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Evalute expression of associated Legendre's polynomial

  1. Nov 12, 2009 #1
    Dear All,

    I am implementing the scattering of dielectric sphere under electromagnetic plane wave. The expression of the field contain [tex]\frac{P_n^1(\cos\theta)}{\sin\theta}[/tex] and[tex]\sin\theta P_n^1'(\cos\theta)[/tex], where the derivative is with respect to the argument.

    These two terms are giving me difficulty when [tex]\theta=0[/tex] or [tex]\theta=\pi[/tex].

    When [tex]\theta=\pi[/tex], in one book (Harrington's Time-Harmonic Electromagnetic Fields), both terms are stated to be
    [tex]\frac{(-1)^n n(n+1)}{2}[/tex] on Page 295. In another book (Balanis' Advanced Electromagnetic Engineering), both terms are equal [tex]-\frac{(-1)^n n(n+1)}{2}[/tex].

    I don't know which one is correct. Could someone tell me how could I evaluate these two expressions at [tex]\theta=0, \pi[/tex]. Thank you.


    kzhu
     
  2. jcsd
  3. Nov 13, 2009 #2
    they are both right, depending only which phase convention do you use for the Legendre Polynomial terms.
    Legendre Polynoms are defined as an orthonormal basis, a phase (as a sign) doesn't affects their physical description, try to understand the physical meaning of what are you doing in order to solve this phase problem...
     
  4. Nov 22, 2009 #3
    Thx for the discussion.

    I was able to derive these two expressions. The method is to use the recursive relation of associated legendre's function
    [tex](m-n-1)P_{n+1}^m(x) + (2n+1)xP_n^m(x) - (m+n)P_{n-1}^m = 0.[/tex]
    and get
    [tex]\frac{P_{n+1}^1}{\sin\theta} = \frac{2n+1}{n}\cos\theta\frac{P_n^1}{\sin\theta}-\frac{n+1}{n}\frac{P_{n-1}^1}{\sin\theta}.[/tex]
    which is another recursive relation. If we substitute
    [tex]P_1^1(\cos\theta) =
    -\sin\theta[/tex][tex]P_2^1(\cos\theta)=-3\cos\theta\sin\theta[/tex], and
    [tex]P_3^1(\cos\theta)=-\frac{3}{2}(5\cos^2\theta-1)\sin\theta[/tex]
    the [tex]\sin\theta[/tex] will cancel out. With the recursive relation, the proof with math induction towards the expressions is straightforward. :)

    I still have difficulty to map the solution with [tex]e^{j\omega t}[/tex] convention to [tex]e^{-j\omega t}[/tex] though. :(
     
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