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Total power radiated from two dipoles (electrodynamics)

  1. May 18, 2015 #1

    CAF123

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    Gold Member

    1. The problem statement, all variables and given/known data
    The retarded vector potential ##\mathbf{A}(\mathbf{r}, t)## in Lorenz gauge due to a current density ##\mathbf{J}(\mathbf{r}, t)## contained entirely within a bounded region of size d is $$\mathbf{A}(\mathbf{r},t) = \frac{1}{4\pi c}\int_V' \frac{\mathbf{J}(\mathbf r', t')}{|r-r'|} dV',$$ where ##t' = t-|r-r'|/c## If this current density is due to a charge density oscillating harmonically with frequency ##\omega \ll c/d,## then at distances ## r \gg d,## ##\mathbf{A}(\mathbf{r}, t) ## the current density is proportional to the real part of ##i\omega \mathbf{p}e^{i\omega(t−r/c)}## , where ##\mathbf p \cos \omega t## is the dipole moment of the charge distribution.

    Now consider two oscillating electric dipoles of equal moment ##\mathbf p## that are placed next to each other on the z axis, a small distance d apart. The two dipoles have the same angular frequency ##\omega##, but are out of phase with each other by ##\pi##. Show that the total power radiated by this system is smaller than that of the single dipole by a factor ##\omega^2 d^2/5c^2##

    2. Relevant equations
    E and B field formulae and the Poynting vector. Total power radiated given by ##\langle P \rangle = \int \langle S \rangle \cdot d \mathbf A##

    3. The attempt at a solution

    I have solved the time averaged power for the single dipole by computing the far field E and B from the given A and then computing the poynting vector before the total time averaged power. Since this second dipole is out of phase by pi relative to the other, then can write ##\mathbf p_1(t) = \mathbf p \cos \omega t## for the first one and ##\mathbf p_2(t) = -\mathbf p \cos \omega t ## for the second. I am just a bit unsure of what to do next. I guess I am to superimpose the E and B fields from the two dipoles and then compute P but how does the superposition work? Thanks!
     
  2. jcsd
  3. May 23, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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