The fields of a moving point charge can be constructed from the Lienard-Wiechert potentials. However could one not just consider a point charge at rest in a frame S' and transform the relevant quantities such as [tex]A^{\mu}[/tex] back to S and then deduce the fields from here. Should not the the two results agree? I considered the problem of a point charge moving along the x-axis with constant velocity v and deduced that(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \vec E = \frac{q}{4\pi \varepsilon_0} \frac{\gamma}{r'^3} \left( \vec r - \vec v t\right)[/tex]

where i just transformed the potential from the charge in S'

[tex]\phi' = \frac{q}{4\pi \varepsilon_0} \frac{1}{r'}, \ \ \ r' = \sqrt{x'^2 + y'^2 + z'^2}.[/tex]

However the result from Griffith's s. 439 eqn (10.68) is that

[tex] \vec E =\frac{q}{4\pi \varepsilon_0} \frac{1 - v^2/c^2}{(1 - v^2 \sin^2 \theta/c^2)^{3/2}} \frac{\hat R}{R^2}, \ \ \ \vec R = \vec r - \vec v t[/tex]

where [tex]\theta[/tex] is the angle between [tex]\vec R[/tex] and [tex]\vec v[/tex].

I do not see how these two results can agree, especially not with the angle.

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# Fields of a moving point charge with constant velocity.

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