# Evalute expression of associated Legendre's polynomial

1. Nov 12, 2009

### 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

2. Nov 13, 2009

### 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...

3. Nov 22, 2009

### 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
$$P_1^1(\cos\theta) = -\sin\theta$$$$P_2^1(\cos\theta)=-3\cos\theta\sin\theta$$, 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. :(