# Fields of a moving point charge with constant velocity.

1. May 30, 2012

### center o bass

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 $$A^{\mu}$$ 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

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

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

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

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

$$\vec E =\frac{q}{4\pi \varepsilon_0} \frac{1 - v^2/c^2}{(1 - v^2 \sin^2 \theta/c^2)^{3/2}} \frac{\hat R}{R^2}, \ \ \ \vec R = \vec r - \vec v t$$
where $$\theta$$ is the angle between $$\vec R$$ and $$\vec v$$.
I do not see how these two results can agree, especially not with the angle.

Last edited: May 30, 2012
2. May 30, 2012

### Bill_K

Take a look at a book like Jackson, which derives the field of a moving point charge both ways and shows that the two results agree.

The angular dependence comes from the fact that E and E transform differently (different factor of γ). Deriving it from φ the way you did, notice that x and z also transform differently. Whichever coordinate is parallel to the motion, it picks up a γ, while the one perpendicular to the motion does not.

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