What is the Four-Vector Potential of a Moving Charge?

[unscientific]

Homework Statement

(a) Find the four-vector potential of a moving charge
(b) Find source time and z-component of electric field
(c) Find electric potential to first order of x and hence electric field[/B]

Homework Equations

The Attempt at a Solution

Part(a)
[/B]
\phi = \frac{q}{4\pi \epsilon_0 r}
\vec A = 0

Consider world line of aritrarily moving charge, where the vector potential is only dependent on what charge is doing at source event. The relevant distance is source-field distance $r_{sf}$. We represent this in 4-vector displacement $R = (ct, \vec r)$ where $t = \frac{r_{sf}}{c}$. Consider $R \cdot U = \gamma(-rc + \vec r \cdot \vec v)$. It gives the right form in rest frame. Thus:

A = \frac{q}{4\pi \epsilon_0} \frac{ \frac{U}{c} }{-R \cdot U}Part (b)

t_s = \frac{r_{sf}}{c} = \frac{\sqrt{z^2+a^2}}{c}
R \cdot U = \gamma (-ct + \vec r \cdot \vec v) = -\gamma r c = -\gamma c \sqrt{z^2 + a^2}

Thus the four-potential is
A = \frac{q}{4\pi \epsilon_0 c^2 \sqrt{z^2 + a^2}} (c, \vec v_s) = (\frac{\phi}{c}, \vec A)
To find electric field,
E_z = -\frac{\partial \phi}{\partial z} - \frac{\partial A_z}{\partial t}
z-component of $\vec A$ is zero, so
E_z = -\frac{\partial \phi}{\partial z} = -\frac{qz}{4\pi \epsilon_0 \left( z^2 + a^2 \right)^{\frac{3}{2}}} \hat kPart (c)

r = \sqrt{ z^2 + \left[ a cos(\omega t) - \Delta x \right]^2 + a^2 sin^2(\omega t) }
r = \approx \sqrt{ z^2 + a^2 - 2a \Delta x cos(\omega t) } \approx \sqrt{z^2 + a^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{z^2 + a^2} \right]

For $R \cdot U$, we have

R \cdot U = \gamma(-rc + \vec r \cdot \vec v) = \gamma \left[ -cr + a\omega \Delta x sin (\omega t) \right]

\phi = -\frac{qc}{4\pi \epsilon_0} \frac{1}{ c\sqrt{a^2 + z^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{a^2 + z^2} \right] + a\omega \Delta x sin(\omega t) }

\phi \approx \frac{-qc}{4\pi \epsilon_0} \frac{1}{c\sqrt{a^2+z^2} + \Delta x \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2}} \right] }

\phi \approx \frac{-qc}{4\pi \epsilon_0} \left[ 1 - \frac{a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2+z^2}} }{c\sqrt{a^2+z^2}} \Delta x \right]

E_x = -\frac{\partial \phi}{\partial x} = \frac{qc}{4\pi \epsilon_0 c^2 (a^2 + z^2) \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2 }} \right] }

Can't seem to match the final expression - what worries me is the lack of $\omega^2$ in the answer.

 

[Orodruin]

What happened to the x-component of the vector potential?