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Homework Help: Poynting vector

  1. Dec 6, 2008 #1
    I am asked to calculate the pointing vector for the following fields:

    [tex]\vec{B}=k^2 \frac{e^{ikr}}{r} \left( 1+\frac{i}{kr} \right) \hat{r} \times \vec{p_{\omega}}[/tex]

    [tex]\vec{E}=\frac{i}{k} (\vec{\nabla} e^{ikr}) \times \left( \frac{k^2}{r} \left(1+\frac{i}{kr} \left) \hat{r} \times \vec{p_{\omega}} \left) + \frac{i}{k} e^{ikr} \vec{\nabla} \times \left( \frac{k^2}{r} \left( i +\frac{i}{kr} \right) \hat{r} \times \vec{p_{\omega}} \right) [/tex]

    We know that:

    [tex]\vec{S} = \frac{c}{4 \pi} Re(\vec{E}) \times Re(\vec{B}) [/tex]

    We know that:

    I can figure out [tex]Re(\vec{B})[/tex] assuming that P_omega points in the z direction:

    [tex]Re(\vec{B})=k^2 p_{\omega} \frac{e^{ikr}}{r} sin \theta \hat{\phi} [/tex]

    since the imaginary term in B vanishes when taking the real part.

    I am not sure how to calculate the real part of E, any thoughts would be appreciated.
     
  2. jcsd
  3. Dec 6, 2008 #2

    turin

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    Homework Helper

    Why do you want to assume a particular direction for P_omega? If I remember correctly, if you play your cross-products right, you should get an expected result ...

    You're forgetting the imaginary part of e^ikr. And this imaginary part will multiply the imaginary part of the other factor in B and result in another real contribution.
     
  4. Dec 6, 2008 #3
    we assume a particular direction for P_omega so that r x p_omega will give the sin(theta) term

    I am having trouble with finding the real part of E because I'm not sure how to find the real parts when imaginary terms are being crossed with real terms, any ideas?
     
  5. Dec 7, 2008 #4

    turin

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    Re x Re = Re.
    Im x Im = (-)Re.
    Re x Im = Im.
    Im x Re = Im.

    You may also use i = e^ipi/2, and add phases to keep the expressions in polar form. In principle, both of these should be possible; however, choosing which way is more convenient comes with experience. Try both, and you will start to develop an intuition for it.

    EDIT: Oh, wait, your expression for S is different than what I'm used to. I use Re(ExB*), or actually Re(ExH*). Sorry for the confusion. Anyway, you can't have Re(something) = something x e^ikr.
     
    Last edited: Dec 7, 2008
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