Lazarus

elfmotat said: Posted Today, 01:34 PM Lazarus, you're under the impression that when the calculations using Newtonian gravity were done, they neglected to include the fact that light always travels at c. I don't know where you got this silly idea from, but it's completely untrue. --------------------------------- Reply: That is the answer that I wanted. Everything I could find talked about the same equations that would be used for a comet. Would you kindly point to a description of the calculation incorperating the velocity of light? All I could find were things like: a = 2GM/rc^2 which is smilar to the equation for a comet but doesn't say anthing about the light not accelerating. α=2GMrc2 Read more: http://www.physicsforums.com

Strange said: No. Newton's equation gives you the wrong result. GR gives you the right result. Reply: There is no question about SR giving good results. The mathematical solution is wondefu and elegantl. It is the interpetation that is the "strange" part. The reason the "Newtonian" result is incorrect is that the wrong equation is being used. It does not take in to account that light will not be accelerated in the forward direction.

Swansont said: What do you mean by the "classical" formula? If you mean Newton's formula then, yes it gives the wrong result. Reply: Take Newton's rules add the requirement that light must travel at a constant velocity. That iis all you need to conclude that light will take a different path by the sun than the equation for a comet passing the sun would generate.

Strange said: What do you mean by the "classical" formula? If you mean Newton's formula then, yes it gives the wrong result. Reply: My typing and spelling is certainly not dependable and I am bad about proofreader. Mea culpa. The whole point is that if the classical calculcation is incorrect, a correct calculation accounting for the costant speed of light should give a correct answer.

Thank yuou for the reply. But still. doesn't it take longer for the light to pass the sun than the clascal fomula would expect?

It appears that the calculoation of the amount of bending of light passing the sun uses the same calculation that is used for a comet passing the sun.. If that is the case there is a problem. The comet accelerates then decelerates while the light maintians a constant velocity. That means it takes longer for the light to pass the sun than it would if it could be accelerated in the forward direction. By allowing for the constant velocity of light the result should match the Special Relativity calculation. Does this reaoning make sense?

Dr Benjamin Crowell pointed out another difficulty in formulating a relationship for the interaction of two electrons. The magnetic field is shaped something like the bow wave of a ship because of the time it takes for the information to travel from one to another. The velocity of the information from a particle is assumed to be the speed of light for electrostatic, electromagnetic and gravitational fields. Those velocities should be independent from the speed of radiation. Astronomers have to use the current position of masses rather than their position when light would have left to get correct answers. Is there experimental verification of any of the 3 velocities of information?

Swansont said: Newtonian gravity is not speed dependent. I see what you are saying. The difference is that the motion of one electron affects the other one, whereas with planets the only consideration is position of the planets. ------------------ Swansont said: It's usually best to do it in the rest frame of one of the particles. It's magnetic field is trivial, since the magnetic moment is almost exactly one Bohr magneton. The magneton is a good clue. If i can relate that to the magnetiic field, B, that gives the answer for the case where the effect electron is on a line from the other electron that is perpendicular to it's velocity vector. Then the problem is to determine the rate that the mangetic field drops away from the perpendicular.

Swansont said: Problem for whom, and how is your equation related to this? And what happened to the planets? A problem for me, at least. I would like to know what the value of the magnetic field from one of the electrons would be perpendicular to its velocity and also, how to calculate the field as the other electron is a distance from the perpendicular. The skew is accounted for in the equation so B is the needed value. Two electrons traveling at different velocities resulting in them being in different frames making it impossible to calculate their interactions implies that the interaction of 2 planets can not be calculated.

I don't think any of the comments are meant to be obnoxious. Only trying to make their point strongly. The vector equation: F = q(E + v x B) describes the effect of one electron on another but how to determine B seems to be a problem.

Swansont said: My objection is that in the rest frame of either particle (v=0 or V=0) there is no interaction, but in any other frame there would be. That seems unphysical. To paraphrase Swansont: My objection is that in the rest frame of either PLANET (v=0 or V=0) there is no interaction, but in any other frame there would be. That seems unphysical.

Please excuse my pathetic attempt to find the interaction of 2 electrons. I have to do calculations a half a dozen time to get the same result twice. That certainly lowers the probability that it is correct. It's just that since we can describe the gravitational interaction of 2 planets it seems that we should be able to describe the interaction of 2 electrons.

Question: Is this a satisfactory formula for the electrical interaction of 2 electrons at speeds << c? K cos(X) Vv/r^2 – Q^2/r^2 Where V is the velocity of the causing electron v is the velocity of the effected electron r is the distance between them Q is the charge of the electrons X is the angle between the direction of v and the plane of the direction of the cause electron and the position of the effect electron. (The alignment with the flux) K is the constant to make the units come out right. I will answer it myself. NO. It at least needs another cosine for the lag behind a perpendecular to a V vector.

An equation for orbits is: RV^2 = GM where R is the semi-major axis, V IS velocity, G is the gravitational constant and M is the mass of the central body. When the system is in motion the centripetal force changes. That is consistant with the change in the actual path of the satelite but is it equal to the compensation required to keep the distances between the satelite and the central body the same. The actual path of the satelite is: A stationary ellipse for system velocity of zero. Like a slinky if the system velocity is slower than the satelite's velocity. Like a spot on a the tread of a rolling tire if the two velocities are equal. Like a sine wave if the system velocity is greater than the satelite's. This does not involve relativity, It is just everyday effects. There should be a mathemetical solution. In fact, riding in a convertable and swinging a ball on a rope probably demonstrates that it is the same.

What does motion do to an orbit? Is the equation for an orbit the same when the whole system is in motion? The centripetal force should be greater when the motion of orbit is in the same direction of the motion of the system and less in the reverse direction but the attraction should not change. That is apposed to the orbit being unaffected by the motion of the system. Which is it?

Fred Champion said: You cannot set up and run the experiment with any degree of certainty without using light. It seems even a blind man could perform this experiment. Put a computer in charge of the F16 and the pilot could be blind, too. Or, just do it in the dark. Cheers

Fred Champion said: I'm sure you will recognize that, for any object, the maximum velocity that can be measured is less than (or possibly just equal to) the velocity of the means of communication. One cannot measure supersonic speeds using only sound, nor super-light speeds using light. Just for fun, here is an example of measuring hypersonic speed with sound. Position yourself with a stopwatch equidistant from 2 mountan peaks a known distance apart.. Have a F16 buzz over the 2 peaks at a couple thousand MPH and fire off a noisy missile as it passes each peak. The time difference of the 2 sounds lets you calculate the F16's speed.

Spyman said: If you click on the [More Reply Options] button next to the button you get a [Preview Post] button that is very helpful. Lazurus: Thank you. Swansont said: None of your derivation includes a magnetic field. There are no predictions you make that include it. Reply: The assumption is that there is a rotating magnetic field around nuclei. There seems to be some physical evidence of that. The field doesn’t have to be very strong to affect the orbiting electrons. The strength of the field does not affect the time of the electron orbits as the electron elliptical orbits have to sync with the rotation time. The time of rotation of the magnetic field was calculated from values of hydrogen, centripetal and electrostatic formulas. That is all that is needed to calculate all the orbits of all the elements. The calculations have to give approximations because to get better results it is necessary to track all the electrons and their interactions. No more is necessary to construct a plausible model.

Swanson said: How do these equations follow from a rotating magnetic field? There is no magnetic field in your derivation. The first question in my last post is still unanswered. Reply: The assumption is that nuclei have a rotating magnetic field. The time of rotation of the magnetic field is derived from the known values of hydrogen and that the time of rotation of the magnetic field has to be the same as the lowest possible orbit of the hydrogen electron. Sorry about the oversize print. I never know how the text will come out when it posts.

Swansont said: How do these equations follow from a rotating magnetic field? The orbital parameters should depend on this, right? Why does the field have to be rotating? What is the source of the magnetic field? ---------------------------------------------------------------------- The rotating magnetic field causes the orbits to be discrete. The elliptical orbits have to sync with a multiple of the time of rotation. Otherwise, the orbits would not be stable. A possible cause of the rotation could be that the magnetic field from the 2 end quarks induce a Larmor rotation in the center quark if the 3 quarks are not in a straight line. Some descriptions of the proton put a bend in the proton. The concept might be testable. Line up some hydrogen atoms and apply a strong magnetic field. That should affect the time of rotation causing a change the orbit of the electron which would cause a change in the spectrum. The steps of calculations: Construct orbital equations involving R, V and T. Solve for R, V and T using known values for H, He and He+. Derive the rotational time of an electron. It is proportional to 1/Z^0.5, Helium has a full shell 1 with 2 electrons but hydrogen only has 1 electron. To find the ratio of the size of hydrogen to a full shell, the ratio of sizes between He and He+ can be used. He is smaller than twice the size of He+ so there has to be overlap of the 2 electron orbits of Helium. The ratio of the sizes of He and He+ is 1.1849. Calculate R for an atom with only 1 electron. Get the size of the atom by multiplying R by the outer shell number squared and 1.1849. DERIVATION OF THE EQUATIONS The symbols are: Z = Atomic number 1 Hydrogen M = Mass of an electron 9.109388e-31 kg Q = Charge of an electron -1.602e-19 coulombs R = Semi- major axis 37 picometers Hydrogen E = Kinetic energy of an electron 2.17896e-18 joules Hydrogen K = Coulombs constant 8.9875e9 V = Velocity of electron 2.187e6 m/s Hydrogen T = Time of orbit 1.4806e-16 seconds Hydrogen (Th) S = Shell number 1 Hydrogen EQUATIONS MV^2 / R = Centripetal force KZQ^2 / R^2 = Electrostatic force MV^2/R = KZQ^2 / R^2 Since they are equal R^3MV^2 = KZQ^2 Reduced R^3 = KZQ^2 / MV^2 Semi-major radius cubed R = (KZQ^2 / MV^2)^.33333 Semi-major axis using velocity V = 2PiR/T Velocity is Distance divided by Time 4Pi^2R^2M / RT^2 = KZQ^2 / R^2 Substituting for V in MV^2/R = KZQ^2 / R^2 4Pi^2R^3M = KZQ^2T^2 T^2 = 4Pi^2R^3M / KZQ^2 T squared T = 2PiR^1.5M^.5/K^.5Z^.5Q Time of orbit T = 1.48e-16 / Z^.5 Period of orbit = Period of H divided by Sq root of Z R = (KZQ^2T^2 / 4Pi^2M)^.33333 Substituting for V in R = (KZQ^2 / MV^2)^.33333 R is the semi-major axis of the atom with only 1 electron. To get the semi-major axis of the whole atom multiply by the shell number squared and the He / He+ ratio. r = 1.1849RS^2 Semi-major axis of the whole atom I haven’t found the reference to nuclei with rotating magnetic fields that I stumbled across in the Handbook of Nuclear properties but I did find 3 other references to rotating magnetic fields or oscillating fields in the book. I was surprised to find that this 18 year old book still sells for over $100. Quotes from pages 29, 59 and 81 follow: On page 29: OSCILLATING TRENEDS IN DEFORMED REGIONS Deeper inside major shell regions, where nuclei are deformed, there are oscillating trends superposed on the quadratic variation of the masses. these are reflected in the systematics of mass differences. On page 59: OSCILLATING TRENDS AND DEFORMATION The oscillating trends (section II.F are accounted for by the configuration interaction part (89)) of the mass equation equation. In particular, the depression of the mass surface and the corresponding humps in separation energy lines in deformed regions result from the mixed terms representing mainly neutron-proton correlations. On page 81: INTRODUCTION In specific regions of the nuclear periodic chart, large quadrupole moments ore observed and the low-lying excitations have a rotational character. ------------------------------------------------------------------------------

Swansont said: So let's have your model. The rules of speculations require it. --------------------------------------------------------------------------------------- Here is a model of the hydrogen atom, the rest of the elements and the formation of molecules. The assumption is that nuclei have a rotating magnetic field with the time of rotation equal to the period of the hydrogen electron orbit divided by the atomic number to the three halves power. (T = Th/Z^1.5) It is the cause of discrete orbits. Using the equation for circular and elliptical orbits, the approximate size of all atoms can be calculated. The form of the equation is 4Pi^2r^3m = kzq^2t^2. The symbols are: r is semi-major axis of the ellipse, m is the mass of an electron, k is Coulombs Constant, z is the atomic number, q is the charge of an electron and t is the period of the orbit. Molecules are formed by electrons orbiting 2 nuclei. Here are the results of the calculations for some of the elements compared to the McGraw-Hill chart. Z is the atomic number Te-17 is the orbit period of a nucleus with only one electron times 10^17 S1 pm is the semi-major axis for a nucleus with one electron, in picometers. Calc is the calculated semi-major axis of the atom, in picometers. Book is the McGraw-Hill value for the radius of the atom. Diff is the difference. Ratio is the ratio of the book value to the calculated value. Z Te-17 S1 pm Calc Book Diff Ratio Shell 1 1 8.887 37.003 43.844 37.000 -6.844 0.844 2 3.142 26.163 31.000 31.000 -0.000 1.000 Shell 2 3 1.710 21.359 101.234 152.000 50.766 1.501 4 1.111 18.503 87.696 112.000 24.304 1.277 5 0.795 16.550 78.439 85.000 6.561 1.084 6 0.605 15.103 71.583 77.000 5.417 1.076 7 0.480 13.986 66.289 75.000 8.711 1.131 8 0.393 13.083 62.008 73.000 10.992 1.177 9 0.329 12.332 58.449 72.000 13.551 1.232 10 0.281 11.697 55.441 70.000 14.559 1.263 Shell 3 11 0.244 11.154 118.949 186.000 67.051 1.564 12 0.214 10.678 113.872 160.000 46.128 1.405 13 0.190 10.263 109.446 143.000 33.554 1.307 14 0.170 9.891 105.476 118.000 12.524 1.119 15 0.153 9.555 101.896 110.000 8.104 1.080 16 0.139 9.250 98.642 103.000 4.358 1.044 17 0.127 8.975 95.713 99.000 3.287 1.034 18 0.116 8.719 92.979 98.000 5.021 1.054 Shell 4 19 0.107 8.487 160.899 227.000 66.101 1.411 20 0.099 8.273 156.849 197.000 40.151 1.256 31 0.051 6.644 125.953 135.000 9.047 1.072 32 0.049 6.540 123.986 123.000 -0.986 0.992 33 0.047 6.442 122.135 120.000 -2.135 0.983 34 0.045 6.345 120.284 117.000 -3.284 0.973 35 0.043 6.253 118.548 114.000 -4.548 0.962 36 0.041 6.168 116.928 112.000 -4.928 0.958 Shell 5 37 0.039 6.082 180.169 248.000 67.831 1.376 38 0.038 6.003 177.818 215.000 37.182 1.209 49 0.026 5.289 156.664 166.000 9.336 1.060 50 0.025 5.234 155.037 140.000 -15.037 0.903 51 0.024 5.179 153.410 141.000 -12.410 0.919 52 0.024 5.130 151.964 143.000 -8.964 0.941 53 0.023 5.081 150.517 133.000 -17.517 0.884 54 0.022 5.032 149.071 131.000 -18.071 0.879 Shell 6 55 0.022 4.990 212.839 265.000 52.161 1.245 56 0.021 4.947 211.017 222.000 10.983 1.052 81 0.012 4.111 175.348 171.000 -4.348 0.975 82 0.012 4.086 174.307 175.000 0.693 1.004 83 0.012 4.062 173.266 155.000 -18.266 0.895 84 0.012 4.037 172.224 164.000 -8.224 0.952 85 0.011 4.013 171.183 142.000 -29.183 0.830 86 0.011 3.989 170.141 140.000 -30.141 0.823

Here is the contention of this model. With the assumptions: 1 All matter, electron, protons, photons, etc, all consists of small positive and negative entities that always travel at the speed of light. 2 Protons and neutrons consist of electrons and positrons. 3 Nuclei are held together by electrostatic forces of electrons and positrons. 4 Nuclei have a rotating magnetic field. 5 That Gauss’ electrostatic and electromagnetic Laws, Faraday’s Law and Ampere’s original law are valid. 6 Newton’s Law of gravity is valid. 7 That Newton’s other Laws are corollaries of the electrical laws. This model shows: The reason for discreet orbits of electrons about the nucleus. The relationship of electron orbits to spectral lines and the orbits of moving atoms to the speed of galaxies The reason for slowing of clocks in gravity and at high speeds That physical matter stretches in the direction of motion eliminating the need for time and distance manipulation. That the expansion of space may not be required because the velocity of the source of old light moving away from where the earth is now can account for the Red Shift of light. ------------------------------------------------------------------------- Comments on Radiation The sources of radiation are: Bremsstrahlung radiation shown by Synchrotron radiation or any charge moving through a magnetic field. The reason an electron radiates when affected by a bending magnetic field is that the magnetic moment reacting with the magnetic field provides forces that try to expand the electron causing part of the electron to break free and become a photon. Consider a charged current loop moving through a magnetic field. In addition to the charge causing a change in the direction of motion, each point on the loop receives an outward force. Atomic shell change radiation. The radiation from an electron changing shells is the energy required to offset the difference in the total energy from one shell to the other. Collision radiation from charged particles colliding with atoms. Radiation from the collision can be caused by the Bremsstrahlug radiation and from an electron changing shells. Electrostatic acceleration. It is possible that electrostatic acceleration does not have any unaccounted for energy to produce a photon when the kinetic and latent energy are offset. At this point in the thread: I haven’t proved that this is a valid model of the real world. You haven’t proved that it’s not.

Swansont said: OK, then. Let's see if your formula accurately predicts the Hydrogen energy levels. The ground state is -13.6 eV. The n=2 level is -3.4 eV. Your equation predicts -1.7 eV. Since R = n3, your formula looks a lot like what physics already predicts, except you have an extra factor of 2. So you need to fix that. More importantly, you need to derive this equation from the underlying physics principles. Just arbitrarily dropping the factor of 2 is an admission that you pulled the equation from … somewhere. Reply: Thank you for pointing out my error. The 2 has no business being in the denominator. This arithmetic stuff gets me in trouble. The only cheating I did was to use the known energy of the lowest hydrogen level. The equation shows how a rotating magnetic field could cause the distinct energy levels of the electron. ------------------------------------------------------------------------------- Swansont said: A photon only has hbar of angular momentum. An electron in a P state has hbar, and if its excited to a D state is now has 2hbar of orbital angular momentum. If the direction changes, then the difference is 3hbar. Your description violates conservation of angular momentum. Let’s save this for entertainment after all the other issues are resolved. --------------------------------------------------------------------------- The Handbook of Nuclear Properties of 1997 Where, exactly, does it say that, and what, exactly, does it say? If it accounts for the energy structure as you describe, it should be easy to notice. You can't have it both ways. The book was in the ASU library. This may be moot now as there is a lot of information on nuclear magnetic moments available. -------------------------------------------------------------------------------- Swansont said: Your digging a deeper hole here, in violating more and more established physics principles. Electrons and photons having structure when none has been observed, positrons being confined inside of protons, etc. Add them to the list of things you need evidence for and new physics to explain. But really, before you pile this any deeper, address the objections that have already been brought up. Reply: This is a recounting of objections to this model: The rotating magnetic field that forces the electrons into discrete orbits. Electrons are required to radiate under acceleration. The mysterious constant, .00030394, in the red shift calculation. The zero angular momentum of hydrogen at level S. The electric dipole moment in the nuclei. Neutrino decay. Structure of protons. Small positive and negative vector like entities traveling at c. Heisenberg theory, Pauli theory and this doesn’t agree with current theories. Nobody has complained yet about the explanation of photons passing the sun. The hbar problem. Now the explainations. The rotating magnetic field of the nucleus to produce the discrete orbits is the most important objection. The Rabi Oscillation and the Lamor Precession are intimately involved with Magnetic Resonance Imaging and are just what we would need to force electrons into discrete orbits. However, it requires a magnetic field to occur. So we need to find a magnetic field to satisfy our needs in the nucleus. The 3 quarks for master Mark, the three hump model of the proton and scattering experiments seem to indicate that there are 3 parts to the proton. We know that the proton has a magnetic moment so it is reasonable that each of the 3 parts have a magnetic moment. Then the center part is in a magnetic field and could exhibit Rabi Oscillation and Lamor Precession. Voila, we have the rotating magnetic field we need. Since a photon is much larger than the diameter of an atom the electron makes several complete orbits while a photon is leaving or arriving. It is far fetched to say that there is impossible for there to be a mechanism that prevents a photon from leaving. The constant .00030394 is determined by the shape of the elliptical orbit of the electron and conversely the known energy of the electron and the electron/proton attraction determines the shape of the orbit The number .00030394 is the ratio of the velocity of the electron at the apogee to the velocity of an electron in a circular orbit in the lowest level of hydrogen. The reason that the ratio is very small is that the elliptical orbit is very skinny, not much fatter than a line. There should not be much angular momentum in a pencil shaped ellipse. Thanks to my favorite physicist’s pointing out problems, the model is much improved, so the current model would not have an electric dipole. Since this model has the proton made up of electrons and positrons they can be combined to create a neutrino. The proton consists of 1198 electron/positron pairs plus one positron. The neutron consists of 1200 electron pairs. The proton has 3 distinct parts. The nuclear binding force is replaced by the extra electrons and the sharing of electron pairs. All electrons, positrons and photon consist of many small positive and negative entities traveling at the speed of light. It is not fair or logical to try to discredit a theory by use of an opposing theory, so Heisenberg Uncertainty, Pauli Exclusion and other theories cannot invalidate another theory. I wish someone would complain about the explanation of the failure of classical equations to correctly show the path of a photon passing the sun. The hbar problem can be the last to get resolved.

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.

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.