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Radiation assumptions

  1. Jun 1, 2012 #1
    Hey I'm taking a course which involves electrodynamics and i am currently reading about radiation. By expanding the relative distance between the current distribution r' and the point we're looking at r in a powerseries and assume that r' << r for all r' and r we obtain the vector potential

    [tex] \vec A = \frac{\mu_0}{4\pi r}( \dot {\vec p} + \frac{1}{c}\dot{\vec m}\times \vec n + \frac{1}{2c} \ddot{\vec D}_n) + \ldots[/tex]
    where p is the dipolemoment m is the magnetic dipole moment and D_n is the quadrupole vector and the derivatives are to be taken at the retarded time tr = t - r/c. The corresponding magnetic field is given by

    (*)[tex]\vec B = \nabla \times \vec A = \frac{\mu_0}{4\pi r^2}( \dot {\vec p} + \frac{1}{c}\dot{\vec m}\times \vec n + \frac{1}{2c} \ddot{\vec D}_n+\ldots)\times \vec n\\
    + \frac{\mu_0}{4\pi cr} \frac{d}{dt} ( \dot {\vec p} + \frac{1}{c}\dot{\vec m}\times \vec n + \frac{1}{2c} \ddot{\vec D}_n+ \ldots)\times \vec n[/tex]

    and now the text that I am reading picks out the terms for the radiation

    [tex]\vec B_{rad}=\frac{\mu_0}{4\pi cr} ( \ddot {\vec p} + \frac{1}{c}\ddot{\vec m}\times \vec n + \frac{1}{2c} \dddot{\vec D}_n+ \ldots)[/tex]
    and states

    "The radiation fields given above are the parts of the total fields that fall off with distance as 1/r. They dominate in the radiation zone, far from the charges and currents. To be more precise there are two conditions that should be satisfied to be in this zone. The first one is r >> a with a as a typical linear size of the charge and current distribution. Our derivation so far has been based on this to be satisfied. If that is not the case there will be fields with a faster fall off with distance (which we have omitted in the expansions) that would compete with the radiation fields in strength. The other is r >>λ , with λ as a typical wave length of the radiation. If that is not satisfied there are contributions to the fields where a smaller number of time derivatives of the multipole momenta could compensate for a higher power in 1/r. In particular this condition is necessary for the second term in (*) to dominate over the first term."

    Now it's the boldface assumption here that I do not quite understand. How does a large λ cause a smaller number of time derivatives of the multipole moments to grow larger than higher number of derivatives?

    I do not see how this is relevant anyhow since radiation in some sence is defined as the energy which does not come back or stays close to the source, i.e it travels to infinity. So if the time derivatives blows up, we could always make r larger and make them negigble. Maybe there are practical and theoretical differences in the definition or radiation?
     
  2. jcsd
  3. Jun 1, 2012 #2

    Jano L.

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    The field and the energy flux can be evaluated anywhere, not just far from the source, but it complicated. The fields and the Poynting flux simplify in great distances, where the terms containing [itex]\ddot p [/itex] are much greater than the terms containing [itex]\dot p[/itex]. This condition implies both inequalitities;

    we can estimate [itex]|\ddot p | \approx \omega^2 p [/itex], where [itex]\omega[/itex] is the frequency of oscillation in the source. Now we insert this estimate into the above expansion of the field. Requiring

    [tex]
    \ddot p /(cr) >> \dot p /r^2

    [/tex]

    implies

    [tex]
    \omega^2/(cr) >> \omega / r^2
    [/tex]

    which implies

    [tex]
    r >> \frac{c}{\omega} = \lambda.
    [/tex]
     
  4. Jun 1, 2012 #3
    Ah! Thank you for that line of reasoning! :)
    I guess cases where this latter assumption is not valid is in the case of standing rather close to sources of radiowaves?
     
  5. Jun 1, 2012 #4
    By the way, in arriving at the two terms in (*) one takes the curl of A considering A as a vector like

    [tex]\vec A = \frac{\mu_0}{4\pi r} \vec B(t_r)[/tex]
    where tr is the retarded time t-r/c. By using the product rule one then gets
    [tex]\vec B = \nabla \times \vec A = \frac{\mu_0}{4\pi r^2} \vec B \times \vec n
    + \frac{\mu_0}{4\pi r} \dot{\vec B}(t_r) \times \vec n[/tex]
    but this has neglected that B is also a function of r trough the
    [tex]\dot{\vec m}\times \vec n[/tex]
    should'nt this also be taken into consideration?
     
  6. Jun 1, 2012 #5

    Jano L.

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    I do not understand your question. Which book are you studying?
     
  7. Jun 1, 2012 #6
    It's some lecture notes located here:
    http://www.uio.no/studier/emner/matnat/fys/FYS3120/v12/LectureNotes3120.pdf [Broken]
    At page 195. Maybe if I reformulate myself; the two terms in (*) are not the only two terms we get by taking the cross product of [tex]\vec A[/tex].
    Or are they? I am especially worrying about the [tex]\dot{\vec m}\time \vec n[/tex] term, since it has a spatial unit vector which i suppose will be affected by the curl operator.
    What I meant by the B vector above was that if one could consider all the vector
    [tex]( \dot {\vec p} + \frac{1}{c}\dot{\vec m}\times \vec n + \frac{1}{2c} \ddot{\vec D}_n) = \vec B(t_r)[/tex]
    as just a function of the retarded time , then the product rule
    [tex]\nabla \times \frac{1}{r}\vec B = \frac{1}{r}\dot{\vec B}(t_r) \times \del t_r + \del(\frac{1}{r}) \times \vec B[/tex]
    would be valid. But does not the
    [tex]\dot{\vec m}\time \vec n[/tex]
    ruin this consideration implying extra terms?
     
    Last edited by a moderator: May 6, 2017
  8. Jun 1, 2012 #7

    It's some lecture notes located here:
    http://www.uio.no/studier/emner/matnat/fys/FYS3120/v12/LectureNotes3120.pdf [Broken]
    At page 195. Maybe if I reformulate myself; the two terms in (*) are not the only two terms we get by taking the cross product of [tex]\vec A[/tex].
    Or are they? I am especially worrying about the [tex]\dot{\vec m}\times \vec n[/tex] term, since it has a spatial unit vector which i suppose will be affected by the curl operator.
    What I meant by the B vector above was that if one could consider all the vector
    [tex]( \dot {\vec p} + \frac{1}{c}\dot{\vec m}\times \vec n + \frac{1}{2c} \ddot{\vec D}_n) = \vec B(t_r)[/tex]
    as just a function of the retarded time , then the product rule
    [tex]\nabla \times \frac{1}{r}\vec B = \frac{1}{r}\dot{\vec B}(t_r) \times \nabla t_r + \nabla(\frac{1}{r}) \times \vec B[/tex]
    would be valid. But does not the
    [tex]\dot{\vec m}\times \vec n[/tex]
    ruin this consideration and implying extra terms.
     
    Last edited by a moderator: May 6, 2017
  9. Jun 1, 2012 #8

    Jano L.

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    The curl of n is zero because it is a gradient field, so the term drops out.
     
  10. Jun 1, 2012 #9
    On the other hand

    [tex]\nabla \times (\dot{\vec m} \times \vec n) = \dot{\vec m} (\nabla \cdot \vec n) - (\dot{\vec m} \cdot \nabla) \vec n[/tex]

    and this does not seem to obviously be zero?
     
  11. Jun 1, 2012 #10

    Jano L.

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    Ah, you are right, the term is not zero. Perhaps he forgot it or dropped it because it is of order 1/r^2 (since grad n ~ 1/r) and hence small in large distances.
     
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