Evalute expression of associated Legendre's polynomial

[kzhu]

Dear All,

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

These two terms are giving me difficulty when \theta=0 or \theta=\pi.

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

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


kzhu

 

[Raghnar]

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...

 

[kzhu]

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
(m-n-1)P_{n+1}^m(x) + (2n+1)xP_n^m(x) - (m+n)P_{n-1}^m = 0.
and get
\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}.
which is another recursive relation. If we substitute
P_1^1(\cos\theta) =<br /> -\sin\thetaP_2^1(\cos\theta)=-3\cos\theta\sin\theta, and
P_3^1(\cos\theta)=-\frac{3}{2}(5\cos^2\theta-1)\sin\theta
the \sin\theta 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 e^{j\omega t} convention to e^{-j\omega t} though. :(