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Electric Current in Special Relativity

  1. Apr 3, 2007 #1
    Consider an electric circuit at rest in the lab inertial reference frame. The circuit carries a steady electric current I. What is the current in the circuit as observed from a different inertial reference frame with a velocity of v relative to the lab frame?

    Thanks in advance for your answers
     
  2. jcsd
  3. Apr 3, 2007 #2
    Special relativity considers the transformation of charge density and current density. Have a look at
    W.G.V. Rosser, Classical electromagnetism via relativity Butterworth London 1968 pp165-173
    For the OX(O'X') components of the current density J(X) in I and J'(X) you find there
    J'(X)=g(V)(J(x)-Vro')
    where g(V) stands for the gamma factor and ro' for rhe charge density in I'.
    The other two components have the same magnitude in all inertial referfence frames in relative motion.
    The transformation for the current depends on the orientation of the conductor through which the current flows.:rolleyes:
     
  4. Apr 3, 2007 #3
    If you write your current and charge density in a 4-vector:

    [tex]j^a = [c\rho, j_x,j_y,j_z][/tex]

    then this object transforms as a 4-vector under boosts and rotations (i.e. in exactly the same way as the displacement 4-vector). The form of transformation is given in, for example, this Wikipedia article section.
     
  5. May 7, 2007 #4
    electrical resistence

    what about the electrical resistance? how does it transform. has it the same magnitude in all inertial reference frames
     
  6. May 7, 2007 #5

    pervect

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    Yep. If we apply this to the particular problem, we see that we need to know the charge density in the lab frame.

    If we assume that the current carrying wire has no charge in the lab frame, and that the direction of the wire and current is the same as the direction of the velocity of the moving observer, we see that in the moving frame the wire acquires a charge, and that both the current and current density are increased by a factor of gamma = 1/sqrt(1-v^2/c^2).

    Note that if we have a complete circuit, some parts of the wire will acquire a positive charge and others will acquire a negative charge, but the total charge on the wire will remain constant at 0.

    Note that this means that moving charge parallel to a wire experiences an electrostatic force in its own rest frame. In the laboratory rest frame, the wire is neutral, so the charge experiences no electric force, but rather a magnetic force.

    This is why it is often said that relativity unifies electrostatics and magnetism.
     
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