harrylin said:
To make it clearer, we can put a positively charged dog in rest with the wire, next to it. Now the cat, using the frame that is co-moving with the electrons as rest frame, has to explain the lack of net force on the dog despite the electric field. The cat can only explain this by the magnetic field of the moving ions and which must exactly compensate the electric field force.
This is a good question. The force acting on the dog should be neutral in both the rest frame of the dog and in the rest frame of the cat when the dog is moving relative to the cat. The analysis of the forces on the dog is a little more complicated than that of the cat so I will take Dalespam's advice and use the transformation of the four-current which can be expressed as:
##[\rho, \mathbf{I_x,I_y,I_z}]## using units where c=1.
I only want to analyse the case where the current and relative motion are always parallel to the wire which in turn remains parallel to the x axis. This simplifies things as ##\mathbf{I_y = I_z} = 0## so the four current can be abbreviated to ##[\rho, \mathbf{I}]## where I have defined ##\mathbf{I_x}## as ##\mathbf{I}##.
After performing a Lorentz boost with velocity v in the x direction:
##\rho' = \gamma_v(\rho - \mathbf{v I})##
##\mathbf{I'}= \gamma_v(-\mathbf{v}\rho +\mathbf{I})##
where ##\gamma_v## is the gamma factor for velocity v.
The Lorentz force acting on a test particle with charge q and velocity ##v_0## is defined as:
##\mathbf{F} = q(\mathbf{E} + \mathbf{v_0 \times B})##
In our simplified case this can be expressed in a non vector form as:
##F = q(E - v_0*B)##
(Note the change of sign when we use ordinary multiplication rather than the cross product.)
The electric and magnetic fields are defined in terms of various constants but we can conveniently use units such that ##E = \rho## and ##B = I## so that
##F = q(p - v_0*I)##.
In the rest frame of the dog, p=0 and ##v_0 = 0## so F =0.
In the general case, when carrying out the Lorentz transformation of the force, we should use the the transformed four velocity of ##v_0## or the relativistic velocity subtraction formula ##v_0' = (v_0 -v)/(1-v_0*v)## For the case of our positively charged dog which is at rest in S, the velocity of the dog in the rest frame of the cat (S') is ##v_0' = -v##.
The force acting on the dog in the cat's reference frame is:
##F' = q\rho' - qv_0'*I'##
##F' = q\rho' + qv*I'##
##F' = q(\gamma_v \rho - \gamma_v vI ) + qv(-\gamma_v v\rho +\gamma_v I)##
##F' = q\gamma_v(\rho - vI ) + q\gamma_v(- v^2\rho + vI)##
In the rest frame of the dog, the boundary conditions specified neutral charge density ##\rho =0## so the above reduces to:
##F' = q\gamma_v( - vI ) + q\gamma_v(v I)##
It can be seen that the electrostatic force on the dog is equal in magnitude and opposite in sign to the magnetic force on the dog as measured in the cat's rest frame, so there is still a total force of zero acting on the dog.