. The +'s are positrons and the -'s are electrons. The three hump model PROTON A side view is something like this: Sequence Layer Count Charge Net 1 * 1 1 +1 +1 2 * * * 2 4 +2 +3 3 * * * * * 3 13 +1 +4 4 * * * * * * * 4 22 -2 +2 5 * * * * * * * * 5 37 +1 +3 6 * * * * * * * * * 6 52 -2 +1 7 * * * * * * * * * * 7 73 +1 +2 8 * * * * * * * * * * * 8 94 -2 +0 9 * * * * * * * * * * * 8 94 +2 +2 10 * * * * * * * * * * * 8 94 -2 0 11 * * * * * * * * * * * 7 73 +1 +1 12 * * * * * * * * * * 6 52 -2 -1 13 * * * * * * * * * 5 37 +1 0 14 * * * * * * * 4 22 -2 -2 15 * * * * * 3 13 +1 -1 16 * * * * * 3 13 -1 -2 17 * * * * * * * 4 22 +2 0 18 * * * * * * * * * 5 37 -1 -1 19 * * * * * * * * * * * 6 52 +2 +1 20 * * * * * * * * * * * * 7 73 -1 +0 21 * * * * * * * * * * * * 7 73 +1 +1 22 * * * * * * * * * * * * * 8 94 -2 -1 23 * * * * * * * * * * * * * 8 94 +2 +1 24 * * * * * * * * * * * * * * 9 121 -1 +0 25 * * * * * * * * * * * * * 8 94 +2 +2 26 * * * * * * * * * * * * * 8 94 -2 0 27 * * * * * * * * * * * * 7 73 +1 +1 28 * * * * * * * * * * * * 7 73 -1 0 29 * * * * * * * * * * * 6 52 +2 +2 30 * * * * * * * * * * 5 37 -1 +1 31 * * * * * * * 4 22 +2 +3 32 * * * * * 3 13 -1 +2 33 * * * * * 3 13 +1 +3 34 * * * * * * * 4 22 -2 +1 35 * * * * * * * * 5 37 +1 +2 36 * * * * * * * * * 6 52 -2 0 37 * * * * * * * * * * 7 73 +1 +1 38 * * * * * * * * * * * 8 94 -2 -1 39 * * * * * * * * * * * 8 94 +2 +1 40 * * * * * * * * * * * 8 94 -2 -1 41 * * * * * * * * * * * 7 73 +1 0 42 * * * * * * * * * * 6 52 -2 -2 43 * * * * * * * * * 5 37 +1 -1 44 * * * * * * * 4 22 -2 -3 45 * * * * * 3 13 +1 -2 46 * * * 2 4 +2 +0 47 * 1 1 +1 +1 The net charge is +1. I contains 2399 particles. It is made up 47 layers of particles arranged in hexagons. There are 9 different size layers. The particles adjacent to each other are of opposite charge. The particle in the center of the proton is negative. The particle on each end is positive. The layers that make up the structure are as follows. Layers 4 through 8 may have the charges reversed. Layer size 1 Number of particles is 1. Charge is +1. + Layer size 2 Number of particles is 4. Charge is +2. + - + + Layer size 3 Number of particles is 13. Charge is +1. - + + - - + + + - - - + + Layer size 4 Number of particles is 22. Charge is +2 or -2 - - - + + + + - - - - + + + - - - + + - - + Layer size 5 Number of particles is 37. Charge is +1 or -1. Charge is +2 Charge is -2 + + - - - - - + + + + + + - - - - - - - + + + + + + + + - - - - - - - - - + + + + + + + + + + - - - - - - - - - + + + + + + + + - - - - - - - + + + The way to get a strong bond is for the protons to share electrons and positrons. That explains why bigger nuclei have less mass than a multiple of hydrogen. The mass loss on of about 10 electrons and 10 positrons matches the heavier atoms. The mass loss for hydrogen 2 is 2 electrons and 2 positrons or 1 pair per proton.. The loss for hydrogen 3 is 9 electrons and 9 positrons or 3 pairs per proton.. The loss for helium 4 is 34 electrons and 34 positrons or 8.5 pairs per proton. The loss for copper is 670 electrons and 670 positrons or 10.65 pairs per proton. The loss for uranium 238 is 2157 electrons and 2157 positrons or 9.06 pairs per proton. The reason the numbers don't come out even is, somewhat like isotopes of atoms, is that there can be more than one way of connecting the electrons and positrons resulting in a different number of lost electrons and positrons. ************************************* Swansont said: And, the next step: calculate the energy of the system, and compare it to the known binding energy of C-12. Reply: The binding energy should match the mass loss. The mass loss of carbon 12 is .093897 AMU compared to the hydrogen 1 atom which equates to 114.28 pairs of electrons and positrons. ************************************* Swansont said: Also, what if you lack the symmetry you have here? What of Li-7, or C-13? What of Deuterium? Shouldn't there be a nuclear electric dipole moment in deuterium, according to your model? Reply: With the strong bond of protons sharing electrons and positrons 3D models can be construted. Lithium 6 Showing the 6 protons 1 1 1 2 3 2 3 2 444 3 5 6 5 6 5 6 For lithium 6 a flat model looks pretty good. Lithium 7 1 1 7 1 2 3 7 2 3 2 444 3 7 5 6 5 6 5 6 Sorry, the 3D model for lithium 7 is hard to follow. I haven't figured out how to put pictures in a post. Carbon 12 and Carbon 13 are built as 3D in a similar manner but I have a problem drawing them here.
swansont Posted August 22, 2013 Posted August 22, 2013 Again, your models imply electric dipole moments which are not observed. The nuclei are not shaped like this. The binding energy of H-2 is 2.224 MeV. The energy of two e-e+ pairs is 2.044 MeV. You're off by 9%. Your other numbers seem to be ad-hoc. In any event, you can't have 114.28 pairs of particles. There has to be an integral number. In any event, this is not the binding energy. I'm talking about the configuration energy of all of these charges — there is an electrostatic potential energy for this configuration. What is it? If you are going to claim a proton is made up of 2399 particles, I must ask for evidence to support the claim. Are you noticing a pattern yet? A claim always instigates a request for evidence. Maybe you could start anticipating this.
Lazarus Posted August 31, 2013 Author Posted August 31, 2013 swansont Posted 22 August 2013 -11:00 AM Swanson said: Again, your models imply electric dipole moments which are not observed. Reply: The charges are pretty evenly distributed so there shouldn't be significant difference in the charge from one side of the nucleus to the other. The instable isotopes may have a slight electric dipole moment. Which model are you referring to? Some light nuclei models are below. (If the picture is visable) ***************************** Swansont said: The nuclei are not shaped like this. Reply: Tell me the shape of a nucleus and I will try to produce a model to match. ***************************** Swansont said: The binding energy of H-2 is 2.224 Me\/. The energy of two e-e+ pairs is 2.044 Iv1e\/. You're off by 9%. Your other numbers seem to be ad-hoc. In any event, you can't have 114.29 pairs of particles. There has to be an integral number. Reply: The reason the numbers don't come out even is that there is more than one way to overlap the protons resulting is a different number of electron/positron pairs missing. Somewhat similar to the way isotopes of elements work. The 2.224 Mev binding energy is an averaged number. ***************************** Swansont said: In any event, this is not the binding energy. I'm talking about the configuration energy of all of these charges — there is an electrostatic potential energy for this configuration. What is it? Reply: That is a very difficult calculation. You have to know the position of each electron and positron in each of the protons then do calculation for each one. However, it may be possible to get an approximation to show that the total comes out negative. Since some of the protons in the nucleus share electron/proton pairs so the multiple interactions between electrons and positrons make a stronger bond than the bond between 1 electron and 1 proton. For instance, the bond between 2 pairs is about 1.4 times as great as 1 pair. + - + as apposed to - + - ***************************** Swansont said: If you are going to claim a proton is made up of 2399 particles. I must ask for evidence to support the claim. Are you noticing a pattern yet? A claim always instigates a request for evidence. Maybe you could start anticipating this. Reply: The fact that the difference in mass between the neutron and the hydrogen atom is divisible into both of them to 7 digits of accuracy is strong evidence of a building block. Since both electrons and positrons can be emitted by decay that should mean that there are electrons and positrons in the nucleus. Even if they are "created on expulsion" this model can accommodate that because all matter consists of the same positive and negative vector like entities. ***************************** Another try at the picture of the models. Still another try with a picture from WORD. Sorry still no picture. Eureka ! The picture posted.
swansont Posted August 31, 2013 Posted August 31, 2013 swansont Posted 22 August 2013 -11:00 AM Swanson said: Again, your models imply electric dipole moments which are not observed. Reply: The charges are pretty evenly distributed so there shouldn't be significant difference in the charge from one side of the nucleus to the other. The instable isotopes may have a slight electric dipole moment. Which model are you referring to? Any of them. Dipole and higher-order multipole moments are known and measured for nuclei. You have to show that your model matches them.
Lazarus Posted August 31, 2013 Author Posted August 31, 2013 Swansont said: Any of them. Dipole and higher-order multipole moments are known and measured for nuclei. You have to show that your model matches them. Reply: That is an excellent point. That needs to be answered but also provides a means of determining the structure of the models. Much better than just picking the pretty ones. How to put the Tinkertoys together to match the dipoles is not immediately obvious. i will head back into my cave and stew about it. Thanks again for sharing your very extensive knowledge of the subject.
Lazarus Posted December 9, 2013 Author Posted December 9, 2013 Swansont said:Any of them. Dipole and higher-order multipole moments are known and measured for nuclei. You have to show that your model matches them. Reply: Sorry for the long delay in replying. Sometimes life gets in the way of fun stuff like this. Your insistence on relating the nuclear moments to the model was very helpful. Having the protons side by side seems to work better in most cases. The moments of two protons with the north south poles in opposite directions cancels out the magnetic moment. The easy part is nuclei that have even Z and even A. Almost any symmetrical configuration would work. The way that a proton and a neutron can become a hydrogen 2 nucleus is described below and in the diagrams that are linked. Since the difference in mass of the proton and the neutron is divisible into both the difference can consist of one positron and 2 electrons. A neutron consists of 1201 electrons and 1201 positrons. A proton consists of 1199 electrons and 1200 positrons. A neutron plus a proton totals 2400 electrons and 2401 positrons. A deuteron has less mass than the sum of the proton plus the neutron. The average loss is 2.84 electron/positron pairs. The loss consists of either 1 pair or 3 pairs. The proton has 3 humps. When a proton and a neutron are forced together to form a deuteron some mass is lost and the center hump shares 3 electrons and 2 positrons with the neutron which now looks just like the proton. The end humps share either 2 electrons and 3 positrons or 1 electron and 2 positrons. The rest of the deuteron has either 2390 electron/positron pairs with a loss of 3 pairs or 2394 pairs with a loss of 1 pair. The proton and neutron are “side by side” with their north/south poles in opposite directions. That allows the unmatched positron to produce the nuclear moment. Because the shared electrons and positrons are between magnetic forces in opposite directions there will be some distortion of the moments. When the atomic number is odd all the matched pairs of protons cancel out except 1 or more of the protons which contribute to the nuclear moment. When the number of neutrons is odd (so electron are odd) 1 or more of the electrons contribute to the moment. Even numbers can have 2 or more contributing to the moment. The tables at the end of this post show mass loss and moments for some of the elements. The following links are to diagrams than I couldn't get to insert for some reason. http://www.flickr.com/photos/111049597@N08/11281648674/ http://www.flickr.com/photos/111049597@N08/11281594223/ The following table shows how the moments can be approximated by the number of protons and electrons that add or subtract moment. Some of the calculated values are fairly close and some are not. The difference in orientation of some of the electrons could be responsible for the disparity. The electrons in the shared portion can't align north/south with both of the protons beside them. Calculated vs Book Moments Name Z e A Book Calc A e H 1 0 1 2.792 2.799 1 0 H 1 1 2 0.860 -0.880 0 -1 He 2 1 3 -2.136 1.919 1 -1 Li 3 3 6 0.820 -0.880 0 -1 Li 3 4 7 3.260 3.678 1 0 Be 4 5 9 -1.180 1.919 1 -1 B 5 5 10 1.800 1.880 0 1 B 5 6 11 2.690 2.799 1 0 C 6 7 13 0.714 0.159 1 -3 N 7 7 14 0.400 -0.880 0 -1 N 7 8 15 -0.282 1.039 1 -2 O 8 9 17 -1.892 1.919 1 -1 Ne 10 11 21 -0.660 0.159 1 -3 Na 11 12 23 2.220 2.799 1 0 Mg 12 13 25 -0.860 0.159 1 -3 Al 13 14 27 3.640 3.678 1 0 Si 14 15 29 -0.563 0.159 1 -3 P 15 16 31 1.133 1.039 1 -2 S 16 17 33 0.640 0.159 1 -3 Cl 17 18 35 0.820 1.039 1 -2 Cl 17 20 37 0.680 1.039 1 -2 K 19 20 39 0.390 1.039 1 -2 K 19 22 41 0.210 1.039 1 -2 Ca 20 23 43 -1.320 1.919 1 -1 Sc 21 24 45 4.765 5.678 1 2 Ti 22 25 47 -0.795 0.159 1 -3 Ti 22 27 49 -1.100 1.919 1 -1 Cr 24 29 53 -0.475 0.159 1 -3 Mn 25 30 55 3.456 3.678 1 0 Fe 26 31 57 0.096 0.159 1 -3 Co 27 32 59 4.636 3.678 1 0 This table shows the mass loss in electron/positron pairs. BOOK DIFFERENCES A Z Calc WT Book WT DIFF WT Per Z Neutrino Per Z ELEC -1 0 0.000000 0.000549 0.000549 0.000000 0.65 100.00 NEU 0 1 1.007825 1.008665 0.000840 0.000840 1.00 1.00 H 1 1 1.007825 1.007825 0.000000 0.000000 0.00 0.00 1 2 2.015650 2.014105 -0.001544 -0.000772 -1.84 -0.92 1 3 3.023474 3.016049 -0.007425 -0.002475 -8.84 -2.95 HE 2 3 3.023474 3.016030 -0.007444 -0.002481 -8.86 -2.95 2 4 4.031299 4.002604 -0.028695 -0.007174 -34.17 -8.54 2 5 5.039124 5.012296 -0.026828 -0.005366 -31.94 -6.39 2 6 6.046948 6.018900 -0.028049 -0.004675 -33.40 -5.57 LI 3 5 5.039124 5.012541 -0.026583 -0.005317 -31.65 -6.33 3 6 6.046948 6.015126 -0.031822 -0.005304 -37.89 -6.32 3 7 7.054773 7.016005 -0.038768 -0.005538 -46.16 -6.59 3 8 8.062598 8.022488 -0.040111 -0.005014 -47.76 -5.97 3 9 9.070423 9.027300 -0.043123 -0.004791 -51.35 -5.71 BE 4 6 6.046948 6.019780 -0.027168 -0.004528 -32.35 -5.39 4 7 7.054773 7.016931 -0.037842 -0.005406 -45.06 -6.44 4 8 8.062598 8.005308 -0.057290 -0.007161 -68.21 -8.53 4 9 9.070423 9.012186 -0.058237 -0.006471 -69.34 -7.70 4 10 10.078248 10.013535 -0.064713 -0.006471 -77.05 -7.71 B 5 8 8.062598 8.024612 -0.037986 -0.004748 -45.23 -5.65 5 9 9.070423 9.013335 -0.057088 -0.006343 -67.97 -7.55 5 10 10.078248 10.012939 -0.065309 -0.006531 -77.76 -7.78 5 11 11.086073 11.009305 -0.076768 -0.006979 -91.41 -8.31 5 12 12.093897 12.014353 -0.079544 -0.006629 -94.71 -7.89 C 6 10 10.078248 10.016830 -0.061418 -0.006142 -73.13 -7.31 6 11 11.086073 11.011433 -0.074640 -0.006785 -88.87 -8.08 6 12 12.093897 12.000000 -0.093897 -0.007825 -111.80 -9.32 6 13 13.101722 13.003354 -0.098368 -0.007567 -117.12 -9.01 6 14 14.109547 14.003242 -0.106305 -0.007593 -126.58 -9.04 6 15 15.117372 15.010600 -0.106771 -0.007118 -127.13 -8.48 N 7 12 12.093897 12.018709 -0.075188 -0.006266 -89.52 -7.46 7 13 13.101722 13.005739 -0.095983 -0.007383 -114.28 -8.79 7 14 14.109547 14.003074 -0.106473 -0.007605 -126.78 -9.06 7 15 15.117372 15.000108 -0.117264 -0.007818 -139.62 -9.31 7 16 16.125196 16.006088 -0.119108 -0.007444 -141.82 -8.86 7 17 17.133020 17.008450 -0.124571 -0.007328 -148.32 -8.72 O 8 14 14.109547 14.008597 -0.100949 -0.007211 -120.20 -8.59 8 15 15.117372 15.003072 -0.114300 -0.007620 -136.09 -9.07 8 16 16.125196 15.994915 -0.130281 -0.008143 -155.12 -9.70 8 17 17.133020 16.999132 -0.133888 -0.007876 -159.42 -9.38 8 18 18.140846 17.999161 -0.141685 -0.007871 -168.70 -9.37 8 19 19.148670 19.003557 -0.145113 -0.007638 -172.78 -9.09 8 20 20.156496 20.004070 -0.152426 -0.007621 -181.49 -9.07 F 9 17 17.133020 17.002098 -0.130922 -0.007701 -155.89 -9.17
Lazarus Posted January 14, 2014 Author Posted January 14, 2014 It turns out that the pick and shovel work for this model was done 100 years ago. Johann Balmer related spectral lines to photon wave length back in 1885. Johannes Rydenberg related the photon wave length to the energy change of an electron in an atom. Neils Bohr introduced his model of electrons in circular orbits 100 years ago. Arnold Summerfield showed that Bohr’s model could work with elliptical orbits. Ralph Konig proposed that electrons could have a magnetic moment. George Uhlebeck and Samuel Goudsmit showed that electron magnet moment (later called spin) could account for the spectral fine lines. Then things came to a screeching halt because of 3 issues: There was no explanation for the discreet orbits of electrons. Accelerated electrons “always” had to radiate. The magnetic moment of an electron was greater than the charge rotating at the speed of light around the circumference of the electron. What they had no way of knowing is that nuclei can have a rotating magnetic field. That is all that is needed to force the orbits to have discreet values. Since the length of a photon is much greater than the circumference of an electron’s orbit, there could easily be a mechanism that would prevent the photon from escaping. (i.e. Electrons can absorb photons) Also, the assumption that accelerating electrons “always” radiate could cloud an exception. This model contends that the electron, and all matter, consist of smaller positive and negative charges traveling at the speed of light so counter flowing charges could create a large magnetic moment. To summarize this model. The difference in mass of a neutron and a hydrogen atom is evenly divisible to 7 digits of accuracy into both of them which implies a “building block”. The building block can be an electron/positron pair. The neutron is equivalent to a proton plus 2 electrons and 1 positron. That removes the need for the bonding force created to hold the nucleus together. An electron can hold 2 protons together. A neutron consists of 1200 electrons and 1200 positrons. A proton consists of 1199 electrons and 1200 positrons. The rotating magnetic field about the nucleus forces electrons to discreet orbits. The energy difference relates to the spectral lines which relate to the velocity of galaxies. All matter, electrons, neutrons, protons, photons etc, consists of positive and negative entities traveling at the speed of light. The entities can change direction but cannot change speed. The red shifted light arriving now was produced by objects moving away from where the earth is now at a velocity great enough to produce the red shift.
swansont Posted January 14, 2014 Posted January 14, 2014 What they had no way of knowing is that nuclei can have a rotating magnetic field. That is all that is needed to force the orbits to have discreet values. Show that this is true. Present a model. Since the length of a photon is much greater than the circumference of an electron’s orbit, there could easily be a mechanism that would prevent the photon from escaping. (i.e. Electrons can absorb photons) Also, the assumption that accelerating electrons “always” radiate could cloud an exception. Ad hoc. Science doesn't like ad hoc. This model contends that the electron, and all matter, consist of smaller positive and negative charges traveling at the speed of light so counter flowing charges could create a large magnetic moment. Now you just have to find experimental evidence of these smaller charges, and that any massive body can move at c.
Lazarus Posted January 16, 2014 Author Posted January 16, 2014 Swansont said: Show that this is true. Present a model. Reply I am so glad you asked. The physical picture of the electrons dancing around the nucleus and the mathematics involved comes out beautifully. The reason shell electrons in an atom conform to discrete energies is there are only certain relationships between the orbital electrons and the rotating magnetic field around the nucleus that are stable. An electron has to approach the nucleus at the minimal magnetic field. With a field rotation time of 1, the lowest energy orbit will have an orbit time of 1 since that will synchronize the orbit and field rotation. Helium has two electrons orbiting synchronized with the rotation of the magnetic field. The electrons in the lowest energy orbit approach the nucleus on opposite sides. Again the time of orbit of both electrons is 1. That becomes shell 1. The time between the electron visits to the nucleus is ½. For shell 2, with room for 8 electrons has orbit times of 8. The time between approaches to the nucleus is 1. The approaches to the nucleus fit in between the shell 1 approaches. For shell 3, a maximum of 18 electrons that have an orbit time of 27. The time between approaches to the nucleus is 1.5. The rest of the shells are in the table below. The units used are: The time of rotation of the nuclear magnetic field is 1. The velocity of the electron in the lowest orbit is 1. The radius of the electron in the lowest orbit is 1. Electron mass is 1. Circular orbits are used. Elliptical orbits are more work. The formulas used are: T=R to the 3/2 power (Time vs Radius) R=1/V*V (Radius vs Velocity) K=V*V/2 (Kinetic energy vs Velocity) In the following table the columns are: A Shell number B Number of electrons in a shell C Radius of the circular orbit D Time of the electron orbit E Velocity of the electron in orbit F Kinetic energy of the electron G Total kinetic energy of the electrons in the shell H Time of next electron to approach the nucleus A B C D E F G H Shell Elec’s Radius Time V KE TKE Next 1 2 1 1 1 1/2 1 .5 2 8 4 8 1/2 1/8 1 1 3 18 9 27 1/3 1/18 1 1.5 4 32 16 64 1/4 1/32 1 2 5 50 25 125 1/5 1/50 1 2.5 6 72 36 216 1/6 1/72 1 3 7 98 49 343 1/6 1/98 1 3.5 8 128 64 512 1/8 1/128 1 4 Interestingly, column H has the same numbers as Spin. Also interesting is that all the shells have the same total kinetic energy. I will try to answer the next question you would have, “What happened to orbit times from 2 to 7?”. Some of the possible explanations are, Mother Nature doesn’t like to do cube roots, She doesn’t like shells with partial electrons or more likely, there is some physical or mathematical result that I have been unable to grasp. ------------------------------------------------------------------------------------------------------------------ Swanson said: Ad hoc. Science doesn't like ad hoc. Reply: Proof that it is impossible for the electron to hang in there without loosing kinetic energy is a hard row to hoe. --------------------------------------------------------------------------------------------------------------------------------------------------------------- Swanson said: :Now you just have to find experimental evidence of these smaller charges, and that any massive body can move at c. Reply: A massive body would have to have all the small vector like entities parallel and that would be difficult to do so electrons aren't supposed to travel at c. Is there a better explanation of the incredable strength of the electron's magnetic moment?
swansont Posted January 16, 2014 Posted January 16, 2014 No, my next question is "Where is your model?" I see a table and a lot of hand-wavy discussion, but no maths. By model, I mean something like a prediction of the energy levels based on a solution to the Schrödinger equation with your mythical rotating magnetic field included, and a model of how you get such a rotating magnetic field in the first place. Matching it all up with observed spectra as a finale.
Lazarus Posted February 6, 2014 Author Posted February 6, 2014 Swansont said: No, my next question is "Where is your model?" I see a table and a lot of hand-wavy discussion, but no maths. By model, I mean something like a prediction of the energy levels based on a solution to the Schrödinger equation with your mythical rotating magnetic field included, and a model of how you get such a rotating magnetic field in the first place. Matching it all up with observed spectra as a finale. Reply: I will start with your request for the grand finale. The equation for the energy of an electron in all the possible electron orbits of hydrogen is: a E = - --------------------- eV 2/3 2R E is the total energy of the electron a is the total energy of the lowest level, for hydrogen it is 13.6 eV R is the number of rotations of the magnetic field of the nucleus, a positive integer. 2/3 1/3 When R is an integer then R is a standard shell number and the E is the energy of an electron in that shell. The time of orbit for the electron has to match the time of rotation of the magnetic field of the nucleus in order to approach the nucleus when the magnetic field of the nucleus is at a minimum For the rest of the elements electron interaction has to be added into the equation. The Fine lines are caused by the magnetic pole reversal of the electron in some orbit changes. The Hyperfine lines are caused by the difference in the interaction of the magnetic moments when an orbit change switches between clockwise and counter-clockwise orbits in relation to the north pole of the nucleus. To make a physical rotating magnetic field, put a wire with a current around a disk then spin the disk about an off center axis. By the way, the reason that a photon passing the sun has a different path from a rock passing the sun is that the rock is accelerated in the forward direction and a photon cannot be accelerated in the forward direction. That means the photon takes longer to pass the sun than it would if it could be accelerated like the rock so its path has to be different. Einstein resolved the difference by jury-rigging time itself.
swansont Posted February 7, 2014 Posted February 7, 2014 I will start with your request for the grand finale. The equation for the energy of an electron in all the possible electron orbits of hydrogen is: a E = - --------------------- eV 2/3 2R E is the total energy of the electron a is the total energy of the lowest level, for hydrogen it is 13.6 eV R is the number of rotations of the magnetic field of the nucleus, a positive integer. 2/3 1/3 When R is an integer then R is a standard shell number and the E is the energy of an electron in that shell. How is R determined? The time of orbit for the electron has to match the time of rotation of the magnetic field of the nucleus in order to approach the nucleus when the magnetic field of the nucleus is at a minimum For the rest of the elements electron interaction has to be added into the equation. The Fine lines are caused by the magnetic pole reversal of the electron in some orbit changes. The Hyperfine lines are caused by the difference in the interaction of the magnetic moments when an orbit change switches between clockwise and counter-clockwise orbits in relation to the north pole of the nucleus. How does an electron change direction like that? Where does the angular momentum go (or come from) when it changes direction? To make a physical rotating magnetic field, put a wire with a current around a disk then spin the disk about an off center axis. There are no wires in a nucleus. How does a proton give you a rotating field, and how come nobody has ever noticed that it has one?
Lazarus Posted February 7, 2014 Author Posted February 7, 2014 Swanasont said: How is R determined? Reply: R is equal to the shell number cubed or you can just count the rotations of the magnetic field. Swansont said:How does an electron change direction like that? Where does the angular momentum go (or come from) when it changes direction? Reply: Both the Fine lines and the Hyperfine lines show up on level changes. The energy is accounted for by the capture or release of a photon. The electron changes from one orbit to another that will have either the same or reversed direction of rotation around the nucleus. The slight difference in energy produces the fine line effect. When an electron is heading for the nucleus it only takes a little jolt to make it pass on the other side of the nucleus but the value of the angular momentum could still be the same. Swansont said: There are no wires in a nucleus. How does a proton give you a rotating field, and how come nobody has ever noticed that it has one? Reply: The Handbook of Nuclear Properties of 1997 mentions nuclei with magnetic forces having a rotational character. Unless it was lying, a rotating magnetic field is an accepted concept. So it could be possible for the hydrogen nucleus to have one. It might not be easy to notice. This model has electrons and positrons in the nucleus and photons, electrons, etc consisting of small positive and negative vector like entities that always travel at the speed of light. The large magnetic moment of an electron is accounted for by the currents created by the movement of the entities. The proton should have the extra positron at its center. One possible means of generating a rotating magnetic field is having entities in an off center oblong path around the positron at the center of the proton which would precess because of the fixed speed of the entities.
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