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   sci.physics.research      Current physics research. (Moderated)      17,516 messages   

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   Message 15,590 of 17,516   
   Roland Franzius to All   
   Re: How to measure a Lorentz contraction   
   09 Mar 17 10:39:07   
   
   From: roland.franzius@uos.de   
      
   Am 08.03.2017 um 09:09 schrieb John Heath:   
   > 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 cylindrical=   
   >>   
   >> 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 Gau=DF, Faraday   
   >> 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   
   >   
   > Yes there is B=0 E=1 such as a Van De Graaff generator. There   
   > is also B=1 E=0 such as a magnet. I will give an example in   
   > applied physics where the same magnet is B=0 E=1.   
   >   
   > This experiment requires a budget of 20 dollars plus a cleared   
   > kitchen table. 1 old school small black and white TV set from   
   > a thrift shop , 5 to 10 bucks , and a strong flat coin type   
   > magnet from a hobby shop 10 to 15 bucks. We now have our   
   > electron particle accelerator , TV , with a phosphor screen   
   > to collect electron trajectory data. Electron velocity is in   
   > the neighborhood of .1 c for a accelerating element of 10 KV   
   > for this type of TV. Place the north end of the magnet facing   
   > the TV screen. You will note the screen raster will turn   
   > clockwise , Maxwell is right. Place the south pole and the   
   > screen will turn counter clockwise , Maxwell is still right.   
   > So far we have B=1 E=0. However if you look closely at the   
   > magnet from the side you will see a black spot. A spot where   
   > electrons with a velocity of .1 c are not hitting the phosphor?   
   > Turning the magnet around will not help as there will still be   
   > a black spot that electron do not care for. A Coulomb force to   
   > cause this would have to be in the range of 15 KV to stop an   
   > electron at .1 c in its tracks. That coulomb force is not in   
   > Maxwell's equations. However it is in the Lorentz contraction   
   > interpretation where effective movement of electrons only in   
   > a magnet will result in a negative Coulomb force. This makes   
   > it B=0 E=1 for a magnet.   
   >   
   > Interesting though. If it were a anti matter magnet then it   
   > would be the positrons that are effectively moving therefore   
   > a strong positive Coulomb force. The black spot would now be   
   > brighter than the rest of the screen. Now that I think of it   
   > the screen plus a 10 city block radius around the screen would   
   > light up. Maybe moving protons to make the positive magnet   
   > would be better.   
   >   
   >   
      
      
   The black spot is the shielding effect of the magentic vector potential   
   of the magnetic field. It is the cause of a centrifugal potential,   
   simply said.   
      
   The kinetic energy equation in a static magnetic field is   
   H = 1/2 ( p-e/c A(x))^2   
      
   The B-field on the axis   
   B = B_0 f(z,r) e_z   
      
   Transformed to cylinder coordinates and written as an antisymmetric rank   
   2 tensor field of flux density through areas dr /\ dphi of the planes   
   perpendicular the z-axis   
      
   F = B_0  f(z,r) dr /\ dphi   
      
   To have div B=0  with this ansatz we must have dF=0-> d/dz f(z,r)=0.   
      
      
   [continued in next message]   
      
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