Another question on radiation of time varying arbitrary source.

[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\;&#039; \; << \;\vec r \;:

\eta \approx \; r- \hat r \cdot \vec r\;&#039; \Rightarrow \rho(\vec r\;&#039;,\;t-\frac {\eta}{c}) \approx \rho (\vec r\;&#039;,\;t-\frac {r}{c} + \frac {\vec r \cdot \vec r\;&#039;}{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\;&#039;,\;t-\frac {\eta}{c}) \approx \rho (\vec r\;&#039;,\; t_0) + \dot{\rho} (\vec r\;&#039;,\; t_0)\left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right ) + \frac 1 {2!} \ddot{\rho} \left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right )^2 + \frac 1 {3!} \rho^{...}_{ } \left ( \frac {\vec r \cdot \vec r\;&#039;}{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

 

[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.

 

[yungman]

Thanks for your time.

Let \rho_{(\vec r\;&#039;, t-k)} = \rho_0\;cos [\omega(t-k)]\;\Rightarrow \;\dot{\rho}_{(\vec r\;&#039;, t-k)} = -\omega \rho_0 sin\; [\omega(t-k)] \;\hbox { and }\; \ddot{\rho}_{(\vec r\;&#039;, t-k)} = -\omega^2 \rho_0 cos\; [\omega(t-k)]

\left|\frac{ \ddot{\rho}_{(\vec r\;&#039;, t-k)} }{\dot{\rho}_{(\vec r\;&#039;, 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\;&#039;, t-k)} }{\dot{\rho}_{(\vec r\;&#039;, t-k)}}\right|

Not equal to \omega

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

Thanks

Alan

 

[yungman]

Anyone please?