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

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

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

 

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

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

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

 

[yungman]

Anyone please?

 

[sardar]

Dear Alan,

Thank you for sharing your question about the radiation of time-varying arbitrary sources. I will do my best to explain the concept mentioned in the book you mentioned.

To begin, the equation provided is known as the retarded scalar potential, which is used in electromagnetism to describe the electric potential generated by a time-varying source. This equation takes into account the time delay in the propagation of the electromagnetic field from the source to the field point.

In the equation, \eta is the distance between the field point and the source point, and it depends on the distance between them, as well as the speed of light. When \vec{r}\;' is much smaller than \vec{r}, the term \vec{r}\;'\cdot\vec{r} can be neglected, resulting in \eta \approx r. This means that the distance between the field point and the source point is approximately equal to the distance between the field point and the origin, which is r.

Now, expanding \rho as a Taylor series in t about the retarded time at the origin, t_0 = t - \frac{r}{c}, we can see that the first term in the expansion is simply \rho(\vec{r}\;',t_0), which is the value of the source at the retarded time. The second term is the derivative of \rho with respect to t at the retarded time, multiplied by the time delay \frac{\vec{r}\;'\cdot\vec{r}}{c}. This is because the source value at the field point is affected by the source value at the retarded time, which is t_0, and the time it takes for the field to propagate from the source to the field point, which is \frac{\vec{r}\;'\cdot\vec{r}}{c}. The third term is the second derivative of \rho, and so on.

Now, the book claims that \left|\frac{\ddot{\rho}}{\dot{\rho}}\right| = \omega. This follows from the fact that the time derivative of a sinusoidal function is also a sinusoidal function with the same frequency. In other words, \dot{\rho} = \omega\rho. This means that the ratio of the second derivative to the first derivative is a constant, which is the frequency, \omega. This is why the book claims that \left|\frac{\ddot