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Potentials of a charge moving with uniform velocity

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A point charge q, moving with uniform velocity [itex]\vec{v}[/itex] = v [itex]\hat{z}[/itex], in the laboratory frame called [itex]\cal{K}[/itex]. The charge is in the origin of the system at time [itex]t=0[/itex].


    2. Relevant equations
    Find scalar potential [itex]\Phi[/itex] and vector potential [itex]\vec{A}[/itex] both in the frame in which the charge is at rest [itex]\cal{K'}[/itex] and the laboratory frame [itex]\cal{K}[/itex]. Verify the potentials to satisfy Lorentz gauge.


    3. The attempt at a solution
    I thought of two methods of proceeding:
    1. Transforming 4-vector-Potential [itex]A'^{\mu}[/itex]: I calculate the easy 4-potential in [itex]\cal{K'}[/itex] and I transformed it back into [itex]\cal{K}[/itex], transforming also the spacetime variables. The 4-vector-Potential [itex]A^{\mu}[/itex] that I found, satisfies Lorentz gauge [itex]\partial^\mu A_\mu = 0[/itex].
    2. Using Lienard Wiechert Potentials: I calculate the 4-potential in [itex]\cal{K'}[/itex] using the same method discussed above. In [itex]\cal{K}[/itex] I calculated the retarded Lienard Wiechert potentials, calculating the retarded time too that is needed to be substituted into the formula. This 4-potential satisfies Lorentz gauge too.

    My question is: these two methods are pretty different as the first does not take into account any "retard" whereas the second is based on the assumption of "retard". Which is correct and why?

    Thank you very much.
     
    Last edited: Feb 19, 2013
  2. jcsd
  3. Feb 19, 2013 #2
    I just managed with more algebra to find that the two results were the same. now my question is: is that always true? is it necessary a uniform motion or whatelse?

    Thanks.
     
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