From: heathjohn2@gmail.com   
      
   On Thursday, December 29, 2016 at 11:21:45 AM UTC-5, Jos Bergervoet wrote:   
   > On 12/29/2016 8:59 AM, John Heath wrote:   
   > > On Tuesday, December 27, 2016 at 10:47:19 PM UTC-5, Jos Bergervoet wrote:   
   > >> On 12/27/2016 11:26 PM, Gregor Scholten wrote:   
   > >>> Hi all,   
   > >>>   
   > >>> I recently observed a problem concerning centripetal force in Special   
   > >>> Relativity.   
   > >>>   
   > >>> Imagine a system of two charges Q > 0 and q < 0, where the first charge   
   > >>> Q has a high mass so that we can assume the second charge q orbitting   
   > >>> it. In some inertial frame S, Q is resting at the spatial origin (x,y,z)   
   > >>> = (0,0,0) and q is performing an orbitting motion in the (y,z)-plane,   
   > >>> yielding the trajectory   
   > >>>   
   > >>> y(t) = R cos(omega t), z(t) = R sin(omega t)   
   > >>>   
   > >>> where R is the orbit radius and omega the angular frequency,   
   > >>> correspondig to orbit period T = 2 pi / omega. Assuming that the   
   > >>> orbitting velocity, given by R omega, is << c, we can calculate the   
   > >>> components of momentum p and centripetal force F as   
   > >>>   
   > >>> py(t) = - m R omega sin(omega t)   
   > >>> pz(t) =   m R omega cos(omega t)   
   > >>>   
   > >>> Fy(t) = - m R omega^2 cos(omega t)   
   > >>> Fz(t) = - m R omega^2 sin(omega t)   
   > >>>   
   > >>> where m is the mass of the charge q. Since the orbitting motion is bound   
   > >>> the the electromagnetic field, the centripetal force can be calculated   
   > >>> from the electric field, too:   
   > >>>   
   > >>> Fy(t) = q Ey = -|q| Ey   
   > >>> Fz(t) = q Ez = -|q| Ez   
   > >>>   
   > >>> So far, everything is fine. Now imagine a seond inertial frame S',   
   > >>> moving with velocity u in x-direction with respect to S. The trajectory   
   > >>> of the charge q in S' becomes   
   > >>>   
   > >>> x'(t') = u t'   
   > >>> y'(t') = R cos(omega' t')   
   > >>> z'(t') = R sin(omega' t')   
   > >>>   
   > >>> where the angular frequency is decreased to   
   > >>>   
   > >>> omega' = sqrt{1 - u^2/c^2} omega = omega/gamma   
   > >>>   
   > >>> with Lorentz factor gamma = 1/sqrt{1 - u^2/c^2} since the orbitting   
   > >>> motion is slowed down in S' by time dilation. The components of the   
   > >>> momentum in S' then become:   
   > >>>   
   > >>> px' = gamma m u   
   > >>> py' = - gamma m R omega' sin(omega' t')   
   > >>> pz' = gamma m R omega' cos(omega' t')   
   > >>>   
   > >>> yielding the following components of the centripetal force in S':   
   > >>>   
   > >>> Fy' = d/dt' py' = - gamma m R (omega')^2 cos(omega' t')   
   > >>>       = - m R (omega^2 / gamma) cos(omega' t')   
   > >>>       = Fy / gamma                                             (1a)   
   > >>> Fz' = d/dt' pz' = - gamma m R (omega')^2 sin(omega' t')   
   > >>>       = - m R (omega^2 / gamma) sin(omega' t')   
   > >>>       = Fz / gamma                                             (1b)   
   > >>>   
   > >>> where we used that omega' = omega/gamma. So, compared to S, the absolute   
   > >>> value of the centripetal force is DECREASED, by factor 1/gamma.   
   > >>>   
   > >>> However, the result for the centripetal force is different if we   
   > >>> calculate it by Lorentz transformation of the electromagnetic field:   
   > >>>   
   > >>> Ex' = Ex = 0, Bx' = Bx = 0   
   > >>> Ey' = gamma(Ey - u Bz) = gamma E_y   
   > >>> By' = gamma(By + u/c^2 Ez = gamma u/c^2 Ez   
   > >>> Ez' = gamma(Ez + u By) = gamma Ez   
   > >>> Bz' = gamma(Bz - u/c^2) Ey = - gamma u/c^2 Ey   
   > >>>   
   > >>> Applying the formula for the Lorentz force:   
   > >>>   
   > >>> \vec F' = q (\vec E' + \vec v' \times \vec B')   
   > >>>   
   > >>> with vx' = u, we get   
   > >>>   
   > >>> Fy' = q (Ey' - u Bz')   
   > >>>       = q gamma Ey (1 + u^2/c^2)   
   > >>>       = gamma (1 + u^2/c^2) Fy              (2a)   
   > >>   
   > >> This plus sign   ^^^  seems to be wrong. For parallel moving   
   > >> charges the magnetic field always acts opposite to the electric   
   > >> field and your charges move in parallel if u becomes big.   
   > >>   
   > >> Take the simpler, well-known case where q is not orbiting Q but   
   > >> is held in fixed position w.r.t Q. You get the electric force   
   > >> of opposite charges (attractive) but the magnetic force of   
   > >> opposite-sign parallel currents and that's *repulsive*.   
   > >>   
   > >> Your derivation of the force would still hold for that case. So   
   > >> with the right B and E you would get two opposite terms for the   
   > >> force. But you don't, so it seems that your By and Bz already   
   > >> have the wrong sign in the earlier equations above.   
   > >>   
   > >> On the z-axis above Q at t=0 you have Ez > 0 and you should   
   > >> have By < 0 for Q moving in the x-direction. That's the opposite   
   > >> of what your By expression gives.   
   > >>   
   > >> --   
   > >> Jos   
   > >   
   > > Two wires from floor to ceiling 1 inch apart with current moving up in   
   > > both wires. The force is attractive.   
   >   
   > Exactly, that's magnetic force. And the electric force   
   > between the parallel moving charges (the electrons) is   
   > repulsive, so that's the opposite.   
   >   
   > > There are as many electrons as   
   > > there are protons in the copper wire. As current starts to flow up in   
   > > both wires the electrons see the other electron as non relativistic as   
   > > they are both moving up . However the electrons see the protons as   
   > > length contracted therefore there is and attractive Coulomb force   
   > > between the wires.   
   >   
   > You now have stationary particles in addition to the   
   > parallel moving charges. The electric force between the   
   > moving charges is still repulsive and my only claim   
   > was that that one is opposite to the magnetic force!   
   >   
   > --   
   > Jos   
      
   I wish this exchange of ideas were closer to 900 MHz vs 2.4 GHz as I   
   could use some help in this area. I suspect you feel my pain. However   
   this is your day where I can give a little. Run two wires in parallel of   
   18 gauge or better and short them with a car battery as a surge source   
   of a good 100 amps or so. This will give you confidence that indeed   
   current in the same dir ection of both wires will lead to an attractive   
   force between the wires not repulsive.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)