- #1

yungman

- 5,755

- 292

[tex]V(\vec r, t) = \frac 1 {4\pi\epsilon_0}\int \frac { \rho(\vec r\;',t-\frac {\eta}{c}) }{\eta} d\;\tau' \;\hbox { where }\;\eta =\sqrt{r^2 + r'^2 - 2 \vec r \cdot \vec r\;' }[/tex]

Where [itex]\;\vec r \;[/itex] point to the field point where V is measured. And [itex]\;\vec r\;' \;[/itex] points to the source point.

For [itex]\;\vec r\;' \;[/itex] << [itex]\;\vec r \;[/itex]:

[tex] \eta \approx \; r- \hat r \cdot \vec r\;' \Rightarrow \rho(\vec r\;',\;t-\frac {\eta}{c}) \approx \rho (\vec r\;',\;t-\frac {r}{c} + \frac {\vec r \cdot \vec r\;'}{c}) [/tex]

Expanding [itex]\rho \;[/itex] as a Taylor series in t about the retarded time at the origin,

[tex]t_0=t-\frac r c [/tex]

We have

[tex]\rho(\vec r\;',\;t-\frac {\eta}{c}) \approx \rho (\vec r\;',\; t_0) + \dot{\rho} (\vec r\;',\; t_0)\left ( \frac {\vec r \cdot \vec r\;'}{c}\right ) + \frac 1 {2!} \ddot{\rho} \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^2 + \frac 1 {3!} \rho^{...}_{ } \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^3 ...[/tex]Then the book go on and claim:

[tex]\left | \frac {\ddot{\rho}}{\dot{\rho}}\right|= \omega[/tex]

I have no idea how this come about. please explain this to me.

Thanks

Alan