# Another question on radiation of time varying arbitrary source.

1. Jun 28, 2011

### yungman

For retarded scalar potential of arbigtrary source around origin:

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

Where $\;\vec r \;$ point to the field point where V is measured. And $\;\vec r\;' \;$ points to the source point.

For $\;\vec r\;' \;$ << $\;\vec r \;$:

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

Expanding $\rho \;$ as a Taylor series in t about the retarded time at the origin,

$$t_0=t-\frac r c$$

We have

$$\rho(\vec r\;',\;t-\frac {\eta}{c}) \approx \rho (\vec r\;',\; t_0) + \dot{\rho} (\vec r\;',\; t_0)\left ( \frac {\vec r \cdot \vec r\;'}{c}\right ) + \frac 1 {2!} \ddot{\rho} \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^2 + \frac 1 {3!} \rho^{...}_{ } \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^3 ........$$

Then the book go on and claim:

$$\left | \frac {\ddot{\rho}}{\dot{\rho}}\right|= \omega$$

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

Thanks

Alan

2. Jun 28, 2011

### Blibbler

It makes sense - think about any simple harmonic oscillator and the relationship between instantaneous velocity and acceleration. Draw it - and then calculate it for yourself using ASinwt as the wave function. The sign of w would change, which is why the ratio is expressed as a modulus in your example, but the absolute value is constant.

3. Jun 28, 2011

### yungman

Let $$\rho_{(\vec r\;', t-k)} = \rho_0\;cos [\omega(t-k)]\;\Rightarrow \;\dot{\rho}_{(\vec r\;', t-k)} = -\omega \rho_0 sin\; [\omega(t-k)] \;\hbox { and }\; \ddot{\rho}_{(\vec r\;', t-k)} = -\omega^2 \rho_0 cos\; [\omega(t-k)]$$

$$\left|\frac{ \ddot{\rho}_{(\vec r\;', t-k)} }{\dot{\rho}_{(\vec r\;', t-k)}}\right| \;=\; \left| \frac {-\omega^2 \rho_0 cos\; [\omega(t-k)]}{ -\omega \rho_0 sin\; [\omega(t-k)] }\right| \;=\; |\omega \;tan\;[\omega(t-k)] |$$

You can still see that

$$\left|\frac{ \ddot{\rho}_{(\vec r\;', t-k)} }{\dot{\rho}_{(\vec r\;', t-k)}}\right|$$

Not equal to $\omega$

I still don't get it. Please explain this.

Thanks

Alan

4. Jun 29, 2011