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|    sci.physics.research    |    Current physics research. (Moderated)    |    17,520 messages    |
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|    Message 15,587 of 17,520    |
|    John Heath to Roland Franzius    |
|    Re: How to measure a Lorentz contraction    |
|    05 Mar 17 19:44:33    |
      From: heathjohn2@gmail.com              [Moderator's note: Please submit non-encoded plain text with unquoted       lines of at most 72 characters so that it displays properly in this       text-based newsgroup. -P.H.]              On Sunday, March 5, 2017 at 2:55:56 AM UTC-5, Roland Franzius wrote:       > Am 03.03.2017 um 10:04 schrieb John Heath:       > > I like it. Simple and do-able. Should       > > be able to swing a 3 foot wire at 3 or       > > 4 Hz by hand with the hand end of the       > > wire clamped to a scope. I have spectrum       > > lab audio analyzer software on a lap top       > > for VLF whistlers , moving sky charges.       > > With this most of what is needed is already       > > in place. As you have already done this I       > > can proceed with confidence. One concern.       > > Standing in the middle of no where spinning       > > a 3 foot wire while looking at a lap top could       > > draw unwanted attention. I am tested for       > > a better WIFI connection officer , that's       > > my story and I am sticking to it.       > >       > > As to your dislike of the Lorentz contraction       > > interpretation of magnetism you are not alone.       > > I can say from my shoes it grows on you       > > over time. How often is there an opportunity       > > to toss a fundamental force in nature such       > > as magnetism straight into the trash bin?       > > One less to worry about.       > >       >       >       > Don't, because there are genuine electric and magnetic fields.       >       > The two quadratic forms E^2-B^2 and E.B are invariant with respect to       > Lorenz transformations. E.B=0 and |E|=|B| is mainly radiation.       >       > So there are static fields with |E|>|B| which are predominantly electric       > fields because you can find a system of reference with B'=0 and E' =       > E/sqrt(1-(v/c)^2).       > This system can be thought to be the rest system where you see charges       > at rest producing their common Coulomb field.       >       > And there are predominantly magnetic fields |B|>|E| where you can find a       > system reference with E'= 0 and B' = B/sqrt(1-(v/c)^2)       >       > Taken cum grano salis of course, because E,B are components of a rank 2       > tensor field, transforming with the product of two Lorentz tranformation       > matrices.       >       > An example of pure magnetic field with E=0 everywhere is the cylindrica=       l=       >       > magnetic field B_phi =1/r around a finite conducting wire with with two       > equal constant currents: one of the negative charges to the left and       > positive charges to the right of identical current and charge density.       >       > So as an experiment, move at constant speed with your synchronized       > clocks/meterstick/ampere-meter/volt-meter equipment along a wire with       > the half velocity v/2 of the electron current.       >       > Now you have two equal currents of negative electrons and positive ions       > with equal but opposite velocities +-v/2.       >       > The wire is assumed somehow to be electrical neutral in the rest system       > of the ions.       >       > Of course, there will be a small longitudinal electrical field E_z to       > drive the electric current against the resistivity and to supply by ExB       > the dissipation energy current density from the field outside into the       > wire.       >       > E_z and dissipation can be made very small or even zero for a       > superconductor. So we can neglect the dissipation problem in a first       > approximation.       >       > Now you have by the Lorenz formula       >       > B'_phi = B_phi /Sqrt(1-(v/2c)^2)       >       > and       >       > E'_r = v/(2c) x B_phi / Sqrt(1-(v/2c)^2)       >       > which means that in this system the current is larger and the wire is       > not neutral but - as an ideal conductor - carries some surface charge       > density D_r that, by definition, is the value of D_r at the surface.       >       > How does this happen?       >       > Lorentz contraction contracts the longitudinal measured distance of the       > ions from rest at v=0 to v/2 using the simultaneity of the moving       > clocks. By the same argument the distance of electrons is enlarged       > because their velocity is reduced from v to v/2. So the net charge       > density varies linearly with v.       >       > This fundamental effect arises at velocities as small as the current       > velocity of electrons of order 10^-11 lightseconds/second for a current       > of 1 A at a density of some 10^27 m^-3.       >       > The gigantic particle density of charges moving freely in metallic       > conductors makes these macroscopic effects easily detectable and       > measureable with high precision. This simple fact enabled Gauß, Farad=       ay       > Maxwell and Einstein to find the laws of electrodynamics.       > These laws, applied by the gauge of the 4-momentum, imply Lorentz       > invariance of all physical laws.       >       > --       >       > Roland Franzius              Two equal + and - charges in opposite directions with equal velocities. Let=        us place two such wires next to each other. The electron of one wire sees =       the other electron in the other wire as non relativistic as they are both m=       oving in the same direction. However the electron see the positive charge i=       n the other wire as length contracted time 2 as they are both moving in opp=       osite directions. This would lead to a strong Coulomb attractive force from=        a Lorentz contraction point of view. Your example of + and - charges has s=       ymmetry but that charge symmetry is lost from the FoR of the individual mov=       ing charge. The Lorentz contraction interpretation of a magnetic force woul=       d not be taught in colleges without first raking it over to make sure it ge=       ts the right answer. To give this balance this does not means it is true. I=       t only means it gets the right answer.              I would add that a current of 1 amp in a copper wire to the right is equiva=       lent to physically moving the same wire to the left at the same speed as th=       e electron current is moving right. Now the electrons are not moving but th=       e protons are in the opposite direction so it all equals out. The reason I =       bring this up is net electron cloud velocity in a 18 gauge wire at 1 amp is=        only a few m meters per second. Surprisingly slow for relativistic effects=       .              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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