# Four-potential of 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]

## 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
$$E_z = -\frac{\partial \phi}{\partial z} = -\frac{qz}{4\pi \epsilon_0 \left( z^2 + a^2 \right)^{\frac{3}{2}}} \hat k$$

Part (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.

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