insane_alien Posted April 4, 2006 Share Posted April 4, 2006 have you tried comparing the equations you have derived against the current ones which have been shown to be accurate? if they are different then your "theory" is a load of bunk Link to comment Share on other sites More sharing options...
sally Posted April 4, 2006 Author Share Posted April 4, 2006 The results are exacltly the same. Link to comment Share on other sites More sharing options...
Norman Albers Posted April 4, 2006 Share Posted April 4, 2006 In both cases you've got the electron spinning like a top, but this view of spin was abandoned long ago for some very good reasons. For one thing there is the behavior of spin 1/2 particles under [imath]2\pi[/imath] rotations: the wavefunction picks up a negative sign. You have to rotate an electron through [imath]4\pi[/imath'] radians to get back to the initial state. That's clearly not ordinary rotation! Perhaps my azimuthal light field can be looked at as a laser ring gyro. What do you get when you give those one rotation? Link to comment Share on other sites More sharing options...
sally Posted April 5, 2006 Author Share Posted April 5, 2006 Could you please give me the reasons why the two spin electron theory was abandoned. ie. What experiment(s) led to its abandon? Is there an actual experiment that outright contradicts the two spin electron theory? Even if you give me an excellent set of reasons I will still try to support my theory.In any case nobody really knows what electron spin is to give it magnetism.I would just include it in my theory. Quote: Originally Posted by Tom Mattson In both cases you've got the electron spinning like a top, but this view of spin was abandoned long ago for some very good reasons. For one thing there is the behavior of spin 1/2 particles under 2\pi rotations: the wavefunction picks up a negative sign. You have to rotate an electron through 4\pi radians to get back to the initial state. That's clearly not ordinary rotation! Perhaps my azimuthal light field can be looked at as a laser ring gyro. What do you get when you give those one rotation? Link to comment Share on other sites More sharing options...
Norman Albers Posted April 6, 2006 Share Posted April 6, 2006 The last line quoted there is mine. Sally, how do you deal with the directional choices here? Also, isn't there a charge imbalance in the conductor, or do you say it is manifest only as what we call B-field? Link to comment Share on other sites More sharing options...
sally Posted April 6, 2006 Author Share Posted April 6, 2006 The last line quoted there is mine. Sally, how do you deal with the directional choices here? Also, isn't there a charge imbalance in the conductor, or do you say it is manifest only as what we call B-field? Over here is an example of magnetic attraction. You do not even need my formula but just Ampere's current law.For repulsion just inverse one of the currents.Note .I am not implying any atomic structure, what I am showing over here is the resultant equivalent current. Also note I have shown the electron flow.To check the north poles etc. either use the left-hand rule or change the direction of flow to the traditional way first. Over here there is probably a nuance between the classical approach and this one. There are no dipoles. The whole of one magnet attracts the whole of the other magnet. The virtual current is along the whole length of the magnet. Now ,lets explain simply the magnetism between two parallel wires Let p be the positive charges and n the negative. Now you must have a look at the *** relative *** velocities between the charges. It is not the velocity w.r.t. us but one charge w.r.t. the other. For a stationary charge there is no resultant charge or force !!!!! We have 4 charge-forces that we have to consider.There are 2 in each wire thus 2*2. Due to the .5*v^2 factor 2*v is 4 times as big. You will have to go through the example carefully to see how simple it is. The values are simplified to make the example clearer. Current in the same direction Charges RelativeVelocities ForceDueToCharges pp 0 -1 pn 1 +1.01 np 1 +1.01 nn 0 -1 (No relative velocity both moving with same direction and speed) Sum +.02 Current in the opposite directions Charges relative velocities Force pp 0 -1 pn 1 +1.01 np 1 +1.01 nn 2 -1.04 !!!!!!!!!! Sum -.02 Link to comment Share on other sites More sharing options...
Norman Albers Posted April 9, 2006 Share Posted April 9, 2006 I looked at the case of an "anti-wire" and you are successful here! I congratulate you at this level. You always get more from the factor of 'v' being squared inside the expression for gamma, when you give it a coefficient.(Is that clear ?) In the case of an anti-wire, the attraction of the two opposite carriers having a relative velocity of '2v' trumps the other terms, so today I will be happy to buy your coffee. I am still suspicious and will work out the questions of Lorentz tranform. You may not use this selectively. Link to comment Share on other sites More sharing options...
Norman Albers Posted April 10, 2006 Share Posted April 10, 2006 Is it so that you describe the interaction of two currents, only? Consider a current-carrying wire next to another conducting mass having no current. Wouldn't your field, where there is 'net excess charge', produce charge separation observable in the mass? Link to comment Share on other sites More sharing options...
sally Posted April 10, 2006 Author Share Posted April 10, 2006 Ok for the force ; The first wire has the current . positive - positive no imbalance positive - negative no imbalance negative-positive .01 negative-negative -.01 As the free current electrons move at the same speed w.r.t. both the positive and negative in the mass we have no resultant force . Now there should be be a slight one time movement of electrons out of the mass but this will be covered up by Lenz's Law which was caused by the changing current. By the time the current is stable the electrons have already left just as they would near a common electrostatic charge . If my theory is right we might still be able to test this.We would have to send exactly the same current down the active wire in both direction and then carefully measure the current in the mass.There might be a small difference. It could also be that this is one of the reasons that near high tension currents there is ionisation. Link to comment Share on other sites More sharing options...
swansont Posted April 10, 2006 Share Posted April 10, 2006 I haven't run the numbers, but it seems to me that the magnetic force is predicted correctly for current flow in a wire, where the drift velocity is very small, but also for particle beams, where the speed is large. How do your equations do with a free charge accelerated through a potential difference, where the classical equation is linear and seems to hold true? Link to comment Share on other sites More sharing options...
sally Posted April 11, 2006 Author Share Posted April 11, 2006 I have not yet worked out cases like this .In general this formula only applies to non-accelerating frames . The real test would be with ordinary magnets . If I am right it would open up a whole new area of magnetism where we could calculate much more exactly magnetic forces and dessign new kinds of motors and generators. We could also use it to create micro-tools for micro-operations etc. If the test is negative this theory is just another useless theory of the thousand that already exist on internet . To start of take two cylindrical magnets and mark the virtual electron flow on them . Move the one around the other from all the angles . It seems to me the that the torques do not correspond exactly to the poles . The case that I feal was the most obvious is the two magnets stuck together as n-s s-n.It seemed to me that the magnets were pulled one towards the other without any pull towards the poles.Also with a small magnet along a large, it did not seem that the pull towards the poles was what I expected. I did all this just by feal so it could all be psychological bias. Link to comment Share on other sites More sharing options...
Norman Albers Posted April 19, 2006 Share Posted April 19, 2006 I predict there will be not so much new found here. We are used to constructing the vector potential field, 'A', as a straightforward image of the current source divided by 'r'. One can in only one further step, construct an energy term of [math]j\cdot{A}[/math]. [Where are latex vector notes?] Link to comment Share on other sites More sharing options...
bombus Posted April 19, 2006 Share Posted April 19, 2006 Hey, I got one - is reality just a relativistic effect? Link to comment Share on other sites More sharing options...
sally Posted April 21, 2006 Author Share Posted April 21, 2006 Reality is not relative.You could call it a dream or virtual but occording to Einstein it it is certainly not relative . The relative effect of magnetism is mentioned in Einstein and Purcell.I am taking what they said on absolute face value.I have gone a little further and changed Purcell's approach a little. When we consider all the "moving" parts of an electric circuit with respect to each other we get a change of charge.The charge increases as the velocity squared of the relative motion.This causes a slight imbalance of the enormous Coulomb charges and has for effect what we call magnetism. Over here the observers are not us but the all charges in the wires themselves.We have to consider the velocities of the charges w.r.t. the other charges. You will have to carefully go through the four charges in two wires to see this. (Each wire has two charges one positive and the other negative thus 2+2=4). The four charges cause a slight imbalance which is called magnetism. If the current is in the same direction for example the positive and negative charges move with respect to each other and are thus slightly stronger. The two negative are stationary as they move with the same speed. The result is that the wires are attracted. Just like the increase of mass with velocity is still considered mass I claim the increase of charge should still be called charge.For my theory static magnetism is a clumsy way of looking at these imbalances. The right-hand rule and Lorentz's Law are round about ways of looking at how charges act directly on other charges. Link to comment Share on other sites More sharing options...
AlienUFO Posted April 22, 2006 Share Posted April 22, 2006 Just like the increase of mass with velocity is still considered massI claim the increase of charge should still be called charge. I woner if two planet move toghether will attract stronger or repel. The right-hand rule and Lorentz's Law are round about ways of looking at how charges act directly on other charges. I don't understand. But I am more concern how's you gonna explain Faraday Law then? Link to comment Share on other sites More sharing options...
gcol Posted April 22, 2006 Share Posted April 22, 2006 Sally; Will your ideas hold for all scales? Link to comment Share on other sites More sharing options...
sally Posted April 24, 2006 Author Share Posted April 24, 2006 The right-hand rule and Lorentz's Law are round about ways of looking at how charges act directly on other charges. I don't understand. But I am more concern how's you gonna explain Faraday Law then? In order to understand how two parallel wires attract or repell you will just have to "break your head over it" untill you actually see how simple it is.Keep in mind the tremendous Coulomb force between the 1 conducting electron per atom of even a tiny wire and also calculate carefully the imbalance of the 4 forces . The positive and negative in one wire to the positive and negative of the other wire. It is very difficult and confusing the first time as we are so used to electricity flowing the other way around. Faraday's Law turns out quite nicely.(I am sorry I can not add a jpeg diagram on the page I have already used up my quota.If you want some jpeg diagrams send me a private mail with your e-mail address and I will send it directly to you.) Faradays law becomes extremely logical , clear and straight forward. We will take a case of a wire moving between the poles of two magnets. A north pole above it and a south pole below it . S >>>>>>>>>>>>>>>>>>> direction of electron >>>>>>>>>>>>>>>>>>> flow. N -------------------------------------------- <--- Wire moving in the field S >>>>>>>>>>>>>>>>>>> direction of electron >>>>>>>>>>>>>>>>>>> flow. N From the top !-------------------<----e----<----------------! ! ........+..................................... ! !.........!.......................................! !.........!.......................................! !.........!.......................................! !.........!.......................................! ! ........! ..--->v..............................! !.........!.......................................! !.........!.......................................! !.........!.......................................! !.........!.......................................! !.........-........................................! !--------------------->--e->-------------------! I have replaced the magnet with a square virtual magnet for simplicity. The electrons go right around the circuit. Ie up on the right hand side and down on the left hand side as well.The dots are there as space keepers. Both the top and bottom magnet have the electrons going around the same way !!!!! I have used the electron flow as there are certain cases where assuming a positive charge flow gives the wrong answer.In order to get the poles , either reverse the flow and use the right hand law or directly use the left-hand law. I am sorry about this confusion of the flow going the "wrong" way but certain cases are too complicated the other way around. Now imgaine an electron in the wire.It is moving parallel to the top and bottom wire.For the top wire it is going in the opposite direction and is so repelled , while for the bottom it is attracted . We thus have our induced emf. The positive charges in the wire cannot move and so do not interfere with our current. Link to comment Share on other sites More sharing options...
sally Posted April 24, 2006 Author Share Posted April 24, 2006 Sally; Will your ideas hold for all scales? Yes it should hold for all scales . Link to comment Share on other sites More sharing options...
Norman Albers Posted May 4, 2006 Share Posted May 4, 2006 This indeed works but there is less here than meets the eye. I am not saying it is not useful, though. The trick we are pulling off is in picking out transverse components of relative velocity. Refer back to my statement about how we construct the integral of magnetic vector potential, in panel #37. There it is manifestly clear that we are dealing with interaction of two currents related to their vector dot product. Link to comment Share on other sites More sharing options...
sally Posted May 23, 2006 Author Share Posted May 23, 2006 I do not exactly agree with you. Governments spend billions of dollars just to look for gravity waves or Higg’s bosons. Just understanding physics is also important.My formula is not the only formula not agreeing exactly with Maxwell’s equations . See Assis over here http://www.ifi.unicamp.br/~assis/wpapers.htm (paper number 55) who brings a different formula from Weber.On the whole if we consider the whole circuit they all give about the same results. On each current element Maxwell’s equation does not obey Newton’s second law of to each action there is an equal and opposite reaction.The others do. I think that it is important to find out which is the right formula or at least which is the most accurate. It seems that this is not as easy as it seems. For my formula it would be fairly simple to test if it is right. In most applications the magnetic forces should be about the same . In all common applications we have coils and so if my simple formula is right the forces are due to the part of the currents running parallel to each other. All motors and generators have the parallel currents and so the differences should not be any big deal. Probably all the real life parasite forces would be much bigger than any differences. There are two cases though that there would be a radical difference. 1.Rail guns. 2.Two perpendicular wires. In these two cases there should be no force on stable currents and the forces should be due to the accelerations involved. This should be fairly easy to verify. If we have a look at the sites talking about rail-guns they claim that the projectile has to be moving for their guns to work. If the projectile starts off at rest it will not move and will end up melted on the rails.It would be a simple matter of one of them to test if the projectile moves at least a little bit with a lower current. They would also have to verify that all the wires leading to the rails would not be parallel to the current running in the projectile. For the two perpendicular wires we would just have to wait for a few seconds and then straighten the wires and see if the wires still bend.It could be that since a lot of current is needed that straight after the big current surge it is switched off to save electricity and stop the wires heating up . The bending of the wires would be due to the change in the current and not the force of stable currents in the wire. These two cases should be easy to verify . If they prove my equation is wrong I would have to add a directional term to it . This would make my whole theory a lot less interesting. If my formula is right ( even if we add on the direction term) it would save some money on research ; Looking for magnetic monopoles would still make no sense. Calculations of very fast moving charges would be simpler and more direct. The calculation of forces on magnets would be a lot more accurate. I am sure that if we could find out which formula is the right one it would help engineers and scientists a lot in all their research . New effects could be found opening new avenues . Link to comment Share on other sites More sharing options...
Norman Albers Posted December 16, 2006 Share Posted December 16, 2006 Yes, a current can be Lorentz-transformed and expansions are useful. Consider just a one-species current flow, then transform yourself to where it seems stationary. You have "lost" the magnetic field, but you will recalculate the charge density per unit length, which has transformed also. My musing asks if this means we do not need the vacuum to have permeability per se. Can vacuum dipole fluctuations, which can be seen to support electric fields, simply be the medium of "magnetism", which is as Sally points out, a certain magnification of co-linear force? Link to comment Share on other sites More sharing options...
Norman Albers Posted December 28, 2006 Share Posted December 28, 2006 In her opening Samurai swing, Sally called B-field mapping "cabalistic". I love this. The reason we do this is that nature manifests many current loops, starting with particle spin, and observed in iron filings. Why, the lines are obvious. "You say potato, and I say potahto." Whatever is useful, just so it's right. Link to comment Share on other sites More sharing options...
Norman Albers Posted January 4, 2007 Share Posted January 4, 2007 SALLY, YOU'RE GONNA LIKE THIS! . You jogged my thinking to where I have created the hypothesis that the vacuum needs no magnetic densities to transmit "magnetism". Nearly everyone hangs on to some such notion, but I say just as electric force is transmitted electrically in the vacuum dipole population, any current source may be analyzed in Lorentz transform locally, and the interacting element may be treated the same. There need be nothing but electric responses from the vacuum. In a one-word response, H.Puthoff just agreed! He had been assuming (ah, the parsing) that the spins of the virtual population were needed. Sally, I would like to give you some credit here. Link to comment Share on other sites More sharing options...
Farsight Posted January 4, 2007 Share Posted January 4, 2007 Hi Albers. Link to comment Share on other sites More sharing options...
Norman Albers Posted January 19, 2007 Share Posted January 19, 2007 The difficult case to vizualize is of a charge approaching a current line or wire. We understand that there should "be a B-field around the current" and that the charge must "cyclotron" about these lines, or at least be deflected. This means a sideways force; how can we pull this out of the hat? I did lay awake for a few spells over this. We may transform to the frame of the approaching charge and see the wire coming at us. The trick is to accurately consider differential current elements in the source. What was moving along the wire now has a perpendicular component added, so that the differential vector velocity is seen as a set of vectors at the resultant angle but laying on that line. Those on the approaching side of center are as an electron coming at you, with a reduced longitudinal field. Those on the other side show you the broadside gamma-squared increase of electric field. Thus there is a net electric side force. Sally, where are you? This could get funny. Link to comment Share on other sites More sharing options...
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