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Another question on radiation of time varying arbitrary source.

  1. Jun 28, 2011 #1
    For retarded scalar potential of arbigtrary source around origin:

    [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
     
  2. jcsd
  3. Jun 28, 2011 #2
    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.
     
  4. Jun 28, 2011 #3
    Thanks for your time.

    Let [tex] \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)][/tex]

    [tex] \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)] |[/tex]

    You can still see that


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

    Not equal to [itex]\omega[/itex]

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

    Thanks

    Alan
     
  5. Jun 29, 2011 #4
    Anyone please?
     
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